Maths Quest Methods Year 12 2v2b6d

[email protected]

Horizontal asymptote y = c, vertical asymptote x=b

x≥b

If a > 0 y≥c If a < 0 y≤c

End point at (b, c)

c x

0 b y

y=

y=a x−b +c or y = a( x − b) + c

Special feature

(b, b, c) c x

y

c 0

Square root

Range

R

x

y 0

Hyperbola

Domain

b

x

y

1 2

0

(b, c)

x

• The equation for any graph y = f (x) above can be written in the general form: y = af (x − b) + c. This form can be used to describe transformations of all of the functions considered. • For all of the above functions: 1. a is the dilation factor: it dilates the graph from the x-axis. 2. When an equation for these types of graphs is put into its general form of y = af (x − b) + c, the horizontal dilation can be described in of a vertical dilation. 3. If a < 0, the basic graph is reflected in the x-axis. 4. f (b − x) or f (−x + b) is the reflection of f (x + b) in the y-axis. 5. b translates the graph b units along the x-axis (to the right if b > 0, or to the left if b < 0). 6. c translates the graph c units along the y-axis (up if c > 0, or down if c < 0).

120

Maths Quest 12 Mathematical Methods CAS for the TI-Nspire

• To put equations into general form: 7. If the coefficient of x is a number other than 1, to find the value of b and a, the equation should be transposed to make the coefficient of x equal to 1. For example, y = (3x + 5)2 + 4 = [3( x + 35 )]2 + 4 = 32 ( x + 35 )2 + 4 = 9( x + 35 )2 + 4 Hence, a = 9, b =

−

5 . 3

The absolute value function

• y = |x| means •

y = x, if x ≥ 0 and y = −x, if x < 0 y y = |x| x

Domain: R Range: R+ ∪ {0} • To sketch y = | f (x)|: 1. Sketch the graph of y = f (x). 2. Reflect the portion of the graph that is below the x-axis in the x-axis. Or: 1. Express the function in hybrid form with specific domains where the absolute value expression is positive and negative. 2. Sketch each rule for the specified domain. • For functions of the form y = a|| f (x)| + c, a and c have the same impact on the graph of the absolute value function, as on the graphs of all other functions discussed in this section. Transformations with matrices

• The use of matrices to map transformations of points and equations can be summarised as follows, where (x′, y′) is the image of the point (x, y) under the transformation.   x  T      y    x  T      y    x  T      y    x  T      y 

 x'   −1 = =  y'   0  x'   1 = =  y'   0

0 x  −x    =   represents a reflection in the y-axis. 1 y  y  0  x

 x =  −  represents a reflection in the x-axis.    y   x'   a 0   x   ax  = =    =   represents a dilation of a factor of a from the y-axis.  y'   0 1   y   y   x'   1 0   x   x  = =    =   represents a dilation of a factor of a from the x-axis.  y'   0 a   y   ay  −1   y 

• Transformations can be combined to represent more than one transformation. For example,  4x + 2   x '  4 0   x   2    = + =     0 1      −y  y' y 3 + 3      2       2  −

∴ x′ = 4x + 2 −y y′ = +3 2

Chapter 2

Functions and transformations

121

1

describes the following: dilation by a factor of 4 from the y-axis, a dilation by a factor of 2 from the x-axis, reflection in the x-axis, a horizontal translation of +2 and a vertical translation of +3. Sum and difference of functions

• For the graph of the sum/difference function, dom ((f (x) ± g(x)) = dom f (x) ∩ dom g(x). The graph of the sum/difference function can be obtained by using the addition of ordinates method. • For the product function, dom ((ff (x)g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: ° the x-intercepts of f (x)g(x) occur where either f (x) or g(x) have their x-intercepts ° f (x)g(x) is above the x-axis where f (x) and g(x) are either both positive or both negative ° f (x)g(x) is below the x-axis where one of the functions f (x) or g(x) is positive and the other is negative. Composite functions and functional equations

• For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. f Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). • Equations involving algebra of functions, for example f (2x (2 ) = 2f 2 (x), are generally tested to determine if they are true for particular functions. ° To determine if an equation is true for a particular function, consider the LHS and RHS of the equation separately to determine if the equation holds true for all values of x. ° Alternatively, you may find a particular x-value for which the equation does not work; that is, a counterexample. ° These types of equations can be investigated by defining the functions on the CAS calculator and then testing the algebraic function equation. Modelling

• Modelling is the process of finding the rule (mathematical model) that fits the given data. • To model: 1. Plot the original data on graph paper or use a CAS calculator. 2. Make an assumption of the model. 3. Transform the data in accordance with your assumption. 4. Check the assumption by plotting the transformed data (if correct, the graph will be a straight or nearly straight line). 5. Draw in a line of best fit. 6. Find the equation of the line (y = mx + c). 1 7. Replace x in the equation with the transformed variable (for example, x2, ). x

122


CHAPTER REVIEW SHORT ANSWER

1 For the function x − 3)2 − 4: a state the coordinates of the turning point b state the domain and range c sketch the graph. a 2 The graph of y = + c has a vertical x−b asymptote at x = 2 and a horizontal asymptote at y = −1. a Find the values of b and c. This graph then undergoes the following transformations: • reflection in the x-axis • dilation by a factor of 3 from the x-axis • horizontal shift of 2 units right. b If the intersection of the two graphs is at (m, 2), find the value of m. c Hence find the equation of the transformed graph. 6x − 5 3 Express the function f ( x ) = in the form 3x + 1 b a+ . Hence describe the transformations x+c required to produce this curve from the graph of 1 y= . x 4 The graph of a cubic function has a stationary point of inflection at (1, 1). It cuts the y-axis at y = 4. Find the equation of the graph. 1 5 The graph of y = was dilated by the factor of x 4 from the x-axis, reflected in the x-axis and then translated 2 units to the left and 1 unit down. a State the equation of the asymptotes. b State the domain and range. c State the equation of the new graph. d Sketch the graph. 6 a State the changes necessary to transform the 1 y graph of y = 2 into x the one shown. b Find the equation of the graph.

−2

−1 − –23

x

7 The domain of a truncus is R \{1}, the range is (−∞, 2) and the graph cuts the y-axis at y = −3. Find the equation of the function. 8 The basic square root curve was reflected in both axes and then translated so that its intercepts at the axes were (0, 1) and (−5, 0). Find the size and the direction of the translations; hence, find the equation of the new graph. 2 − 2, 9 a Sketch the graph of y = 2 − ( x + 2)2 clearly showing the coordinates of the cusps, the intercepts with the axes and the position of the asymptotes. b State the domain and range of the graph in a. 10 The graph of ff: [−5, 1] → R where f (x ( ) = x3 + 6x 6 2 + 9x is shown. y

−3

0

x

a On the same set of axes sketch the graph of y = /f (x)) /. b State the range of the function with the rule y = /f (x)) / and domain [−5, 1]. [© VCAA 2006]

11 The point (−1, 3) undergoes a translation given by a  the matrix   to (2, 0). Find a and b and describe b  the transformations involved. 12 The point (1, 2) undergoes a series of a 0   transformations given by the matrices   and 0 b   1  then   to (−7, 4). 2 a Find the values of a and b. b Find the image under the transformations of: i y=2 x ii y = x3 + x 13 A point on a curve ((x, y) undergoes a a 0  transformation descibed by   to (x', y'), 0 2  where a is a real constant such that a > 0.

Chapter 2


123

a Find the values of x and y in of a, x' and y'. b If the point is on the curve y = 2x 2 2 − x, find the image of the curve in of a under this transformation. c If the point (3, 6) is on the transformed curve, find the value(s) of a and hence the rule of this image. 14 The graph of the function f : (−2, 1) → R, f (x ( ) = x3 + 2x 2 2 is shown below. y

f x) = x3 + 2x f( 2 2

2 1

(−2, 0)

−2 −1.5 −1 −0.5 0 −1

0.5

1

x

Let g(x) = f (x) + 1, and sketch this graph on the same set of axes. Hence, sketch ((f + g)(x). 15 The graph of the function f:( f −2, 2), f (x ( ) = /x / is shown below. (−2, 2)

f x) =x y f( 2 (2, 2) 1

0 −2 −1 −1

1 2 x

−2

Let g(x) = −f (x) + 1 and sketch the graph of this function on the same set of axes. Hence, sketch f (g(x)). 16 Let f (x ( ) = x2 determine which of the following relationships are true. a f (x) − f (−y) = f (x) − f (y) b −f (x) − f (y) = f (x) + f (y) c f (−x) + f (−y) = f (x) + f (y) d f (x) − f (−y) = f (x) + f (y) 17 The data in the table below exactly fit one of these a models: y = ax3, y = 2 or y = a x . x x

2

y

25

4

5

10

6.25

4

1

20 0.25

25 0.16

a Plot the values of y against x and use the scatterplot to choose a suitable model. b Plot the values of y against either x3, 1 or x x2 (depending on your choice in part a ). Did you choose the right model? Explain your answer. c Find the value of a.

124

1 The equation of a parabola is given by y = m − 2(x 2( + 3)2, where m > 0. The increase in m will result in: A the graph being thinner B the graph being wider C the increase of the domain D the increase of the range E the graph being shifted further to the right 2 The coordinates of the turning point of the parabola y = 2(3x + 6)2 − 3 are: A (6, −3) B (−6, −3) C (2, −3) − − − D ( 2, 3) E ( 2, 3)

(1, 3)

3

MULTIPLE CHOICE

bx − 3)3 + 1 is dilated in the 3 The graph of y = 23 (bx y direction by the factor of: 3 C 2b B 23 b A 23 3 2 3 D b E 3b3 4 The graph of y = 2 − (3 + 4x 4 )3 has a stationary point of inflection at: A ( 43 , 2)

B (

−

3 − , 2) 4 − 3 , 2) 4

C (−3, 2) D ( E (−3, −2) 2 5 If f ( x ) = + 1, then f (x ( ) + 2 will have: x A the horizontal asymptote y = 2 B the horizontal asymptote y = 1 C the horizontal asymptote y = 3 D the vertical asymptote x = 2 E the vertical asymptote x = 1 6 The equation of the graph shown is likely to be: −2 y A y= −1 x−2 −2 B y =1 x+2 −2 x −2 −1 C y= −2 x +1 2 −1 D y= x+2 −2 E y= −1 x+2 7 If the graph of y = 1 is reflected in the y-axis, x translated 43 units to the right and 2 units up, the resulting graph would have the equation: 1 3 A y= + x−2 4 3 +2 B y= 4x


1 3 − 4x −1 +2 D y= 4x − 3 1 +2 E y= 3 − 4x C y = 2−

8 Which of the following is not true for the graph at right? y A The vertical asymptote is x = 2. B The horizontal asymptote is y = −4. x 2 C The domain is R\{2}. D The range is R\{−4}. −4 E The value of the y-intercept is greater than −4. 9 To obtain the graph shown, we need to: A translate the graph of y 1 y = 2 one unit to the left x and reflect in the x-axis x −1 B translate the graph of 1 y = 2 one unit to the left x and reflect in the y-axis 1 C translate the graph of y = one unit to the x2 left and reflect in both axes 1 D translate the graph of y = 2 one unit to the x left and dilate it in the x direction E none of the above 10 The equation of this graph is of the form: A y = a x − m + n, a > 0 B y = a m − x + n, a > 0

y (m, n) x

C y = a x − m + n, a < 0 D y = a m − x + n, a < 0 E y = a x + m + n, a < 0 11 The equation of this graph could be: A y=d− a−x

y d

B y= x−a +d C y=d− x−a D y = c− a − x E y= a− x −c

b a c

x

12 The domain of the function f ( x ) = 2 x − 1 + 3 is: A [1, ∞) B [3, ∞) C [0.5, ∞) D [2, ∞) E [−1, ∞) 1 13 The range of the function y = − 3 − − 2 is: x2 A (1, ∞) B (−∞, −5] C [3, ∞) D (−∞, 2) − − E ( ∞, 2] 14 The equation of the graph shown in the diagram below is best described by: y A y = |x + 2| + 2 2 B y = 2 − |x + 2| C y = |2 − x| + 2 x −2 D y = 2 − |2 − x| E y = |x + 2| − 2 15 The value(s) of k for which |2k + 1| = k + 1 are: − −2 2 B A 0 only C 0 or 3 3 − 1 2 E or 0 D 0 or 2 3 [© VCAA 2006] 16 Under the transformation T : R2 → R2 of the plane   x   2 0    x   1  defined by T    =      +   , the   y   0 1    y   − 3  image of the curve y = | x | has the equation: A B C D E

y=

1 2

x −1 + 3

1 2

x −1 − 3

y = | 22x + 1| − 3 y = 2 | x + 1| − 3 y = 2 | x − 1| + 3 y=

17 If g:[2, 4] → R, where g(x ( ) = x2(x (x ( − 3), and h:(0, 3] → R, where h(x ( ) = 3 − x, then the (x function f (x ( ) such that f (x ( ) = g(x ( ) × h(x (x ( ) is defined (x by the rule: A f: f R → R, where f (x) = −x2(x − 3)2 B ff: (0, 3] → R, where f (x) = −x2(x − 3)2 C ff: [2, 3] → R, where f (x) = −x2(x − 3)2 D ff: (0, 4] → R, where f (x) = −x2(x − 3)2 E ff: [2, 3] → R, where f (x) = −x2(x − 3) (3 + x) 18 Given the function f ( x ) = x − 3 then f ° h could exist if h(x ( ) was defined as: (x A h(x) = −x + 2 B h(x) = x2 − 3 C h(x) = x3 D h(x) = x2 + 3 E h( x ) = x − 2

Chapter 2


125

19 The data in the following table exactly fit one of these models: y = ax2, y = ax3 or y = a x . x

1

2

3

y

0.3

2.4

8.1

The value of a is: A 2.4 C 2.7 E 0.3

4 19.2

B 1.2 D 0.9

20 For certain data the values of y were plotted against 1 and the line of best fit was drawn as seen on x the diagram below. The model that relates the variables x and y is: y A y = 20x 20 − 1 (1, 19) B y = 19x + 1 x −1 C y= 20 19 D y = −1 x (0.1, 1) 20 1 −1 E y= x– x

EXTENDED RESPONSE

1 The graph of y = f (x ( ) is shown at right. a Sketch the graph of each of the following functions on the same set of axes with the original graph and give the coordinates of the points A, B, C and D. i y = −f (x) ii y = f (−x) iii y = f (xx − 2) iv y = f (x) + 3 v y = 2f 2 (x) vi y = 1 − f (x + 1) b Maya, a fabric designer, wishes to use the curve of y = f (x) (red) to create a ‘wavy’ pattern as shown in the diagram at right. If she wants the waves to be 2 units apart vertically, suggest the best way she could alter the equation of y = f (x). ( a fabric has a fixed width!)

y

D(4, 6)

B C(2, 3) A 2 −3 −2 7 x

2 units apart

2 Consider the function f:R → R, f( f x) = (x − 1)2 (x − 2) + 1. a The coordinates of the turning points of the graph of y = f( f x) are (a, 1) 25 and (b, 27 ). Find the values of a and b. b Find the real values of p for which the equation f( f x) = p has exactly one solution. c For the following, k is a positive real number.  x i Describe a sequence of transformations that maps the graph of y = f (x) onto the graph of y = f   − 1.  k  x y = f − 1 ii Find the x-axis intercepts of the graph of in of k.  k  d Find the real values of h for which only one of EXAM TIP the solutions of the equation f (x + h) = 1 is positive. Students must ensure that they show [© VCAA 2004]

3 The graph of the function f: 1 f ( x ) = 2 is shown below. x (−0.5, 4)

y 4

(–2, –0.5)

∪ (0.5, 2) → R,

(0.5, 4)

3

[Assessment reports 1 and 2 2007 © VCAA]

f x) = 12 f(

2 (−2, 0.25)

126

x

1

−2.5 −2 −1.5 −1 −0.5 0

their working — if a question is worth more than one mark, students risk losing all available marks if only the answer is given and it is incorrect. The instructions at the beginning of the paper state that if more than one mark is available for a question, appropriate working must be shown. When students present working and develop their solutions, they should use conventional mathematical expressions, symbols, notation and terminology.

(2, 0.25) 0.5

1 1.5

2 2.5 x


The graph of y = f (x) is to be transformed to become the graph of y = f (2x (2 ) + 1. a Describe these two transformations. b Create matrices to represent these transformations. c Use these matrices to find the images of the points (1, 1) and (2, 0.25) under these transformations, and use these values to deduce the images of the points (−1, 1) and (−2, 0.25). d On the same axes, sketch this transformed function, showing the coordinates of the four points from c above. e Using any method, find a rule for this transformed function. 4 A proposed section of a ride at an amusement park is to be modelled on the curve y = 1 (600 x + 25 x 2 − x 3 ), 500 where y is the height (in metres) of the ride above ground level and x is the horizontal distance (in metres). The x-axis represents ground level. It will travel through a tunnel from A to C; B is the lowest point in the tunnel and D is the highest point on the ride. y

A

D

Ex

C B

a Find the horizontal distance from A to E. b Find the greatest depth below ground level and the maximum height above ground level that the rollercoaster will reach in this section (correct to 2 decimal places). c Describe the impact that a dilation by a factor greater than 1 from the x-axis would have on: i the maximum depth and maximum height from b ii the point at which the rollercoaster would emerge from the tunnel iii the gradient of the slope at this point. 5 Lena and Alex are planning to buy a new house. They’ve been watching the prices of 3-bedroom houses in a specific area, where they want to live, for the whole year. During each month they collected the data and then, at the end of the month, they calculated the average price for that month. The results of their calculations are shown in the table below. (The prices given are in thousands of dollars.) Month Price

1

2

3

4

5

6

7

8

9

10

11

12

240 248 255 261 266 271 273 274 275 274 272 270

a Plot the prices against the months. What model does the graph suggest? b If the model of the form y = a(x − b)2 + c is to be used for these data, what is (judging from the graph) the most suitable value for h? c Plot the values of y (the prices) against (x − b)2, where b is the value you’ve selected in part b . Comment on the shape of the graph. d Draw a line of best fit and find its equation. Hence, state the values of a and c in the model. e Write the equation of the model. f According to the Real Estate Institute, the property market is on a steady rise (that is, the prices are going up and are likely to rise further). Do the data collected by Lena and Alex this theory? g Use the model to predict the average price for the next 2 months.

Chapter 2


127

h Lena and Alex were planning to spend no more than 250 000 for their new house. Several months ago the prices were in their range, but they could not find what they wanted. If the prices are going to behave according to our model, how long do they have to wait until the prices fall back into their range? 6 An eagle soars from the top of a cliff that is 48.4 metres above the ground and then descends towards unsuspecting prey below. The eagle’s height, h metres above the ground, at time t seconds can be modelled by a the equation h = 50 + , where 0 ≤ t < 25 and a is a constant. t − 25 a Find the value of a. b Find the eagle’s height above the ground after i 5 seconds ii 20 seconds. c After how many seconds will the eagle reach the ground? d Comment on the changes in speed during the eagle’s descent. e Sketch the graph of the equation. After 24 seconds, the eagle becomes distracted by another bird and reaches the ground exactly 2 seconds later. For this second part of the journey, the relationship between h and t can be modelled by the equation h = a(t − 24)2 + c. f Find the values of a and c. g Fully define the hybrid function that describes the descent of the eagle from the top of the cliff to the ground below. eBoo k plus eBook Digital doc

Test Yourself Chapter 2

128


eBook plus

ACTIVITIES

Chapter opener Digital doc

• 10 Quick Questions: Warm up with ten quick questions on functions and transformations. (page 57) 2A

Transformations of the parabola

Digital docs

• Spreadsheet 132: Investigate transformations. (page 59) • Spreadsheet 108: Investigate the quadratic function in power form. (page 61) 2B

The cubic function in power form

Digital docs

• Spreadsheet 015: Investigate the cubic function in power form. (page 65) • Spreadsheet 236: Investigate graphs of functions. (page 70) 2C

The power function (the hyperbola)

Tutorial

• WE 9 int-0522: Watch a worked example on sketching the graph of a hyperbola. (page 74) Digital docs

• Spreadsheet 051: Investigate the hyperbola. (page 72) • Spreadsheet 236: Investigate graphs of functions. (page 71) • WorkSHEET 2.1: Find the domain, range, coordinates of turning points and equations of asymptotes of various graphs. (page 78) • History of Mathematics: Investigate the history of major curves. (page 78) 2D

The power function (the truncus)

Digital doc

• Spreadsheet 236: Investigate graphs of functions. (page 85) 2E

The square root function in power form

Tutorial

• WE 16 int-0523: Watch a worked example on implied domain and range. (page 88) Digital doc

• Spreadsheet 236: Investigate graphs of functions. (page 92) 2F

The absolute value function

Tutorials

• WE 20 int-0524: Watch a worked example on sketching the graph of an absolute value function. (page 93) • WE 21 int-0525: Watch a worked example on expressing an absolute value function as a hybrid function. (page 94)

Digital docs

• WorkSHEET 2.2: Identify transformations, state domain and range, sketch graphs of power functions and absolute value functions. (page 96) • Spreadsheet 002: Investigate graphs of absolute value functions. (page 96) 2G

Transformations with matrices

Interactivity

• Transformations with matrices int-0247: Consolidate your understanding of using matrices to transform functions. (page 96) Tutorial

• WE 25 int-0526: Watch how to use matrices to determine the resultant equation after transformations. (page 100) 2H

Sum, difference and product functions

Tutorial

• WE 27 int-0257: Watch how to use addition of ordinates to sketch the sum of two functions. (page 104) 2J

Modelling

Digital docs

• Spreadsheet 076: Investigate modelling with functions. (page 117) • Investigation: Goal accuracy. (page 119) Chapter review Digital doc

• Test Yourself Chapter 2: Take the end-of-chapter test to test your progress. (page 128) To access eBookPLUS activities, log on to www.jalus.com.au

Chapter 2


129

EXAM PRACTICE 1 SHORT ANSWER

15 minutes

1 The functions f g are graphed below. On the same axes sketch a graph of f + g. f

g

CHAPTERS 1 TO 2

1 −a x−b 1 C y= +b x+a 1 E y= +a x+b A y=

B y=

−1

+b a−x −1 D y= −a x−b

2 The graph sketched below is best represented by the rule:

0

1 mark

y

2 Write down the maximal domain of f : X → R where f ( x ) = x 2 − 3 .

a

1 mark

3 For y = (2x (2 − 3)2 − 1: a write down the y-coordinate of the turning point b determine the equation of the axis of symmetry. 2 marks

4 a Using the factor theorem, show that x + 2 is a factor of x3 − 7x − 6. b Given x3 − 7x − 6 = (x + 2)Q(x) where Q(x) is a quadratic factor, determine Q(x). 2 marks −2x 2

5 For what values of c does the graph of y = +c intersect the graph of y = −x2 + x − 2 at two distinct points? 2 marks

6 Sketch the graph of the function f :[−2,5) − R, f (x ( ) = | 3(x 3( − 3)2 − 10 |. 2 marks

MULTIPLE CHOICE

10 minutus

Each question is worth 1 mark. 1 Given a, b ∈ R, this graph could have the rule:

A B C D E

b

c

x

y = (xx – a)(xx – b)2(xx – c)2 y = (xx – a)(xx – c)2 y = (xx – a)3(xx – c)2 y = (xx – a)(xx – c) y = (xx – a)(xx – b)(xx – c)

3 The coefficient of the term in x4 in the expansion of 3 (2 x 2 − )5 is: x A −240 B −72 C −60 D 72 E 720 4 What is the equation of the horizontal asymptote of the graph of y = A y = −4 3 D y= 4

3x − 2 ? 4−x B y = −3

C y = −1

E y=3

5 The quadratic function f : D → R, f (x ( ) = 2(x 2( + 1)2 − 5 has a domain, D,, of [0, 3]. The range of f is: A [−1, 3] C [−5, 27] B R D [−3, 27] E [−5, ∞)

y

y=b x=a 0

130

x

6 The simultaneous linear equations 3x + ay = 12 and ax + 3y = 4a have infinitely many solutions for: A a ∈ [−3,3] B a=3 C a = 3 or a = −3 D a=9 E a=0


EXTENDED RESPONSE

40 minutes

1 Severe tropical cyclone Vance devastated the town of Exmouth in northern WA in 1999. It produced a measured wind gust of 267 km/h, which is the highest wind speed ever recorded on mainland Australia. In order for it to be first categorised as a cyclone, its wind speed needed to exceed 119 km/h. Several years later, it is a peaceful day, with no wind detectable at 1 pm. By 3 pm, however, the wind speed is gusting to 200 km/h, and the residents know they are in trouble. a Let t be the time in hours after noon and v be the wind speed. Establish a linear model of the form v = at + b to represent the relationship between the wind speed and time. 2 marks b Using this model, determine to the nearest minute when the wind speed will be high enough for classification as a cyclone. 1 mark c Predict to the nearest minute when the cyclone will break the record for the highest wind speed ever recorded. 1 mark

d Explain why the linear model is unsatisfactory as a model for the cyclone’s behaviour. e

1 mark

i The wind speed actually peaks at 256 km/h at 5 pm. Use this data and the wind speed at 1 pm to help to create a quadratic model relating wind speed (v) to hours after noon (t). How well does this model match all the data provided?

2 marks

ii The wind speed actually returns to zero at 1 am the following morning, 12 hours later. Discuss how well this result matches the quadratic model. f

1 mark

i Establish a second quadratic model that exactly matches the following data: Time

Wind speed

1 pm

0 km/h

5 pm

256 km/h

1 am

0 km/h 1 marks

ii Evaluate how well this model represents the relationship between time and wind speed. g It has been suggested that a cubic function would be a better model for the data. Use all the data provided in order to establish a cubic model.

1 mark 2 marks

1020 1010 1000 990 980 970 960 950 0 km

Eye

Pressure Wind speed

175 150 125 100 75 50 25

Wind speed (km/h)

Pressure (hPa)

h The graph below shows how air pressure in hectopascals (P hPa) and wind speed (v km/h) relate to 1 distance (x) across a cyclone. It has been suggested that a truncus (general form v ( x ) 2 ) could represent x the shape of the relationship between speed and distance across the cyclone.

500 km

i What is the equation to the vertical asymptote for this relationship? ii Determine a rule for the relationship between speed (v) and distance (x). iii Determine a rule for the relationship between air pressure (P) and distance (x).

1 mark 2 marks 2 marks

eBoo k plus eBook Digital doc

Solutions Exam practice1

Exam practice 1

131

3

3A 3B 3C 3D 3E 3F

The index laws Logarithm laws Exponential equations Logarithmic equations using any base Exponential equations (base e) Equations with natural (base e) logarithms 3G Inverses 3H Literal equations 3I Exponential and logarithmic modelling

Exponential and logarithmic equations arEas oF sTudy

• Logarithm and exponent laws • Identification of appropriate solution processes for solving logarithmic and exponential equations • Numerical solutions of exponential and logarithmic equations • Solution of literal equations such as emx + n = k

• Application of algebraic and logarithmic properties to the simplification of expressions and the solution of equations • Finding the rule of an inverse function involving logarithmic and exponential equations • Application of logarithmic and exponential functions in modelling practical situations eBook plus

3a

Digital doc

The index laws

10 Quick Questions

A number in index form has two parts, the base and the index, power, exponent or logarithm. A number in index form is represented like this: Index, power, exponent or logarithm

ax Base

The index laws are summarised below. ax = x a

ax ÷ ay = ax - y

a y = ax =

(ax)y = axy

(ab)x = axbx

a0 = 1 (a ≠ 0)

ax  a  b  = b x , b ≠ 0

a

132

1

ax × ay = ax + y

−x

=

1 (a ≠ 0) ax

maths Quest 12 mathematical methods Cas for the Ti-nspire

x

y

x

( a) y

x

Worked Example 1

Simplify

( 2 x 2 y3 ) 3 × 3( xy4 )2 . 6 x 4 × 2 xy4

Think

Write

1

Remove the brackets by multiplying the indices.

2

Add the indices of x and add the indices of y. Simplify 23 to 8 and multiply the whole numbers.

3

Subtract the indices of x and y. Divide 24 by 12.

(2 x 2 y 3 )3 × 3( xy 4 )2 23 x 6 y 9 × 3 x 2 y8 = 6 x 4 × 2 xy 4 12 x 5 y 4 =

24 x8 y17 12 x 5 y 4

= 2x3y13

For negative indices and fractional or decimal indices, the same rules apply.

Worked Example 2

Write in simplest form: 2

− 0.4

a 64 3 b 32 −2

c 125

3

using a CAS calculator.

THINK

WRITE/display x

a

b

2 y

1

Rewrite using the index law a y = a x .

2

Rewrite using a x = ( y a ) x .

= ( 3 64 )2

3

Simplify by taking the cube root of 64.

= 42

4

Square 4.

a 64 3 =

y

3

64 2

= 16 −0.4

1

Write as a fraction with a positive index.

2

Change 0.4 to 104 .

3

Simplify the fractional index.

4

Rewrite using the index law a y = a x .

5

Simplify by taking the 5th root of 32.

6

Square 2.

b 32

= = =

1 320.4 1 4

3210 1 2

32 5 x

y

=

1 5

( 32 )2 1 = 2 2 1 = 4

Chapter 3 Exponential and logarithmic equations

133

−2

c

1

c

To simplify 125 3 with a CAS calculator, open a Calculator page. −2

Complete the entry line as: 125 3 . Then press ENTER ·. Note: If the calculator is set to Approximate, the answer will be displayed as a decimal.

2

−2 3

Write the answer.

125

1

= 25

Worked Example 3

Simplify, leaving your answer with positive indices: a

−

−

a 2 b4 × ( a 3 b 4 )

 1 −1  a2 b  b  − 1 2 3 b   

−1

−1

using a CAS calculator.

THINK a

1

Remove the brackets by multiplying the indices.

2

Add the indices of a and of b. Place a5 in the denominator with a positive index.

3

b

WRITE/display

1

On a Calculator page, complete the entry line using the fraction template  12 1  a b  as:  3 1 b 2   

-

- -1

a a 2b4 × (a3b 4)

-

-

= a 2b4 × a 3b4 -

= a 5b8 =

b8 a5

b

−1

−

−

Then press ENTER ·.

2

Write the answer. The answer is expressed with positive powers.

 12 −1  a b   −1 2   3 b 

−1

=

b

3

3 a

If the expression contains different numbers that do not have the same base, write each number as a product of prime factors.

134


WorkEd ExamplE 4

Simplify

3 n × 6 n + 1 × 12 n − 1 : 32 n × 8 n

a showing working, and

Think a

WriTE/display

1

Write each number as the product of prime factors.

2

Remove the brackets.

3

In the numerator, add the indices of numbers with base 3 and add indices of numbers with base 2. Subtract the indices.

a

3n × 6 n + 1 × 12 n − 1 32 n × 8 n 3n × (3 × 2) n + 1 × (22 × 3) n − 1 = 32 n × ( 2 3 ) n 3n × 3n + 1 × 2 n + 1 × 2 2 n − 2 × 3n − 1 32 n × 2 3 n 3 n 3 n − 1 3 ×2 = 2n 3 × 23 n =

-1

5

Write the term with a negative index in the denominator with a positive index.

= 3n × 2 1 = 3n × 2

6

Simplify.

=

1

On a Calculator page, complete the entry line using the fraction template as:

4

b

b with a CAS calculator.

3n 2

b

3n × 6 n + 1 × 12 n − 1 32 n × 8 n Then press ENTER ·.

2

Write the answer.

3n × 6 n + 1 × 12 n − 1 3n = 2 32 n × 8 n

WorkEd ExamplE 5

eBook plus

Simplify, leaving your answer with positive indices. 3 3 1 − a x 2− − b − + −1 2 1 x x +2 x −2 Think a

1

Tutorial

int-0528 Worked example 5

WriTE

Rewrite the expression with positive indices.

a x

3 − x 2 1 = 2 − 3x 2 x

−2

−

Chapter 3

Exponential and logarithmic equations

135

2

Find the lowest common denominator.

b

=

1 3x 4 − 2 x2 x

1 − 3x 4 x2 1 3 b + = 1  1   x + 2  x − 2

Simplify.

1

Rewrite the question using positive indices.

2

Find the common denominator for the in the brackets and simplify.

=

Follow the process for division of fractions (change the division sign to a multiplication sign and invert the second fraction).

4

Find the common denominator.

5

Expand the brackets on the numerator.

6

Simplify the numerator by adding like .

Factorise the numerator and write the answer.

7

1 x2 − 3x 2 × 2 2 x x

=

3

3

=

3 x +  1 + 2x   1 − 2x   x   x 

=

3x x + 1 + 2x 1 − 2x

=

3 x (1 − 2 x ) + x (1 + 2 x ) (1 + 2 x ) (1 − 2 x )

=

3x − 6 x 2 + x + 2 x 2 (1 + 2 x ) (1 − 2 x )

=

4x − 4x2 (1 + 2 x ) (1 − 2 x )

∴

4 x (1 − x ) (1 + 2 x ) (1 − 2 x )

1. axay = ax + y ax 2. ax ÷ ay = ax − y or y = a x − y a 3. (ax)y = axy 4. a0 = 1, a ≠ 0 1 1 5. a x = x and − x = a x a a −

1

x

y

y

y

6. a y = a and a y = a x = ( a ) x Exercise

3A

The index laws 1 WE1

136

Simplify: ×

a

x3

x4

e

( x 2 )3 × x 5 ( x 5 )2

b x7 ÷ x2 f

5x 2 y 4 × 4 x 5 y 22 x 3 y 2


c (x2)5 2 3 4 2 g (2 xy ) × 5( x y ) 4 x 5 y 3 × 3x 2 y 3

-

d (x 3)2

2 WE2 a

Simplify: −3

2 27 3

b 25 2

 27  3 d    64 

c 810.25

2

 243  e   32 

−3 5

3 Simplify: a

x4 × x5 x3

3

c 361.5

b 16 4

1

e  256   81 

d  9  2  49  4 WE3

−3 4

Simplify, leaving your answer with positive indices. 1

a

3x 3y2

-4

× (x2y)

−3  − 1 2 x 2y2

c  

 

b 5

 3 2 ×  9x 2 y2   

1 2 2x 2 y 3

 3  2  a2b c  d  − 1   3a 2 bc 2 

 −1 −1  2 ×  9x 5 y 2   

−2

−

 2  a 3 b3 ÷ 3 1 2  a c    −

−

5 Simplify: 4y-1

a x

c

1 16 2

×

(x 2y3) 1

−1  2 −1  2 x5y 4

 

 

b

6 WE4 Simplify: a 2n × 4n + 1 × 8n - 1 d

32 × 2

−3

3 92

 1 2 −1  ×  83 x 3 y 2   

 −3 3 a 2b4 d   ab 2 

1  2 12 × 4x 5 y 2

 

−1 3 5x 3 y 4

 

b 3n × 9n - 1 × 27n + 1

   

3

2

−2

1

 9a − 3 b  2 ÷ 2 3  4a b 

c 2n × 3n + 1 × 9n

× 16

7 Simplify: a 2n - 1 × 3n × 6n + 1

b

52 × 3

−1

125 × 9

−2

÷

27 5

8 WE5 Simplify, writing your answer as a single fraction with positive indices. −

a x 1 +

1

-

-

b (x 1 + x 2) 2

−1

x 1 1 + − c −1 x +1 x 1 −1

-

-1

d 2x(x2 - y2) 1 - (x - y)

−

9 MC 3 x + 3x is equal to: A 1

B

1 + 32 x 3x

C 3

− x2

D 6

x E 1 + 3 3x


137

3B

Logarithm laws If a > 0, then N = ax ⇔ loga (N) = x. For example, an expression in index form can also be rewritten in logarithmic form. 8 = 23 ⇔ log 2 (8) = 3 • Since ≠ 0 then loga (0) is undefined. • a0 = 1 ∴ loga (1) = 0 • a1 = a ∴ loga (a) = 1 Let m = ax ⇔ loga (m) = x and n = ay ⇔ loga (n) = y. • mn = ax × ay ∴ loga (mn) = x + y = ax + y = loga (m) + loga (n) x m   m a • = ∴ log a   = x − y  n n ay = ax - y = loga (m) − loga (n) • mp = (ax)p ∴ loga (mp) = px = axp = p loga (m) ax

Change-of-base rule Suppose b = ax, then loga (b) = x. Consider N = by, then logb (N) = y. But N = by = (ax)y = axy. Therefore, loga (N) = xy = loga (b) logb (N). Thus, log b ( N ) =

log a ( N ) . log a (b)

This is called the change-of-base rule.

Worked Example 6

Evaluate: a log2 (1) b log5 (5). Think

Write

a

Log of 1 to any base is equal to zero: loga (1) = 0.

a log2 (1) = 0

b

If the number and base are equal the answer is 1: loga (a) = 1.

b log5 (5) = 1

Worked Example 7

Write in index form: a log2 (8) = 3 b logx (81) = 4. Think

138

Write

a Use ax = y ⇒ loga (y) = x.

a log2 (8) = 3 ⇔ 23 = 8

b Use ax = y ⇒ loga (y) = x.

b logx (81) = 4 ⇔ x4 = 81


WorkEd ExamplE 8

Simplify: a log10 (5) + log10 (2)

b log4 (20) - log4 (5)

c log2 (16)

Think a

b

c

d

d log 5

( x ). 5

WriTE a log10 (5) + log10 (2) = log10 (5 × 2)

1

Rewrite using loga (mn) = loga (m) + loga n.

2

Simplify.

= log10 (10)

3

Simplify using loga (a) = 1.  m Rewrite using log a   = log a (m) − log a ( n).  n

=1

1

 20   5 

b log 4 (20) − log 4 (5) = log 4 

2

Simplify.

= log 4 (4)

3

Simplify using loga (a) = 1.

=1

1

Rewrite 16 as a number with base 2.

2

Rewrite using loga (mp) = p loga (m).

= 4 log2 (2)

3

Simplify using loga (a) = 1.

=4

1

Rewrite using y a = a y .

2


1

c log2 (16) = log2 (24)

d log 5



( x ) = log  x 5

5

1 5

 

1

= 5 log 5 ( x )

WorkEd ExamplE 9

Simplify log3 (27) + log3 (9) - log3 (81). Think

WriTE/display

1

On a Calculator page, complete the entry line as: log3 (27) + log3 (9) + log3 (81) Then press ENTER ·.

2

Write the answer.

log3 (27) + log3 (9) +log3 (81) = 1

WorkEd ExamplE 10

eBook plus

Simplify: a 2 + log10 (3)

b 3 log3 (6) - 3 log3 (18)

log 3 ( 9 ) c . log 3 ( 27 )

Chapter 3

Tutorial



139

Think a

b

1

c

Write

Write 2 as 2 log10 (10) because log10 (10) = 1. (mp)

a 2 + log10 (3) = 2 log10 (10) + log10 (3)

= log10 (102) + log10 (3)

= p loga (m).

2

Rewrite using loga

3

Rewrite using loga (mn) = loga (m) + loga (n).

= log10 (102 × 3)

4

Write 102 as 100.

= log10 (100 × 3)

5

Multiply the numbers in the brackets. (mp)

= log10 (300) b 3 log3 (6) − 3 log3 (18) = log3 (63) − log3 (183)

= p loga (m).

1

Rewrite using loga

2

Rewrite using  m log a   = log a (m) − log a ( n).  n

3

Write 63 as 6 × 6 × 6 and 183 as 18 × 18 × 18.

4

Simplify.

 1 = log3  3  3 

5

Write the numbers with the base 3.

= log3 (3 3)

6


= −3 log3 (3)

7

Simplify using loga a = 1.

= −3 × 1 = −3

1

Write the numbers with the same base. It is not possible to cancel the 9 and the 27 because they cannot be separated from the log.

2


=

2 log3 (3) 3 log3 (3)

3

Cancel the logs because they are the same.

=

2 3

 63  = log3  3   18   6×6×6  = log3   18 × 18 × 18 

−

c

log3 (9) log3 (32 ) = log3 (27) log3 (33 )

Worked Example 11

Calculate the value of log2 (18), correct to 2 decimal places. Think

140

1

On a Calculator page, complete the entry line as: log2 (18) Then press ENTER ·. Pressing Ctrl / ENTER · will give a decimal approximation.

2

Write the answer.

Write/display

log2 (18) = 4.17, correct to 2 decimal places.


For a > 0: 1. loga (1) = 0 2. loga (a) = 1 3. loga (0) is undefined. 4. loga (x) does not exist when x ≤ 0. 5. loga (mn) = loga (m) + loga (n) Exercise

3B

 m 6. log a   = log a (m) − log a ( n)  n 7. loga (mp) = p loga (m) log a ( N ) 8. log b ( N ) = log a (b)

Logarithm laws 1 WE6 Evaluate the following. a log3 (1) b log5 (1)

c log2 (2)

2 WE7 Write the following in index form. a log2 (16) = 4 b logx (25) = 2 d log3 (x) = 5 3

c log5 (125) = x

 1 e log 5  5  = −1

Write the following in logarithmic form. a 23 = 8 b 34 = 81 − d 5x = 125 e 2 1 = 12

4 WE8

c 43 = x f x3 = 27

Simplify:

a log6 (3) + log6 (2) d log3 (81) 5

d log6 (6)

b log2 (10) − log2 (5)

c log2 (32)

log 5  1  5

f log3 

b 3 log3 ( 3 x )

c log 2 

e

1  27 

Simplify: a log 2 ( x )

6 WE9 Simplify: a log4 (10) + log4 (2) − log4 (5) c 1 log10 (16) + log10 (52 )



 x4   y 2 

b log5 (25) + log5 (125) − log5 (625) d log3 (2) − log3 (10) + log3 (15)

2

e log2 (16) + log2 (8) + log2 (4) 7 WE10 Simplify: a 4 log2 (12) − 4 log2 (6) d

log 2 (64) log 2 (8)

b 2 + log5 (10) − log5 (2) e

log a ( x ) log a ( x )

8 WE11 Evaluate correct to 3 decimal places. a log10 (3) b log5 (4) d log2 (0.8) 9

c 1 + log2 (5)

e log4 (20)

c log10 (0.5) f log3 (60)

Simplify: a 5 log3 (x) + log3 (x2) − log3 (x7)

b 4 log2 (x) + log2 (x3) − log2 (x6)

c 3 log4 (x) − 5 log4 (x) + 2 log4 (x) e log10 (x2) + 3 log10 (x) − 2 log10 (x)

d 4 log6 (x) − 5 log6 (x) + log6 (x) f 4 log10 (x) − log10 (x) + log10 (x2)

g log5 (x + 1) + log5 (x + 1)2

h log4 (x − 2)3 − 2 log4 (x − 2) Chapter 3 Exponential and logarithmic equations

141

10 MC 2 log10 (5) − log10 (20) + log10 (8) is equal to: B −log10 (2) C 1 A log10 (2)

D −1

E log10 (4)

11 MC If loga (b) = 2, then b is equal to: A 0 B 1 C 2 D a E a2 12 If y = a log10 (x), find x when a = 2 and y = 3. Give your answer correct to 3 decimal places.

3C

Exponential equations The equation ax = b is an example of a general exponential (or indicial) equation and 2x = 32 is an example of a more specific exponential equation. To solve one of these equations it is necessary to write both sides of the equation with the same base if the unknown is an index or with the same index if the unknown is the base.

Worked Example 12

Solve for x in each of the following. 1 a 2x = 32 b 3 x = c 2 × 3x = 162 d 2(1 − x) = 16 27 Think a

1

b

The indices are equal because the base is 2 on each side of the equation.

1

Write 27 with base 3.

3 1

d

Write 32 with base 2, the same as the left-hand side.

2

2

c

Write a 2x = 32

2x = 25 x=5

1 27 1 = 3 3 − 3x = 3 3

b 3x =

Write 1 as a number with base 3. 33 Equate the indices. Divide both sides by 2 to leave 3x on the left-hand side.

2

Write 81 as a number with base 3.

3

Equate the indices.

1


2

Equate the indices.

3

Solve for x.

x = −3 c 2 × 3x = 162

3x = 81

3x = 34 x=4 d

= 16 21 − x = 24

21 − x

1−x=4 x = −3

Worked Example 13

Solve 5x × 252x − 3 = 625 for x: a using index laws b with a CAS calculator. Think a

1

Write/display

Write all numbers with the same base.

5x × 252x − 3 = 625 5x × (52)2x − 3 = 54

2

142

Simplify.

5x × 52(2x − 3) = 54


5x × 54x - 6 = 54

3

Remove the brackets in the index.

4

Add the indices on the left-hand side.

55x - 6 = 54

5

Equate the indices.

5x - 6 = 4

6

Solve the equation.

5x = 10 x=2

b

1

On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(5x × 25(2x - 3) = 625,x) Then press ENTER ·.

2

Write the solution.

Solving 5x × 25(2x - 3) = 625 for x gives x = 2.

Sometimes it is possible to use the methods for solving quadratic equations to help solve indicial equations. that 22x = (2x)2. WorkEd ExamplE 14

eBook plus

Solve for x in the following. b 32x - 12 × 3x + 27 = 0

a (2x - 16)(2x + 4) = 0 Think a

b

1

c 4x - 2x + 3 + 16 = 0 WriTE

Use the Null Factor Law to solve by making each bracket equal to zero.

(2x - 16) = 0 or (2x + 4) = 0

Solve each equation.

2x = 16 or 2x = -4

3

Write 16 as a number with base 2 but -4 can not be written with base 2.

2x = 24 or no real solution

4

Solve by equating the indices.

1

Write

as

(3x)2.


a (2x - 16)(2x + 4) = 0

2

32x

Tutorial

x=4 b 32x - 12 × 3x + 27 = 0

(3x)2 - 12 × 3x + 27 = 0

2

Let 3x = a to make a simpler quadratic equation to solve.

a2 - 12a + 27 = 0, where a = 3x

3

Factorise.

(a - 3)(a - 9) = 0

4

Use the Null Factor Law by making each bracket equal to zero.

a - 3 = 0, a - 9 = 0

5

Solve for a.

6

Substitute back a = 3x.

3x = 3, 3x = 9

7

Write numbers with base 3.

3x = 31, 3x = 32

a = 3, a = 9

Chapter 3


143

c

8

Equate the indices.

1

4x

Rewrite

as

(2x)2

x = 1, x = 2 and

2x + 3

as

2x

×

23.

4x – 2x + 3 + 16 = 0 − 2x × 23 + 16 = 0

c

(2x)2

(2x)2 − 2x × 8 + 16 = 0

2

Rewrite 23 as 8.

3

Let 2x = a to make a simpler quadratic equation to solve.

4

Replace a × 8 with 8a because the coefficient precedes the pronumeral.

5

Factorise.

(a − 4)(a − 4) = 0

6

Use the Null Factor Law.

a − 4 = 0, a − 4 = 0

7

Solve for a.

a = 4 and a = 4

8

The two factors are equal because a2 − 8x + 16 = 0 is a perfect square.

9

Substitute back a = 2x.

2x = 4

10

Write 4 as a number with base 2.

2x = 22

11

Solve by equating the indices.

a2 − a × 8 + 16 = 0 where a = 2x a2 − 8a + 16 = 0

x=2

to always make the right-hand side equal to zero when solving quadratic equations. It is a good idea to substitute your answer back into the original equation to check the accuracy of your work. If the base is not the same and the numbers cannot be written with the same base, then logarithms can be used. It is possible to take the logarithm of both sides of an equation provided the same base is used.

Worked Example 15

Solve for x in the following. Give your answers in exact form using base 10 and correct to 3 decimal places. a 5x = 10 b 2(x + 1) = 12 Think a

b

144

Write a

5x = 10 log10 (5x) = log10 (10)

1

Take the logarithm of both sides to base 10.

2

Use loga (mp) = p loga (m) and loga (a) = 1.

x log10 (5) = 1

3

Divide both sides by log10 (5).

x=

4

Use a calculator to simplify.

x ≈ 1.431, correct to 3 decimal places.

1

Take the logarithm of both sides to base 10.

2

Use loga (mp) = p loga (m) to simplify.

3

Divide both sides by log10 (2).

1 (exact form) log10 (5)

2(x + 1) = 12 log10 (2(x + 1)) = log10 (12)

b

(x + 1) log10 (2) = log10 (12)


( x + 1) =

log10 (12) log10 (2)

4

Use a calculator to simplify the right-hand side.

5

Solve for x.

∴ x=

log10 (12) − 1 (exact form) log10 (2)


Note: Logarithms in bases other than 10 may be used. Inequalities are worked in exactly the same way except that there is a change of sign when dividing or multiplying both sides of the inequality by a negative number. Worked Example 16

Solve the following equations for x, giving your answers both in exact form and correct to 3 decimal places. a 2x > 5 b 0.5x ≤ 1.4 Think

Write/display

1

On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry lines as: solve(2x>5,x) solve(0.5x≤1.4,x) Press ENTER · after each entry.

2

Write the answers in exact form. The calculator defaults to base e when solving exponential equations in exact form. Note: ln (A) ⇔ loge (A). This will be discussed later in this chapter. Note: In part b , the inequality ≤ has been 1 changed to ≥ because log e 2 < 0.

a Solving 2x > 5 for x gives

Press Ctrl / ENTER · after solving each equation to obtain a decimal approximation. Write the answers correct to 3 decimal places.

a Solving 2x > 5 for x gives x > 2.322,

3

x>

log e (5) . log e (2)

b Solving 0.5x ≤ 1.4 for x gives −

x≥

 7 log e    5 log e (2)

.

correct to 3 decimal places.

b Solving 0.5x ≤ 1.4 for x gives x ≥ −0.485, correct

to 3 decimal places.

1. The equations ax = y and 2x = 32 are exponential equations. 2. Write numbers with the same base to help simplify problems. The most commonly used bases are 2, 3 and 5. 3. If the base is the same, equate the indices. 4. If the indices are the same, equate the bases.


145

5. Use the Null Factor Law to solve quadratic equations. 6. A negative number cannot be expressed in index form, for example, -4 cannot be expressed with base 2. 7. a2x = (ax)2 8. Take the logarithm of both sides of an equation or inequation using the same base. 9. Change the sign of an inequality when multiplying or dividing by a negative number. 10. loga (x) > 0 if x > 1 11. loga (x) < 0 if 0 < x < 1 ExErCisE

3C

Exponential equations 1

a

eBook plus

Index form

3x

b 10-x = 1000

= 81

d 7x = 1 49

Digital doc

SkillSHEET 3.1

Solve for x in each of the following.

WE12

2

Digital doc

c 2x × 4x - 1 = 16 4

×

Digital doc

b 5x × 52x + 1 = 625

243

d

6

7

Solve for x in the following. a - 1) = 0 b 22x - 6 × 2x + 8 = 0 2x x c 6 -7×6 +6=0 d 4x - 6 × 2x - 16 = 0 x x e 9 =2×3 +3 b 42x - 20 × 4x = -64

WE15 Solve for x correct to 3 decimal places.

a 2x = 5

b (0.3)x - 1 = 10

c (1.4)2 - x = 6

d 3 × 5x = 27

e 5 × 7x = 1

f 2x × 3x + 1 = 10

WE16 Solve for x correct to 3 decimal places. a 3x > 5 b 22x ≤ 7

d 7x ≥ 0.5 8

33 x + 1 = 81 9x − 2

9)(3x

5 Solve for x in the following. a 25x + 4 × 5x - 5 = 0

SkillSHEET 3.3 Solving indicial equations by equating the bases

3x - 1 =

WE14

(3x

eBook plus

mC The value of x for which 5 × 2x = 1255, correct to 3 decimal places, is:

A 7.971

B 897.750

Digital doc

D 7.972

E 2.059

Solving liner inequations

146

9

c (0.2)x > 3

e (0.4)x > 0.2

eBook plus SkillSHEET 3.4

d 22x - 6 = 1


WE13

3x

c 52 x − 1 = 1 125

b 6x - 2 = 216

a

Solving equations

= 32


eBook plus SkillSHEET 3.2

1 2x

e 243x = 3

a 3 × 2x = 48 3

c

C 897.749

mC The solution to the equation 102x = 3 × 10x + 4 is:

A log10 (-1), log10 (4)

B

-1,

D 0, 0.602

E log10 (4)

4


C 10x + 1, 10x - 1

3d

logarithmic equations using any base The equation loga (y) = x is an example of a general logarithmic equation. Laws of logarithms and indices are used to solve these equations.

WorkEd ExamplE 17

Solve for x in the following equations. a log2 (x) = 3 b log3 (x4) = -16

c log5 (x - 1) = 2

Think a

1

b

c

WriTE

Rewrite using ax = y ⇔ loga (y) = x.

23 = x

x=8

2

Rearrange and simplify.

1

Rewrite using loga

(mp)

2

Divide both sides by 4.

3


4

Rearrange and simplify.

1


2

a log2 (x) = 3

= p loga (m).

b

log3 (x4) = -16 4 log3 (x) = -16 log3 (x) = -4 -

3 4 = x 1 34 1 = 81

x=

c log5 (x - 1) = 2

52 = x - 1

x - 1 = 25 x = 26

Solve for x.

The base of a logarithmic function and the base of an exponential function must be a positive real number other than 1. In the expression loga (x), a ∈ R+\{1}.

WorkEd ExamplE 18

eBook plus

Solve for x in each of the following: a logx (4) = 2

without technology

Think a

 1   125 

b log x 

= − 3 with a CAS calculator.

Tutorial

int-0531

Worked example 18

WriTE/display

1


2

Solve the quadratic equation.

logx (4) = 2 x2 = 4

a

x2 - 4 = 0 (x - 2)(x + 2) = 0 x - 2 = 0 or x + 2 = 0 x = ± 2

Chapter 3


147

3

Check to see if solutions are valid.

x = 2 is the only solution.

This is the only solution. The solution x = −2 is not valid because the base of a logarithmic function must be a positive real number other than 1. b

1

On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(log x 

1   125 

= − 3, x )


2

Solving log x 

Write the solution.

1   125 

= − 3 for x gives x = 5.

Worked Example 19

Solve for x in the following. a log2 (16) = x b log 3  1  = x c log9 (3) = x 3

Think a

b

1

Write

Rewrite using

ax

= y ⇔ loga (y) = x.

2


3

Equate the indices.

1


a log2 (16) = x

2x = 16 2x = 24 x=4

b log 3  1  = x 3

1 3 1 = 1 3

3x =

c

148

−1

3x = 3

1

2


3

Equate the indices.

1


x = −1 c log9 (3) = x

9x = 3

(32)x = 3

2


3

Remove the brackets.

32x = 31

4

Equate the indices.

2x = 1

5

Solve.


x=

1 2

WorkEd ExamplE 20

Solve for x in the following. a log2 (4) + log2 (x) - log2 (8) = 3

b log10 (x) + log10 (x - 3) = log10 (4)

Think a

b

1

WriTE

Simplify the left-hand side. Use loga (mn) = loga (m) + loga (n) and  m log a   = log a (m) − log a ( n) .  n

2


3

Solve.

1

Simplify the left-hand side by using loga (mn) = loga (m) + loga (n).

a log2 (4) + log2 (x) - log2 (8) = 3

 4 × x log 2  =3  8   x log 2   = 3  2 23 =

x 2

x = 2 × 23 =2×8 = 16 b log10 (x) + log10 (x - 3) = log10 (4)

log10x (x - 3) = log10 (4) x(x - 3) = 4

2

Equate the logs.

3

Expand.

4

Solve the quadratic equation.

x2 - 3x - 4 = 0 (x - 4)(x + 1) = 0 x = 4 or x = -1

5

x > 0, x - 3 > 0 because it is not possible to take the logarithm of a negative number, x > 3.

x = 4 is the only solution.

x2 - 3x = 4

1. In a logarithmic equation, the unknown can be: (a) the number, log2 (x) = 5 (b) the base, logx (8) = 3 (c) the logarithm, log2 (4) = x. 2. In the expression loga (x), a is a positive real number other than 1. 3. Use laws of logarithms and indices to solve the equations. 4. Check that all solutions are valid. ExErCisE

3d eBook plus Digital doc

WorkSHEET 3.1

logarithmic equations using any base 1

WE17 Solve for x in the following.

a

i log5 (x) = 2

i i i log10 (x2) = 4 v log4 (2x - 3) = 0

i i log2 (x) = -3 i v log3 (x + 1) = 3 vi log2 (-x) = -5

v i i log5 (1 - x) = 4

Chapter 3


149

i i log4 (x) = −2

i log3 (x) = 4

b

i i i log2 (x3) = 12

i v log5 (x − 2) = 3 vi log3 (−x) = −2

v log10 (2x + 1) = 0 v i i log10 (5 − 2x) = 1

2 WE18 Solve for x in the following. a i logx (9) = 2

2 3

i i i log x  1  = − 3  

i v logx (62) = 2

i logx (16) = 4

i i log x (125) =

8

b

i i log x (25) =

i i i log x  

1  64 

3 4

i v logx (43) = 3

= −2

3 WE19 Solve for x in the following. a

i log2 (8) = x

i i log 5  1  = x  

i i i log4 (2) = x

i v log6 (1) = x

5

v log 1 (2) = x 2

b

 1  16 

i log3 (9) = x

i i log 4 

i i i log8 (2) = x

=x

i v log8 (1) = x

v log 1 (9) = x 3

4 WE20 Solve for x in the following. a

i log2 (x) + log2 (4) = log2 (20) i i i log3 (x) − log3 (2) = log3 (5)

i i log5 (3) + log5 (x) = log5 (18) i v log10 (x) − log10 (4) = log10 (2)

v log4 (8) − log4 (x) = log4 (2) b

i log3 (10) − log3 (x) = log3 (5) i i i log2 (x) + log2 (5) = 1

i i log6 (4) + log6 (x) = 2 i v 3 − log10 (x) = log10 (2)

v 5 − log4 (8) = log4 (x)

5 Solve for x in the following. i log2 (x) + log2 (6) − log2 (3) = log2 (10) i i log2 (x) + log2 (5) − log2 (10) = log2 (3) a i i i log3 (5) − log3 (x) + log3 (2) = log3 (10) i v log5 (4) − log5 (x)+ log5 (3) = log5 (6) v log5 (x) + log5 (x − 2) = log5 (3) b

i log3 (x) + log3 (x + 2) = log3 (8) i i i log5 (x) + log5 (x + 20) = 3

i i log4 (x) + log4 (x − 6) = 2 i v log5 (x + 1) + log5 (x − 3) = 1

v log6 (x − 2) + log6 (x + 3) = 1

6 MC If loga (x) = 0.7, then loga (x2) is equal to: A 0.49

B 1.4

D 0.837

E 0

C 0.35

7 MC If log10 (x) = (a), then (log10 x)2 + log10 (x) − 6 becomes:

150

A (log10 (a))2 + log10 (a) − 6

B a2 + a + 6

D (a − 2)(a + 3)

E log10 (106x3)


C log10 (x3) − 6

8 Solve for x in the following. a (log10 (x))2 + log10 (x) − 2 = 0 (Hint: Let a = log10 (x)) c (log2 (x))2 − 2 log2 (x) = 8 e (log3 (x))2 − log3 (x4) + 3 = 0 g log2 (x4) = (log2 (x))2 i log10 (x2 + 2x − 5) = 1 log10 ( x ) = 4, find x. 9 If log10 (2)

3E

b d f h j

(log10 x)2 − 2 log10 (x) − 3 = 0 (log2 (x))2 + 3 log2 (x) = 4 (log5 (x))2 − log5 (x3) + 2 = 0 log3 (x3) = (log3 (x))2 log3 (x2 − 3x − 7) = 1

Exponential equations (base e)

Euler’s number e, named after an 18th century Swiss mathematician, is a very important number used in problems involving natural growth and natural decay. Like π, it is irrational and has to be approximated: e = 2.718 281 828 459 . . . The number e can be used to find the value of an investment after a period of time, or the temperature of a liquid after it has been cooling. n  1 To find the value of e, take the expression  1 +  and evaluate it for increasing values of n.  n n 1 1  1  n=1  1 + n  =  1 + 1 = 2 n

2

n

3

n

5

n = 2

1  1   1 + n  =  1 + 2  = 2.25

n = 3

1  1   1 + n  =  1 + 3  = 2.370 37

n = 5

1  1   1 + n  =  1 + 5  = 2.48832

n = 10

1 1    1 + n  =  1 + 10 

n = 100

1 1     1 + n  =  1 + 100 

n

n

n

10

= 2.59374 100

1 1    n = 1000  1 +  =  1 +  1000   n n

= 2.70481 1000

 1 1   n = 10 000  1 +  =  1 +  n  10 000 

= 2.716 92 10 000

= 2.71815

n

1  As n increases,  1 +  becomes closer and closer to 2.718 281 or e, or  n n 1  e = lim  1 +  n→ ∞  n An answer given in of e is an exact answer. The laws of indices apply in the same way if e is the base, that is: • ex × ey = ex + y • ex ÷ ey = ex − y • (ex)y = exy • e0 = 1 1 −x • e = x e •

x ey

y

= ex


151

Worked Example 21

Solve for x in e3x = e. Think

Write

1

Write the equation.

e3x = e

2

Write e with a power of 1.

e3x = e1

3

Equate the indices.

4

Solve for x.

3x = 1 1 x= 3

CAS calculators have an ex function, which is treated in the same way as any other number.

Worked Example 22

Solve for x, showing working and with a CAS calculator. Express your answers in exact form and correct to 3 decimal places. a ex = 3 b ex − 3e−x = 2 Think

Write/display

Method 1: Technology-free a

b

1

Write the equation.

2

As loge (e) = 1, take loge of both sides of the equation.

3

Rewrite using loga (xp) = p loga (x).

4

Solve for x.

1

Write the equation. 1 − Write e x as . ex

2

152

ex = 3

a

loge (ex) = loge (3) x loge (e) = loge (3) x = loge (3) ≈ 1.099, correct to 3 decimal places. −x

b ex − 3e

ex

=2 3 − x =2 e

3

Multiply every term by ex.

(ex)2 − 3 = 2ex

4

Make the right-hand side equal to zero.

(ex)2 − 2ex − 3 = 0

5

Let ex = a.

a2 − 2a − 3 = 0 where a = ex

6

Factorise and solve for a.

(a − 3)(a + 1) = 0 a − 3 = 0 or a + 1 = 0 a = 3 or a = −1

7

Substitute ex for a.

ex = 3 or ex = −1

8

Solve for x by taking the log of both sides to base e.

loge (ex) = loge (3)

9

ex = 3 is the only solution because ex = −1 has no real solution.



Method 2 : Technology-enabled 1

On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry lines as: solve(ex = 3,x) solve(ex − 3e−x = 2,x) Press ENTER · after each entry.

2

Write the answers in exact form.

a Solving ex = 3 for x gives x = loge (3). b Solving ex − 3e−x = 2 for x gives

x = loge (3).

3

Write the answers in approximate form, correct to 3 decimal places.

a Solving ex = 3 for x gives x = 1.099, correct to

3 decimal places.

b Solving ex − 3e−x = 2 for x gives x = 1.099,

correct to 3 decimal places.

1. Euler’s number e is an irrational number, which is approximated to 2.718 (correct to 3 decimal places). 2. Evaluate e by using the ‘ex’ button on the calculator. 3. The number e is an exact answer. Use the calculator to give an approximation if required. 4. The laws of indices apply in the same way if e is the base. 5. On a calculator, use the LN button to take the log of a number to base e. The LOG button defaults to base 10 if not specified. 6. loge (x) = ln (x). 7. ex > 0, that is, ex = −1 has no real solution. Exercise

3E

Exponential equations (base e) 1 Evaluate the following, giving your answer correct to 3 decimal places. 1

1

4

a e2

b e4

c e 2

d e 3

e

f

g ln (4)

h ln (5)

i loge (1.5)

j loge (3.6)

5

e

2 WE21 a ex = e

Solve for x in each of the following. b ex = e2 c ex − 2 = e4 1 − 1 g e3 x + 6 = e e e x + 1 = f e x − 2 = 2 e e

e

−1

d e2x = e

h e 2 x − 1 = e3


153

3 WE22 Solve for x in each of the following, giving your answer correct to 3 decimal places. a ex = 2

b ex = 5

c e x =

e ex = 1.3

f ex = 2.6

g 2ex = 6

1 2

d e x =

1 4

h 3ex = 12

4 Solve for x in each of the following, giving your answer correct to 3 decimal places. a (ex − 1)(ex + 2) = 0 b (e−x − 2)(e2x − 3) = 0 c (3e−x − 2)(2ex − 1) = 0 d (ex)2 − ex = 0

e (ex)2 − e × ex = 0

g 6 − 11ex + 3e2x = 0

h 18 − 23ex + 7e2x = 0

f (ex)2 − 7ex + 10 = 0

5 Solve for x in each of the following, giving your answer correct to 3 decimal places. a ex − 4e−x = 0

b ex − 15e−x − 2 = 0

c 5ex − 12e−x − 11 = 0

d 3ex + 6e−x − 11 = 0

e 4ex + 6e−x − 11 = 0

f ex + 2e−x = 3

6 Solve for x in each of the following, giving your answer correct to 3 decimal places. a ex > 1

b ex < e

c ex < 2

d e2x ≥ 4

e ex + 1 ≤ 6

f e1 − x ≤ 10

−x

g e > 0.75 7 If y = Ae−kt, and y = 19.6 when t = 2, and y = 19.02 when t = 5, find the value of the constants A and k. Give your answers correct to 2 decimal places. 8 For a body that has a higher temperature than its surroundings, Newton’s Law of Cooling is given by the formula θ = θ0e−kt, where θ is the difference between the temperature of the body and its surroundings after t minutes and θ 0 is the difference between the original temperature of the body and its surroundings. If the temperature of a freshly poured cup of coffee is 90 °C in a room with a constant temperature of 18 °C, and it cools to 65 °C after 10 minutes, find the value of k. Give your answer correct to 2 decimal places.

3F

Equations with natural (base e) logarithms CAS calculators have an LN function that can be used to find the log to base e in the same way that the LOG key is used to find the log of a number to base 10. CAS calculators default to base 10 when using the log key. In general, aloga(x) = x, where a ∈ R+\{1}. The laws of logarithms apply in the same way for base e as they do for base 10. ex = y ⇔ loge (y) = x.

Worked Example 23

Solve for x, giving your answer both in exact form and correct to 3 decimal places, given that loge (x) = 3. Think 1

154

Rewrite using ex = y ⇔ loge (y) = x.

Write/display

loge (x) = 3 e3 = x


2

Write the answer in exact form.

3

On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(ln (x) = 3,x) Then press ENTER ·.

4

Write the solution in approximate form, correct to 3 decimal places.

∴ x = e3

Solving loge (x) = 3 for x gives x = 20.086, correct to 3 decimal places.

Worked Example 24

Solve for x giving your answer correct to 3 decimal places where appropriate. a loge (3) = loge (x) b loge (x) + loge (3) = loge (6) Think a

b

Write a loge (3) = loge (x)

Since the base is the same, equate the numbers. 1 2 3

x=3

Rewrite using loge (mn) = loge (m) + loge (n).

b loge (x) + loge (3) = loge (6)

loge (3x) = loge (6) 3x = 6

Equate the number parts.

x=2

Solve for x.

Exercise

3F

1. ex = y ⇔ loge (y) = x

2. e log

4. loge (1) = ln (1) = 0

5. loge (e) = ln (e) = 1

e

( x)

=x

3. 10 log

10

( x)

=x

6. loge (0) = ln (0) is undefined.

7. loge (mn) = loge (m) + loge (n) or

ln (mn) = ln (m) + ln (n)

 m 8. log e   = log e (m) − log e ( n)  n

or

 m ln   = ln (m) − ln ( n)  n

9. loge (mp) = p loge (m)

or

ln (mp) = p ln (m)

Equations with natural (base e) logarithms 1 WE23 Solve for x in each of the following giving, exact answers. a loge (x) = 1 b loge (x) = 2 c loge (x) = −2

d loge (x) = −1


155

2 Solve for x, giving exact answers when appropriate, otherwise, correct to 3 decimal places. a ln (2x) = 2 b ln (3x) = 1 c ln (x3) = 3 d ln (x2) = 2

e ln (x2) = 0.4

f ln (x3) = 0.9

g ln (x - 1) = -1

h ln (2x + 1) = -2

3 WE24 Solve for x, giving exact answers when appropriate, otherwise, correct to 3 decimal places. a loge (x) = loge (2) b loge (x) = loge (5) c loge (x) + loge (3) = loge (9)

d loge (x) + loge (2) = loge (8)

e loge (x) - loge (5) = loge (2)

f 1 + loge (x) = loge (6)

4 Solve for x, giving exact answers. a loge (x) + loge (5) - loge (10) = loge (3)

b 2 loge (3) + loge (x) - loge (2) = loge (3)

c 3 loge (2) + loge (x) - loge (4) = loge (5)

d loge (4) + loge (3) - loge (x) = loge (2)

e loge (x) + loge (x + 1) = loge (2)

f ln (x + 1) + ln (2x - 1) = ln (5)

5 mC If ln (y) = ln (x) + ln (a), then an equation relating x and y that does not involve logarithms is: x a A y=x+a B y = ax C y=x-a D y= E y= a x 6 mC In the equation 2 loge (x) - loge (3x) = a, x = A 3ea B -a C 3a

D loge (6a)

E no solution

7 Write the following equation without logarithms and with y as the subject. 2 loge (x) + 1 = loge (y) 8 If loge (x) = a and y = ea, express y in of x. 9 Solve for x the equation eln (x) = 2. eBook plus Digital doc

WorkSHEET 3.2

3G eBook plus Interactivity

int-0248 Inverses

156

10 Five grams of a radioactive substance is decaying so that the amount, A grams, that is left after t days, is given by the formula A = 5e-kt. a Find the value of A when the number of grams of the radioactive substance has been halved. b Rewrite the equation with the new value of A. c Rearrange the equation so that t is the subject of the equation. d If k = 0.005, find how long it will take for the number of grams of the radioactive substance to be halved. Give your answer correct to the nearest day.

inverses Inverse operations are opposite operations. Addition and subtraction are inverse operations and multiplication and division are inverse operations. Squaring and taking the square root are also inverse operations. The equation of the inverse of the function y = ex can be found by interchanging the x and y so that y = ex becomes x = ey. Using ax = y ⇔ loga (y) = x, x = ey becomes loge (x) = y or y = loge (x). Therefore y = ex and y = loge (x) are the equations of inverse functions. Thus two important properties follow: aloga(x) = x, x ∈ R+ and loga a(x) = x, x ∈ R.


Worked Example 25

Calculate the inverse of y = 3ex + 1: a without the use of technology b using a CAS calculator. Think a

b

Write/display

1

Interchange x and y to write the inverse equation.

2


3

In order to make y the subject, begin by rewriting the equation using ax = y ⇔ loga (y) = x.

4

Make y the subject.

1


2

On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(x = 3ey + 1,y) Then press ENTER ·.

3

Write the equation of the inverse.

y = 3ex + 1 Inverse is x = 3ey + 1 ey + 1 =

x 3

 x log e   = y + 1  3  x y = log e   − 1  3 x = 3e y + 1

The inverse of y = 3ex + 1 is  x y = log e   − 1 where x > 0  3

Worked Example 26

Calculate the inverse of f(x) = 2 log10 (x − 1) + 1: a without the use of technology b using a CAS calculator. Think a

1 2

Write

Interchange x and y to write the inverse equation. In order to make y the subject, begin by subtracting 1 from both sides.

3


4


Let y = 2 log10 (x − 1) + 1 Inverse is x = 2 log10 (y − 1) + 1 2 log10 (y − 1) = x − 1 log10 ( y − 1) = 10

x −1 2

x −1 2

= y −1


157

b

5

Add 1 to both sides.

6

Write the answer.

f −1 ( x ) = 10

1


x = 2 log10 (y - 1) + 1

2

On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(x = 2 log10 (y - 1) + 1,y) Then press ENTER ·.

y = 10

x −1 2

x

Write the equation of the inverse.

4

The solution from the CAS can be simplified to obtain the solution found in a . Rewrite the answer with a negative power.

y = 10

5

Add the powers when multiplying indices of the same base.

y = 10

6

Simplify the powers and write the answer.

1 2

+

10 2 ∴y = 10

3

∴y

−

x −1 2

× 10 2

−

1+

+1

+1 x

1

x −1 = 10 2

f −1 ( x )

+1

x 1 + 2 2

+

1 2

+1

+1

+1

x −1 = 10 2

+1

1. The equation y = loge (x) is an inverse function of y = ex. 2. To find the inverse of a function, interchange x and y and then make y the subject. ExErCisE

3G

inverses 1

eBook plus Digital doc

SkillSHEET 3.5 Inverses

158

WE25 Find the inverse of the following. a y = 2ex b y = ex + 1 c y = ex - 1

d y = e2x - 1

2 Find the equation of the inverse of the following. a y = 2 + ex b y = 2 - ex d y = 2 + ex + 1 e y = 3 - 2ex - 2 3

WE26 Find the inverse of the following. a f (x) = 2 loge (x) b f (x) = loge (x + 1) d f (x) = loge (2x - 1) e f (x) = loge (2 - x)


e y = e2 - x

f y = e2 - 3x

c y = 1 - 2ex f y = 2 - 3ex + 1 c f (x) = loge (x - 1) f f (x) = loge (2 - 3x)

4 Find the equation of the inverse of the following. b y = 2 - loge (x) a y = 2 + loge (x) d f (x) = 2 - ln (x - 1) 5

D

7

e y = 3 + 2 ln (x - 1)

x −1 5

3

x −1 e 5

B e

+2 −2

E

3

x −1 5

3

x+2 e 3

5

C e

−2

x −1 5

+2

3

−1

mC If y = 5e 2x + 1 - 1, the equation of the inverse is:

A

1  x + 1 +1 ln 2  5 

B

1  x + 1 −1 ln 2  5 

D

1  x + 1 1 + ln 2  5  2

E

1  x + 1 1 − ln 2  5  2

(2x)

mC If eloge

A x2

C 5 ln ( x − 1) + 1 2

= y , then y equals: B loge (2x)

C e2x

E 2ex

D 2x

3h

f f (x) = 1 - 3 ln (x + 2)

mC If y = 5 loge (3x - 2) + 1, the equation of the inverse is:

A e

6

c y = 2 + 3 loge (x)

literal equations An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. It does not have a numerical solution. The solution will be expressed in of the other pronumerals, in this case a and k, often called parameters. For this equation, the solution is: kx = ln (a) ∴ x =

1 ln (a), k ≠ 0, a ∈ R + k

WorkEd ExamplE 27 -kx

Solve ekx = 5 + 2e

eBook plus

for x, where k ∈ R\{0}.

Think

Tutorial

WriTE/display

2 e kx

1

Rewrite the equation with positive powers.

e kx = 5 +

2

Multiply both sides by ekx.

(ekx)2 = 5ekx + 2

3

Let y = ekx and make the right-hand side zero to obtain a quadratic equation in of y.

y2 - 5y - 2 = 0

4

Solve for y using the quadratic formula.

y= y=


− ( − 5) ±

( − 5)2 − 4 × 1 × − 2 2 ×1

5 ± 33 2

∴y =

Chapter 3

5 + 33 or 5 − 33 y= 2 2


159

5

Substitute ekx for y.

e kx =

5 + 33 or kx 5 − 33 e = 2 2

6

Only the first solution is valid as ekx > 0.

e kx =

5 + 33 2

7

Re-write the exponential equation in logarithmic form using ax = y ⇔ loga (y) = x or take logs of both sides. Divide both sides by k.

8

Write the solution and state the restriction for k.

9

On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: − solve(e k × x = 5 + 2e k × x,x) Note: You must put in a multiplication sign between the k and the x. Then press ENTER ·.

10

Write the answer.

 5 + 33  kx = log e  2   ∴x =

 5 + 33  1 log e  , k ∈ R \ {0}. k 2  

∴x =

 33 + 5  1 log e  k 2  

Worked Example 28

1 Solve for x, given that log 2 ( x ) − 5 log 2 ( p) = log 2 ( 6 ) where p > 0, 2 a manually b with a CAS calculator. Think a

160

1

Write/display

Simplify the left-hand side using p loga (m) = loga (mp) and  m log a (m) − log a ( n) = log a    n

 1 log 2  x 2  − log 2 ( p5 ) = log 2 (6)    x log 2  5  = log 2 (6)  p  x =6 p5

2

Equate both sides.

3

Multiply both sides by p5.

x = 6 p5

4

Square both sides to obtain x.

x = (6p5)2

5

Write the solution and state the restriction for p.

∴ x = 36p10 where p > 0


b

1

On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: 1  solve log 2 ( x ) − 5 log 2 ( p) = log 2 (6),x  2  Then press ENTER ·.

2

Write the answer. Note that the CAS gives the incorrect restriction p ≥ 0. The correct restriction is p > 0 for log2 (p) to be defined.

∴ x = 36p10

Some equations can only be solved using a CAS calculator, that is, no algebraic method exists. The following example demonstrates this. Worked Example 29

Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ex = x3 b loge (x) = x − 2 Think

Write/display

1

On a Calculator page, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry lines as: solve(ex = x3,x) solve(ln (x) = x − 2,x) Press ENTER · after each entry.

2

Write the solutions correct to 3 decimal places.

a Solving ex = x3 for x gives x = 1.857 or x = 4.536. b Solving loge (x) = x − 2 for x gives x = 0.159 or

x = 3.146.

1. An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. 2. Literal equations do not have numerical solutions. 3. The solution of a literal equation is expressed in of the other pronumerals, in this case a and k, often called parameters.


161

ExErCisE

3h

literal equations 1 Solve 5e2x = a for x, where a ∈ R+. 2 If log3 (D) = cy + log3 (Z), solve for D. 3 Solve emx + n = 3k for x, where m ∈ R\{0} and k ∈ R+. 4 Solve for q given that 2 log3 (p + 5q) = 4.  x2   z2 x5   y4   y  5 Prove that log10  3 2  + log10  3  − log10  3 4  − log10  4  = 0. z x  y z   z   z x 6 Consider the exponential equation 9xb × 273a = 81. a Find x in of a and b, where a ∈ R, b ∈ R\{0}. b Hence find the value of x if a = 2 and b = -3. 7 Solve 42x - b = 20 for x, where b ∈ R. 8 Solve 2x - 1 = 3x + a for x, where a ∈ R. 9 If y = m + Rebx, solve for x. 10 Solve for a given that (log2 (5a))2 = 16b2. a 11 WE27 Solve e kx = 3 + kx for x, where a ≥ 0, k ∈ R\0. e 1

12 WE28 Solve for x given that 2 log 4 ( x ) − 3 log 4 ( y ) = log 4 (3). 13 Solve for b given that 2 loge (a) - 5 loge (b) - 2 = 0, where a, b ∈ R+. 14 WE29 Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ex = 3x b x+2=e x c x2 - 1 = e2x 15 WE29 Solve the following equations using a CAS calculator. Give your answers correct to 3 decimal places. a ln (x) = 2 - x b ln (x - 2) = x - 4 c x2 - 1 = ln (2x) 16 Solve for x given that log7 (x) = log4 (p).

3i eBook plus eLesson

eles-0091 Exponential and logarithmic modelling

162

Exponential and logarithmic modelling Exponential and logarithmic functions can be used to model many real situations involving natural growth and decay. Continuous growth and decay can be modelled by the equation A = A0ekt, where A0 represents the initial value, t represents the time taken and k represents a constant. For continuous growth, k is positive, but for continuous decay, k is negative. Logarithms to base 10, often called common logarithms, are used in scientific formulas for measuring the intensity of earthquakes, the acidity of solutions and the intensity of sound.


WorkEd ExamplE 30

eBook plus

In the town of Ill Ness, the number of cases of a particular disease, Tutorial D, can be modelled by the equation D = D0ekt, where t is the time in years. int-0533 Using available medication the number of cases is being reduced Worked example 30 by 20% each year. There are 10 000 people with the disease today. a How many people will have the disease after one year? b Find the value of k correct to 3 decimal places. c Write the equation substituting values for k and D0. d Find how long it would take for the number of people with the disease to be halved. Give your answer correct to the nearest year. e How long would it take for the number of people with the disease to be reduced to 100? Give your answer correct to the nearest year. Think a

b

d

a (100 - 20)% = 80%

1

Find the percentage of people with the disease after one year.

2

Find 80% of the original number.

80% of 10 000 = 8000

3

Write a sentence.

Therefore, 8000 people will have the disease after one year.

1

Substitute t = 0 and D = 10 000 into the given equation.

2

c

WriTE

Substitute t = 1 and D = 8000 into [1], and solve for k.

Use the given equation D = D0ekt. 1

Substitute 5000 for D.

2

Simplify by dividing both sides by 10 000.

3

Take loge of both sides.

4

Solve for t.

b

D = D0 ekt When t = 0 and D = 10 000, 10 000 = D0 ek × 0 10 000 = D0 × 1 10 000 = D0 So D = 10 000ekt

[1]

When t = 1 and D = 8000, 8000 = 10 000ek × 1 8000 = ek 10 000 0.8 = ek loge 0.8 = loge (ek) -0.223 = k log (e) e -0.223 = k × 1 k = -0.223 -0.223t

c D = 10 000e

-0.223t

d D = 10 000e

When D = 5000, 5000 = 10 000e 0.223t -0.223t

0.5 = e

loge (0.5) = -0.223t t=

log e (0.5) − 0.223

≈ 3.108 (3 decimal places) 5

Write a sentence.

It would take about 3 years.

Chapter 3


163

e

−0.223t

D = 10 000e

e

1

Write the equation.

2

Substitute 100 for D.

3

Simplify by dividing by 10 000.

4


5

Solve for t.

6

Write the answer in a sentence.

When D = 100, − 100 = 10 000e 0.223t −0.223t

0.01 = e

loge (0.01) = −0.223t t ≈ 20.651 (3 decimal places) It would take about 21 years.

1. Make a note of all the information that has been given. 2. Give your answers to the correct number of decimal places. 3. Continuous growth and decay is modelled by the equation A = A0ekt, A0 represents the initial value (that is, when t = 0) and k represents a constant.

Exercise

3I

Exponential and logarithmic modelling 1 WE30 Changing δ-gluconolactone into gluconic acid can be modelled by the equation − y = y0e 0.6t, where y is the number of grams of δ-gluconolactone present t hours after the process has begun. Suppose 200 grams of δ-gluconolactone is to be changed into gluconic acid. a Find the value of y0. b Write the equation replacing y0 with your answer. c How many grams of δ-gluconolactone will be present after 1 hour? Give your answer correct to the nearest gram. d How long will it take to reduce the amount of δ-gluconolactone to 50 grams? Give your answer correct to the nearest quarter of an hour. −

2 The decay of radon-222 gas is given by the equation y = y0e 0.18t, where y is the amount of radon remaining after t days. When t = 0, y = 10 g. Give all answers to the nearest whole number. a Find the value of y0. b Write the equation substituting your value of y0. c What will be the mass after 1 day? d How many days will it take for the mass to reach 1 g? 3 The equation y = A + B loge (x) relates two variables x and y. The table below shows values of x and y. x y a b c d

164

1 3

2 4.386

3 m

Find the value of A and B correct to the nearest whole number. Write the equation relating x and y substituting values for A and B. Using your new equation, find the value of m correct to 3 decimal places. If y = 7.6, find x correct to the nearest whole number.


4 An amount of $1000 is invested in a building society where the 5% p.a. interest paid is compounded continuously. The amount in the after t years can be modelled by the equation A = A0ert, where r is the continuous interest rate. a Find the value of A0 and r. b Write the equation substituting values of A0 and r. c Find the amount in the bank after i 1 year ii 10 years. Give your answer correct to the nearest dollar. d How long will it take for the investment to double in value? Give your answer to the nearest year. 5 The number of people living in Boomerville at any time, t years, after the first settlers arrived can be modelled by the equation P = P0ekt. Suppose 500 people arrived on 1 January 1850, and by 1 January 1860 there were 675 people. a What is the value of P0? b Find the value of k correct to 2 decimal places. c Write the equation substituting values for P0 and k. d What will be the population on 1 January 1900? Give your answer to the nearest 10 people. e When will the population be 2000? 6 A cup of soup cools to the temperature of the surrounding air. Newton’s Law of Cooling can be − written as T − TS = (T0 − TS)e kt, where T is the temperature of the object after t minutes, and TS is the temperature of the surrounding air. The soup cooled from 90 °C to 70 °C after 6 minutes in a room with an air temperature of 15 °C. a Find the values of TS, T0 and k correct to 2 decimal places. b Write the equation substituting the values for TS, T0 and k. c Find the temperature of the soup after 10 minutes. Give your answer to the nearest degree. d How long would it take for the soup to be 40 °C? Give your answer to the nearest minute. e If the soup is placed in a refrigerator in which the temperature is 2 °C, how long will it take for the soup to reach 40 °C? Use the same value of k and give your answer to the nearest minute. 7 The diameter of a tree for a period of its growth can be modelled by the equation D = D0ekt, where t is the number of years after the beginning of the period. The diameter of the tree grew from 50 cm to 60 cm in the first 2 years that measurements were taken. a Find the values of D0 and k. b Write the equation using these values. c How much will it have grown in the first 5 years? Round to the nearest centimetre. d How long will it take for the tree’s diameter to double? Round to the nearest year. −

8 The decay of a radioactive substance can be modelled by the equation M = M0e kt, where M grams is the mass of the substance after t years. After 10 years the mass of the substance is 98 grams and after 20 years the mass is 96 grams. a What was the mass of the substance initially? Give your answer to the nearest gram. b Find the value of k. Give your answer to 3 decimal places. c Write the equation using these values. d Find the mass of the substance after 50 years. e How long would it take for the mass to be halved? 9 The number of bacteria present in a culture at any time, t hours, can be modelled by the equation N = N0ekt. a If the original number is doubled in 3 hours, find k correct to 2 decimal places. b Write the equation substituting the value of k. c Find the original number of bacteria if there were 2500 bacteria after 4 hours. Give the answer correct to the nearest thousand.


165

d Write the equation substituting your value for the original population. e Find the number of bacteria present after 8 hours. Give your answer correct to the nearest thousand. 10 The intensity of light d metres below the surface of the sea can be modelled by the equation − I = I0e kd. Divers in the Sea of Loga have found that the intensity of light is halved when a diver is 5 metres below the surface of the water. a Find the value of k correct to 4 decimal places. b Write the equation substituting the value of k. c Find the percentage of light available at a depth of 10 metres. d If artificial light is necessary when the intensity of light is less than 0.1 of the intensity at the surface (I < 0.1I0), find how deep a diver can go before artificial light is necessary.

166


Summary Index laws

• ax × ay = ax + y • ax ÷ ay = ax − y x y xy • (a ) = a • a0 = 1 1 1 − • a x = x and − x = a x a a x

1

• a y = y a and a y = y a x • ax = y ⇔ loga (y) = x Logarithm laws

loga (1) = 0 loga (a) = 1 loga (0) is undefined. loga (mn) = loga (m) + loga (n) m • log a   = log a (m) − log a ( n)  n • loga (mp) = p loga (m) log a ( N ) • log b ( N ) = log a (b) • • • •

Exponential equations

• To solve exponential equations: 1. write all with the same base, write with the smallest possible base or take the logarithm of both sides of the equation 2. then solve the equation. • A negative number cannot be expressed in index form. • If 0 < x < 1, then loga (x) < 0 and if x > 1 then loga (x) > 0. • It is not possible to take the logarithm of a negative number. Exponential equations (base e) n

 1 • Euler’s number e = lim  1 +  = 2.718 281 828459 ... h →∞  n • The laws of indices and logarithms apply in the same way when using e. Equations with natural (base e) logarithms

• To solve logarithmic equations use the laws of logarithms and indices. Inverses

• • • •

The equation y = loge (x) is an inverse function of y = ex. To find an inverse, interchange x and y. a log ( x ) = x, x ∈ R + log a (a x ) = x, x ∈ R a

Literal equations

• An equation such as ekx = a, where k ∈ R and a ∈ R+, is called a literal equation. • Literal equations do not have numerical solutions. • The solution of a literal equation is expressed in of the other pronumerals, in this case a and k, often called parameters.


167

chapter review 2 2 log3 (x) + 4 log3 (x) − log3 (x6) is equal to:

Short answer −2

3 2  − 4 1 Simplify 4 x 5 y 3 ×  2 x 3 y 5  , leaving your   answers with positive indices.

A 0

 6x  B log3  x 6 

2 If log2 (5) = 2.321 and log2 (9) = 3.17, find

C log3 (6x - x6)

 8x  D log3  x 6 

E 6log3 (x - x6)

log 2  95  . 3 Solve 3 × 2x − 7 = 17 for x. log 2 (32) . log 2 (8) 1 5 Solve log x (2) = for x. 3 6 If 4e(2 − x) = 128, find x, giving your answer in exact form. 4 Evaluate

7 Solve for x in loge (5) + loge (x) − loge (2) = loge (10). 8 Find the rule of the inverse function to y = 3e2x − a. 9 Find the rule of the inverse function of y = loge (1 − x) + 3. 10 Solve the equation loge (3x + 5) − loge (2) = 2 for x.

Exam tip Common mistakes include cancelling out the logarithms, adding to obtain 3x + 7, or multiplying to get 6x + 5. [Assessment report 1 2007]

[© VCAA 2007]

11 Solve 6e3x = k for x, where k ∈ R+. 12 Solve 3eax + b − 6k = 0 for x and state the necessary restrictions for the parameters a, b and k. 13 Solve for x given that 4 log2 (ax + b) = 12. 14 Solve for x given that log2 (x) = y + log2 (z). State the restrictions for the parameters. Multiple choice

1 If a > 1, the solution of x for the equation x = a2 is: A 1 B a negative number less than 1 C a positive number less than 1 D a negative number greater than 1 E a positive number greater than 1

168

3 The solution of the equation 3e2x = 4 is closest to: B −0.405 A −0.406 C 0.143 D 0.144 E 0.575 [© VCAA 2005] 4 Evaluated to 3 decimal places, log3 (24) is: A 2.892 B 2.893 C 0.345 D 0.346 E 1.380 5 The solution set of the equation (2x − 1)(22x − 4) = 0 over R is: A {0, 1} B {0, 2} C {1, 2} D {1, 4} E {2, 4} −

6 The solution set of the equation ex − 12e x = −4 over R is: A loge (2), loge (6) B ex − 2, ex + 6 − C 2, 6 D loge 2 E loge 6 7 If loge (x) = a, then e2a + 3ea − 2e−a is equal to: 2 a 2 B x 2 + 3 x − x

A a 2 + 3a −

C 2 log e (a) + 3 log e (a) −

2 log e (a)

d log e ( x 2 ) + 3 log e ( x ) −

2 log e ( x )

E (ea + 2)(e + 1) 8 If loge (2x) = a, then x is equal to: A 2ea ae D 2


B 2ae E e2a

C

ea 2

9 If ex + 4 = e2x − 1, then x is equal to: − 5 B 5 A 3 D -5

C e5 −

14 If y = loga (7x − b) + 3, then x is equal to: 1 A 1 a y − 3 + b B (a y − 3) + b 7 7 b 1 y−3 + b) D a y − 3 − C (a 7 7 y−3 E log a (7 − b) [© VCAA 2007]

5

E e 3 10 The solution(s) to the equation 2 ln (x) = ln (x + 4) + ln 2 is/are: B 2, −4 A -2, 4 C 1 D 2 E 4 11 If 2a =

x2 , then y is equal to: y

x 2a x2 C a 2 A

 x B    2a  x D 22 a

2

x2 4a 12 The equation which is the inverse of y = ex − 1 is: A y = loge (x) − 1 B y = loge (x − 1) C y = loge (x) + 1 D y = loge (x + 1) − E y = e x − 1 E

13 The air pressure P cm of mercury at h km above sea level can be modelled by the equation − P = 76e 0.13h. One kilometre above sea level the pressure has: A increased by approximately 9 cm B decreased by approximately 9 cm C increased by approximately 41 cm D decreased by approximately 41 cm E neither increased nor decreased significantly

15 If 2 loge (x) − loge (x + 2) = 1 + loge (y), then y is equal to: x2 A 10( x + 2) 2x −1 B x+2 x2 C x+2 2x D x+2 x2 E e( x + 2) [© VCAA 2003] 16 If 3log ( x + 4 ) = y , then y equals: A log (x + 4) B x + 4 C (x + 4)3 D 3 log (x + 4) E −4 3

17 The solution set of the equation e4x − 5e2x + 4 = 0 over R is: A {1, 4} B {−4, −1} C {−2, −1, 1, 2} D {−loge (2), 0, loge (2)} E {0, loge (2)} [© VCAA 2007]

Extended response

 a 1 log e   = x . If loge (a) = 0.6932, find the value of x, giving your answer correct to 2 decimal places.  3 x x 2 x3 + + ... 2 e x = 1 + + 1! 2! 3! a Write the next 3 . b Substitute x = 1 in the equation using the first 7 . c Show that e ≈ 2.7182. 3 The apparent brightness of a star can be found using the formula B = 6 - 2.5 log10 A, where A is the actual brightness of that star. Find the apparent brightness of a star with actual brightness of 3.16.  a 4 Earthquake magnitude is often reported on the Richter scale. The magnitude, M, is given by M = log10   + B, T where a is the amplitude of the ground motion in microns at the receiving station, T is the period of the seismic wave in seconds, and B is an empirical factor that allows for the weakening of the seismic wave with the increasing distance from the epicentre of the earthquake.


169

Find the magnitude of the earthquake if the amplitude of the ground motion is 10 microns, the period is 1 second and the empirical factor is 6.8. 5 Five grams of a radioactive substance is decaying so that the amount, A grams, that is left after t days, is given by the formula A = 5e kt. a Find the value of A when the number of grams of the radioactive substance has been halved. b Rewrite the equation with the new value of A. c Rearrange the equation so that t is the subject of the equation. d If k = 0.005, find how long it will take for the number of grams of the radioactive substance to be halved. Give your answer correct to the nearest day. 6 A school in the suburb of Bienvenue opened with 30 students in February 1995. It has been found for the first years after opening that the number of students enrolled in the school t years after opening can be modelled by the equation N = N0ekt. There were 45 students enrolled in February 1996. a Find the values of N0 and k. b Write the equation substituting the values for N0 and k. c How many students will there be 5 years after the opening? d How many years will it take for the school to have 1000 pupils? Another school in the suburb of Enbaisse has a declining student population. The number of students enrolled at any one time can be modelled by the equation E = E0e rt. There are 1000 students enrolled in February 1995 and 900 in February 1996. e Find the values of E0 and r. f Write the equation substituting the values for E0 and r. g How many students will be enrolled after 5 years? h How many years will it take for the two schools to have approximately the same number of pupils? i What will the population be then? Use the calculator value in the working and do not round off until the final answer. -kx

7 Solve ekx = 4 + ke

for x, where k ∈ R+.



170


eBook plus

aCTiviTiEs


• 10 Quick Questions: Warm up with ten quick questions on exponential and logarithmic equations. (page 132) 3A

The index laws

Tutorial

• WE 5 int-0528: Watch a worked example on writing expressions with positive indices. (page 135) 3B

Logarithm laws

Tutorial

• WE 10 int-0529: Watch a worked example on simplifying logarithmic expressions. (page 139) 3C

Exponential equations

Tutorial

• WE 14 int-0530: Watch a worked example on solving exponential equations. (page 143) Digital docs

• SkillSHEET 3.1: Practise writing expressions using index form. (page 146) • SkillSHEET 3.2: Practise solving equations. (page 146) • SkillSHEET 3.3: Practise solving indicial equations by equating the bases. (page 146) • SkillSHEET 3.4: Practise solving linear inequations. (page 146) 3D

Logarithmic equations using any base

Tutorial

• WE 18 int-0531: Watch a worked example on solving logarithmic equations using a CAS calculator. (page 147) Digital doc

• WorkSHEET 3.1: Simplify exponential and logarithmic expressions, and solve logarithmic and exponential equations. (page 149) 3F

Equations with natural (base e) logarithms

3G

Inverses

Interactivity

• Inverses int-0248: Consolidate your understanding of inverses using the interactivity. (page 156) Digital docs

• SkillSHEET 3.5: Practise finding inverses. (page 158) 3H

Literal equations

Tutorial

• WE 27 int-0532: Watch a worked example on solving literal equations. (page 159) 3I

Exponential and logarithmic modelling

eLesson

• Exponential and logarithmic modelling eles-0091: Learn about how exponential and logarithmic modelling is used. (page 162) Tutorial

• WE 30 int-0533: Watch a worked example on exponential modelling. (page 163) Chapter review Digital doc

• Test Yourself Chapter 3: Take the end-of-chapter test to test your progress. (page 170) To access eBookPLUS activities, log on to www.jalus.com.au

Digital doc

• WorkSHEET 3.2: Solve logarithmic and exponential equations and application questions. (page 156)

Chapter 3


171

4

4A Graphs of exponential functions with any base 4B Logarithmic graphs to any base 4C Graphs of exponential functions with base e 4D Logarithmic graphs to base e 4E Finding equations for graphs of exponential and logarithmic functions 4F Addition of ordinates 4G Exponential and logarithmic functions with absolute values 4H Exponential and logarithmic modelling using graphs

Exponential and logarithmic graphs AREAS OF STUDY

• Graphs and identification of key features of graphs of the following functions: – exponential functions: exponential equations, y = ax – logarithmic functions, y = loge (x) and y = log10 (x), the relationship a = ek, where k = loge (a) • Key features of functions, including: – axis intercepts – domain (including maximal domain) and range – asymptotic behaviour – symmetry • Transformations for y = f (x) to y = Af (n(x + b)) + c, where A, n and c ∈ R and f is an exponential or logarithmic function and relation between the graph of the original

• •

• •

•

function and the graph of the transformed function Families of transformed functions for a single transformation parameter Graphs of sum, difference, product and composite functions of f and g, where f and g are exponential or logarithmic equations such as y = x2ekx, k ∈ R Graphical and numerical solutions of equations Recognition of the general form of possible models for data presented in graphical or tabular form using exponential and logarithmic functions Applications of simple combinations of exponential or logarithmic functions and interpretation of features of the graphs of these functions in modelling practical situations eBook plus

4A

Graphs of exponential functions with any base

Digital doc

10 Quick Questions

The function f (x) = ax is an exponential function where a is a positive, real number which is not 1 (that is, a ∈ R+\{1}). In this chapter, a will take the values of 2, 10 and Euler’s number e. An exponential function f (x) can be written f: R → R where f (x) = ax and a ∈R+\{1}. In the previous chapter it was seen that if x has a positive coefficient, f (x) is an increasing function and may be used to describe physical growth. Examples of this include population and bacterial growth, and increases in investment values, light intensity and temperature. If x has a negative coefficient, f (x) is a decreasing function and may be used to describe physical decay. Examples of this include population and bacterial decline, radioactive decay, temperature cooling and decreases in light intensity and vehicle values. Graphs with a positive coefficient of x are considered first.

172


Graphs of f (x) = 2x and f (x) = 10x The graphs of f (x) = 2x and f (x) = 10 x are shown below. y 12 10 8 6 (0, 1) 4 Asymptote 2 (1, 2) y= 0 x 0 −3 −2 −1 1 2 3

y 12 (1, 10) 10 8 6 4 Asymptote 2 (0, 1) y= 0 x −3 −2 −1 0 1 2 3

f (x) = 2x f (x) = 10x These graphs have a number of common features: 1. They both cross the y-axis at the point (0, 1) because a0 = 1. 2. The graph does not cross the x-axis; therefore, there are no x-intercepts. 3. There is a horizontal asymptote along the x-axis (y = 0). 4. For f (x) = a x, another point on the graphs is (1, a). 5. The maximal domain is R. 6. The range is R+. 7. They are both increasing functions. That is, as x → ∞, y → ∞. 8. It can be seen that the greater the value of a, the steeper the graph.

Dilation A dilation changes the shape of the graph, making it wider or narrower.

Dilation from the x-axis If the coefficient of ax is changed to a positive real number greater than 1, the graph is stretched vertically and is said to be dilated from the x-axis. This could be written f (x) = Aax where A is the dilation factor. The graph is stretched vertically (along the y-axis) away from the x-axis because each y-value is being multiplied by the constant A. In mapping notation, a dilation factor of A from the x-axis, where A > 0, is given by (x, y) → (x, Ay). If A = 3 and a = 2 the function becomes f (x) = 3 × 2x. A comparison can be made more easily if graphs are drawn on the same axes. The graphs of f (x) = 2x, g(x) = 3 × 2x and h(x) = 4 × 2x are shown. The dilation factor, A, is 1, 3 and 4 respectively. f (x ( ) = 2x

g(x ( ) = 3 × 2x (x

h(x ( ) = 4 × 2x (x

x-intercept

—

—

—

y-intercept

(0, 1)

(0, 3)

(0, 4)

R

R

R

R+ = (0, ∞)

R+ = (0, ∞)

R+ = (0, ∞)

y=0

y=0

y=0

1

3

4

(1, 2)

(1, 6)

(1, 8)

Domain Range Horizontal asymptote Dilation factor from the x-axis A point on the graph Mapping

y (0, 4)

h(x) = 4 × 2x g(x) = 3 × 2x

4 3 (0, 3) f( f x) = 2x 2 1 (0, 1) Asymptote x y= 0 −3 −2 −1 0 1 2 3

(x, y) → (x, 3y) (x, y) → (x, 4y)

Chapter 4

Exponential and logarithmic graphs

173

If A is a real number between 0 and 1, the y-values are multiplied by a constant which is less than 1 and the graph becomes less steep. A comparison can be made more easily if the graphs are drawn on the same axes. The graphs of f (x) = 2x, g(x) = 12 × 2x and h(x) = 14 × 2x are drawn below. y (1, 2)

2

The dilation factor is 1,

1 2

and

1 4

(0, 1–2) −1

1– 2

× 2x

h(x) = 1–4 × 2x

(1, 1)

(0, 1) Asymptote y= 0

f x) = 2x f( g(x) =

–1

(0, 1–4) (1, 2 ) x 0 1

respectively.

f (x ( ) = 2x

g( x ) = 12 × 2 x

h( x ) = 14 × 2 x

x-intercept

—

—

—

y-intercept

(0, 1)

 1  0, 2 

 1  0, 4 

Domain Range

R R+

Horizontal asymptote

= (0, ∞)

R R+

R R+

= (0, ∞)

= (0, ∞)

y=0

y=0

y=0

Dilation factor from the x-axis

1

1 2

1 4

A point on the graph

(1, 2)

(1, 1)

Mapping

(

(x, y) → x, 1 y 2

(1, ) 1 2

)

(

(x, y) → x, 1 y 4

)

For any positive real value of A, as A increases, the graph of f (x (x) = A × 2x becomes steeper and closer to the y-axis. Similarly, as A decreases the graph becomes less steep and further from the y-axis. The domain, range and horizontal asymptote stay the same as for f (x (x) = 2x. The mapping is ((x, y) → (x ( , Ay).

Dilation from the y-axis If the coefficient of x changes, the graph is stretched horizontally and is said to be dilated from the y-axis. This can be written f (x) = akx where 1 is the dilation factor. The dilation factor is the k reciprocal of the coefficient of x. In mapping notation, a dilation factor of 1 from the y-axis, where k > 0, is given by (x, y) → ( 1 x, y). k k To show the effect of a dilation from the y-axis, the three graphs x 2x and h(x) = 2 are drawn at right. The dilation f (x) = 2x, g(x) = 22x 2 1 factors are 1, 2 and 2, respectively.

174


y 4 2 (0, 1) −3 −2 −1

2 g(x) = 22x f x) = 2x –x f( h(x) = 22

Asymptote y=0 01 2 3 x

x

f (x ( ) = 2x

2 g(x ( ) = 22x (x

h(x ( ) = 22 (x

x-intercept

—

—

—

y-intercept

(0, 1)

(0, 1)

(0, 1)

R

R

R

R+ = (0, ∞)

R+ = (0, ∞)

R+ = (0, ∞)

y=0

y=0

y=0

1

1 2

2

(2, 4)

(1, 4)

(4, 4)

Domain Range Horizontal asymptote Dilation factor from the y-axis A point on the graph

(x, y) →

Mapping

(

1 2

x, y

)

(x, y) → (2x (2 , y)

For all the graphs discussed so far (that is, of the form f (x (x) = A × akx, a ∈ R+\{1}, k > 0), the + maximal domain is R, the range is R , the x-axis is the horizontal asymptote and they are all increasing functions. The dilations have affected the steepness of the graphs. The mapping is ( , y) → ( 1 x, Ay). (x k WORKED EXAMPLE 1 x

Sketch the graph of f (x (x) = 2 × 2 2 , showing the intercepts and the asymptote, and stating the domain and the range. THINK

WRITE/DRAW x

1

Write the rule.

2

State the basic shape and transformations.

An exponential curve with basic shape f (x) = 2x. Dilation of 2 units from the x-axis and 2 units from the y-axis.

3

Find the y-intercept. Either let x = 0 or use the fact that the y-intercept is A in the function

If x = 0, then

x

f (x) = A × a k .

f (x) = 2 × 2 2

0

y = 2 × 22 = 2 × 20 =2×1 =2 so the y-intercept is 2.

4

Find the horizontal asymptote.

The horizontal asymptote is the x-axis. There are no x-intercepts.

5

Find another point on the graph.

If x = 2, y = 2 × 2 2 = 4.

6

Sketch the graph.

2

y 4

(2, 4)

(0, 2)

x

x −3 −2 −1 0 1 2 3 7

State the domain and the range.

f (x) = 2 × 2 2 Asymptote y = 0

The domain is R and the range is R+.

Chapter 4


175

Translation Vertical translation If a constant is added to the function, the graph is moved up or down and is said to be translated vertically. In mapping notation, a vertical translation of B units is given by (x, y) → (x, y + B). The graph of g(x) = 2x + 1 is shown below with the graph of f (x) = 2x. The graph of f (x) = 2x has been moved up 1 unit. Every y-value has been increased by 1. The line y = 1 is the horizontal asymptote. f (x ( ) = 2x

g(x ( ) = 2x + 1 (x

x-intercept

—

—

y-intercept

(0, 1)

(0, 2)

R

R

Range

(0, ∞)

(1, ∞)


y=0

y=1


(1, 2)

(1, 3)

Domain

y

g(x) = 2x + 1 f x) = 2x f(

4 3 (0, 2)

(1, 3)

(0, 1) 0 1

−1

(1, 2) Asymptote y=1 2

x

(x, y) → (x, y + 1)

Mapping

The graph of g(x) = 2x − 1 is the graph of f (x) = 2x translated 1 unit down. This means that the horizontal asymptote is now the line y = −1 and the graph crosses the x-axis. Therefore, there is an x-intercept. Every y-value has been decreased by 1. The graph of y = 2x − 1 is shown with the graph of y = 2x. y

f (x ( ) = 2x

g(x ( ) = 2x − 1 (x

x-intercept

—

(0, 0)

4

y-intercept

(0, 1)

(0, 0)

R

R

2 (0, 1)

Range

(0, ∞)

(−1, ∞)


y=0

y = −1


(1, 2)

(1, 1)

Domain

−33 −2 −11

(1, 2)

g(x) = 2x − 1

(1, 1) x 0 1 2 3 Asymptote y = −1

(x, y) → (x, y − 1)

Mapping

Horizontal translation In mapping notation, a horizontal translation of b units is given by (x, y) → (x + b, y). The graph of g(x) = 2x − 1 is the graph of f (x) = 2x translated 1 unit to the right. The graph of h(x) = 2x + 1 is the graph of f (x) = 2x translated 1 unit to the left. The graph of g(x) = 2x − 1 is shown at right with the graph of f (x) = 2x. 176

f x) = 2x f(


y

h(x) = 2x + 1 f(x) = 2x f(

4 2 (0, 1) −3 −2 −1

(1, 1) (0, 1–2 )

01 2 3

g(x) = 2x − 1 Asymptote y=0 x

f (x ( ) = 2x

g(x ( ) = 2x − 1 (x

h(x ( ) = 2x + 1 (x

x-intercept

—

—

—

y-intercept

(0, 1)

(0, )

(0, 2)

R

R

R

R+ = (0, ∞)

R+ = (0, ∞)

R+ = (0, ∞)


y=0

y=0

y=0


(1, 2)

(2, 2)

(0, 2)

(x, y) → (x + 1, y)

(x, y) → (x − 1, y)

Domain Range

Mapping

1 2

Horizontal and vertical translations have the same effect regardless of the base. g(x ( ) = 2 + 10x + 1 is f (x (x (x) = 10 x translated 1 unit to the left and 2 units up. Horizontal translations keep the same asymptote, domain and range as the original function; vertical translations keep the same domain, but the range changes. It is interesting to note that f (x) = 0.5 × 2x can be written as f (x) = 2x − 1 because f (x) = 0.5 × 2x 1

= 2 × 2x −

= 2 1 × 2x = 2x − 1 A dilation of 0.5 from the x-axis has the same effect on f (x) = 2x as a translation of 1 unit to the right. WORKED EXAMPLE 2

eBoo k plus eBook

Sketch the graph of f (x (x) = 2x − 1 − 2, showing intercepts and asymptotes, and stating the domain and range. THINK

WRITE/DRAW

Tutorial


f (x) = 2x − 1 − 2

2

Write the rule. State the basic shape.

3

State the translations.

Horizontal translation of 1 unit to the right Vertical translation of 2 units down

4

Find the horizontal asymptote by translating y = 0 down 2 units.

The horizontal asymptote is y = −2.

5

Find the y-intercept. Make x = 0.

y-intercept: If x = 0, then

1

An exponential curve with the same shape as f (x) = 2x

−

y=2 1−2 = 12 − 2 = −1 12

6

Find the x-intercept. Make y = 0.

7

Equate the indices.

8

Solve for x.

x-intercept: If y = 0, then

Chapter 4

2x − 1 − 2 = 0 2x − 1 = 21 x−1=1 x=2


177

9

Sketch the graph.

y 1 −3 −2 −1 −1 −2

10

(2, 0) 0 1 2 3 x

f (x) = 2x − 1 − 2

1 (0, −11–2 ) Asymptote y = −2

The domain is R and the range is (−2, ∞).


For all the graphs of the form f (x (x) = ax + b + B, where b, B ∈ R and a ∈ R+\{1}, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B and they are all increasing functions. The graph remains exactly the same shape as f (x (x) = ax. The mapping is ((x, y) → (x ( − b, y + B).

Reflections If a negative sign is in front of the a the graph is reflected in the x-axis. ( a > 0.) The mapping is (x, y) → (x, −y). The graph of g(x) = −2x is shown with the graph of f (x) = 2x. f (x ( ) = 2x

g(x ( ) = −2x (x

y

x-intercept

—

—

4

y-intercept

(0, 1)

(0, −1)

R

R

R+ = (0, ∞)

R− = (−∞, 0)


y=0

y=0


(1, 2)

(1, −2)

Domain Range

f x) = 2x f(

2

Asymptote (1, 2) y = 0 −3 −2 –1 0 1 2 3 x (1, −2) −2 (0, 1)

(0, −1) g(x) = −2x

(x, y) → (x, −y)

Mapping

If there is a negative sign before the x term, the graph is reflected in the y-axis. The graph of g(x) = 2−x is shown with the graph of f (x) = 2x. f (x ( ) = 2x

g(x ( ) = 2 −x (x

x-intercept

—

—

y-intercept

(0, 1)

(0, 1)

R

R

R+ = (0, ∞)

R+ = (0, ∞)


y=0

y=0


(1, 2)

(−1, 2)

Domain Range

Mapping

f x) = 2x f(

– y g(x) = 2–x

3 (−1, 2) 2 1 −3 −2 −1

(1, 2) (0, 1) 0

Asymptote y= 0 x 1 2 3

(x, y) → (−x, y) For all the graphs of the form g(x) = −ax, where a ∈ R+\{1}, the maximal domain is R and the range is R−. The horizontal asymptote is y = 0 and the functions are all decreasing. It is the reflection of f (x) = ax in the x-axis. The mapping is (x, y) → (x, −y).

178


For all the graphs of the form g(x) = a−x, where a ∈ R+\{1}, the maximal domain is R and the range is R+. The horizontal asymptote is y = 0 and the functions are all decreasing. It is the reflection of f (x) = ax in the y-axis. The mapping is (x, y) → (−x, y). WORKED EXAMPLE 3

Sketch the graph of f (x (x) = 2 − 2x − 1, showing intercepts and asymptotes, and stating the domain and the range: a showing all working b with a CAS calculator. THINK a

WRITE/DRAW

1

Write the rule.

f (x) = 2 − 2x − 1

2

State the basic shape.

An exponential curve with basic shape f (x) = 2x

3

State the transformations.

Horizontal translation of 1 unit to the right Vertical translation of 2 units up Reflection in the x-axis

4

Find the horizontal asymptote by translating The horizontal asymptote is y = 2. y = 0 up 2 units. − If x = 0, f (x) = 2 − 2 1 Find the y-intercept by making x = 0.

5

=2−

1 2

= 1 12 The y-intercept is 1 12. 6

Find the x-intercept by making y = 0.

7

Sketch the graph.

2 − 2x − 1 = 0 2x − 1 = 21 x−1=1 x=2 The x-intercept is 2. If y = 0,

y 2

Asymptote y=2 (0, 3–2 )

1 0

−2 −1

1

(2, 0) x 2

f (x) = 2 − 2x − 1

−1

b

8


1

To graph y = 2 − 2x − 1 on a Graphs page, complete the function entry line as: f1(x) = 2 − 2x − 1 Press ENTER ·. Note: The horizontal asymptote at y = 2 is not displayed.

The domain is R and the range is (−∞, 2).

Chapter 4


179

2

To locate the x-intercept on a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(f 1(x) = 0, x) Press ENTER ·.

3

Write the x-intercept in coordinate form.

∴ (2, 0) are the coordinates of the x-intercept.

4

To calculate the y-intercept, complete the entry line as: ff1(0) and press ENTER ·. Write the y-intercept in coordinate form.

∴ 0,

( ) are the coordinates of the y-intercept. 3 2

1. Graphs of the form f (x) = ax, where a ∈ R+\{1} (a) The maximal domain is R. (b) The range is R+. (c) The x-axis is the horizontal asymptote. (d) The y-intercept is 1. (e) They are all increasing functions. 2. Dilation (a) Graphs of the form f (x) = Aax, where a, A ∈ R+\{1}, have a y-intercept of A. A changes the steepness and proximity of the graph to the y-axis when compared with the graph of f (x) = ax. As A increases, the graph becomes steeper. So f (x) = ax is dilated by a factor of A from the x-axis. (b) Graphs of the form f (x) = akx, where a ∈ R+\{1} and k > 0, have a different steepness than graphs of the form f (x) = ax. As k increases, the graph becomes more steep (that is, it gets closer to the y-axis). The y-intercept remains the same. So f (x) = akx is dilated by a factor of 1 from the y-axis. k 3. Translation (a) Graphs of the form f (x) = ax + B, where a ∈ R+\{1}, B ∈ R, have a y-intercept of 1 + B, a horizontal asymptote of y = B and a range of (B, ∞). If B < 0 there is one x-intercept. If B > 0 there are no x-intercepts. (b) Graphs of the form f (x) = ax + b, where a ∈ R+\{1}, b ∈ R, have a y-intercept of ab. They are the same shape as f (x) = ax but have been translated b units to the left if b > 0 and to the right if b < 0. 4. Reflection (a) Graphs of the form f (x) = −ax, where a ∈ R+\{1}, have a y-intercept of −1, a range of R− and they are decreasing functions. They are the same shape as f (x) = ax but have been reflected in the x-axis. (b) Graphs of the form f (x) = a−x, where a ∈ R+\{1}, have the same key features as f (x) = ax but are decreasing functions; they have been reflected in the y-axis. 5. Combinations of transformations For all the graphs of the form f (x) = A × ax + b + B, where b, B, A ∈ R and a ∈ R+\{1}, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B and they are all increasing functions.

180


EXERCISE

4A eBoo k plus eBook Digital doc

Spreadsheet 034 Exponential graphs

Graphs of exponential functions with any base 1 WE1 Sketch the graph of each of the following functions, showing the intercepts and the asymptote, and stating the domain and range of each one. a f (x) = 2x b f (x) = 10x c f (x) = 2 × 10x x x d f (x) = 3 × 2 e f (x) = 0.3 × 2 f f (x) = 0.5 × 10x 3x 2 2x 2 g f (x) = 2 h f (x) = 10 i f (x) = 5 × 32x x x 3x j f (x) = 4 × 2 k f (x) = 2 × 10 2 l f (x) = 2 × 2 3 2 WE2 Sketch the graph of each of the following functions, showing intercepts and asymptotes, and stating the domain and range of each one. a f (x) = 2x + 1 b f (x) = 3x + 2 c f (x) = 3x − 3 x x + 2 d f (x) = 2 − 4 e f (x) = 2 f f (x) = 10x + 1 x − 4 x − 3 g f (x) = 3 h f (x) = 2 i f (x) = 2x + 1 − 8 x − 2 j f (x) = 10 +1 3 WE3 Sketch the graph of each of the following functions, showing intercepts and asymptotes, and stating the domain and range of each one. a f (x) = 2x b f (x) = −10x −x c f (x) = 10 d f (x) = 2−x x e f (x) = 1 − 3 f f (x) = 10 − 10x −x g f (x) = 2 + 10 h f (x) = 1 + 2−x 1 − x i f (x) = 2 − 2 j f (x) = 1 − 32 − x 4 Sketch the graph of f (x (x) = 2 × 31 − x, showing the intercepts and asymptotes and stating the domain and range. 5 Sketch the graph of the function f: f R → R where f (x (x) = −3 × 2x − 1, showing the intercepts and asymptotes and stating the domain and range. 6 Sketch the graph of the function f: f R → R where f (x (x) = 5 − 4 × 31 − x showing the intercepts and asymptotes and stating the domain and range. 7 State the transformation of y = 2x needed to sketch the graphs of the following functions. Give details of each transformation. a y = 23x b y = 24x c y = 2 × 2x x − x d y=3×2 e y= 2 f y = 2−x x x g y=2 +1 h y=2 −3 i y = 2x − 1 x + 5 j y=2 8 Each of the following functions is a translation of f (x (x) = 10x. State how far each graph is translated, and in which direction. a f (x) = 10 x + 4 b f (x) = 10 x − 2 c f (x) = 10 x + 2 x − 3 x d f (x) = 10 e f (x) = 10 − 3 f f (x) = 2 + 10 x x − 1 2 + x g f (x) = 10 h f (x) = 10 i f (x) = 5 + 10 x + 1 x − 4 x − 3 j f (x) = 10 +2 k f (x) = 10 −4 l f (x) = 10 x + 2 − 3 9 Write down the domain and range for each of the following graphs and the equations for the horizontal asymptotes. a

y

b

y

c

y

6 4 2 (0, 2) y=0 x 0 −3 −2 −1 1 2

6 4 4 (0, 4) 2

(0, 2) y=1

−3 −2 −1

0 1

Chapter 4

2

x

y=3

2 –3 −2 −1

0 1

2

x


181

y 2

d

y

e y=1

−3 −2 −1 0

1

2

2

6

x

−2

−1 0

2

−3 −2 −1

y=1

(1, 0)

4 (0, 4) y=2

y

f

1 2

3

x

−2 (0, −2) 0

1 2 3

x

10 If the graph of f (x (x) = 2x is translated 1 unit up and reflected in the x-axis, what is the new equation? 11 If the graph of f (x (x) = 1 − 2x + 3 is translated 2 units to the right and dilated by factor 3 from the x-axis, what is the new equation? 2 − 1 + 3 are respectively: 12 MC The domain and range of the graph of f (x (x) = 102x − A R, R B R, [ 1, ∞) C [3, ∞), R D [1, ∞), R E R, (3, ∞)

13 MC The y-intercept of the graph of f (x (x) = 10−x + 1 is: A 1 B 2 C −1

D −2

E 11

2x − 1

14 MC When the graph of f (x (x) = − 4 is translated 3 units down and 2 units to the left, it becomes: A f (x) = 2x − 3 − 7 B f (x) = 2x − 3 − 1 x + 1 C f (x) = 2 −7 D f (x) = 2x + 1 − 1 x − 2 E f (x) = 2 −6 15 Under certain conditions a mathobacillus bacterial colony doubles its numbers every minute. The population can be modelled by: P = 50 × 2t where P is the number of bacteria t minutes after counting has begun. a Find the number of bacteria when t = 0. b Find the number of bacteria after 3 minutes. c Sketch the graph of the population as a function of time.

4B

Logarithmic graphs to any base The function f (x) = loga (x) is a logarithmic function where a is a positive, real number which is not 1 (that is, a ∈ R+\{1}). Logarithms to base 10 appear in many scientific formulas. An example of this is the intensity of earthquakes which is measured on the Richter scale. that loga (x) does not exist when x < 0. In this course only graphs of functions where a > 1 are considered. Graphs with a positive coefficient of x will be considered first.

Graphs of f (x) = log2 (x) and f (x) = log10 (x) The graphs of f (x) = log2 (x) and f (x) = log10 (x) are shown below. Asymptote y x= 0 (2, 1) 2 (1, 0) 1 −1 0 −1 −2

1

2 3

f (x) = log2 (x) 182

y

Asymptote x= 0

1

x

0

(10, 1)

(1, 0) 2 4 6 8 10

−1

f (x) = log10 (x)


x

These graphs have a number of common features: 1. They are both increasing functions; that is, as x → ∞, y → ∞. 2. Both graphs cross the x-axis at the point (1, 0), as loga (1) = 0 3. The y-axis is a vertical asymptote, so there is no y-intercept. This is because loga (0) is undefined. 4. There are no negative values of x because the log of a negative number does not exist. 5. Another point on the graph is (a, 1), where a is the base because loga (a) = (1). 6. The domain is R+. 7. The range is R. 8. It can be seen that the smaller the value of a, the steeper the graph or the more rapidly the graph rises.

Dilation Dilation from the x-axis If the coefficient of loga (x) is changed to a positive real number greater than 1, the graph is stretched along the y-axis and is said to be dilated from the x-axis. This could be written f (x) = A loga (x) where A is the dilation factor. The graph is stretched along the y-axis because each y-value is being multiplied by the constant, A. In mapping notation, a dilation factor of A from the x-axis, where A > 0, is given by Asymptote (x, y) → (x, Ay). y x= 0 (10, 2) g(x) = 2 log10 (x) The asymptote, x-intercept, domain and range are the same for all graphs of the form f (x (x) = A loga (x ( ). Graphs with different values of A can be more easily compared by drawing them on the same axes. The graphs of g(x) = 2 log10 (x), f (x) = log10 (x) and h(x) = 12 log10 (x) are drawn at right. Notice that, as A increases, the graph becomes steeper.

1 0

f x) = log10 (x) (10, 1) f( h(x) = 1–2 log10 (x) (10, 1–2 ) x 2 4 6 8 10

−1

1

f (x ( ) = log10 (x ( )

g(x ( ) = 2 log10 (x (x ( )

h(x ( ) = 2 log10 (x (x ( )

x-intercept

(1, 0)

(1, 0)

(1, 0)

y-intercept

—

—

—

R+ = (0, ∞)

R+ = (0, ∞)

R+ = (0, ∞)

R

R

R

x=0

x=0

x=0

1

2

1 2

(10, 1)

(10, 2)

(10, )

Domain Range Vertical asymptote Dilation factor from the x-axis A point on the graph Mapping

(x, y) → (x, 2y)

1 2

(

(x, y) → x, 12 y

)

Note: h(x) = 12 log10 (x) can be written as h(x) = log10 ( x ) and is drawn in the same way. 1 2

Chapter 4


183

Dilation from the y-axis If the coefficient of x is changed to a positive real number not equal to 1, the graph is stretched along the x-axis and is said to be dilated from the y-axis. This could be written f (x) = loga (kx), where 1 is the dilation k factor. When k > 1, the graph moves away from the x-axis. To see the effect of a dilation from the x-axis the graphs g(x) = log10 (2x (2 ), f (x) = log10 (x) and h(x) = log10 ( 12 x) are drawn at right.

y Asymptote x= 0 1

0

1

h(x) = log10 (–2 xx) 2 (2, 0) (1, 0)

(1, 0)

( , 0)

= (0, ∞)

h(x ( ) = log10 (x

— R+

= (0, ∞)

= (0, ∞) R

x=0

x=0

x=0

Dilation factor from the y-axis

1

1 2

2

Another point on the graph

(10, 1)

(5, 1)

(20, 1)

(x, y) →

1 2

— R+

R

Mapping

( x)

(2, 0)

R


x

6

1 2

— R+

4

−1

g(x ( ) = log10 (2x (x (2 )

y-intercept Range

f x) = log10 (x) f(

( 1–2 , 0)

f (x ( ) = log10 (x ( ) x-intercept

Domain

2 ) 2x g(x) = log10 (2x

( x, y ) 1 2

(x, y) → (2x (2 , y)

For all graphs discussed so far (that is, of the form y = A loga (kx), A ∈ R+\{1}, k > 0), the maximal domain is R+, the range is R, the y-axis is the vertical asymptote and they are all increasing functions. The graphs have varied in their x-intercepts. The dilation factor from the x-axis is A and the dilation factor from the y-axis is 1 . In mapping notation, k Asymptote ( , y) → 1 x, A (x Ay . y x=0 k f x) = 2 log2 (3x) f( If the base is 2, the graph is steeper than when the base is 10 but the vertical asymptote, intercept, domain and range 2 stay the same as for f (x) = A log10 (kx). f x) = 2 log10 (3x) f( 1 The graphs of f (x) = 2 log10 (3x) and f (x) = 2 log2 (3x) are ( –3 , 0) x shown at right. 0 1 2 3 1. The vertical asymptote is the y-axis. −2 2. The x-intercept is 13 . 3. The domain is R+ and the range is R. 4. The basic graph is dilated by factor 2 from the x-axis and factor 13 from the y-axis.

(

184

)


WORKED EXAMPLE 4

Sketch the graph of f (x (x) = 3 log10 (2x (2 ), showing the intercepts and the asymptote, and state the domain, range and the kind of transformation. THINK

WRITE/DRAW

1

Write the rule.

f (x) = 3 log10 (2x (2 )

2


A logarithmic graph with basic shape f (x) = log10 (x)

3

Find the vertical asymptote (log10 (0) is undefined). The vertical asymptote is the y-axis.

4

Find the x-intercept. that log10 (1) = 0.

x-intercept: Let y = 0, 3 log10 (2x (2 ) = 0 2 =1 2x x = 12 The x-intercept is 12.

5

Find the y-intercept.

There are no y-intercepts.

6

Mark another point on the graph. Choose one which makes 22xx equal the base.

Let x = 5, y = 3 log10 (10) =3×1 =3 Coordinates are (5, 3).

7

Sketch the graph.

Asymptote y x=0 3

f x) = 3 log10 (2x f( (2 )

1

0

(5, 3)

( –2 , 0) 1

2

3

4

5

x

−3 8


The domain is R+ and the range is R.

9


The dilations are of factor 3 from the x-axis and factor 12 from the y-axis.

Translation Translations may be either vertical or horizontal.

Vertical translation If a constant is added to the function, the graph is moved up or down and is said to be translated vertically. The graphs of f (x (x) = log10 (x ( ), g(x ( ) = log10 (x (x ( ) + 1 and h(x ( ) = log10 (x (x ( ) − 1 are drawn and compared. y Asymptote x=0 2

g(x) = log10 (x) + 1 f x) = log10 (x) f(

1 0 −1

h(x) = log10 (x) − 1 2

4

6

8

10

12 x

−2 −3

Chapter 4


185

f (x ( ) = log10 (x ( )

g(x ( ) = log10 (x (x ( )+1

(

1 , 10

0

h(x ( ) = log10 (x (x ( )−1

)

x-intercept

(1, 0)

y-intercept

—

—

—

R+ = (0, ∞)

R+ = (0, ∞)

R+ = (0, ∞)

R

R

R

Vertical asymptote

x=0

x=0

x=0

Vertical translation

—

1 unit up

1 unit down

(1, 0)

(1, 1)

(1, −1)

(x, y) → (x, y + 1)

(x, y) → (x, y − 1)

Domain Range

A point on the graph Mapping

(10, 0)

Horizontal translation If a constant is added to x, the graph of f (x) = log10 (x) is translated horizontally. The graphs of f (x) = log10 (x), g(x) = log10 (x + 1) and h(x) = log10 (x − 1) are drawn. f (x ( ) = log10 (x ( )

g(x ( ) = log10 (x (x ( + 1)

h(x ( ) = log10 (x (x ( − 1)

(1, 0)

(0, 0)

(2, 0)

—

—

x-intercept y-intercept Domain

— R+

(−1,

= (0, ∞)

Range

R

Horizontal translation A point on the graph

(1, ∞)

R

x=0

Vertical asymptote

∞)

x=

R

−1

x=1

—

1 unit to left

1 unit to right

(10, 1)

(9, 1)

(11, 1)

(x, y) → (x − 1, y)

(x, y) → (x + 1, y)

Mapping y 2

g(x) = log10 (x + 1)

1 −2 −1

−1

0 1

3

4

5

x

f x) = log10 (x) f( h(x) = log10 (x − 1)

−2 x = −1

2

x=1

For all graphs of the form f (x (x) = loga (x ( + b) + B, where b and B ∈ R and a ∈ R+\{1}, − the maximal domain is ( b, ∞), the range is R, the vertical asymptote is x = −b and they are all increasing functions. The graph has exactly the same shape as f (x (x) = loga (x ( ). The horizontal translation is −b and the vertical translation is B. In mapping notation, ( , y) → (x (x ( − b, y + B).

186


WORKED EXAMPLE 5

eBoo eBook k plus

Sketch the graph of f (x (x) = 1 + log10 (x ( + 2), showing intercepts and asymptotes, and stating the domain, range and transformations. Give intercepts to 1 decimal place. THINK

Tutorial


WRITE/DRAW

1

Write the rule.

f (x) = 1 + log10 (x + 2)

2


Logarithmic graph, basic shape f (x) = log10 (x)

3

Find the vertical asymptote (log10 (0) is undefined). x + 2 = 0 x = −2 is the vertical asymptote.

4

Find the x-intercept.

If y = 0,

5

Use the law ax = y ⇔ loga(y) = x.

x + 2 = 10 1 x = 0.1 − 2 = −1.9 − The x-intercept is 1.9.

6


If x = 0, y = 1 + log10 (0 + 2) (exact answer) ≈ 1 + 0.301 03 ≈ 1.3, correct to 1 decimal place The y-intercept is 1.3.

7

Sketch the graph.

1 + log10 (x + 2) = 0 log10 (x + 2) = −1 −

Asymptote y x = –2 2

(0, 1.3)

(−1.9, 0) −2

0

2

x 4 6 8 f x) = 1 + log10 (x + 2) f(

8


The domain is (−2, ∞) and the range is R.

9

State the translations.

The horizontal translation is 2 units to the left and the vertical translation is 1 unit up.

Reflections If there is a negative sign in front of the loga (x) term, the graph is reflected in the x-axis. The mapping is (x, y) → (x, −y). The graphs of f (x) = log10 (x) and g(x) = −log10 (x) are shown at right.

Asymptote y x=0 f x) = log10 (x) f( 1 (10, 1) (1, 0) 0 −1

Chapter 4

2

4

g(x) =

6 8 10

x

(10, –1) 10 (x))

–log


187

f (x ( ) = log10 (x ( )

g(x ( ) = −log10 (x (x ( )

(1, 0)

(1, 0)

x-intercept y-intercept

— R+

Domain

= (0, ∞)

Range

= (0, ∞)

R

R

x=0

x=0

(10, 1)

(10, −1)

Vertical asymptote A point on the graph

— R+

(x, y) → (x, −y)

Mapping

If there is a negative sign in front of the x term, the graph is reflected in the y-axis. The mapping is (x, y) → (−x, y). The graphs of f (x) = log10 (x) and g(x) = log10 (−x) are shown below right. f (x ( ) = log10 (x ( )

g(x ( ) = log10 (−x) (x

(1, 0)

(−1, 0)

x-intercept y-intercept Domain Range Vertical asymptote A point on the graph

— R+

— R−

= (0, ∞)

= (−∞, 0)

R

R

x=0

x=0

(10, 1)

(−10, 1)

Mapping

Asymptote y x=0 f x) = log10 (x) g(x) = log10 (−x) f( 1 (−10, 1) (10, 1) (−1, 0) −10

−2

(1, 0) 2

−1

(x, y) → (−x, y)

WORKED EXAMPLE 6 a Sketch the graph of f (x (x) = 2 log10 (3 − x) − 2, showing intercepts and asymptotes, and stating

the domain, range and transformations. Give exact values or round to 3 decimal places.

b Sketch the same graph using a CAS calculator. THINK a

188

WRITE/DRAW

1

Write the rule.

f (x) = 2 log10 (3 − x) − 2

2


Logarithmic graph, f (x) = log10 (x)

3

Find the vertical asymptote (log10 (0) undefined).

For the vertical asymptote, 3−x=0 x = 3 is the vertical asymptote.

4

Find the x-intercept.

If y = 0,

2 log10 (3 − x) − 2 = 0 2 log10 (3 − x) = 2 log10 (3 − x) = 1 3 − x = 101 x = −7 − The x-intercept is 7.


10

x

5


6

Sketch the graph.

If x = 0, y = 2 log10 (3) − 2 = −1.046 (to 3 decimal places) The y-intercept is −1.046. (–7, 0)

y

x −6 −44 −2 0 2 Asymptote (0, 2 log10 (3) − 2) x=3 −22

f (x (x) x) = 2 log10 (3 − x) x −2 7 8

b

State the domain and the range. State the transformations.

1

To graph y = 2 log10 (3 − x) − 2 on a Graphs page, complete the function entry line as: f1(x) = 2 log10 (3 − x) − 2 Press ENTER ·. Note: The vertical asymptote at x = 3 is not displayed.

2

To locate the x-intercept on a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve (f 1(x) = 0, x) Press ENTER ·.

3

Write the x-intercept in coordinate form. To calculate the y-intercept, complete the entry line as: ff1(0) and press ENTER ·. Write the y-intercept in coordinate form.

4

(−∞,

The domain is 3) and the range is R. Reflection in the y-axis, dilation 2 units from the x-axis, vertical translation 2 units down, horizontal translation 3 units to the right

The coordinates of the x-intercept are (−7, 0). The coordinates of the y-intercept are (0, −1.046) (correct to 3 decimal places).

1. The function f (x) = loga (x) is a logarithmic function where a ∈ R+\{1}. (a) The vertical asymptote is the y-axis, so there are no y-intercepts. (b) The graph crosses the x-axis at (1, 0) because loga (1) = 0. (c) The domain is R+; that is, there are no negative values of x. (d) The range is R.

Chapter 4


189

2. Dilation (a) The function f (x) = A loga (x) dilates the graph of f (x) = loga (x) by a factor of A units from the x-axis. The vertical asymptote, x-intercept, domain and range remain the same. As A increases, the graph becomes steeper. (b) The function f (x) = loga (kx) dilates the graph of f (x) = loga (x) by a factor of 1 k from the y-axis. The vertical asymptote, domain and range stay the same, but the x-intercept is 1 . As k increases, the graph becomes steeper and the x-intercept k becomes smaller. 3. Translation (a) The function f (x) = loga (x) + B translates the graph of f (x) = loga (x) vertically B units. The vertical axis, domain and range remain the same but the x-intercept changes. There is no change in the shape of the graph. (b) The function f (x) = loga (x + b) translates the graph of f (x) = loga (x) horizontally −b units. The shape and the range remain the same but the vertical asymptote, the x-intercept and the domain change: (i) The vertical asymptote becomes x = −b. (ii) The graph crosses the x-axis at (1 − b, 0). (iii) The domain is (−b, ∞). 4. Reflection (a) The function f (x) = −loga (x) reflects the graph of f (x) = loga (x) in the x-axis. All key features remain the same but the graph is a decreasing function instead of an increasing function. (b) The function f (x) = loga (−x) reflects the graph of f (x) = loga (x) in the y-axis. The vertical asymptote and the range remain the same but the x-intercept and the domain change. (i) The graph crosses the x-axis at (−1, 0). (ii) The domain is (−∞, 0).

EXERCISE

4B

Logarithmic graphs to any base 1 a c e g

Find the vertical asymptote of each of the following functions. f (x) = log2 (x) b f (x) = log10 (x) f (x) = 2 log10 (x) d f (x) = 5 log2 (x) f (x) = log2 (3x) f f (x) = log10 (4x) f (x) 3 log10 (2x (2 ) h f (x) = 2 log2 (3x)

2 WE4 Sketch the graph of each of the functions in question 1, showing the intercepts and asymptotes, and stating the domain and range of each one. 3

State the horizontal and vertical translations required to transform f (x (x) = log2 (x ( ) into the following functions. a f (x) = log2 (x + 2) b f (x) = log2 (x + 1) c f (x) = log2 (x − 3) d f (x) = log2 (x − 4) e f (x) = 1 + log2 (x − 5) f f (x) = log2 (x − 3) + 2 g f (x) = 2 + log2 (x + 1) h f (x) = 3 + log2 (x + 1) i f (x) = log2 (x + 3) − 2 j f (x) = log2 (x + 1) − 2

4 WE5 Sketch the graph of each of the functions in question 3, showing the intercepts and asymptotes, and stating the domain and the range of each one.

190



Spreadsheet 069 Logarithmic graphs

5 Sketch the graph of each of the following functions showing intercepts and asymptotes, and stating the domain, range and the transformation of f (x (x) = log10 (x). a f (x) = −log10 (x) b f (x) = −2 log10 (x) c f (x) = log10 (−x) d f (x) = 3 log10 (−x) e f (x) = 1 − log10 (x) f f (x) = 1 + log10 (x) g f (x) = 2 + log10 (−x) h f (x) = 2 − log10 (−x) 6 Sketch the graph of f (x (x) = log2 (2x (2 − 1), showing intercepts and asymptotes, and stating the domain, range and transformations of the graph f (x (x) = log2 (x ( ). 7 WE6 Sketch the graph of each of the following functions, showing intercepts and asymptotes, and stating the domain and range of each one. Give exact values or round to 1 decimal place. a f (x) = 2 log2 (x) + 3 b f (x) = 3 log10 (x) + 1 c f (x) = log10 (1 − x) d f (x) = log10 (3 − x) e f (x) = 1 − 2 log2 (x) f f (x) = 2 − 5 log10 (x) g f (x) = 3 log10 (x + 1) − 2 h f (x) = 4 log2 (x − 2) − 1 i f (x) = 1 + 2 log10 (3x) j f (x) = 3 − 4 log10 (2x (2 ) 8 Each graph in question 7 is a transformation of the graph of f (x (x) = loga (x), a = 2, 10. State the base of each graph and the kinds of transformation that have taken place. 9 Sketch the graph of f: f R → R where f (x (x) = 3 log2 (2 − x) + 1, showing all key features and stating the domain and range. 10 Sketch the graph of ff: [0, 10] → R where f (x (x) = 3 log10 (x ( + 1) − 2, showing key features.


WorkSHEET 4.1

4C

11 MC When the function f (x (x) = log2 (x ( + 2) is translated 2 units up and 3 units to the left, the function becomes: A f (x) = log2 (x + 4) + 3 B f (x) = log2 (x + 5) + 2 C f (x) = log2 (x − 1) + 2 D f (x) = log2 (x + 5) − 3 E f (x) = log2 (x − 1) + 2

Graphs of exponential functions with base e Graphs of exponential functions with base e are drawn in exactly the same way as with any other base. The graph of f (x) = ex can be dilated, translated and reflected in the same way. The graphs of f (x) = ex, g(x) = 2x and g(x) = 10x are shown at right. The graph of f (x) = ex is shown in red. It can be seen that h(x) = 10x is steeper than f (x) = ex and h(x) = 2x is less steep than f (x) = ex.

Features common to all three graphs

y 4

−2

h(x) = 10x f x) = ex f(

g(x) = 2x 3 (1, e) 2 (1, 2) Asymptote (0, 1) y=0 x 0 2 −1 1

1. The graphs all cross the y-axis at the point (0, 1). 2. The horizontal asymptote is the x-axis or the line y = 0. 3. The domain is R. 4. The range is R+. 5. They are all increasing functions.

WORKED EXAMPLE 7 3 − 1. State the transformations of f (x (x) = ex needed to form the graph of f (x (x) = 2 e3x

THINK 1

Write the rule.

WRITE/DRAW

f (x) = 2e3x − 1

Chapter 4


191

2

3

State the dilation. The coefficient of ex gives the dilation from the x-axis. The reciprocal of the coefficient of x gives the dilation from the y-axis.

Dilation by a factor of 2 units from the x-axis

State the translation. The vertical translation is given by the constant added to the ex term.

The graph is translated 1 unit down.

Dilation by a factor of

1 2

units from the y-axis

WORKED EXAMPLE 8

Sketch the graph of f (x (x) = ex. On the same set of axes sketch the graph of f (x (x) = ex − 2, marking the asymptote and y-intercept, and state the transformation, the domain and the range. Give exact answers. THINK

WRITE/DRAW

1

Write the rule for the first graph.

f (x) = ex

2


Exponential curve

3

State the horizontal asymptote.

The horizontal asymptote is the x-axis.

4

Find the y-intercept by making x = 0.

If x = 0, y = e0 y = 1 is the y-intercept.

5

Draw the graph.

y

f x) = ex f(

2

Asymptote y = 0

(0, 1) −3 −2 −1 0

1

x

6

Write the rule for the second graph.

f (x) = ex − 2

7

State the transformation. The horizontal translation is given by the constant added to the x term.

The horizontal translation is 2 units to the right.

8

State the horizontal asymptote, which is the same as for f (x) = ex.


9

Find the y-intercept by making x = 0.

If x = 0, y = e0 − 2 − y=e 2 =

10

Sketch the graph on the same set of axes.

y 2 (0, 1)

1 e2 f x) = ex f( f x) = e (x − 2) f( (0, e−2) Asymptote y = 0. (2, 1)

−3 −2 −1 0

192

2 3

1

2 3

x

11

Check the accuracy of your graph by transforming the point (0, 1) and marking it on your graph.

(0, 1) → (2, 1)

12

State the domain and the range which are the same for both graphs.



For graphs of y = Aekx, k > 0 the coefficient of the ex term shows the dilation of A from the x-axis and the reciprocal of the coefficient of the x term shows the dilation of 1 from the y-axis. k The mapping is (x, y) → ( 1 x, Ay). k WORKED EXAMPLE 9 2 , marking the asymptote and intercept, and state the transformations, Sketch the graph of y = 3e2x domain and range.

THINK

WRITE/DRAW

1

Write the rule.

2 f (x) = 3e2x

2


Exponential curve

3


Dilation by a factor of 3 units from the x-axis and dilation by a factor of 12 unit from the y-axis

4

Find the asymptote, recognising that it is unchanged by dilation.


5

Find the y-intercept by making x = 0 or by multiplying y-values of f (x) = ex by 3. (0, 1) → (0, 3)

If x = 0,

6

Sketch the graph.

y = 3e2 × 0 =3×1 =3 so y-intercept is 3. y 4 2

(0, 3)

−3 −2 −1 0 1 2 3 7

State the domain and range.

2 f (x) = 3e2x Asymptote y = 0

x


The vertical translation is given by the constant added to the ex term. The horizontal translation is given by the constant added to the x term. WORKED EXAMPLE 10

Sketch the graph of f (x (x) = ex − 2 + 1, marking the asymptote and intercept, and state the transformations, domain and range. Find the y-intercept correct to 2 decimal places. THINK

eBoo k plus eBook Tutorial


WRITE/DRAW

1

Write the rule.

f (x) = ex − 2 + 1

2


Exponential curve

3


The horizontal translation is 2 units to the right and the vertical translation is 1 unit up.

4

Find the asymptote by translating the asymptote of f (x) = ex up one unit.

The horizontal asymptote is y = 1.

5


If x = 0, y ≈ 1.14 (to 2 decimal places) so the y-intercept is 1.14.

6

Find another point by translating the y-intercept of (0, 1) → (2, 2) f (x) = ex to the right by 2 units and up by 1 unit.

Chapter 4


193

7

y

Sketch the graph.

(2, 2)

(0, 1.14) 2 −1 0 8


1

Asymptote y=1

2

f (x) = ex − 2 + 1

x

3

The domain is R and the range is (1, ∞).

A reflection in the x-axis is shown by a negative sign before the ex term. A reflection in the y-axis is shown by a negative sign before the x term. WORKED EXAMPLE 11

Sketch the graph of f (x (x) = 2 − e−x, marking the asymptote and intercepts. State the transformations, domain and range. Give exact answers. Check using a CAS calculator. THINK

WRITE/DRAW

1

State the rule.

f (x) = 2 − e−x

2


Exponential curve

3


A reflection in the x-axis and a reflection in the y-axis. The vertical translation is 2 units up.

4

Find the horizontal asymptote by translating the asymptote of f (x) = ex up 2 units.

The horizontal asymptote is y = 2.

5

Find the y-intercept by making x = 0 or by reflecting (0, 1) in the x-axis and translating it up 2 units.

If x = 0,

6

Find the x-intercept by making y = 0 and solving the equation. 1 e−x = x e loge (eex) = x loge (e) =x×1

If y = 0,

−

y=2−e 0 =2−1 =1 or (0, 1) → (0, −1) → (0, 1) The y-intercept is 1. 2 − e−x = 0 − =2 e−x 1 =2 ex ex = 12 loge (eex) = loge  1  2

x=

loge  1  2

so the x-intercept is loge  1  . 2

7

Sketch the graph.

y (loge( 1–2 ), 0)

2

Asymptote y=2 (0, 1) 0 1

−1

f (x) = 2 − e−x x

−2 8

194


The domain is R and the range is (−∞, 2).


9

To graph y = 2 − e−x, on a Graphs page, complete the function entry line as: f1(x) = 2 − e−x Then press ENTER ·. Note: The horizontal asymptote at y = 2 is not displayed.

2 , g(x) = e2x 2 − 1 and h(x) = e2x 2 −2 The graphs of f (x) = e2x are drawn at right.

y

2 f x) = e 2x f( 2 −1 g(x) = e 2x

2

2 −2 h(x) = e 2x

(0, 1) 1 (0, e−1) 0

−1 2 f (x ( ) = e2x

2 −1 g(x ( ) = e2x (x = e 2  x −

x-intercept

—

—

y-intercept

(0, 1)

Domain Range Horizontal asymptote

(

−  1 1  0, e  = 0, e



2 −2 h(x ( ) = e2x (x 2( 2(x = e − 1)

—

)

(

−  1 2  0, 2  = 0, e e

R

R

R

R+ = (0, ∞)

R+ = (0, ∞)

R+ = (0, ∞)

y=0

y=0

y=0

1 2

Horizontal translation A point on the graph

1 2

(0, 1)

unit to right

( , 1) 1 2

(1, 1) ( —12 , 1) Asymptote (0, e−2) y = 0 x 1

)

1 unit to right (1, 1)

In summary: Transformations can be represented in the following way: 1. Dilation by factor A from the x-axis can be written as y = Af (x), A > 0. 2. (a) Dilation by factor 1 from the y-axis can be written as y = f (kx), k > 0. k (b) Dilation by factor k from the y-axis can be written as y = f 1 x , k > 0. k 3. (a) Translation of B units up can be written as y = f (x) + B, B > 0. (b) Translation of B units down can be written as y = f (x) − B, B > 0. 4. (a) Translation of b units to the right can be written as y = f (x − b), b > 0. (b) Translation of b units to the left can be written as y = f (x + b), b > 0. 5. (a) Reflection in the x-axis can be written as y = −f (x). (b) Reflection in the y-axis can be written as y = f (−x).

( )

Chapter 4


195

For all the graphs of the form f (x) = Aex + b + B, where A ∈ R+, b, B ∈ R, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B. When A ∈ R− the range changes to (−∞, B). 1. A is the dilation factor from the x-axis. As A increases, the graph becomes steeper. 2. B is the vertical translation. When B > 0 the graph is translated up B units. When B < 0 the graph is translated down B units. 3. b is the horizontal translation. When b > 0 the translation is to the left. When b < 0 the translation is to the right. 4. When A < 0 the graph is reflected in the x-axis. 5. When f (x) = e−x the graph is reflected in the y-axis. 6. When f (x) = ekx the graph is dilated by factor 1 from the y-axis. k 7. Consider transformations in the following order: reflection, dilation, translation. EXERCISE

4C

Graphs of exponential functions with base e 1 WE7 State the transformations of f (x (x) = ex needed to form each of the following functions. 2 a f (x) = e3x b f (x) = e2x c f (x) = 4 e d f (x) = 2e e f (x) = 1 + ex − 2 f f (x) = 2 + ex + 5 2 2x g f (x) = 3 − e h f (x) = 1 − e3x 2 i f (x) −1 x 2

x 3


Spreadsheet 034 Exponential functions

2 WE8 For each of the following examples sketch the graph of f (x (x) = ex. On the same set of axes, sketch the graph of the given functions, marking the asymptote and y-intercept, and state the transformation, the domain and the range. Give exact answers. a f (x) = ex − 1 b f (x) = ex − 3 c f (x) = 2eex d f (x) = 3eex e f (x) = 1 + ex f f (x) = ex − 2 g f (x) = e2 − x h f (x) = e1 − x i f (x) = 3 − ex 3 WE9 Sketch the graph of each of the following functions, marking the asymptote and intercept, and stating the transformations, domain and range. For the intercepts give exact answers or correct to 1 decimal place where appropriate. 2 a f (x) = e3x b f (x) = e2x c f (x) = 4eex 1 1 4x x d f (x) = 2ee e f (x) = 2 e f f (x) = 4 e 2 x g f (x) = 2e3x h f (x) = 3e4x 4 WE10 Sketch the graph of each of the following, marking the asymptote and intercepts, and stating the transformations, domain and range. Mark the y-intercepts as exact values. a f (x) = ex + 3 b f (x) = ex − 3 c f (x) = ex + 1 x x + 1 d f (x) = e + 2 e f (x) = e +2 f f (x) = ex + 2 + 1 x − 1 x − 2 g f (x) = e +3 h f (x) = e +5 i f (x) = ex + 2 − 1 5 WE11 Sketch the graph of each of the following functions, marking the asymptote and intercepts, and stating the transformations, domain and range. For intercepts, give exact answers or correct to 1 decimal place where appropriate. a f (x) = e−x b f (x) = −ex c f (x) = 1 − ex −x −x d f (x) = 2 + e e f (x) = 1 + e f f (x) = 1 + ex −x −x g f (x) = 3 − e h f (x) = 5 − e i f (x) = −2e−x 196


2 − 3 − 1, marking all the key features and stating the domain 6 Sketch f: f R → R where f (x (x) = e2x and range. Write answers correct to 2 decimal places where appropriate.

7 Sketch f: f [0, ∞) → R where f (x (x) = ex − 2 − 1, marking all the key features and stating the domain and range, giving answers correct to 2 decimal places where appropriate. 8 If f (x (x) = ex, sketch the following functions. a y = f (x) + 1 b y = f (x) − 2 1 d y = 2 f (x) e y = 1 − f (x) g y = f (x + 1) h y = f (x − 2) j y = 1 + f (−x)

c y = 2f 2f (x) f y = 2 − f (x) i y = f (−x)

2 + 3 is translated so that the new graph has a horizontal 9 MC The graph with equation y = e2x asymptote of y = −1. The new graph has undergone a translation of: A 1 unit down B 2 units down C 3 units down D 4 units down E 5 units down

10 MC An increasing exponential function has a horizontal asymptote of y = 2 and a y-intercept of 4. A possible equation for this exponential is: 2 +4 2 −4 A y = e2x B y = e4xx + 2 C y = e2x x x y D y = 2ee + 4 E y = 2ee + 2 −

11 The graph shown is modelled by the equation y = 2eex b + B. a Find the values of B and b. b If the graph was translated down 1 unit, translated 3 units to the left and dilated by factor 12 from the x-axis, write the equation of the new graph.

4D

4 3 2 −2 −1

(3, 3) Asymptote y=1 0 1 2 3

4

x

Logarithmic graphs to base e Graphs of logarithmic functions with base e are drawn in exactly the same way as with any other base. The function f (x) can be dilated, translated and reflected in the same way. The graphs of f (x) = log2 (x), g(x) = log10 (x) y Asymptote f x) = log2 (x) f( x=0 and h(x) = loge (x) are shown at right. The graph of 2 h(x) = loge (x) h(x) = loge (x) is in blue. (1, 0) 1 f (x) = log2 (x) and h(x) = loge (x) are steeper than g(x) = log10 (x) g(x) = log10 (x). x h(x) = loge (x) −1 0 1 2 3 4 5 −1 ⇔ h(x) = ln (x)

Common features 1. The graphs all cross the x-axis at (1, 0) because loga (1) = 0. 2. The vertical asymptote is the y-axis (x = 0) because loga (0) is undefined. 3. The domain is R+. 4. The range is R. 5. They are all increasing functions. WORKED EXAMPLE 12

State the transformations of f (x (x) needed to form the graph of f (x (x) = 2 loge (x ( − 3) + 1. THINK

WRITE

1

State the rule.

f (x) = 2 loge (x − 3) + 1.

2

State the dilation. The coefficient of loge (x) gives the dilation from the x-axis.

Dilation is 2 units from the x-axis.

Chapter 4


197

3

State the translations. (a) Horizontal translation is given by the constant added to the x term.

Horizontal translation is 3 units to the right.

(b) Vertical translation is given by the constant added to the log term.

Vertical translation is 1 unit up.

Dilations do not change the vertical asymptote, the domain or the range of a logarithmic graph. The x-intercept does change, however. WORKED EXAMPLE 13

Sketch the graph of f (x (x) = 3 loge (2x (2 ), marking all key features and stating the domain and range. THINK

WRITE/DRAW

1

Write the rule.

f (x) = 3 loge (2x (2 )

2

State the basic shape of the curve.

A log graph with basic shape f (x) = loge (x).

3

State the vertical asymptote, which occurs at loge (0). Vertical asymptote is the y-axis.

4

Find the x-intercept by letting y equal 0. Solve for x. Divide both sides by 3. Use loge (x) = a ⇔ ea = x. Use e0 = 1. Divide both sides by 2.

5

Sketch the graph of f (x) = 3 loge (2x (2 ).

If y = 0, 3 loge (2x (2 ) = 0 loge (2x (2 ) = 0 e0 = 2x 2 2 =1 2x x = 12 The x-intercept is 12. y Asymptote x=0 6 4 2

6


−1 0 −2

f (x) = 3 loge (2x (2 ) ( 1–2 , 0) 1

2 3

x

The domain is R+, and the range is R.

Translations do not change the shape of the basic graph, only the position. The asymptotes, intercepts and domain may change but the range stays the same. WORKED EXAMPLE 14

Sketch the graph of f (x (x) = ln ((x + 1) − 2, marking the vertical asymptote and the intercepts. THINK

198

WRITE/DRAW

1

State the rule.

f (x) = ln (x + 1) − 2

2


Log graph with shape f (x) = ln (x).

3

Find the vertical asymptote by making (x + 1) = 0.

x + 1 = 0 for asymptote so x = −1 is the vertical asymptote.

4

Find the y-intercept by making x equal to 0. that ln (1) = 0.

If x = 0, y = ln (0 + 1) − 2 = ln (1) − 2 − = 2 − y-intercept = 2.


5

Find the x-intercept by making y equal to 0. Use ln (x) = a ⇔ ea = x.

6

Sketch the graph.

If y = 0,

ln (x + 1) − 2 = 0 ln (x + 1) = 2 e2 = x + 1 x = e2 − 1 x ≈ 6.39 x-intercept ≈ 6.39 (to 2 decimal places) Asymptote x = −1 y −2

0 2 −2 (0, −2)

(e2 − 1, 0) 4 6 x

f (x) = ln (x + 1) − 2

4

A reflection of f (x) = loge (x) in the x-axis does not change the graph’s asymptote, x-intercept, domain or range. If there is a y-intercept, the sign changes. A reflection in the y-axis does not change the vertical asymptote or the range, but the x-intercept and the domain change.

WORKED EXAMPLE 15

eBoo k plus eBook

Sketch the graph of f (x (x) = 2 − 3 loge (1 − x), marking the asymptote and intercepts. State the domain and range and check using a CAS calculator. THINK

Tutorial

int-0537

Worked example 15

WRITE/DRAW

1

State the rule.

f (x) = 2 − 3 loge (1 − x)

2

Find the vertical asymptote by translating the line x = 0 one unit to the right or by making 1 − x = 0.

Vertical asymptote is x = 1.

3

Find the y-intercept by making x equal to 0 and solving the equation.

If x = 0, y = 2 − 3 loge (1) =2

4

Find the x-intercept by making y equal to 0.

If y = 0, 2 − 3 loge (1 − x) = 0 3 loge (1 − x) = 2 loge (1 − x) = 23 2

e3 = 1 − x

2

x = 1 − e3 x ≈ −0.95 (to 2 decimal places) 5

Sketch the graph, ing that there is a reflection in both the x- and the y-axes.

y 4 (0, 2) 2– (1 − e 3 , 0)

f (x) = 2 − 3 loge (1 − x)

x −22 −1 0 1 −2 Asymptote x=1 6


The domain is (−∞, 1) and the range is R.

Chapter 4


199

7

To graph f (x) = 2 − 3 loge (1 − x) on a Graphs page, complete the function entry line as: f1(x) = 2 − 3 ln (1 − x) Then press ENTER ·. Note: The vertical asymptote at x = 1 is not displayed.

To sketch a graph by using transformations it is necessary to dilate, reflect and then translate.

1. The function f (x) = loge (x) is an increasing logarithmic Asymptote y x=0 function: y = loge (x) (a) The vertical asymptote is the y-axis so there are no y-intercepts. (b) The graph crosses the x-axis at (1, 0) because loga (1) = 0. x 0 (1, 0) (c) The domain is R+; that is, there are no negative values of x. (d) The range is R. (e) After transformations, the function becomes f (x) = A loga k(x + b) + B. 2. Dilation (a) Dilations from the x-axis are given by the coefficient of loge (x). Then, y = f (x) becomes y = Af (x). (b) Dilations from the y-axis are given by the reciprocal of the coefficient of x. Then, y = f (x) becomes y = f (kx). (c) Dilations do not change the vertical asymptote, the domain or the range of a logarithmic graph. Dilations from the y-axis change the x-intercept. 3. Translation (a) Vertical translation is given by the term added to loge (x). Then, y = f (x) becomes y = f (x) + B. If B > 0 the graph is moved up, and if B < 0 the graph is moved down. (b) Horizontal translation is given by the term that is added to x. y = f (x) becomes y = f (x + b). If b > 0 the graph is translated to the left, and if b < 0 the graph is translated to the right. (c) Translations do not change the shape of the basic graph, only the position. The asymptotes, intercepts and domain may change but the range stays the same. 4. Reflection (a) Reflection in the x-axis occurs if the coefficient A is negative. Then, y = f (x) becomes y = −f (x). (b) Reflection in the y-axis occurs if the coefficient of x is negative. Then, y = f (x) becomes y = f (−x). (c) When following a sequence of transformations to sketch a graph, it is necessary to dilate, reflect and then translate.

200


EXERCISE

4D

Logarithmic graphs to base e 1 WE12 State the transformations of f (x (x) needed to form the graph of each of the following. a f (x) = 5 loge (2x (2 ) b f (x) = 2 loge (4x) c f (x) = loge (x + 1) − 3 d f (x) = loge (x − 2) + 1 e f (x) = −loge (−x) f f (x) = −loge (2x (2 ) g f (x) = 1 − loge (x − 2) h f (x) = 2 − loge (x + 3) i f (x) = 3 loge (x + 4) − 1 j f (x) = 1 − loge (x − 4) k f (x) = 2 loge (1 − x) + 3 l f (x) = 3 loge (2 − x) − 1


Spreadsheet 069 Logarithmic graphs

2 WE13 Sketch the graph of the following functions, marking all key features and stating the domain and range. Give exact values. a f (x) = loge (2x (2 ) b f (x) = loge (3x) c f (x) = 3 loge (x) x d f (x) = 2 loge (x) e f (x) = 2 loge f f (x) = 3 loge (2x (2 )

(3)

3 WE14 Sketch the graph of the following functions, marking the vertical asymptote and the intercepts. Give exact answers for a – f, otherwise round to 1 decimal place. a f (x) = 1 + loge (x) b f (x) = 2 + loge (x) c f (x) = loge (x − 1) d f (x) = loge (x − 2) e f (x) = loge (x + 2) f f (x) = loge (x + 3) g f (x) = loge (x − 3) + 2 h f (x) = loge (x − 1) + 1 i f (x) = loge (x + 3) − 2 j f (x) = loge (x + 1) − 2 4 WE15 Sketch the graph of the following functions, marking the asymptote and intercepts, and stating the domain and range. Give intercepts correct to 2 decimal places where appropriate. a f (x) = loge (−x) b f (x) = −loge (x) − c f (x) = 2 loge (x) d f (x) = loge (−2x 2 ) − − − e f (x) = 3 loge ( 2x 2 ) f f (x) = −2 loge ( 4x) g f (x) = loge (1 − x) h f (x) = loge (2 − x) i f (x) = −loge (2 − x) j f (x) = −loge (3 − x) 5 The graph of f (x (x) = loge (x ( ) undergoes the following transformations. Find the equation of the image of f (x (x) in each case. a Translation of 1 unit to the right. b Translation of 2 units down. c Dilation of 5 units from the x-axis. d Dilation of 3 units from the y-axis. e Reflection in the x-axis. f Reflection in the y-axis. g Reflection in the x-axis and translation of 3 units to the left. h Dilation of 4 units from the y-axis and translation 1 unit up. i Reflection in the x-axis, dilation of 2 units from the x-axis and translation of 3 units down. 6 Sketch the graphs of the equations found in question 5, showing the asymptote and intercepts, and stating the domain and the range. Give intercepts correct to 2 decimal places where appropriate. 7 State the transformations of f (x (x) needed to form the graph of f (x (x) = 2 loge (3x + 6) − 1 and sketch the graph, showing the asymptote and intercepts. Give intercepts correct to 2 decimal places. −

8 Sketch the graph of ff: ( ∞, 1] → R where f (x (x) = −3 loge (2 − x), showing key features. 9 MC The graph of y = loge (x) is transformed into the graph of y = 5 loge (2x (2 ) by: A A dilation of factor 5 from the x-axis and a dilation of factor 2 from the y-axis B A dilation of factor 2 from the x-axis and a dilation of factor 5 from the y-axis C A dilation of factor 15 from the x-axis and a dilation of factor 2 from the y-axis D A dilation of factor 5 from the x-axis and a dilation of factor 12 from the y-axis E A dilation of factor 12 from the x-axis and a dilation of factor 5 from the y-axis. Chapter 4


201

10 MC For the function f (x) = 3 loge (x − 2), the vertical asymptote and the coordinates of the x-intercept are respectively: A x = 3, (2, 0) B x = 2, (3, 0) C y = 3, (2, 0) D y = 2, (3, 0) E x = 3, (3, 0) a 11 MC The function ff: [0.5, 6] → R where f (x) = is undefined when x is equal to: logg e ( x ) A 0 B 0.5 C 1 D 6 E a 12 a When the function f (x (x) = a + b loge (x ( ) is reflected in the x-axis and translated 2 units to the right, find g(x ( ), the equation of the image of f (x (x (x). b State the domain and range of g(x) and write it using correct function notation. c If g(x) is reflected in the y-axis and dilated by factor 4 from the x-axis, write in correct function notation the equation of h(x), the image of g(x).

4E


WorkSHEET 4.2

Finding equations for graphs of exponential and logarithmic functions As with other functions it is sometimes necessary to be able to find the equation of an exponential or logarithmic function from a graph. If we know points on the curve, we can substitute the values into the most suitable general equation: 1. For an exponential graph the general equation is y = Ae(x + b) + B. 2. For a logarithmic graph the general equation is y = A loge (x + b) + B. Both of these examples are written with base e but they could be written with any base, for example y = A × 2(x + b) + B or y = A log10 (x + b) + B. If there are two unknowns, two pieces of information are necessary. The coordinates of two points, substituted into a general equation, will give two equations and enable two unknowns to be found.

WORKED EXAMPLE 16 y 4

The equation of the graph shown is of the form f (x (x) = Aex + B. Find the values of A and B correct to 2 decimal places and hence find the equation of the function.

(0, 2) (−2.44, 0) −4 −3 −2 −1

THINK 1

2 3

202

Use the point on the y-axis and substitute values into the given equation. Substitute the coordinates of another point into the given equation. Solve simultaneous equations by subtracting [1] from [2].

0 1 x

WRITE

For (0, 2): 2 = Ae0 + B 2=A+B (−2.44, −2.44

For 0 = Ae

[2] − [1]:


[1]

0): +B

[2] −2

−2

−2.44

= Ae −A − = A(e 2.44 − 1)

4

Find A, rounding the answer correctly.

−2

A=

−

2.444

Substitute in [1]: 2 =

−

2.444

e −1 ≈ 2.19 (to 2 decimal places)

5

Find B by substituting the exact value of A in equation [1] or [2] to find B.

−2

e

B=2−

e − ≈ 0.19 x f (x) = 2.19ee − 0.19

6

−

−1 −2 2.444

+B −1

Rewrite the original equation, substituting values for A and B. Note: You can use a CAS calculator to solve equations [1] and [2]. 1 On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Press /r to obtain the expression template and choose the simultaneous equation template for two unknowns. Complete the entry line as:  a + b = 2  solve   , a  −2.44 +b=0   a × e Press ENTER ·. 2

Write down the solutions correct to 2 decimal places.

∴ A = 2.19 and B = −0.19

The horizontal asymptote of an exponential equation gives the vertical translation and hence the value of B in the general equation f (x) = Aex + B. WORKED EXAMPLE 17

The equation of the graph shown is of the form f (x (x) = aex + b. Find the values of a and b and hence find the equation of the function.

y 10 (0, 5)

THINK

WRITE

−3 −2 −1

Asymptote y=2 0 1

2

x

1

Use the horizontal asymptote to find the value of b.

The graph of ex has been translated up 2 units, so b = 2.

2 3

Substitute the value of b in the equation. Use the y-intercept to find a by substituting in the given equation.

4

Write the equation, using the values of a and b.

f (x) = aex + 2 For (0, 5): 5 = ae0 + 2 ae0 = 3 a=3 f (x) = 3eex + 2

If there are three unknowns, three pieces of information are necessary to solve the equation.

Chapter 4


203

The vertical asymptote of a logarithmic graph gives the horizontal translation and hence the value of b in the equation y = A loge (x + b) + B. A translation to the left gives a positive value of b and a translation to the right gives a negative value of b. WORKED EXAMPLE 18

The equation of the graph shown is of the form y = A loge (x ( + b) + B. Find the values of A, b and B and hence find the equation.

eBoo k plus eBook

y

x = −1 5 (0, 5) (2, 0) 1

2

3

Use the vertical asymptote to find the value of b.

2 3

Substitute the value of b into the equation. Use the y-intercept to find an equation.

4

Simplify using loge (1) = 0.

5

Substitute the value of B in the equation. Use the x-intercept to find A.

x

The graph of f (x) = loge (x) is translated 1 unit to the left, so b = 1. f (x) = A loge (x + 1) + B. For (0, 5): A loge (0 + 1) + B = 5 A loge (1) + B = 5 =5 f (x) = A loge (x + 1) + 5 For (2, 0): A loge (2 + 1) + 5 = 0 A loge (3) + 5 = 0 A loge (3) = −5

Substitute values back into the original equation.

−

5 log e (3) A ≈ −4.55 (to 2 decimal places) f (x) = −4.55 loge (x + 1) + 5 A=

7

4

WRITE

1

6


0 THINK

Tutorial

1. The general equation for an exponential graph is f (x) = Aek(x + b) + B, where A is the dilation factor from the x-axis, 1 is the dilation factor from the y-axis, b is the k horizontal translation and B is the vertical translation. 2. The general equation for a logarithmic graph is f (x) = A loge k(x + b) + B, where A is the dilation factor from the x-axis, 1 is the dilation factor from the y-axis, b is the k horizontal translation and B is the vertical translation. 3. The coordinates of points on the curve can be substituted into the general equation to find the values of unknowns in the equation. 4. One piece of information is required for each unknown in the general equation. 5. The horizontal asymptote of an exponential graph gives the vertical translation and hence the value of B. If the asymptote is above the x-axis B is positive and if it is below the x-axis the value of B is negative. 6. The vertical asymptote of a logarithmic graph gives the horizontal translation and hence the value of b. If the asymptote is to the left of the y-axis b is positive and if it is to the right of the y-axis b is negative.

204


EXERCISE

4E

Finding equations for graphs of exponential and logarithmic functions 1 WE16 The equation of the graph shown is of the form f (x (x) = Aex + B. Find the values of A and B correct to 2 decimal places and hence find the equation of the function. State the equation of the horizontal asymptote.

y 4 2

(1, 4) (0, 3) 0 1

2 Find the values of b and B if the general equation of the graph shown is y = ex + b + B. Give your answers correct to 2 decimal places.

2

3

y 6

4

x

(1, 6)

4 2 (0, 2) −1−0.5 0 0.5 1 1.5

3 Find the values of A and k if the general equation of the graph shown is y = Aekx. Give your answers correct to 2 decimal places.

x

y 0 −2 −1 (−1, −4) −5

1

2

x

−10 (−2, −10)

4 Find the values of A and B if the general equation of the graph shown is y = A × 2x + B. Give exact answers.

y 4 (1, 3) 2 x −33 −2 −1 0 1 2 Asymptote (−3, −1) −2 –– y = −19 15

5 WE17 The equation of the graph shown is of the form f (x (x) = a × 2x + b. Find the values of a and b and hence find the equation of the function. Give exact answers.

4 2

(0, 11–2 ) 1 Asymptote y=1 0 1 2 x −3 −2 −1

6 MC If the horizontal asymptote is y = 1 and the y-intercept is −2, the equation for the exponential function of the form y = −e x + b + B is: A y = −ex + 1 − 2 B y = −ex + 2 + 1 − x + 1 C y= e −2 D y = −ex − 1 − 2 − x + 1.1 E y= e +1

Chapter 4


205

7 Find the values of b and B if the general equation of the graph shown is y = e x + b + B. Give exact answers.

y 6 4

(2, 4) Asymptote y=3

2 0 1

2

8 Find the values of A and B if the general equation of the graph shown is y = Aex + B . Give exact answers.

3

4

x

4 (0, 3) 2 0 1 2 x Asymptote −2 y = −1

−3 −2 −1

9 WE18 The equation of the graph shown is of the form y = A loge (x ( + b) + B. Find the values of A, b and B and hence find the equation. Give answers correct to 1 decimal place.

y Asymptote 6 − x= 2

10 Find the values of A and k, given that the graph of y = A log10 kx es through the points (1, 1) and (3, 2). Give your answers correct to 4 decimal places where appropriate.

4

11 If the horizontal translation is 2 units to the left, the vertical translation is 3, the graph es through the point (2, 6), and the equation is of the form y = A log2 (x ( + b) + B, find the values of A, b and B.

4F

Addition of ordinates

(2, 6)

2 −3 −2 −1

0 1

2

eBoo k plus eBook Interactivity

Sometimes we need to sketch the graph of a function that can be thought int-0249 of as the sum of two functions. For example, the function y = x2 + ex can Addition of ordinates be thought of as the sum of the functions y = x2 and y = ex. Such a graph can be drawn by sketching the two individual functions on the same set of axes, then adding the y-values (ordinates) for each x-value and plotting the resulting points. This method is convenient to use when we know the basic shape of the individual functions but cannot recognise the basic shape of the given function. We can sketch both graphs on the same set of axes, then plot the resulting points by adding the y-values for each x-value, and hence sketch the new curve. For example, the graph of y = x2 + ex can be sketched using the addition of ordinates technique, since the basic shape of the function is not known, but the two individual functions are a basic positive parabola and a basic exponential curve. The graph of y = ex + 1 can also be thought of as the sum of two functions, but since we can recognise its shape as the basic exponential curve translated 1 unit up, there is no need to use the addition of ordinates method. Note that the ‘sum function’ can only be defined for the domain over which both of the individual functions are defined. So the domain of the ‘sum’ function is obtained by finding the intersection of the domains of the individual functions. That is, if h(x) = f (x) + g(x), domain h(x) = domain f (x) ∩ domain g(x). Sometimes a function is defined as the difference between two individual functions. We can still use the method of addition of ordinates to graph such a function, because the second function can be expressed as the addition of a negative function. That is, h(x) = f (x) − g(x) can be written as h(x) = f (x) + (−g(x)). 206


x

So if, for example, we need to graph y = x2 − ex, we could think of it as y = x2 + (−ex). We then would sketch the basic parabola and the basic negative exponential curves on the same set of axes and use the addition of ordinates technique to obtain the graph of the original function. WORKED EXAMPLE 19

State the domain of f (x (x) = 3x 3 2 + loge (x ( ). THINK

WRITE

1

Write f (x) as the sum of the two individual functions g(x) and p(x).

Let f (x) = g(x) + p(x), where g(x) = 3x2 and p(x) = loge (x).

2

State the domain of g(x).

Domain g(x) = R

3

State the domain of p(x).

Domain p(x) = R+

4

State the domain of f (x) by finding the intersection of the domains of individual functions.

Domain f (x) = domain g(x) ∩ domain p(x) = R ∩ R+ = R+

When sketching the graph by hand, it is important to select wisely the values of x for which the ordinates should be added. As a guide, good points to select are: 1. the end points of the graph 2. the points of intersection of the graphs 3. the x-intercepts of either of the graphs. WORKED EXAMPLE 20

Using the same scale and axes, sketch the graphs of y1 = ex and y2 = e−x over the domain [−2, 2). Hence, sketch the graph of y = ex + e−x, rounding coordinates to 1 decimal place. THINK 1

Sketch y1 = ex by finding the horizontal asymptote, the y-intercept, and the end points.

WRITE/DRAW

The horizontal asymptote is the x-axis. x = 0, y = e0, y = 1 x = 2, y = e2, y = 7.4 − x = −2, y = e 2, y = 0.1 The y-intercept is 1. y 8

(2, 7.4)

6

y1 = ex

4 (−2, 0.1)

2

(0, 1) 0

−2 −1 2

On the same axes, sketch y2 = e−x by reflecting y1 in the y-axis and finding the horizontal asymptote, the y-intercept, and the end points.

x

2

1

The horizontal asymptote is the x-axis. The y-intercept is 1. − y2 = e−x (−2, 2, 7.4)

y1 = ex

y 8

(2, 7.4)

6 4 (−2, 0.1) −2 −1

Chapter 4

2

(0, 1) 0

1

2

(2, 0.1) x


207

−

3

Add ordinates at the end point with the lowest x-value (y = y1 + y2 = ex + e−x).

When x = −2, y = e 2 + e2 ≈ 7.5 Point (−2, 7.5)

4

Add ordinates when the graphs intersect.

When x = 0,

y=1+1 =2

Point (0, 2) When x = 2, Point (2, 7.5)

5

Add ordinates at the other end point.

6

Plot the points (which were obtained by adding ordinates), and them to sketch the graph. It is necessary to label only end points and intercepts.

−x

y2 = e

(−2, 7.5)

y ≈ 7.5 y 8 6 4 2

−2 −1

y1 = ex

(2, 7.5)

y = ex + e − x

(0, 2)

0 1

2

x

WORKED EXAMPLE 21 y

Given the graphs of f (x (x) and g(x ( ), (x sketch the graph of h(x ( ) = f (x (x (x) + g(x ( ). (x

f x) f( g(x) x

0 THINK 1

2

Add the ordinates at the point where g(x) has the x-intercept and mark the resulting point on the set of axes. Note that g(x) = 0 at this point; therefore f (x) + 0 = f (x) (that is, if one of the functions cuts the x-axis, the sum is equal to the y-value of the other function).

3

Add the ordinates at the first point of intersection of the 2 functions. Note that at the point of intersection the value of y is the same for both functions, so the resulting point is double the y-value. Mark the point on the set of axes.

4

208

Add the ordinates at the LHS end points of the graph: a large positive value plus a smaller negative value should give a smaller positive value. Mark this point on the axes.

WRITE/DRAW 1

0

3

Add the ordinates at the point where f (x) has the x-intercept and mark the result on the set of axes.


2

y

y

f x) f(

f x) f(

g(x)

g(x)

x

y

0

4

y

f x) f(

f x) f(

g(x) 0

x

x

g(x) 0

x

5

6

7

Add the ordinates at the second point of intersection of the 2 graphs and mark the resulting point on the set of axes.

y

5

Add the ordinates at the RHS end points of the graph: 2 positive values together give an even larger positive value.

the points with a smooth curve to produce the graph of h(x). Note that the turning point of h(x) is between the y-axis and the point with the same y-value. The graph begins to turn at the left end point because the graph of g(x) turns there.

y

6

f x) f(

f x) f(

g(x)

g(x)

0

x

y

h(x)

0

x

f x) f( g(x) 0

x

WORKED EXAMPLE 22

Sketch the graph of f (x (x) = x + x , using addition of ordinates. THINK 1

State the two individual functions.

2

State the domain of f (x) by finding the intersection of the domains of individual functions.

3

On the set of axes sketch the graph of g(x) (a straight line, ing through the origin and bisecting the first quadrant) and p(x) (a basic square root curve). Note that only the first quadrant is needed, since the domain is R+ ∪ {0}.

WRITE/DRAW

Let f (x) = g(x) + p(x), where g(x) = x and p(x) = x Domain g(x) = R Domain p(x) = R+ ∪ {0} Domain f (x) = domain g(x) ∩ domain p(x) Domain= R ∩ R+ ∪ {0} Domain= R+ ∪ {0} 1

y g(x) p(x)

0 4

5

6

The first point of intersection of g(x) and p(x) is at the origin, f (x) = 0 + 0 = 0 (that is, f (x) will also start at the origin). Mark this point on the set of axes. Add the ordinates at the second point of intersection of the two graphs and mark the resulting point on the set of axes.

2

x

y

3

g(x)

y g(x)

p(x)

0

x

p(x)

0

x

Add the ordinates at the RHS end points of the graph and mark the resulting point on the set of axes.

Chapter 4


209

7

Sketch the graph of f (x) by ing the points. Note, as x increases, f (x) approaches g(x) because p(x) is increasing slowly.

f x) f( y g(x) p(x)

0

x

Further graphs The example below shows how to draw the graph of a product function. WORKED EXAMPLE 23

Sketch the graph of y = x2ex using a CAS calculator. Show all axis intercepts and any asymptotes. THINK

WRITE/DISPLAY

1

To graph y = x2ex on a Graphs page, complete the function entry line as: f1(x) = x2ex Then press ENTER ·. Note: y = x2ex is asymptotic to the negative x-axis, as when x → −∞, x2ex → 0.

2

Substitute x = 0 to locate the y-intercept. Write the y-intercept in coordinate form (which is also the x-intercept).

y = 02e0 y=0 ∴ (0, 0) y = 0 is an asymptote.

1. A graph of the sum of 2 functions can be drawn by sketching the 2 functions on the same set of axes and adding the y-values for each value of x. 2. The domain of the ‘sum’ function is obtained by finding the intersection of the domains of individual functions: if h(x) = f (x) + g(x), domain h(x) = domain f (x) ∩ g(x). 3. Suitable points at which to add ordinates are: (a) left end points and right end points (b) the points of intersection of the 2 graphs (c) the x-intercepts of either of the 2 functions. 4. Exact values of y can be obtained by substituting the required x-values into the given function. 5. The ‘difference’ function can be rewritten as the ‘sum’ function as follows: h(x) = f (x) − g(x) = f (x) + [−g(x)], so that addition of ordinates can be used.

210


EXERCISE

4F

Addition of ordinates 1 WE19 State the domain of each of the following functions. 1 a y=x+ b y=3 x −x x d y= x+2 −x g y=

e y = x2 − 3x3

2 − x2 ( x + 1)2

h y = 3x − 2 1 − x

c y = x2 + 1 x − 1 2 1 f y = 2x 2 3− x− 3 2 i y = 3(2 − x)2 + 3x

j y = 2 3− x − 3 x +1 2 MC If f (x) = g(x) + p(x), and if f (x) is defined over the domain [−2, 7] and g(x) is defined for x ≤ 7, the domain of p(x) could not be: A [−2, 7] B [−2, ∞) C [−2, 7) D [−2, 7.5) E [−2, 15] 3 WE20 Using the same scale and axes, sketch the graphs of y1 and y2 over the given domain. Hence, sketch the graph of y = y1 + y2, rounding coordinates to 1 decimal place as this is accurate enough for most sketch graphs. Check all graphs using a CAS calculator. a y1 = 2eex, y2 = 2e−x, [−1, 1] b y1 = 3e−x, y2 = 3eex, [−1, 1] −x − c y1 = e , y2 = x, ( 2, 2] d y1 = ex, y2 = x, (−2, 2] x 2 − e y1 = e , y2 = x , ( 2, 2) f y1 = e−x, y2 = x2, (−2, 3) 4 Using the same scale and axes, sketch the graphs of f (x (x) and g(x ( ). Hence, sketch the graph of (x h(x ( ) = f (x (x (x) + g(x ( ), rounding coordinates to 1 decimal place as this is accurate enough for most (x sketch graphs. State the domain and range of h(x ( ). Check all graphs using a CAS calculator. (x a f (x) = 2 loge (x), g(x) = x b f (x) = 3 loge (x), g(x) = x c f (x) = loge (x), g(x) = 2x 2 d f (x) = loge (x), g(x) = 12 x − − e f (x) = loge (x), g(x) = x f f (x) = 2 loge (x), g(x) = x 5 Using the addition of ordinates, sketch the graph of ff: [−2, 2] → R, where f (x (x) = 12 ex + 12 e−x. State the domain and the range, giving answers to 1 decimal place where rounding is necessary. 6 MC If the domain of y1 is (−2, 2] and the domain of y2 is (0, ∞), the domain of y1 + y2 is: A (−2, ∞) B [2, ∞) C (0, ∞) D (−2, 0) E (0, 2] 7 WE21 Given the graphs of f (x (x) and g(x ( ), sketch the graph of h(x (x ( ) = f (x (x (x) + g(x ( ). (x y y y a b c f x) f(

0

d

g(x)

x

0

g(x)

f x) f(

x y

x f x) f( g(x)

g(x)

e

y

0

f

f x) f(

y

f x) f(

0

g(x)

x 0

x

x

0

f x) f( g(x)

8 On the same set of axes sketch the graphs of f (x (x) = x2 and g(x ( ) = 5x + 6. Use the addition-of(x 2 ordinates method to sketch the graph of y = x + 5x + 6.

Chapter 4


211

9 On the same set of axes sketch the graphs of f (x (x) = x3 and g(x ( ) = x2 − 1 and hence sketch the (x graph of y = x3 + x2 − 1, using the addition-of-ordinates technique. 10 WE22 Sketch the graph of each of the following functions, using addition of ordinates. Check your answers using a CAS calculator. a y= x + 2− x

b y = 2x 2 − x

c y = x − 3 − x2

d y= x+5 + 5− x y

11 MC The graph at right is likely to represent the sum of which 2 functions?

y

A

B

0

C

x

y

0

D 0

0

y

x

x

y

E

y

x 0

x

0

x

12 Sketch the graph of y = loge x + x by sketching the 2 individual functions on the same set of axes and then adding the ordinates. State the domain of the function. your answer with a CAS calculator. 13 If f (x (x) = 3x − 3 and g(x ( ) = loge x, sketch the graph of h( x ) = f ( x ) + g( x ), using addition of (x ordinates. Check your answer using a CAS calculator. 14 WE23 For each of the following, sketch the graph using a CAS calculator. Show all axis intercepts and any asymptotes. State the domain and range. 2 a y = x2e2x − b y = x2e−x c y = ex 2

15 For each of the following, sketch the graph using a CAS calculator. Show all axis intercepts and any asymptotes. State the domain and range. a y = x2 loge (x) b y = −x2 loge (3x) c y = x loge (x)

4G

Exponential and logarithmic functions with absolute values  x for x ≥ 0 The modulus, or absolute value, function is defined as f ( x ) = x =  − .  x for x < 0 To obtain the graph of y = f x , the graph of y = f (x), where x ≥ 0, is reflected in the y-axis.

212


y

The rule of the composite function y = f x , where f (x) = loge (x), x > 0, can be written as: logg e ( x ) y = log e x =  − logg e ( x )

2 1

for x > 0 for x < 0

−2

−1

0 −1

2 x

1

−2 −3

To obtain the graph of y = f ( x ) , negative y-values of y = f (x) are reflected in the x-axis. The rule of the composite function y = f ( x ) , where f (x) = loge (x), x > 0, can be written as:  − logg e ( x ) y = log e ( x) x) =   logg e ( x )

y 4 3 2 1

for 0 < x < 1

0

for x ≥ 1

1

2 y

The rule of the composite function y = f x , where f (x) = ex, can be written as: e x for x ≥ 0 y = ex =  − x for x < 0 e

(0, 1) −2

y=

−1

0

1

2 x

y

The rule of the composite function y = f ( x ) , where f ( x ) = e x − k, k ∈ R+, can be written as: ex

4 x

3

y=k

 − (e x − k ) for x < log e k −k =  for x ≥ log e k  e x − k

(0, |1 − k|)

−2

−1

0

1

2 x

(loge(k), 0) WORKED EXAMPLE 24

For the function y = 2 log e x + 2 − 3: a sketch the graph of y = 2 log e x + 2 − 3, showing any asymptotes b calculate all axis intercepts both in exact form and correct to 2 decimal places c state the domain and range. THINK a

1

WRITE/DRAW y

3 Sketch the graph of y = 2 log e x + 2 − 3, x > −2. The graph is not defined when x = −2, so there is a vertical asymptote at x = −2.

2 1 −7 −6 −5 −4 −3 −22 −1−10

1 2 3 4 5 x

−2 −3 Asymptote x = −2

Chapter 4


213

2

y

Reflect y = 2 log e x + 2 − 3 in the vertical asymptote at x = −2.

2 1 −7 −6 −5 −4 −3 −22 −1−10

1 2 3 4 5 x

−2 −3 Asymptote x = −2

3

To graph y = 2 log e ( x + 2) − 3 with a CAS calculator, open a Graphs page and complete the function entry line as:

(

)

f1(x) = 2 log e x + 2 − 3 Then press ENTER ·. Press /r and select the absolute value template.

b

1

To locate the x-intercepts, express y = 2 log e x + 2 − 3 as a hybrid function.

2

Substitute y = 0 and solve each equation.

3

Write the equation in exponential form and solve for x.

2 log e ( x + 2) − 3 y= 2 log e ( − ( x + 2)))) − 3 2 loge (x + 2) − 3 = 0 3 logg e ( x + 2) = 2

for x > − 2 for x < − 2

3

4

Repeat for the other x-intercept.

x + 2 = e2 3

x = e2 − 2 ∴ x = 2.481 69 2 loge (−(xx + 2)) − 3 = 0 3 logg e ( − ( x + 2)) 2)) = 2 − ( x + 2)

3

= e2

3

x = −e2 − 2 ∴ x = −6.481 69 5

Write down the coordinates of the x-intercepts in exact form and correct to 2 decimal places.

 3  3   −  e 2 − 2, 0 and  e 2 − 2, 0     − (2.48, 0) and ( 6.48, 0)

6

Substitute x = 0 to obtain the y-intercept and simplify.

y = 2 log e 0 + 2 − 3

y = 2 log e 0 + 2 − 3

y = 2 log e 2 − 3

y = 2 log e 2 − 3

∴y = −1.613 71 7

214

Write down the coordinates of the y-intercept (0, 2 loge (2) − 3) in exact form and correct to 2 decimal places. (0, −1.61)


c

1

State the domain.

x ∈ R\{−2}

2

State the range.

y∈R

WORKED EXAMPLE 25

eBoo k plus eBook

a Sketch the graph of y = e − 1 showing all axis intercepts and asymptotes. b State the domain and range. x

THINK a

1

2

b

3

Note that the graph of y = − 1 now has a horizontal asymptote at y = 1.

1

Substitute x = 0 to obtain the y-intercept and simplify.


WRITE/DRAW

Sketch the graph of y = e x − 1 . The graph has a horizontal asymptote at y = −1. Reflect the negative part (below the x-axis) of y = e x − 1 and the asymptote about the x-axis.

Tutorial

y 2 1 −2

−1

0 −11 − −2

1

Asymptote y=1 x 2 Asymptote y = −1

ex

y = e0 − 1 = 1−1 =0

2

Write the coordinates of the y-intercept (which is also the x-intercept).

The coordinates of the y-intercept are (0, 0).

3

State the domain.

x∈R

4

State the range.

y ∈ R+∪{0}

 x for x ≥ 0 1. The modulus, or absolute value, function is defined as f ( x ) = x =  − .  x for x < 0 2. For y = f x , the graph of y = f (x), where x ≥ 0, is reflected in the y-axis. 3. For y = f ( x ) , negative y-values of y = f (x) are reflected in the x-axis. EXERCISE

4G

Exponential and logarithmic functions with absolute values 1

WE24 Sketch the graphs of the following functions, showing all axis intercepts and asymptotes in exact form. For each graph, state the maximal domain and the range of the function.

b y = − log e ( x + 1) − 2 c y = log10 3 − x + 1 y = 3 log e x − 2 + 3 2 Sketch the graphs of the following showing all axis intercepts and asymptotes in exact form. For each graph, state the maximal domain and the range of the function. y = log e ( x − 2)

b y=

− log

e (1 −

x ) − 2) x)

Chapter 4

c y = log10 ( x + 3) + 1


215

3 Sketch the graphs of the following showing all axis intercepts in exact form. For each graph, state the maximal domain and the range of the function. a

x 1 y = e x−1 +4

c y = −e 4

− x −1

+2

WE25 Sketch the graphs of the following showing all axis intercepts and asymptotes in exact form. For each graph, state the maximal domain and the range of the function.

y = ex − 3 c y= e

4H

b y = e 4− x − 3

−x

b y = − ex − 2 + 4

−1 + 2

Exponential and logarithmic modelling using graphs As seen in chapter 3, exponential and logarithmic functions can be used to model real situations. Graphs of these functions can be used to illustrate the model and make predictions for future changes. In most cases when modelling real life situations, the domain is restricted to [0, ∞) because t = 0 when the model begins.

WORKED EXAMPLE 26

The population of wombats in Snubnose Gully is increasing according to the equation: W = 100e0.03t where W is the number of wombats t years after 1 January 1998. a Find the initial size of the population. b Find the population 2 years and 10 years after the number of wombats was first recorded. Give answers to the nearest whole wombat. c Graph W against t for 0 ≤ t ≤ 30. d Find the expected size of the population in the year 2020. e Find the year in which the wombat population reaches 250. WRITE/DRAW

THINK a

b

216

a W = 100e0.03t

1

State the rule.

2

Find W when t = 0.

When t = 0, W = 100

3


The initial size of the population is 100 wombats.

1

Find W when t = 2.

b When t = 2, W = 100e0.03

×2

= 100 × 1.0618 ≈ 106 (nearest whole number)

2


After 2 years there are 106 wombats.

3

Find W when t = 10.

When t = 10, W = 100e0.03 10 = 100 × 1.3499 ≈ 135 (nearest whole number)

4


After 10 years there are 135 wombats.

×


c

1

To graph f (x) = 100e0.03x on a Graphs page, complete the function entry line as: f1(x) = 100e0.03x Then press ENTER ·. Note the viewing window settings: Xmin:0, Xmax:30, Ymin:0 and Ymax:250 When drawing your graph, label the axes with the given variables.

c

d

1

Convert the year 2020 to the correct number of years. Find W by substituting t = 22 into the equation.

d t = 2020 − 1998

2

e

= 22 years × When t = 22, W = 100e0.03 22 = 193.479 = 193 (nearest whole number) In the year 2020, there are approximately 193 wombats.

3

Write the answer as a sentence.

1

Let W = 250.

2


2.5 = e0.03

3

Take natural logs of both sides.

loge (2.5) = loge (e0.03 t) 0.03t = loge (2.5)

4

Divide both sides by 0.03.

t=

5

Evaluate the answer using a CAS calculator.

t = 30.543 ∴ t = 31 (nearest year)

6

Express t = 31 as a year and write the answer as a sentence.

There will be 250 wombats in the year 2029.

e 250 = 100e0.03

×t

×t ×

1 logg ( 2.5) 0.003 e

1. In general the number of unknowns indicates the number of points required to substitute into the equation. 2. The equation for exponential growth is A = A0ekt, where A0 represents the initial value, t represents the time taken and k represents the rate constant. This is a rapidly increasing function. 3. The equation for exponential decay is A = A0e−kt, where A0 represents the initial value, t represents the time taken and k represents the rate constant. This is a rapidly decreasing function. EXERCISE

4H

Exponential and logarithmic modelling using graphs 1

WE26 The population of a species of koala found on Eucalyptus Island is increasing according to the equation K = 50e0.04t, where K is the number of koalas t years after 1 January 1998.

Chapter 4


217


Spreadsheet 041 Function grapher

a Find the initial size of the population. b Find the population 2 years and 10 years after the number of koalas was first recorded. Give answers to the nearest whole number. c Plot a graph of K against t. d Use the graph to find the size of the population after 15 years. 2 The population of a species of wallaby found on a reserve is increasing according to the equation W = 150 × 1.08t, where W is the number of wallabies t years after records were first kept. a Find the initial size of the population. b Find the population 1 year and 5 years after records were first kept. Give answers to the nearest whole number. c Plot a graph of W against t. d Use the graph to find the size of the population after 15 years. e Use the graph to find how long it would take for the population to double. 3 A student invests $500 with a company that pays interest of 6% compounded continuously. (Interest paid according to the formula A = A0ert is said to be compounded continuously and r is called the continuous interest rate.) a How much money made up the initial investment? b How much did the student have with the company after 1 year? Give your answer correct to the nearest 5 cents. c How much interest did the student have after 5 years? Give your answer correct to the nearest 5 cents. d Plot a graph of A against t. e Use your graph to find how much the student would have in the after 8 years. 4 The decay of a radioactive element, E, is given by the equation E = E0 e−kt, where E is the number of radioactive nuclei present t days after the experiment begins. a If 200 radioactive nuclei are present in the element at the beginning of the experiment and there are 33 radioactive nuclei present after 10 days, find the value of k correct to 2 decimal places. Use this rounded value for all working in the other parts of this question. b Find the number of radioactive nuclei, E, present after 1 day and after 5 days. c Plot a graph of E against t. d Use the graph to find how long it would take before 50 radioactive nuclei are left. e Would there ever be no radioactive nuclei left? Give reasons for your answer. f The half-life of a radioactive element is the time taken for half of the radioactive nuclei to decay. Use your graph to find the half-life of this element. 5 A hard-boiled egg is placed in water to cool. The cooling process can be modelled by the equation T − T1 = (T T0 − T1)e−kt, where T is the temperature of the egg t minutes after it was placed in the water and T1 is the temperature of the water. Assume that the temperature of the egg is 98 °C when it is first placed in the 18 °C water and it takes 5 minutes for it to cool to 38 °C. a Substitute the values of T0 and T1 into the equation and simplify, making T the subject. b Find the value of k to 3 decimal places. c Substitute it into the equation. d Find how long it would take to reach a temperature of 25 °C. Give your answer to the nearest minute. e Draw a graph of T against t, marking in asymptotes. f Use the graph to find the temperature after 15 minutes. g Assuming that the water does not become significantly warmer, use your graph to determine whether the egg will ever reach the temperature of the water.

218


6 The population of a species of fish in Lake Underwater is declining due to an unknown cause. The number of fish t weeks after the first dead fish was found on the surface of the water can be modelled by the equation P = A loge (t) + B. One week after the first dead fish was discovered, the population was 10 000. The population was 8000 after 5 weeks. a Find the values of A and B. Give answers correct to 2 decimal places where appropriate. b Find how many fish there are after 10 weeks. c Find how long it would take for the population to be 3000. Give your answer to the nearest week. d Sketch a graph of P against t, clearly indicating the asymptote. e Use the graph to find how many fish there would be after 20 weeks. 7

MC The relationship between m and n is modelled by the equation m = log10 (an + b). The graph is shown at right. The values of a and b are: A 3 and 2 B 2 and 3 C 3 and −2 − D 1 and 2 E 1 and 3

m 2 1 −1

(4, 1)

(1, 0) 0 1

2

3

4

n

8 It costs a clothing company $20 to produce a jacket. Production costs are proportional to the number of jackets produced. a If the company produces n jackets, write an equation for the company’s production costs for the jackets, $C. As the company produces more jackets they find that they have to sell them at a lower price. The company’s revenue, $R, is modelled by the equation − R(n) = 2000(1 − e 0.1n) b Show that R(0) = 0. c According to this model, revenues plateau as costs increase. What is the value that the revenue approaches? d On the same axes, sketch the graphs of the cost equation and the revenue equation for 0 ≤ n ≤ 55. e Use addition of ordinates to sketch the graph of the profit the company can expect to make from selling the jackets (profit = revenue − cost). f Write an equation for this profit. g Use a CAS calculator to find the number of jackets that must be sold in order to maximise the profits and what that profit would be. h Use a CAS calculator to find how many jackets the company could make before they began to make a loss. i Is this a reasonable model for a company to use? Explain your reasons.

Chapter 4


219

SUMMARY Exponential functions y • For graphs of the form f (x) = ax, where a ∈ R+\{1}: 1. The maximal domain is R. 4 2. The range is R+. 3. The x-axis is the horizontal asymptote. 2 4. The y-intercept is 1. (0, 1) 5. They are all increasing functions. x 0 1 2 −2 −1 • Reflection: 1. If f (x) = ax is reflected in the x-axis the result is the graph of f (x) = −ax. The graph is a decreasing function instead of an increasing function. The y-intercept changes to (0, −1) and the range becomes R−. 2. If f (x) = ax is reflected in the y-axis the result is the graph of f (x) = a−x. All key features stay the same but the graph is a decreasing function instead of an increasing function. • Translation: For all of the graphs of the form f (x) = ax + b + B, where b, B ∈ R, and a ∈ R+\{1}, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B and they are all increasing functions. b translates the graph horizontally, B translates the graph vertically. • Dilation: If f (x) = Aakx, where A, k ∈ R+, the graph of f (x) = ax is dilated by factor A from the x-axis and by factor 1 k from the y-axis. • Combinations of transformations: For all the graphs of the form f (x) = A × ak(x + b) + B, where b, B ∈ R, A, k ∈ R+ and a ∈ R+\{1}, the maximal domain is R, the range is (B, ∞), the horizontal asymptote is y = B and they are all increasing functions. b translates the graph horizontally, B translates the graph vertically and A dilates the graph by factor A from the x-axis, k dilates the graph by factor 1 from the y-axis. If A, k < 0 the graphs are reflected in the x- and k y-axes, respectively. Logarithmic functions

• For graphs of the form f (x) = loga (x), where a ∈ R+\{1}: y 1. The maximal domain is R+; that is, there are no negative values of x. 2. The range is R. 3. The vertical asymptote is the y-axis so there are no x 0 (1, 0) y-intercepts. 4. The graph crosses the x-axis at (1, 0) because loga (1) = 0. 5. They are all increasing functions. • Reflection: 1. If f (x) = loga (x) is reflected in the x-axis the result is the graph of f (x) = −loga (x). All key features remain the same but the graph is a decreasing function instead of an increasing function. 2. If f (x) = loga (x) is reflected in the y-axis the result is the graph of f (x) = loga (−x). The vertical asymptote and the range remain the same but the x-intercept and the domain change. (a) The graph crosses the x-axis at (−1, 0). (b) The domain is (−∞, 0). • Dilation: 1. The function f (x) = A loga (x) dilates the graph of f (x) = loga (x) by a factor of A from the x-axis. The vertical asymptote, x-intercept, domain and range remain the same. As A increases, the graph becomes steeper.

220


2. The function f (x) = loga (kx) dilates the graph of f (x) = loga (x) by a factor 1 from the y-axis. The vertical k asymptote, domain and range stay the same, but the x-intercept is 1 . As k increases, the graph becomes k steeper and the x-intercept becomes smaller. • Translation: 1. The function f (x) = loga(x) + B translates the graph of f (x) = loga (x) vertically, B units. The vertical axis, domain and range remain the same but the x-intercept changes. There is no change in the shape of the graph. 2. The function f (x) = loga (x + b) translates the graph of f (x) = loga (x) horizontally b units. The shape and the range remain the same but the vertical asymptote, the x-intercept and the domain change: (a) The vertical asymptote becomes x = −b. (b) The graph crosses the x-axis at (1 − b, 0). (c) The domain is (−b, ∞). • Combinations of transformations: The function f (x) = A loga (x + b) + B has dilation factor A from the x-axis and is translated b units horizontally and B units vertically. If A < 0 there is a reflection in the x-axis and if −x is used there is a reflection in the y-axis. Addition of ordinates

• A graph of the sum of two functions can be drawn by sketching the two functions on the same set of axes and then adding the y-values for each value of x. • If h(x) = f (x) + g(x), domain h(x) = domain f (x) ∩ domain g(x). • Suitable points at which to add ordinates are: 1. the end points of the graph 2. the points of intersection of the two graphs 3. the x-intercepts of the two graphs. • The technique can be used for the difference of the two functions, if it is rewritten as a sum: h(x) = f (x) − g(x) = f (x) + [−g(x)]. Further graphs

• Use a CAS calculator to graph the product of two functions or composite functions. • Obtain the equation of any asymptote of the function by considering asymptotic behaviour of the individual functions. • On your graph, clearly label: 1. asymptotes 2. axis intercepts. Absolute value graphs

 x for x ≥ 0 • The modulus, or absolute value, function is defined as f ( x ) = x =  − .  x for x < 0 • For y = f x , the graph of y = f (x), where x ≥ 0, is reflected in the y-axis. • For y = f ( x ) , negative y-values of y = f (x) are reflected in the x-axis.

Chapter 4


221

CHAPTER REVIEW −

SHORT ANSWER ( + 1) − 4, showing 1 Sketch the graph of x) = 2(x intercepts and asymptotes, and stating the domain and range.

2 Sketch the graph of f (x (x) = 3 log10 (2x (2 ), showing intercepts and asymptotes, and stating the domain and range. 3 Find the equation of the graph below, given that it is of the form f (x (x) = Aex + B. y 4 2 x −4 −3 −2 −1 0 1 2 (0, −1) −2 y = −4 −44 −5

4 The graphs of f (x (x) and g(x ( ) are shown below. (x Sketch the graph of ((f + g)(x )( ), using the addition)(x of-ordinates technique. y a g(x) f x) f( x

b

g(x)

y

b determine the rule for g 1(x), and hence state − the domain and range of g 1(x). 8 If h(x ( ) = f (x (x (x) + g(x ( ) where f (x (x (x) = x2 + 1 and g(x ( ) = loge (x (x ( ), sketch the graph of h(x ( )= (x f (x (x) + g(x ( ). State the domain and range (x of h(x ( ). (x 9 Describe a sequence of transformations that maps the graph of y = f (x (x) on to the graph of y = 2f 2f (3 – x) + 5. 10 After a protection program for tigers was introduced in a province of India on 1 January 2006, the population of the tigers is modelled by T t) = 50 × 20.4t, where t is the number of months T( after the start of the program. At the same time, the number of elephants in the province is modelled by − E(t) = 400 × 4 0.1t. Find: a the number of tigers and elephants in the province on 1 January 2006 b which of the two animals has the highest numbers in the province on 1 April 2007 c the date when the population of the tigers will equal the population of elephants in the province. If the number of elephants falls below 25, they are at risk of extinction in this province. d According to the model, will this happen? If so, when will it happen?

f x) f( x

5 a Use the law loga (m mp) = p loga (m) to simplify 2 f (x (x) = log10 (x ( ) and hence sketch the graph of the function ff: (0, ∞) → R where f (x (x) = log10 (x ( 2). b Sketch the graph over the domain, R, noting that it is now possible to take negative values of x. 6 P = P0e−t. If P = 120 when t = 0, find the exact value of P when t = 20. 7 For the function g(x ( ) = 2eex + 1 – 4: (x a sketch the graph of g(x) by finding the equation of any asymptotes and the coordinates of all intercepts

222

11 Sketch the graph of y = − log e ( x + 3) , showing all intercepts and asymptotes in exact form. State the maximal domain and the range.


)

(

12 Let f : D → R, f ( x ) = 2 log log e x + 3 + 1, where D is the maximal domain of f. f a State D. b Find the exact coordinates of the points where the graph of y = f (x) intersects the x- and y-axes. c Sketch the graph of y = f (x). Indicate any asymptote with its equation.

6 Which of the following graphs best represents the function f (x (x) = log10 (x ( − 2) + 1? A y x=2

2 −6 −4 −2 0 −2

2 4

6

2 4

6

2 4

6

x

−4

EXAM TIP

R/{−3}

R\{−3}.

a Some students wrote instead of Some also wrote R−3, or just −3, which are insufficient when specifying the domain. b Some students did not give exact values and some missed the second x-intercept, because they were not aware of the absolute value function inside the natural logarithm. Instead of writing loge (3), some students appeared to write ‘3’. c Some students only drew the right-hand branch of the function. Some did not show asymptotic behaviour correctly, particularly by having branches suddenly stop at a circle somewhere on the asymptote. The function should not curve away from the asymptote, nor touch it.

B

x = −2 y 2 −6 −4 −2 0 −2 − −4 −

C

x = −2 y 2

[Assessment report 1 2003]

[© VCAA 2003]

−6 −4 −2 0 −2 −

MULTIPLE CHOICE

1 The horizontal asymptote for the graph of f (x (x) = 2x − 1 is: A x=2 B x=1 C y=0 D y = −1 E y = −2

4 When the function f (x (x) = log10 (x ( − 1) is translated 2 units up and 1 unit to the left the function becomes: A log10 (x − 2) + 2 B log10 (x − 3) + 3 C log10 (x) + 2 D log10 (x + 1) + 2 E log10 (x + 2) + 1 5 The function f (x (x) = log2 (x ( + 1) has as its domain: A (1, ∞) B [1, ∞) C (−1, ∞) − D [ 1, ∞) E R

x

−4 −

D

2 The domain and range of the graph of f (x (x) = 3 × 10x + 1 are respectively: A R, [1, ∞) B R, (1, ∞) C R, [3, ∞) D R, (3, ∞) E R, R 3 The y-intercept for the graph of f (x (x) = 1 − 2x is: A 0 B 1 C 2 D 3 E 4

x

y 2

0

x=2

1 2 3 4 5 6 7

x

−2

E

x = −2 y

2

−3 −2 −1

0 1 2 3 x

7 The graph of f (x (x) = 2eex + 1 is obtained from the graph of f (x (x) = ex by: A a dilation of 2 units from the y-axis and a translation of 1 unit to the left

Chapter 4


223

B a dilation of 2 units from the y-axis and a translation of 1 unit to the right C a dilation of 2 units from the x-axis and a translation of 1 unit to the left D a dilation of 2 units from the x-axis and a translation of 1 unit to the right E a dilation of 1 unit from the x-axis and a translation of 2 units to the right 8 If f (x (x) = ex, the function in the graph below is: y

14 If f (x (x) = loge (x ( + 2) + 1, then f (1) and f (0) are respectively (correct to 2 decimal places): A 2.10, 1.69 B 2.09, 0.69 C 2.098, 1.693 D 1.10, 1.69 E 2.10, 0.69 15 The domain of {{ff (x (x) + g(x ( )} is: (x A dom f ∪ dom g B dom f ∩ dom g C dom f D dom g E R −

16 If f (x (x) = g(x ( ) + p(x (x ( ), and dom g(x (x ( ) = ( ∞, 5] and (x − dom p(x ( ) = ( 2, ∞), then the domain of f (x (x (x) is: − − A ( 2, 5] B [ 2, 5) − − C [ 2, 5] D ( 2, 5) E none of the above

2 (0, 1) x −3 −2 −1 0 1 2 3 Asymptote –2 y = −2

17 Which of the following shows the graph of y=

A 3f 3f (x) + 2 C 2f 2f (x) + 3 E 2 + 3f 3f (x)

B 3f 3f (x) − 2 D 2 − 3f 3f (x)

x+4 + y

A

10 The vertical asymptote for the graph of f (x (x) = 3 loge (x ( − 2) + 1 is: − A x= 1 B x=0 C x=1 D x=2 E x=3

4

−4

4 x

4

13 The general equation of the graph shown y is y = Aex + B. The values of 4 A and B are: − − (0, 3) A 3, 1 B 2, 1 − − 2 C 4, 1 D 1, 3 E 4, 1

4 x

−4

4 x

y 4

−

12 If f (x (x) = loge (x ( + 1), the y-intercept -intercept of 22ff (x (x) + 3 is: A 0 B 1 C 2 D 3 E 4

y

2

E

4 x

–4

D

y

C −4

11 For the function ff: [ 1, ∞] → R where f (x (x) = 3 loge (x ( + 2), the domain and range are respectively: − A ( 2, ∞), (0, ∞) B (2, ∞), R − − D [ 1, ∞), [0, ∞) C ( 1, ∞), R+ E R+, R

−2

2

x

18 The decay of uranium-235 is modelled by the equation U = U0ekt, where U is the number of grams of uranium-235 after t million years. If a 1000 g mass of uranium-235 decays to 907 g in 100 million years, the values of U0 and k are respectively: − A 100, 0.000 976 B 1000, 0.000 976 − − C 10 000, 0.000 907 D 1000, 0.000 907 E 100, 907

19 The function g has the rule g(x ( ) = logex − b, (x where b is a real constant. The maximal domain of g is: A R+ B R\{b} 0 2 x C R D (−b, b) Asymptote [© VCAA 2006] E (b, ∞) −

−2 y = 1

224

y

B

4

9 If the horizontal asymptote is y = 2 and the y-intercept is 1, a possible equation for the graph is: A y = ex + 2 B y = 2eex + 1 C y = 2eex − 1 D y = ex − 2 x E y=2−e

−6 −4 −2

4−x ?


20 The maximal domain D of the function f: f D→R with rule f (x ( ) = loge (x − 3) + 6 is: B (3, ∞) A R\{3} D (−3, ∞) C R − [© VCAA 2007] E ( ∞, 3) 21 The function f satisfies the functional equation  x + y   f ( x) + f ( y)  f =  where x and y are any  2   2 non-zero real numbers. A possible rule for the function is: A f ( x ) = log e x B f ( x) = 1 x C f ( x) = 2 x D f ( x) = 2 x E f ( x ) = sinn (2 x )

22 The function f has the property that f (x (x) + f (y (y) = (x (x + y) f (xy (xy) for all non-zero real numbers x and y. A possible rule for f is: A f (x (x) = 2x 2 B f (x (x) = 5 − 2x 2 1 C f ( x) = x D f (x (x) = ex E f ( x ) = log e x [© VCAA 2005]

23 Let f (x) = ex. For all positive real numbers x and y, f (x + y) is equal to: A f (x (x) + f (y (y) B f (x (x) f (y (y) C f (xy (xy) D (f (f (x (x)) y y [© VCAA 2003] E f (x (x )

[© VCAA 2006]

EXTENDED RESPONSE

1 Sketch the graph of f (x (x) = 2eex − 1 + 1, showing all key features. State the domain and the range. 2 Sketch the graph of f (x (x) = 1 − loge (2 − x), clearly showing intercepts and asymptotes, and state the domain and range. 3 By adding ordinates, sketch the graph of f (x (x) = log10 (2x (2 ) + log10 (x ( ). State the domain and the range. 4 N is the number of bacteria in a culture where N = 10 000e0.04tt is the formula for the number of bacteria in the culture after t hours. Find: a the initial number of bacteria b the number of bacteria after 10 hours. Give your answer to the nearest thousand. 5 Sketch the graph of y = 4 − e x − 1 showing all axis intercepts in exact form and any asymptotes. State the maximal domain and the range of the function. 6 Part of the graph of the function f : R → R, f (x (x) = a + be−x is shown at right. The line with equation y =1 is an asymptote and the graph es through the origin. a Explain why a = 1 and b = −1. Let h be the function h:[0, 2] → R, h(x) = f (x). b State the range of h using exact values. − c Find the inverse function h 1.

y y=1

0

1

x

EXAM TIP

a Most students were able to explain why a = 1 by discussing the asymptote; however, many of them were unable to explain why b = −1. The most common error occurred when students tried to discuss reflections and translations but failed to indicate which graph they were transforming. − b This question was generally well done. Some students were unable to obtain full marks because they rounded up 1 − e 2 to 0.8647. Others used round brackets instead of square brackets. c When the inverse function is asked for, the domain must be given. Students will be penalised if the domain is left out. If only the rule for the inverse function is asked for, the domain does not have to be given. Some students gave the incorrect −

answer h 1 ( x ) = − log e 1 − x .


[© VCAA 2007]

Chapter 4


225

log e (5 (5 − x) x ) + 1. 7 A function f is defined by the rule f ( x ) = log a Sketch the graph of f over its maximal domain. Clearly label any intersections with the axes with their exact coordinates and any asymptotes with their equations. − b Find the rule for the inverse function f 1. EXAM TIP

The graph on a calculator screen is of limited use. Many students did not give exact intercepts. Some tried to ‘cheat’ by finding numerical approximations and then writing these as fractions (which were incorrect). Many students failed to identify the vertical asymptote at all, while some identified a horizontal asymptote. If they failed to identify the vertical asymptote, the right-hand end of the graph frequently ended abruptly on or close to the x-axis. Many students were also unable to represent asymptotic behaviour correctly. A few students still wrote a circle where their graph ended on the asymptote — this should not be encouraged, as students will lose marks for doing it. [Assessment report 1 2005]

[© VCAA 2005]

8 Kerri invested $5000 with a company that pays interest of 5% compounded continuously. To answer the questions below, use the formula A = A0ert, where A is the amount of the investment, A0 is the original investment, r is the continuous interest rate and t is the number of years since the money was originally invested. Round the answer to the nearest 5c. a How much money did Kerri first invest with the company? b How much money did Kerri have with the company after 1 year? c How much did she have after 5 years? If the company had been paying interest compounded quarterly, the formula used would have been

( )

4t

A = A0 1 + r . 4 d Using this system, how much would Kerri have after 5 years of the investment? e How long would it take to double the investment? Give your answer in years. f Which is the better investment? Give reasons. g What is the difference in the amount of interest after 5 years? h What would be the difference in 5 years if Kerri had invested $10 000? 9 A local council decided to build a new road along the coast. To make it safer it was decided to design the road so that it followed the curve represented by the equation y= 2 log10 (2x (2 − a) + 3 where a > 0. y The grid at right shows the road. In each direction, 1 unit represents Bridge 1 kilometre. Ship a If the new section of road goes 4 over the bridge marked on the graph, find the value of a. b Find the x-coordinate of the point where the road begins. Give the answer correct to 2 2 decimal places. How far would this be from the vertical axis to the nearest metre? c What is the shortest distance x 0 2 4 from the beginning of the road 6 8 1 3 5 7 9 to the coastline if the coastline is on the vertical asymptote at that point? Give the answer correct to the nearest metre. d The main highway is along the x-axis. How far is the road from the main highway when it is in line with the ship marked on the grid? Give the answer in kilometres, correct to 1 decimal place.

226


10 The air pressure P in kilopascals (kPa) at a height of x kilometres (km) above sea level may be modelled by the equation P = aebx. A mountain climber uses an altimeter to record air pressures at known heights on a climb of Mount Kosciuszko (height 2.228 km). These pressures are shown in the table below. x

0

0.5

1.0

1.5

2.0

P

101.3

95.2

89.4

84.0

78.9

−

−

a If b is between 0.1 and 0.2, find the value for b (to 3 decimal places) which produces the best fitting model of the form P = aebx for the above data. b Use your model to predict the air pressure at the top of Mount Kosciuszko. 2 − bex + c is shown below. 11 The graph of the function f (x (x) = e2x

y y=6

(0, 2) 0

a b c d e f g h

x

Find the values of b and c. Show that the exact values of the x-intercepts are x = loge (2) and x = loge (3). Use a CAS calculator to find the coordinates of the turning point. Round answers to 2 decimal places. Find the exact values of the coordinates of the point of intersection of the function and the horizontal asymptote. − eBoo k plus eBook If the function is reflected in the x-axis, fully define the new function g(x) = f (x). If the function is reflected in the y-axis, fully define the new function, h(x). Digital doc If the function is reflected in both the x- and the y-axis, sketch the graph of Test Yourself the new function, k(x), write its equation and state the domain and the range. Chapter 4 Find the equation of f (2 − x) + 1. State the domain and range, rounding to 2 decimal places where appropriate.

Chapter 4


227

eBook plus

ACTIVITIES


• 10 Quick Questions: Warm up with ten quick questions on exponential and logarithmic graphs. (page 172) 4A

Graphs of exponential functions with any base

Tutorial

• WE 2 int-0534: Watch how to sketch a graph of an exponential function. (page 177) Digital doc

• Spreadsheet 034: Investigate exponential graphs. (page 181) 4B

Logarithmic graphs to any base

Tutorial

• WE 5 int-0535: Watch how to sketch the graph of a logarithmic function. (page 187) Digital docs

• Spreadsheet 069: Investigate logarithmic graphs. (page 191) • WorkSHEET 4.1: Sketch graphs of exponentials and logarithms, identify transformations and determine rules for graphs. (page 191) 4C

Graphs of exponential functions with base e

Tutorial

• WE 10 int-0536: Watch how to sketch the graph of an exponential function and state the transformations. (page 193) Digital doc

• Spreadsheet 034: Investigate the exponential function. (page 196) 4D

Logarithmic graphs to base e

Tutorial

• WE 15 int-0537: Sketch the graph of a logarithmic function stating the domain and range using a CAS calculator to check. (page 199) Digital docs

• Spreadsheet 069: Investigate logarithmic graphs. (page 201) • WorkSHEET 4.2: Sketch graphs of exponentials and logarithms and determine points of intersection between graphs. (page 202)

228

4E

Finding equations for graphs of exponential and logarithmic functions

Tutorial

• WE 18 int-0538: Watch how to find the equation of an exponential given the graph. (page 204) 4F

Addition of ordinates

Interactivity int-0249:

• Addition of ordinates: Consolidate your understanding of addition of ordinates using the interactivity. (page 206) 4G

Exponential and logarithmic functions with absolute values

Tutorial

• WE 25 int-0539: Watch a worked example on how to sketch the graph of an absolute function. (page 215) 4H

Exponential and logarithmic modelling using graphs

Digital doc

• Spreadsheet 041: Invesigate graphs using a function grapher. (page 218) Chapter review Digital doc

• Test Yourself: Take the end-of-chapter test to test your progress. (page 227) To access eBookPLUS activities, log on to www.jalus.com.au


5

5A 5B 5C 5D

Relations and their inverses Functions and their inverses Inverse functions Restricting functions

inverse functions AREAS oF STuDy

• Graphs and identification of key features of graphs of functions studied • Graphs of inverse functions • Functions and their inverses, including conditions for the existence of an inverse function, and use of inverse functions to solve equations involving exponential, logarithmic and power functions

• The concept of an inverse function, connection between the domain and range of the original function and its inverse and the conditions for the existence of an inverse function, including the form of the graph of the inverse function for specified functions • Finding the rule of an inverse function and giving its domain and range eBook plus

5A eBook plus Interactivity

int-0248 Relations and their inverses

Relations and their inverses

Digital doc

10 Quick Questions

You will recall from Maths Quest 11 Mathematical Methods CAS that a relation is a set of ordered pairs that can be graphed or may be described by a rule. As we have seen, the inverse of a relation can be found by: 1. interchanging the x- and y-coordinates of an ordered pair; for example, (−1, −2) becomes (−2, −1) 2. reflecting the relation in the line y = x y 3. interchanging x and y in the rule and rearranging y=x 3 the equation to make y the subject; for example, (1, 2) y = x − 1 becomes x = y − 1, giving y = x + 1. 2 (2, 1) (0, 1) The domain of a relation becomes the range of its inverse 1 and the range of a relation becomes the domain of its inverse. (−1, 0) (1, 0) x 0 1 2 3 The diagram at right shows a set of ordered pairs −3 −2 −1 −1 (0, −1) − − − (−2, −1) A = {( 1, 2), (0, 1), (1, 0), (2, 1)}, the line y = x and the −2 inverse B = {(−2, −1), (−1, 0), (0, 1), (1, 2)}. (−1, −2) − −3 The domain of A is { 1, 0, 1, 2} and the range of A is {−2, −1, 0, 1}. The domain of B is {−2, −1, 0, 1} and the range of B is {−1, 0, 1, 2}.

using matrices to describe a reflection in the line y = x In chapter 2, we looked at how matrices can be used to find the image of a point that is transformed on a plane, or to find the rule for the graph of a relation that undergoes a transformation or series of transformations.

Chapter 5

inverse functions

229

As the graph of an inverse is the reflection in the line y = x of the original, we can use a matrix to describe this transformation and find the image of a point on f (x), or the rule of the inverse − f 1(x) given f (x). 0 1   x  The matrix operation that produces a reflection in the line y = x is    . 1 0   y  Finding the image of a point using this method is trivial, as we know that the points on the graph of an inverse are found by interchanging the (x, y) values of any point on the original. This can be shown as follows: The graph of a relation es through the point (3, −4). Find the image of this point on the graph of the inverse. 0 1   3   − 4    −  =   1 0   4   3  That is, (3, −4) maps to (−4, 3). A more general application will be to find the rule of the inverse of a relation. This can be done using matrices but in practice this is unnecessary as it simply involves swapping the variables and rearranging the new equation. Again, it is not time effective to employ this method, but it is shown as a demonstration of the relationship of the inverse graph to its original.   x   x ′   0 1   x   y  T    =   =    =    y   y ′   1 0   y   x  That is, (x, y) maps to (y', x' ). Therefore, we can write y' = x and x' = y, which relates to the practice of reversing the variables to find the rule of the inverse. WoRkED ExAMplE 1

Sketch the graph of each of the following relations. State the domain and range of each. a {(−3, −1), (−1, 1), (1, 3), (3, 5)} b y = x2 + 2 Think a

b

WRiTE/DRAW

1

Plot each coordinate pair on a set of axes.

2

The domain is the set of first elements of the ordered pairs.

3

The range is the set of second elements of the ordered pairs.

1

The relation is a parabola with a minimum turning point (0, 2).

2

Sketch the parabola.

3

The domain is the set of real numbers.

4

The range is any real number greater than or equal to 2.

a

y 5 4 3 2 1 −3−2−1 0 −1

b

Domain = {−3, −1, 1, 3} 1 2 3

y

x

Range = {−1, 1, 3, 5}

y = x2 + 2

(0, 2) 0

x

Domain = R Range = [2, ∞)

WoRkED ExAMplE 2

Find the inverse of each relation in worked example 1. Sketch the graph of each inverse relation, stating its domain and range.

230

Maths Quest 12 Mathematical Methods CAS for the Ti-nspire

Think a

WRiTE/DRAW a {(−3, −1), (−1, 1), (1, 3), (3, 5)}

1

Write the original relation.

2

Interchange the x and y elements to obtain the inverse. Plot these points on a set of axes.

3

The inverse is {(−1, −3), (1, −1), (3, 1), (5, 3)} y 3 2 1 −1 0 1 2 3 4 5 x −1 −2 −3

4 5

b

1 2 3 4 5

6

Domain = {−1, 1, 3, 5}

State the domain, which is the set of first elements of the ordered pair. State the range, which is the set of second elements of the ordered pair. Write the original relation. Interchange x and y in the rule.

Range = {−3, −1, 1, 3} b y = x2 + 2

The inverse is: x = y2 + 2 y2 = x − 2

Subtract 2 from both sides. Take the square root of both sides to make y the subject.

y=± x−2 y

The graph of y = ± x − 2 is a ‘sideways’ parabola with a turning point (2, 0). (use a CAS calculator to this graph.)

+ x− 2 y =−

x

2

0

7

Sketch the graph of the relation. State the domain.

Domain = [2, ∞)

8

State the range.

Range = R

Note that the domain of the original relation is the range of its inverse, and the range of the original relation is the domain of its inverse. WoRkED ExAMplE 3

eBook plus

For each relation graphed below, sketch the graph and its inverse on the same axes. Draw in the line y = x. a

b

y

0

(0, −5)

y

(2, 0) x

c

4 (−2, 2)

−2

(−2, 4)

(2, 2)

0

2

y

x

4

Tutorial


(1, 4)

(−3, 0) 0

Chapter 5

x

inverse functions

231

Think a

b

c

DRAW

1

Copy the graph shown.

2

Interchange the points (2, 0) and (0, −5) to (0, 2) and (−5, 0), respectively, and mark them on a set of axes.

3

Draw a straight line through each pair of points.

4

Draw the line y = x, noting that the inverse is a reflection of the original function in y = x.

1


2

Interchange the points (−2, 2), (0, 0), (0, 4) and (2, 2) to (2, −2), (0, 0), (4, 0) and (2, 2) and mark on a set of axes.

3

Draw a circle through the second set of 4 points.

4

Draw the line y = x, noting that the inverse is a reflection of the original relation in y = x.

1 2

3

4


a

y

(0, 2) (−5, 0)

x

(2, 0)

0

(0, −5)

y=x b

y (0, 4)

(−2, 2)

y=x (2, 2)

2

(4, 0) x

0 −2 c

Interchange the points (−3, 0), (−2, 4), (0, 0) and (1, 4) to (0, −3), (4, −2), (0, 0) and (4, 1) and mark them on the same set of axes. Draw a ‘sideways cubic’ through the second set of points. Draw the line y = x, noting that the inverse is a reflection of the original function in y = x.

(−2, 4)

(2, −2) y 4

(1, 4)

y=x

2 (4, 1) (−3, 0) −2

0 −2 (0, −3)

2

4

x

(4, −2)

Note: This example shows that the graph of the inverse relation can be obtained by reflecting the graph of the original relation in the line y = x. Some specific points can be marked and used as a guide; intercepts are particularly useful. Note also that a relation and its inverse intersect on the line y = x.

How to visualise the graph of an inverse The following activity can help you to quickly see the shape of an inverse from the graph of a relation. Make a sketch of the graph of a relation in the right-hand bottom corner of a blank page. This should be 4 or 5 cm square. You will need to use a pen or a dark pencil for best results. Fold the corner in to the centre at 45°. Keeping the page orientation the same, view the axes through the paper and you will see the graph of the inverse!

232


Looking ‘through’ paper

Fold in along this line at 45°

1. 2. 3. 4.

A relation is a set of ordered pairs which may be described by a rule. The inverse of a relation can be found by interchanging the x and y elements. The graph of a relation and its inverse are reflected in the line y = x. If the rule for a relation is known, then the rule for its inverse is obtained by interchanging x and y in the rule and then making y the subject. 5. The domain of a relation is the same as the range of its inverse and the range of a relation is the same as the domain of its inverse. ExERCiSE

5A eBook plus Digital doc

SkillSHEET 5.1 Domain and range



Relations and their inverses 1 WE 1 Sketch the graph of each of the following relations. State the domain and range of each. a {(0, 1), (1, 2), (3, 2), (3, 5)} b {(−8, 7), (−5, 2), (−2, −1), (1, 1)} c y=x d y = 2x − 5 e 2x + 4y = 8 f y = x2 + 4x 3 g y = x2 − 1 j y= h y = (x + 1)2 i x2 + y2 = 4 x k y=2 l x = −4 m y = 2x3 2 WE 2 Find the inverse of each relation in question 1. Sketch the graph of each inverse relation, stating its domain and range. (Check your graphs using a CAS calculator.)

ExAM Tip The best procedure for determining the range of a function is through sketching a graph for the given domain.

3 MC The graph of a relation es through the points (0, 1), (1, 2) and (3, 3). The graph of the inverse of this relation must through the points: A (2, 1) and (3, 3) B (0, 1) and (3, 3) C (0, 1) only D (1, 2) and (3, 3) E (3, 3) and (4, 10)

Chapter 5

inverse functions

233

4 MC A relation has x-intercepts 2 and −3. The y-intercepts of the inverse of this relation are: B −2 and −3 C 2 and −3 A −2 and 3 D 2 and 3 E cannot be determined 5 WE 3 For each relation, sketch the graph and its inverse on the same axes. Draw in the line y = x. a

b

y

d

y

−2

0

f

y

x

3

−6

j

k

y

l

y

y

4 3

x

−4

(−1, −1)

4

0

x

−4

m

n

y

−3

3

p

y

x

(−2, 3)

5b

−4

−2

0

2 x

0

x

(1, −2)

Functions and their inverses Recall that a function is a relation which has only one y-value for each x-value. The graph of a function can be crossed only once by any vertical line. To find the inverse of a function, we use the same procedures that we used for relations in the previous exercise. 1. The rule for the inverse of a function is obtained by either: (a) interchanging the x- and y-values of its ordered pairs or (b) interchanging x and y in the rule for the function, then making y the subject.

234

y

2 −4

x

x

0

x

0

o

y 0

2

0

x

−1 0

x

−2

(1, 1) 0

y

2 x

0

x

y

x

0

−4

h

y

0

i

2 x

0

g

y −2

0

y

(1, 3)

x

0

e

c

y


y y = f(x)

0

Vertical lines cross the curve only once.

x

2. The graph of the inverse of a function is obtained by either: (a) interchanging the x- and y-values of its ordered pairs, or (b) reflecting the graph of the function through the line y = x, or (c) finding the rule of the inverse and graphing the new function. The domain of a function is the range of its inverse. The range of a function is the domain of its inverse. WoRkED ExAMplE 4

State whether or not each of the following is a function by using the vertical line test. a y = (x + 1)2 − 3 b x2 + y2 = 1 4 +1 c y= x−2 Think a

WRiTE/DRAW

1

Write the equation.

2

State the type of graph.

3

Do a rough sketch and use the vertical line test to determine if the relation is a function.

a y = (x + 1)2 − 3

A parabola with minimum turning point (−1, −3) y

x

0 (−1, −3)

b

4

State whether or not it is a function.

1

Write the equation.

2


3

Do a rough sketch and use the vertical line test to determine if the relation is a function.

(0, −2)

y = (x + 1)2 − 3 is a function. b x2 + y2 = 1

A circle, centre (0, 0), radius 1 y (0, 1) (−1, 0)

(1, 0) 0

x

(0, −1)

c

4


1

Write the equation.

2


x2 + y2 = 1 is not a function. c y=

4 +1 x−2

A hyperbola with asymptotes x = 2 and y=1

Chapter 5

inverse functions

235

3

Do a rough sketch and use the vertical test to determine if the relation is a function.

y (x = 2)

(−2, 0) 0 (0, −1)

4


y=

eBook plus

For each of the following functions, sketch, on the same set of axes, the graph of the function, its inverse and the line y = x. State the domain and range for the function and its inverse. a y = x2 − 2x b y = loge (x + 1) a

(y = 1)

4 + 1 is a function. x−2

WoRkED ExAMplE 5

Think

x

Tutorial


WRiTE/DRAW a y = x2 − 2x

1

Write the original function.

2

Change it to turning point form by completing the square.

= x2 − 2x + 12 − 12 = (x2 − 2x + 1) − 1 = (x − 1)2 − 1

3

Determine the turning point of the parabola.

This is a parabola with turning point (1, −1).

4

Find the y-intercept by letting x equal 0 and substituting in the original equation.

If x = 0, y = 02 − 2(0) y=0 The y-intercept is 0.

5

Find the x-intercepts by letting y equal 0.

If y = 0, x2 − 2x = 0

6

Solve the equation by factorising the left-hand side.

x(x − 2) = 0 x = 0 and x = 2 The x-intercepts are 0 and 2.

7

Sketch the graph of the function.

y

0 −1

x 1 2 (1, −1) y = x2 − 2x

8

Interchange x and y to find the equation of the inverse.

x = (y − 1)2 − 1

9

Rearrange to make y the subject.

x + 1 = (y − 1)2 y −1= ± x +1 y = 1 ± x + 1 which is a sideways parabola.

236


10

Interchange the x- and y-coordinates of the turning point and intercepts in the original equation to find the turning point and intercepts of the inverse.

11

On the original axes, sketch the graph of the inverse and the line y = x.

(1, −1) becomes (−1, 1), (2, 0) becomes (0, 2) and the origin remains the same. y

y=x y = x2 – 2x

x y=1± x+1

b

12

State the domain and range of the function.

The domain is R and the range is [−1, ∞).

13

State the domain and range of the inverse.

The domain of the inverse is [−1, ∞) and the range is R. b y = loge (x + 1)

1

Write the function.

2

Find the vertical asymptote using the fact that loge (0) is undefined.

Vertical asymptote occurs where x+1=0 x = −1 is the vertical asymptote.

3

Find the x-intercept using loge (1) = 0.

x-intercept occurs where y = 0 or x+1=1 x = 0 so the x-intercept is 0.

4

Sketch the graph of the function.

x = −1

y

x

0

−1

y = loge (x + 1)

5

Interchange x and y to find the equation of the inverse.

x = loge (y + 1)

6

Rearrange to make y the subject.

ex = y + 1 y = ex − 1

7

Find the horizontal asymptote and intercepts from the original function.

x = −1 is the original vertical asymptote, so y = −1 is the horizontal asymptote for the inverse. (0, 0) is on both graphs.

Chapter 5 Inverse functions

237

8

On the original axes, sketch the graph of the inverse and the line y = x.

x=

−1

y

y = ex − 1

y=x

y = loge (x + 1) x y = −1 9

State the domain and range of the function.

The domain is (−1, ∞) and the range is R.

10

State the domain and range of the inverse.

The domain of the inverse is R and the range is (−1, ∞).

Worked Example 6

If f(x) = ln (x + 1) + 1, use a CAS calculator to: a find f 1(x) b draw the graph of f(x) and its inverse f 1(x). Think a

1 2 3

4

b

238

WRITE

Let y = f (x). Interchange x and y. To make y the subject, press: • Menu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(ln(y + 1) + 1 = x,y) (This means to solve the equation with respect to y). Then press ENTER ·.

a y = ln (x + 1) + 1

x = ln (y + 1) + 1

-

Write the answer in the form f -1(x). -

To draw the graphs of f (x) and f 1(x), open a Graphs page. Complete the function entry line as: f 1(x) = ln(x + 1) + 1 Then press ENTER ·. Complete the function entry line as: f 2(x) = ex - 1 - 1 Then press ENTER ·.

f 1(x) = ex - 1 - 1 b


1. A function is a relation which has only one y-value for each x-value. The graph of a function can be crossed only once by a vertical line. 2. The graph of the inverse is obtained by: (a) interchanging the x- and y-coordinates in the ordered pairs (b) reflecting the graph in the line y = x (c) finding the rule of the inverse and then graphing the inverse. 3. A horizontal asymptote y = a is the vertical asymptote x = a of the inverse, and vice versa. 4. The domain of a function is the same as the range of its inverse. The range of a function is the same as the domain of its inverse. ExERCiSE

5b

Functions and their inverses 1 WE 4 State whether or not each of the following is a function by using the vertical line test. a y = 2x − 1 d y=

−

1 −3 (2 x + 1)2

Digital doc

SkillSHEET 5.2 Matching graphs with equations

c y=

e x2 + y2 = 4

f y = 3e(x −2) + 1

h y= x +2

g y = loge (x + 1) − 2 eBook plus

2 Sketch the graph of each of the following functions and state the domain and range of each. ( that it is correct using a CAS calculator.) a (−4, −2), (−2, 0), (0, 1), (2, 4), (3, 6)} c f (x) = 5 − 2x d f (x) = x2 − 9 4 f f ( x) = g f (x) = x2 + 8x x

b 3x + 4y = 12 e f (x) = (x + 2)2 x3 h f ( x) = 2

i f (x) = 2ex

k f ( x) = 4 − x 2

j f (x) = loge (2x)

3 WE 5 Find the inverse of each function in question 2. Sketch the graph and state the domain and range of each inverse. ( using a CAS calculator.)

ExAM Tip When the inverse function is asked for, the domain must be given. Students will be penalised in the future if the domain is left out. If only the rule for the inverse function is asked for, the domain does not have to be given.

4 For each function graphed below: i copy the graph of each function and sketch its inverse on the same axes ii state the domain and range of f (x) iii state the domain and range of the inverse of f (x). eBook plus Digital doc

Spreadsheet 059 Inverse graphs

4 −1 x+3

b y = 3(x − 1)2 + 2

a

b

y

c

y

y=x

f(x)

[Assessment report 2007]

f(x)

f(x)

y

4

1 0

2

x

0

x 0

y=x

x

y=x

Chapter 5

inverse functions

239

d

e

y

f

y

f(x)

y x=1

3

f(x)

3 f(x)

y=x

g

y=x

y=x y

h

f(x)

y=x

f(x) −5

1

5x

0

x

0

−4

x

0

y y=x

(−4, 1) –3

i

y

f(x)

y=x

3

x

2

3 x

0

−3

x

0

0

−5

j

y

k

y=x

0

2

y x = −3

f(x) 6

−3 −2


WorkSHEET 5.1

3

−2 y=x

Questions 5 to 8 relate to the following function. f : R + → R, f ( x ) =

(3, 4)

x

0

−2

5 MC The range of f (x) is: A R+ B R

y 4

f(x)

2 x

l

y=x

−

0

3

x D R−

E (−∞, 0]

6 MC The domain of the inverse of f (x) is: A R− B R C R+

D [0, ∞)

E (−∞, 0]

7 MC The range of the inverse of f (x) is: C [0, ∞) B R A R−

D (−∞, 0]

E R+

C [0, ∞)

x

8 MC The inverse of f (x) can be defined by: A y = x2, where x ∈ [0, ∞) B y = x2, where x ∈ R− 2 + D y = x , where x ∈ R E y = x, x where x ∈ [0, ∞)

C y = x2 where x ∈ R

9 WE 6 For each of the following functions, fully define the inverse. a f : [−2, ∞) → R, f (x) = (x − 2)2 − 3 b f : R → R, f (x) = 3e x − 1 + 2

5C

inverse functions A one-to-one function is a function where for each x-value there is only one y-value and vice versa. The graph of a one-to-one function can be crossed only once by any vertical or horizontal line. A function that is not one-to-one is many-to-one.

240


A function will have an inverse that is also a function if and only if it is a one-to-one function. − If a function, f , is one-to-one, then its inverse function is denoted by f 1. Furthermore: − dom f 1 = ran f −1 ran f = dom f −1 The graph of f is obtained by reflecting the graph of f through the line y = x. (Note that if f − crosses the line y = x at any point, f and f 1 will intersect at that point.) The maximal domain of a y function is the largest domain y = f(x) for which its rule is defined. If a function is given without its domain specified, then it is understood that the maximal or implied domain is Horizontal lines x 0 intended. cross the curve only once.

Vertical lines cross the curve only once. eBook plus

WoRkED ExAMplE 7

1 . x+2 State whether or not it is a one-to-one function, and hence determine if the inverse function exists. If the inverse function exists: on the same set of axes, sketch f(x), its inverse and the line y = x − use a CAS calculator to find the points of intersection of f(x) and f 1(x) find the maximal domain and range of the original function find the domain and range of the inverse function.

Tutorial

Consider the function f ( x ) = a b c d e

WRiTE/DRAW

Think a


a

1 , a basic hyperbola x+2 translated 2 units to the left. f ( x) =

1

Write the function.

2

Find the vertical asymptote by making the denominator of the fraction equal to 0.

Vertical asymptote occurs where x+2=0 x = −2

3

Find the horizontal asymptote ing there is no vertical translation.

Horizontal asymptote occurs where y = 0.

4

Sketch the graph of f (x) and use vertical and horizontal line tests to decide whether or not f (x) is a one-to-one function.

x = −2

y (0, 1–2)

−2

x

0

1 f(x) = x + 2 5

State whether or not the inverse function exists.

−

f (x) is a one-to-one function, so f 1(x) exists.

Chapter 5

inverse functions

241

b On the same axes sketch f (x), f

−1

b The horizontal asymptote is y = −2.

(x) and y = x.

The vertical asymptote is x = 0. 1 0, 2 becomes 1 , 0

( )

( ) 2

The equation is y = 1x − 2. x = −2 1) (0, — 2 1 y = —— x+2

y

y=x

0 1 , 0) (— 2

1 x− 2 y= — x y = −2

c The points of intersection are (−2.414,

c Find the points of intersection.

−2.414)

and (0.414, 0.414). Note that these points both lie on the line y = x.

d The domain is R\{−2} and the range is

d Find the domain and range of f (x).

R \{0}. e The domain of the inverse is the range of the original

e The domain of the inverse is R \{0} and

the range is R \{−2}.

function. The range of the inverse is the domain of the original function.

Functions are either one-to-one or many-to-one. If a function, f, is not one-to-one but is many-to-one, it is possible to restrict its domain so that the limited domain is one-to-one and hence the inverse function exists. WoRkED ExAMplE 8 −

State the largest negative domain of f(x), shown in the figure, so that f 1(x) exists. y f(x) (0, 9)

−3

0

Think

242

1

Decide whether f (x) is a one-to-one function.

2

The parabola can be divided through its axis of symmetry, x = 3, so that 2 one-to-one functions are formed.


x

WRiTE/DRAW

f (x) is not one-to-one, because horizontal lines cut the parabola in 2 places.

3

State the domain of the left part of the parabola that is one-to-one.

f (x) is one-to-one if the domain is restricted to (−∞, −3].

4

State the domain of the right part of the parabola that is one-to-one.

f (x) is also one-to-one if the domain is restricted to [−3, ∞).

5

State the largest negative domain of f (x) so that − f 1(x) exists.

The largest negative domain for f 1(x) to exist is (−∞, −3].

−

−

A function f (x) has an inverse function f 1(x) if and only if the function is one-to-one. − The domain of a function f (x) is the range of its inverse f 1(x). − The range of a function f (x) is the domain of its inverse f 1(x). −1 The graph of f (x) is obtained by reflecting f (x) in the line y = x. The maximal or implied domain is used when the domain of a function is not specified. It is the largest domain for which the rule can be defined. 6. If a one-to-one function intersects with its inverse, the points of intersection lie on the line y = x. 7. If a function is not one-to-one it is possible to restrict the domain so that the new function is one-to-one and hence an inverse function exists.

1. 2. 3. 4. 5.

ExERCiSE

5C

inverse functions 1

WE 7 Consider the following functions. For each function: i state whether or not it is a one-to-one function and hence determine if the inverse function exists. If the inverse function exists: ii on the same set of axes, sketch f (x), its inverse and the line y = x − iii use a CAS calculator to find the points of intersection of f (x) and f 1(x) iv find the maximal domain and range of the original function v find the domain and range of the inverse function. a f (x) = 4x + 1 b f (x) = 6x c f (x) = 5 d f (x) = x2 + 2 e f (x) = (x − 3)2 f f (x) = (x + 1)3 2 g f ( x) = i f (x) = x2 − 6x + 3 h f ( x ) = − 16 − x 2 x 4x − 2 k f (x) = 2 loge (x − 1) j f (x) = e

2 Find: i which of the functions below − has an inverse function f 1(x) −1 ii f (x), if it exists.

ExAM Tip When asked for the domain of an − inverse, simply stating that the domain of f 1 is the same as the range of f was not sufficient to gain the mark. Students were expected to explicitly state the domain as a set. [Assessment report 2006]

a f (x) = 4x

b f ( x) =

1 −2 x2

c f (x) = 5 − x2

d f (x) = (x − 1)2

e f ( x) =

x3 2

f f (x) = (x + 5)2 − 28

Chapter 5

inverse functions

243

g f ( x) = x − 2 j f (x) = 5 − ex − 2

h f ( x ) = 16 − x 2 k f (x) = 3e−x − 2

i f (x) = 2ex + 1 l f (x) = loge (3x)

m f (x) = 2 loge (x − 4)

n f (x) = 1 + 2 ln (x)

o f (x) = 3 − loge (2x + 3)

3 Copy and complete the following table. Function

Inverse of the function

Domain

Range

Domain

Range

a

R

R−

b

[1, ∞)

R

c

[−3,

d

[0, 3] R+

R

e

[−10, ∞)

R

f

3]

R−

(0, ∞)

g h

[−5, 5]

[0, 8] R+

R

4 MC Consider the function in the figure shown at right. a The function would be one-to-one if the domain were restricted to: A [−3, 5] B [2, 10] C [1, 3] D [0, 4] E [0, 2.2] b The largest domain that restricts it to a one-to-one function is: B [0, ∞) A (−∞, 4] C R+ − − E ( ∞, 2] D R

y

4

2 + 1. 5 MC Consider the function f ( x ) = ( x − 3)2 a The function would be one-to-one if the domain were restricted to: A (−3, ∞] B [−3, ∞) C (3, ∞) D (1, ∞) E [3, ∞) b The inverse would be a function if the domain of f (x) were: A [0, 4] B (1, 4] C [2, 4] D [0, 3) E [1, ∞) 6 MC Consider the function f : S → R, f (x) = x2 + 1. − a The maximal domain, S, for f 1(x) to exist is: + C [1, ∞) A R B R− + − E (−∞, 1] D R ∪ {0} or R ∪ {0} − b For f 1 to exist, the largest possible positive set, S, is: C [0, ∞) D R+ A R B R− 7 WE8 exists. a

2

b

y x

f(x)

c

y f(x)

(−3, 3)

y f(x)

−3


0

x

x

E [1, ∞)

For each function graphed below, state the largest possible domain of f so that f

0

244

0

−1 0

x

−1

d

y 0

f(x) 2

e

y

f

f(x)

y 9

x

f(x)

1

g

h

y

−9

x

0

i

y

y

f(x) 0 1 −3 −7

5D

5

x=5

x

0

x

5

0

9 x f(x)

9x

0

f(x)

Restricting functions As we have seen in the previous exercise, functions which are not one-to-one can have their domains restricted so that they become one-to-one. As a result their inverses will then be − functions, that is, f 1 will exist.

WoRkED ExAMplE 9

eBook plus

Consider the function fully defined as follows: f : [ − 3, 3] → R, f ( x ) = 9 − x 2 − a Find the largest possible positive domain so that f 1 exists. −1 b Use this restricted domain to fully define f (x).

1


WRiTE

Think a

Tutorial

a Let y = f (x)

Write the function.

y = 9 − x2 y2 = 9 − x2

2

Square both sides.

3

Add x2 to both sides.

x2 + y2 = 9

4

State the shape and key features of the graph.

This is the equation of a circle with centre (0, 0) and radius 3.

5

Describe f (x), ing that a positive square root implies the upper part of a semicircle. Sketch the graph of f (x).

The graph of f ( x ) = 9 − x 2 is an upper semicircle.

6

y 3

−3 7

State the possible domains if f (x) is one-to-one.

8

State the largest positive domain for f 1(x) to exist.

−

0

3

x

f (x) is one-to-one if the domain is [−3, 0] or [0, 3]. The largest positive domain for f exist is [0, 3].

Chapter 5

−1

to

inverse functions

245

b

1

b If dom f = [0, 3], then ran f = [0, 3].

Find the range of f (x). −1

−1

= ran f.

dom f

−1

= ran f = [0, 3]

ran f

−1

= dom f = [0, 3]

2

Find the domain of f

3

Find the range of f

4

Write the function.

y = 9 − x2

5

Find the inverse by interchanging x and y.

Inverse is x = 9 − y 2

6

Make y the subject.

7

Select the appropriate inverse function.

8

State the function f 1.

−1

using dom f

using ran f

−1

= dom f.

x2 = 9 − y2 x2 + y2 = 9 y2 = 9 − x2 y = ± 9 − x2 Since ran f

−

−1

= [0, 3], use y = 9 − x 2 .

Therefore the inverse of f with the largest positive domain is: − − f 1: [0, 3] → R, f 1 ( x ) = 9 − x 2

WoRkED ExAMplE 10

If f: S → R, f(x) = 3(x − 1)2 − 2: − a find the largest positive set, S, for which f 1(x) exists −1 b fully define f and sketch its graph. WRiTE/DRAW

Think a

a y = 3(x − 1)2 − 2

1

Write the function.

2

State the shape and key features of the graph.

3

Sketch the graph of f (x).

A parabola with turning point (1, −2), y-intercept (0, 1) and x-intercepts (0.184, 0) and (1.816, 0) y

(0, 1) (0.184, 0) (1.816, 0) x 0 (1, −2)

b

4

State whether or not it is a one-to-one function.

5

State the possible domains if f (x) is to be one-to-one.

6

State the largest positive domain, set S, for f to exist. Find the range of f (x).

1 2 3

246

Find the domain of f Write the function.

−1

using dom f

−1

−1

= ran f.


The parabola is not one-to-one so the domain must be restricted. The domain could be (−∞, 1] or [1, ∞). S = [1, ∞) b The range is [−2, ∞). −

The domain of f 1 is [−2, ∞). y = 3(x − 1)2 − 2

4

x = 3(y − 1)2 − 2 x + 2 = 3(y − 1)2 x+2 = ( y − 1)2 3

Interchange x and y and make y the subject.

x+2 Reject the negative 3 solution because the domain is [−2, ∞) x+2 y = 1+ 3 y −1=

5

−

−

f 1:[−2, ∞) → R, f 1 ( x ) = 1 +

Fully define the inverse function.

x+2 3

key feature of inverse functions A feature of inverse functions is that by taking the inverse function of a function, or vice versa, the independent variable (x) is always obtained. That is: − − f 1 [ f (x)] = f [ f 1(x)] = x. The following example illustrates how this works. WoRkED ExAMplE 11

eBook plus

For f : R → R, f(x) = x3: − a find f 1(x) and state its range − − − − b find f [ f 1(x)] and f 1[f(x)] and show that f [ f 1(x)] = f 1[ f(x)] = x. Think a

1 2 3

b


WRiTE

Write the function. Interchange x and y. Make y the subject by writing both sides to the power of 13 and simplifying.

a f(x) = y = x3

Its inverse is x = y3. 1

Write the rule for f 1(x).

5

Find the range of f 1(x), ing that ran − f 1 is the same as dom f.

6

Write your answer in the same form as the question.

1

Find f [ f 1(x)] by rewriting f 1(x) as x 3 .

1

y = x3

2

Find f ( x 3 ) by replacing x by x 3 in f(x) = x3.

3

Simplify.

4

Find f

−1

5

Find f

−1

6

Simplify.

dom f = R − ran f 1 = R f

1

−

b

1

1

1

−

f 1 ( x) = x 3

−

−

1

(y3 ) 3 = x 3

−

4

7

Tutorial

−1

1

−

: R → R, f 1 ( x ) = x 3 1

−

f [ f 1 ( x )] = f ( x 3 )  1 =  x3   

3

=x − f 1[ f ( x )] =

x3.

[f(x)] by rewriting f(x) as

(x3) by replacing x by x3 in f 1 ( x ) = x 3.

−

1

−

f 1 (x3 ) 1

= (x3 ) 3 =x

−1

Compare f [ f (x)] to f

−1

[ f(x)].

− f [ f 1 (x ( x )] =

f −1[ f ( x)] x )] = x

Chapter 5

inverse functions

247

WoRkED ExAMplE 12

eBook plus

a Sketch the graph of f(x) = x2 − 3x + 3, showing the turning point and relevant Tutorial intercept(s). int-0545 b Find the rule of the inverse by an algebraic method and sketch this graph Worked example 12 on the same set of axes together with the line y = x. c Is the inverse a function? d The inverse is a reflection in the line y = x of the original function f(x). Use this information to find any points of intersection between the original curve and its inverse. − e Find the maximum value of a for f: (−∞, a] → R, f(x) = x2 − 3x + 3 so that f 1(x) exists. WRiTE/DRAW

Think a

use a CAS calculator to help in drawing a graph of f (x) including the relevant points.

a

y 4 3 2 1

f(x) = x2 − 3x + 3

(1.5, 0.75)

0 b

c

248

−

1 2 3 4

x

b x = y2 − 3y + 3

1

To find the equation of f 1(x), let y = x2 − 3x + 3. Interchange x and y.

2

To make y the subject, on a Calculator page, press: • MeNu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(y2 − 3y + 3 = x,y) Then press eNTeR ·.

3

Write the answer in the form of f 1(x).

4

use the calculator to draw the graphs − of f(x), f 1(x) and y = x.

−

The inverse is a one-to-many relation and therefore is not a function.

f 1 ( x) = ± −

4x − 3 + 3 2

y 4 3 2 1

f(x) = x2 − 3x + 3 y=x

0

1 2 3 4 5 6 7

x

c The graph of the inverse does not

the vertical line test, as it is a one-tomany relation, and therefore it is not a function.


d

e

The points of intersection between f (x) and − f 1(x) will occur at the intersections of the graph of f (x) and y = x. To locate these points on a Calculator page, press: • MeNu b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(x2 − 3x + 3 = x, x) Then press eNTeR ·. The points of intersection are at (1, 1) and (3, 3). Draw the graphs of f(x) and y = x.

d

For the left-hand branch of the parabola, we need f (x) = x2 − 3x + 3, since the domain is (−∞, a). It is a one-to-one function. Locate the turning point of the graph and the x-coordinate will be the value of a. To locate the minimum of f (x) = x2 − 3x + 3 on a Calculator page, press: • Menu b • 4:Calculus 4 • 7:Function Minimum 7 Complete the entry line as: f Min(x2 − 3x + 3, x) Then press eNTeR ·. 3 The x-value of the turning point is 2 , which 3 implies that a = 2 .

e

f(x) = x2 − 3x + 3

y 4 3 2 1

y=x

(1.5, 0.75)

1 2 3 4

0

x

The points of intersection are at (1, 1) and (3, 3).

3

f : ( − ∞, 2 ) → R, f ( x ) = x 2 − 3 x + 3 a=

3 2

1. Functions that are not one-to-one can have the domain restricted so that they become one-to-one. Then their inverses will be functions. 2. A fully defined function is written in the form f : [domain] → R, f (x) = rule. − 3. The graph of f 1(x) is a reflection of f (x) in the line y = x. − 4. The points of intersection of f (x) and f 1(x) lie on the line y = x. 5. By taking the inverse function of a function or vice versa, the independent variable (x) − − is always obtained: f 1[f (x)] = f [f 1(x)] = x. ExERCiSE

5D eBook plus Digital doc

SkillSHEET 5.3 Function notation

Restricting functions 1 WE 9a Find the largest possible domain(s) of the following functions so that the inverse function exists. (use a CAS calculator to sketch the graph if necessary.) a f : R → R, f (x) = x2 + 3 b f : R → R, f (x) = 3x2 − 1 1 c f : R → R, f (x) = (x + 3)2 − 2 d f : R \{0} → R, f ( x ) = 2 − 3 x 1 1 f f : R \ {2} → R, f ( x ) = e f : R \{4} → R, f ( x ) = +1 2 x−2 ( x − 4)

Chapter 5

inverse functions

249

g f : [− 5, 5] → R R,, f ( x) x) = i

−

25 − x 2

R,, f ( x) x) = 1 − x 2 h f : [− 1, 1] → R j f : R → R, f (x) = x2 − 2x + 5

f : [4, ∞) → R R,, f ( x) x) = x − 4

k f : R → R, f (x) = e x + 2

l f : (5, ∞) → R, f (x) = 2 loge (x − 5) −

2 WE 9b For each function in question 1 use the restricted domain to fully define f 1(x). (If there are two possible domains, use the one which is to the right.) 3 MC use the function f : [0, ∞) → R, f (x) = x2 to answer the following questions. − a Any points of intersection of f (x) and f 1(x) must lie on the line: A y=0 B y=x C y = 2x D y=x+1 E x=0 − b f 1(x) is correctly defined by: − − − − A f 1 : [0, ∞) → R, f 1(x) = x B f 1 : [0, ∞) → R, f 1(x) = −x −

−

−

−

R, f 1 ( x) x = x D f 1 : [0, ∞) → R, f 1 ( x ) = − x C f 1 : [0, ∞) → R, 1 − − R, f 1 ( x) x = 2 E f 1 : (0, ∞] → R, x − c The range of f 1(x) is: −∞, 0] D [0, ∞) A ( B R C R+ E R− −1 d The graphs of f and f intersect at the point(s): A (0, 0) and (1, 1) B (1, 1) only C (0, 0) and (1, −1) 1 D (0, 0) and (3, 3) E (1, 1) and (2, 4 ) 4 Copy each of the following graphs, then on the same set of axes sketch the graphs of y = x and − y = f 1(x). b a y y f(x)

3 5

c

f(x)

0

6

f(x)

y

x

−1

d

x

0

y 3 f(x)

x

0

e

f

y 1

250

0

x

3 y

f(x) x

0

f(x)

−1

0

x


(2, −5)

g

h

y

y 2

4 f(x)

0

1

x

0

3

x

5

f(x) −4

5 WE 10 If f : S → R, f ( x ) = 3 + x − 1, find: − a the largest set, S, for which f 1(x) exists −1 b f and sketch its graph. 6 If f : [−2, a] → R, f (x) = (x − 4)2 − 5, find: a the largest possible value of a so that f will have an inverse which is a function − b f 1 and sketch its graph. − 1 7 WE 11 For f : R+ → R, f ( x ) = x − a find f 1 and state its range −1 b find f [ f (x)] − c find f 1[ f (x)] − − d show that f [ f 1(x)] = f 1[ f (x)] = x. 1 : 8 For f : R → R, f ( x ) = 2 x +2 a sketch the graph of f (x) − b state the largest positive domain for f (x) so that f 1(x) exists −1 c sketch the graph f (x) on the same set of axes as f (x), using the positive domain − − d show that f [ f 1(x)] = f 1[ f (x)] = x. 9 WE 12 If g : [b, 8] → R, g(x) = 9 − x2, find: a the smallest value of b so that g(x) has an inverse that is a function − b g1 − c the range of g 1(x). 10 Given that g( x ) = 3 + 4 − x 2 , fully define two inverse functions, − g 1, using maximal domains.

Chapter 5


WorkSHEET 5.2

inverse functions

251

Summary Functions and their inverses

• A function is a relation which has only one y-value for each x-value. The graph of a function can be crossed only once by a vertical line. • A one-to-one function is a function which has only one x-value for each y-value. The graph of a one-to-one function can be crossed only once by any vertical or horizontal line. • A function which is not one-to-one is many-to-one. • The rule for the inverse of a relation, or a function, can be obtained by either: 1. interchanging the first and second elements of the ordered pairs, or 2. interchanging x and y in the rule and making y the subject. • Graphing the inverse — The graph of the inverse of a relation, or a function, can y be obtained by either: −1(x) f 1. interchanging the first and second elements of the ordered pairs of the relation or function, or 2 (−5, 2) 2. reflecting the graph of the relation or function through the line y = x, or x 0 3. using the rule of the relation or function to find the rule of the inverse and −5 f(x) then graphing the inverse. y=x • The domain and range of a function and its inverse are interchanged. (2, −5) • The graphs of a function and its inverse intersect on the line y = x. Inverse functions

• • • • •

−

The inverse of a one-to-one function, f (x), is also a function and is denoted by f 1(x). − dom f 1 = ran f −1 ran f = dom f − − f 1[ f (x)] = f [ f 1(x)] = x The implied (or maximal) domain of a function is the largest domain for which the function has meaning. When the domain is not mentioned, use the maximal domain. Restricting functions

−

• A function, f (x), which is not one-to-one can have its domain restricted so that f 1(x) exists. − • The points of intersection of f (x) and f 1(x) lie on the line y = x. • A function is fully described using the notation f : X → Y, f (x) = rule where X is the domain and Y is the co-domain. − • The graph of f 1(x) is a reflection of f (x) in the line y = x. −1 − • f [ f (x)] = f [ f 1(x)] = x

252


ChApTER REViEW c

ShoRT AnSWER

1 Sketch the graph of the inverse of each of the following relations. y a

y

8 x

0

2

f(x) (4, −4) x

0

−3

–3

b

y (−4, 4)

2 −6

−2

0

x

2 Sketch the graph of each of the following functions and state its domain and range. a y = − 36 − x 2

b f (x) = 2e x − 2

3 using the graph of the functions given below sketch the graph of the inverse of each function. y a x = −2 1 −2

−1 0

b

f(x)

x

y f(x)

0

(−1, −4)

x

4 State the largest possible domain for each of the following functions to be one-to-one. a f (x) = 3 − loge (x + 4) b g(x) = (x − 2)2 1 c f ( x) = 5 − x2 5 For the following functions: i Sketch their graphs and, by inspection, define them as one-to-one or many-to-one functions. ii For the functions defined in i as many-to-one, define the maximal domain for which the inverse is a function, stating their domains in formal function notation. If there is more than one option, choose the ‘right-hand’ option. iii Find the rules of their inverses. iv For all of the one-to-one functions from i and the redefined functions from iii, sketch all these inverse functions. a f (x) = 2x − 1 b f (x) = 2(x − 1)3 + 1 c f (x) = x2 + x  d f ( x) = 2 − x − 3 e f (x) = 3ex + 1 2 −1 f f ( x) = 3( x − 2) − 1 +1 g f ( x) = ( x + 3)2 6 a Sketch the graph of f (x) = loge (x − 2) + 1, marking the intercepts and asymptotes and stating the domain and the range. b Find the rule for the inverse function. − c If f 1(m) = 3, find the value of m. − d Draw the graphs of y = x and f 1(x) on the same axes as f (x) = loge (x − 2) + 1, marking the point (m, 3) and checking that it is on − f 1(x). State the domain and the range. e Mark the image of (m, 3) on the original graph.

Chapter 5

inverse functions

253

7 If g : (−∞, b) → R, g(x) = x2 + 5x − 1, find the − largest value of b for g 1 to exist.

A

B

y

8 Part of the graph of the function f : R → R, − f (x) = a + be x is shown below.

−1

y

0

x

0

−1

y y=1 0

1

y

C

x

3 The graph of the function with the equation y = f (x) is shown below. y 1

2

The graph of the inverse function f be: y

2, 4, −2 −2, 4 0, 0

B

D

y

0

1

x

1

x

−1

is most likely to y 1 0

x

−1 0

C

−4

x

0

1 A relation has an x-intercept of −4 and y-intercept of 2. The inverse of this relation has an x-intercept and y-intercept, respectively, of: −4,

x

x

0 −1

A

MulTiplE ChoiCE

1

y

E

[© VCAA 2007]

[© VCAA 2008]

x

1

0

The line with the equation y = 1 is an asymptote and the graph es through the origin. a explain why a = 1 and b = −1. Let h be the function h: [0, 2] → R, h(x) = f (x). b State the range of h using exact values. − c Find the inverse function h 1. d Sketch and label the graph of the inverse − function h 1 on the graph axes shown above. 9 Let f : R → R, f (x) = e2x − 1. a Find the rule and domain of the inverse − function f 1. − b Sketch the graph of y = f (f 1(x)) for its maximal domain. ax − c Find f (−f 1(2x)) in the form where a, b bx +c and c are real constants.

y

D

0

A B C D E

x

x

y

−1 0

x

2 The graph that best represents the inverse of the relation shown in the figure below is: E

y

−1

0

x

y

0

[© VCAA 2005]

254


4 The relation x2 + (y + 1)2 = 16 has an inverse that is: A a one-to-one function B a many-to-one relation C a many-to-many relation D a one-to-many function E a one-to-one relation −

-intercept of the inverse function f 1(x), 5 The x-intercept where f ( x ) = 3 − x − 3 , is closest to: A 1.27 B 6 C −3 D −1.27 E −6 6 The y-intercept of the inverse function f 2 + 3, is closest to: where f ( x ) = ( x − 2) 4 A 3 B no x-intercept −4 C 3 D 2 E −2

−1 (x),

7 Asymptotes exist on the graph of the inverse 2 − + 3, at: function f 1(x), where f ( x ) = ( x − 2) A x = 1 only B y = 1 only C x = 3, y = 2 D x = 2, y = 1 E x = 3, y = −2 4 8 The inverse of the relation y = 2 is: x 2 2 B y= A y=± x x −2 2 D y= C y= x x −2 E y= x 9 A function has a domain of [0, ∞) and a range of R+. The domain of its inverse must be: A R+ B R D [0, ∞) C R+ E (−∞, 0] 1 10 The implied domain of the function, f ( x ) = 2 x −1 is: A R C R \{0} E R+

B R \{1} D R \{−1, 1}

Questions 11 to 13 can be answered by considering the function f (x) = 4 ln (x + 3) − 2. 11 The maximal domain for f (x) is: A [3, ∞) B (−∞, −3) − D (−3, ∞) C R \{ 3} E R+ −

12 dom f 1 is equal to: A R+ C R E (−3, ∞) 13 The rule for f

−1

B [0, ∞) D (−∞, −3)

is completely described by:

−

−

−

−

x

A f 1 : R → R, f 1 ( x ) = e 4 − 3 1 1 B f : R → R, f ( x ) = e −

x+2 4

−3

−

1 − − R, f 1 ( x) x) = e C f : ( ∞, 3) → R, −

−

1 1 D f : R → R, f ( x ) = e

E

−

−

f 1 : R + → R, f 1 ( x ) =

x+4 3

x+4 2

−3

−2

x−4 e 2

−3

14 At which one of the following points is it possible for a function and its inverse to intersect? A (2, 2) B (1, 2) C (2, 1) D (0, 1) − − E ( 1, 2) −

15 The value of f [ f 1 (− 13 )] is equal to: A −3 1 B 3 C x 1 D 3 E 3 −

16 The function shown in the figure is one-to-one if the domain is restricted to: A B C D E

y

[2, ∞) [−3, ∞) [−3, 2] (−∞, 2] [0, ∞)

f(x) −3

0

2

x

17 If f : S → R, f (x) = x2 − 10x + 18, then the largest − possible set, S, for f 1 to exist is: A B C D E

(−∞, 0] [0, ∞) [0, 5] [5, 100] (−∞, 5]

Chapter 5

inverse functions

255

ExTEnDED RESponSE

1 using a CAS calculator: a Find the rule for the inverse of the function f (x) = 0.213x2 + 1.127x − 2.124. b explain why this inverse is not a function. c Find the value of a, correct to 2 decimal points, for [a, ∞), the maximal domain of f (x) so that the inverse is a function. Let this new function be g(x). − d Find the domain and range of f 1(x). −1 e Solve g(x) = g (x) for x, correct to 2 decimal places. 2 The function, f (x), in the figure is a parabola with a turning point at (0, 4). a Find the function which describes f (x). b Find the coordinates of the point, A. c State the domain and range of f (x). d Sketch the graph of f (x) and its inverse on the same set of axes. e What shape is enclosed between f (x) and its inverse? f i State the domain for which the inverse of f (x) is a function. ii Give the rule for the inverse of f (x) over this domain. g i State the domain for which the inverse of f (x) is not a function. ii Give the rule for the inverse of f (x) over this domain. 3 The graph of f (x) = exx2 is shown at right. a use a calculator to find the turning points, rounding answers to 2 decimal places as appropriate. b State the domain and range of f (x). c On the same set of axes, sketch f (x), its inverse and the line y = x. Mark all turning points and asymptotes. d State the domain and range of the inverse of f (x). e use a CAS calculator to solve exx2 = x to find the points of intersection of f (x) and its inverse. − f Find the largest possible negative domain of f (x) so that f 1(x) exists (that is, the inverse is a function). g Find the domain of the inverse function.

y 4

y=x

A −2

0

2

x

–2

y

f(x) = exx2 x

4 a use a CAS calculator to draw the graph of f (x) = 6 loge |x − 3|. − b Find the value of a for (a, ∞), the maximal domain of f (x), such that f 1(x) exists. Let this new function be g(x). − c Solve g(x) = g 1(x) for x, correct to 2 decimal places. − d Consider the function f (x) = a loge (x − 3), where a > 0. Let f 1(x) = h(x). Given that x a 1 a f ′( x ) = and h ′( x ) = e , find the value of a (correct to 2 decimal places) for which the graph of a x−3 f (x) and its inverse intersect only once.



256


eBook plus

ACTiViTiES


• 10 Quick Questions: Warm up with ten quick questions on inverse functions. (page 229) 5A

Relations and their inverses

Interactivity int-0248:

• Inverses: use the interactivity to consolidate your understanding of graphs of the inverses of relations. (page 229) Tutorial

• WE 3 int-0540: Watch a worked example on how to sketch relations and their inverses. (page 231) Digital docs

• SkillSHeeT 5.1: Practise identifying domain and range. (page 233) • Spreadsheet 041: Investigate graphs of functions. (page 233)

5D

Restricting functions

Tutorials

• WE 5 int-0541: Watch a worked example on sketching functions and their relations. (page 236)

• WE 9 int-0543: Watch a worked example on defining inverse functions. (page 245) • WE 11 int-0544: Watch a worked example on identifying that the composite functions − − f [ f 1(x)] = f 1[ f (x)] = x. (page 247) • WE 12 int-0545: Watch a worked example on restricting a function to define an inverse function. (page 248)

Digital docs

Digital docs

5B

Functions and their inverses

Tutorial

• SkillSHeeT 5.2: Practise making graphs with equations. (page 239) • Spreadsheet 059: Investigate inverse graphs. (page 239) • WorkSHeeT 5.1: Sketch relations, functions and their inverses. (page 240) 5C Tutorial

Inverse functions

• WE 7 int-0542: Watch a worked example on sketching graphs of inverse functions. (page 241)

• SkillSHeeT 5.3: Practise using function notation. (page 249) • WorkSHeeT 5.2: Recognise types of functions and sketch polynomials and power functions. (page 251) Chapter review Digital doc


Chapter 5

inverse functions

257

6 Circular (trigonometric) functions areaS oF STudy

• Graphs and identification of key features of graphs of the circular functions: y = sin (x), y = cos (x), y = tan (x) • Transformation from y = f (x) to y = Af (n(x + b)) + c, where A, n, b and c ∈ R, and f is one of the circular functions specified above and the relation between the graphs of the original and transformed functions for functions such as: y = −10 sin (π (x − c)) − 5, c ∈ R • Graphs of sum, difference, product and composite functions of f and g, where f and g are functions such as y = sin ((x) + 2x 2x and y = |cos cos (2 (2x)| • Graphical and numerical solutions of trigonometric equations

6a 6b 6c 6D 6E 6F

Revision of radians and the unit circle Symmetry and exact values Trigonometric equations Trigonometric graphs Graphs of the tangent function Finding equations of trigonometric graphs 6G Trigonometric modelling 6H Further graphs 6I Trigonometric functions with an increasing trend

• Recognition of the general form of possible models for data presented in graphical or tabular form, using circular functions • General solutions of equations such as 1 cos ( x ) + cos cos (3 x ) = 2 , x ∈ R and the specification of exact solutions or numerical solutions, as appropriate, within a restricted domain • Applications of simple combinations of functions and interpretation of features of the graphs of these functions in modelling practical situations, for example, y = ax + b + m sin (nx) as a possible pattern for economic growth cycles


nit c

1

P

nit

revision of basic concepts

y

1u

revision of radians and the unit circle

1u

6a

10 Quick Questions

Angles are measured in degrees or radians. O S x To define a radian we can use a circle which has a radius of one unit. This circle is called the unit circle. If we take a piece of string which is the same length as the radius and place it along the circumference of the circle from S to P to form an arc, then A unit circle the angle formed by ing S and P to O, the centre of the circle, measures one radian. The radius of the circle can be any length and can still be regarded as a unit. As long as the arc is the same length as the radius, the angle will always measure one radian. In general, therefore, a radian is the angle formed at the centre of any circle by radii meeting an arc which is the same length as the radius of the circle. Note the following.

258

maths Quest 12 mathematical methods CaS for the Ti-nspire

y 1. One radian is written as 1c (or 1 radian can be written as 1). A 2. The circumference of a circle is 2π r units in length. P 3. If the radius is one unit, as in the case of the unit circle, then the r r circumference is 2π units, and the angle at the centre of the circle c is 2π radians. 1 r 4. 2π radians = 360° B S x 5. The length of the semicircle from S through A to B is half the circumference and is π units. 6. π radians = 180° 7. An arc length of r units subtends an angle of 1 radian. A radian 8. An arc length of 2π r units subtends an angle of 2π radians. πr   2π r 9. An arc length of a quarter of a circle is units  that is, and subtends an angle of  2  4 π radians. 2

Finding the number of degrees in one radian π c = 180° 180° 1c = = 57.296° (correct to 3 decimal places) π

Since we have

Converting radians to degrees Radians are converted to degrees using the following equation. 180° 1c = π Worked Example 1

Convert the following to degrees, giving the answer correct to 2 decimal places. a 2c b 6.3c c 9π

10

Think a

1 2

b

Write

180 Multiply the number of radians by . π Simplify where possible.

180° π 360° = π = 114.59°, correct to 2 decimal places.

a 2c = 2 ×

3

Write the answer correct to 2 decimal places.

1

b On a Calculator page, type: 6.3 Note: The radian sign does not need to be entered if the calculator is already in radian mode. To convert 6.3c to degrees, press Catalog k, press D to get to the items beginning with the letter D. Then select ¢DD. Then press ENTER ·.

Chapter 6 Circular (trigonometric) functions

259

2

Then press ENTER ·. Then write the answer.

6.3c = 360.96° correct to 2 decimal places. c

1

Multiply the number of radians 180 by . π

2

Simplify by cancelling.

c c 9π = 9π × 180°

10

10

π

= 162°

Converting degrees to radians Degrees are converted to radians using the following equation. 180° = π c 1° = Worked Example 2

Convert the following to radians. a 2° b 36.35° c 150° Think a

260

1

Write

On a Calculator page, type: 2 The calculator is in radian mode so you will need to type the degree symbol before converting to radians. You will find the degree symbol in the Maths expression template by pressing Ctrl / Catalog k.


πc 180

2

Complete the entry line as: 2° Then press ENTER ·. Then write the answer.

π or 0.035c correct to 3 decimal places. 90 π b 36.35° = 36.35 × 180 2° =

b

1

Multiply the number of degrees by π . 180

2

Simplify.

= 0.634c

Note: In this example it is not appropriate to leave your answer in exact form. c

1

Multiply the number of degrees by π . 180

2

Simplify, leaving your answer in exact form.

c

150° = 150 ×

π 180

150π 180 5π = 6 =

Special cases (degrees to radians) Note the following special cases. 180° = π π 90° = 2 π 60° = 3 π 45° = 4 π 30° = 6

Divide both sides by 2. Divide both sides by 3. Divide both sides by 4. Divide both sides by 6.

Basic definitions of sine, cosine and tangent Sine and cosine In the unit circle the vertical distance PR is defined as sine (θ ) or sin (θ ) and the horizontal distance OR is defined as cosine (θ ) or cos (θ ). The coordinates of the point P are (cos (θ ), sin (θ )) where θ can be in radians or degrees. The x-coordinate of P is cos (θ ) and the y-coordinate of P is sin (θ ).

y P (cos (θ ), sin (θ )) sin (θ) θ x O cos (θ) R

sin (q ) and cos (q )


261

Special cases (sin, cos) Note the following special cases for sin and cos. sin (0) = 0 π sin   = 1  2 sin (π) = 0

y

 3π  sin   = − 1  2

π

cos (0) = 1 π cos   = 0  2 cos (π) = −1

π – 2

sin (π–2 ) cos (π ) cos (0) 0

sin (2π) = 0 sin (0°) = 0 sin (90°) = 1 sin (180°) = 0 sin (270°) = −1 sin (360°) = 0

3π ) sin (— 2

 3π  cos   = 0  2

x

cos (2π) = 1 cos (0°) = 1 cos (90°) = 0 cos (180°) = −1 cos (270°) = 0 cos (360°) = 1

3π — 2

Special cases

y

Tangent

T

Using the unit circle, the vertical distance TS is defined as tan θ. TS is the tangent to the circle which intersects with the x-axis and ∠TOS = θ. Using Pythagoras’ theorem in triangle OPR (figure below), PR2 + OR2 = OP2. 2 So, sin (θ ) + cos2 (θ ) = 1 From the diagram below, ∆OPR is similar to ∆OTS (angle, angle, angle).

Special cases (tan)

x

T

it

P un

sin (θ ) tan (θ ) = cos (θ )

S

tan (q )

y

1

TS PR Therefore: = OS OR tan (θ ) sin (θ ) = 1 cos (θ )

O

tan (θ )

θ

O

tan (θ ) sin (θ )

θ cos (θ ) R 1 unit Identities

S x

Note the following special cases for tan. tan (0) = 0

π tan   is undefined  2

tan (0°) = 0

tan (90°) is undefined

tan (π) = 0

 3π  tan   is undefined  2

tan (2π) = 0

tan (180°) = 0

tan (270°) is undefined tan (360°) = 0 sin (90o ) 1 = , which is undefined. It can be seen that tan (90°) is undefined because tan (90o ) = cos (90o ) 0

1. Radians: (a) 1c = the size of the angle formed where the length of an arc is equal to the radius of the circle. (b) π c = 180° (c) Angles are in radians unless a degree symbol is shown.

262


180 . π π 3. To change degrees to radians, multiply by . 180 4. Identities:

2. To change radians to degrees, multiply by

(b) tan (θ ) =

(a) sin2 (θ ) + cos2 (θ ) = 1

exerCiSe

6a eBook plus Digital doc

Spreadsheet 142 Unit circle

eBook plus

revision of radians and the unit circle 1 We1 Convert the following to degrees, giving answers correct to 2 decimal places. a 3c 7π c e 20

b 5c c 4.8c d 2.56c 5π c 5π c 3π c h f g 4 6 10 2 We2 Convert the following to radians. Give exact answers for a , b , c and d . Write other answers correct to 2 decimal places.

Digital doc

SkillSHEET 6.1 Changing degrees to radians

sin (θ ) cos (θ )

a 5°

b 15°

c 120°

d 130°

g 235° h 260° i 310° f 78.82° 3 Evaluate using a calculator. Give answers correct to 3 decimal places.

e 63.9° j 350°

a sin (0.4)

b sin (0.8)

c cos (1.4)

d cos (1.7)

e tan (2.9)

f tan (2.4)

g sin (75°)

h sin (68°)

i cos (160°)

j cos (185°)

k tan (265°)

l tan (240°)

4

Evaluate the following. a sin (0)  3π  e tan    2

b sin (π) π f tan    2

c cos (2π)

d cos (π)

g sin (90°)

h sin (360°)

i cos (180°)

j cos (0°)

k tan (270°)

l tan (240°)

5

Evaluate without using your calculator. a sin2 (20°) + cos2 (20°)

b cos2 (50°) + sin2 (50°) π π e sin 2   + cos 2    2  2

d sin2 (2.5) + cos2 (2.5) g 2 sin2 (α) + 2 cos2 (α)

c sin2 (π) + cos2 (π) θ θ f sin 2   + cos 2    2  2

h 5 sin2 (β) + 5 cos2 (β)

6 Write the following in order from smallest to largest. a sin (35°), sin (70°), sin (120°), sin (150°), sin (240°) b cos (0.2), cos (1.5), cos (3.34), cos (5.3), cos (6.3) 8

15

7 If sin (θ ) = 17 and cos (θ ) = 17 , find tan (θ ). 8 If sin (A) = 0.6 and cos (A) = 0.8, find tan (A). Draw a triangle marking in the position of angle A and possible lengths of the sides.

π radians is equal to: 3 a 0° b 30° c 45°


9 mC

SkillSHEET 6.2 Tangent ratios

D 60°

E 90°

Chapter 6

Circular (trigonometric) functions

263

10 MC The expression 1 − sin2 (α) is equal to: c cos (α) D tan (α) a 1 b cos2 (α)

E tan2 (α)

11 The temperature T °C inside a shop t hours after 2 am is given by π  T = 15 − 3 cos  t   12  Calculate the exact temperature after 4 hours and the temperature to the nearest tenth of a degree at 9.00 am.

6B

Symmetry and exact values Exact values Using the equilateral triangle (of side length 2 units) shown at right, the following exact values can be found. π sin (30°) = sin   =  6

1 2

π cos (30°) = cos   =  6 π tan (30°) = tan   =  6

3 2 1

=

3

3 3

30° 2

π sin (60°) = sin   =  3

3 2

π cos (60°) = cos   =  3

1 2

π tan (60°) = tan   = 3  3

60° 1 1 Exact values of sine, cosine and tangent of 30° and 60°

Using the right isosceles triangle shown, the following exact values can be found. π sin (45°) = sin   =  4

1 2

π cos (45°) = cos   =  4

=

1 2

=

2 2

2

3

45° 1

2

45° 1 Exact values of sine, cosine and tangent of 45°

2 2

π tan (45°) = tan   = 1  4

The unit circle and symmetry properties The unit circle is symmetrical so that the magnitude of sine, cosine and tangent at the angles shown are the same in each quadrant but the sign varies. y

π–

θ

S

θ

θ

θ θ

θ θ

π+

A

T

C

x

2π –

Symmetrical properties

In the first quadrant sin, cos and tan are all positive. In the second quadrant only sin is positive.

264


θ

In the third quadrant only tan is positive. In the fourth quadrant only cos is positive. This can be ed as All Students To Class (ASTC). 1st quadrant

2nd quadrant

3rd quadrant

4th quadrant −sin

sin (2π − θ ) = −sin (θ )

sin (θ )

sin (π − θ ) = sin (θ )

sin (π + θ ) =

cos (θ )

cos (π − θ ) = −cos (θ )

cos (π + θ ) = −cos (θ )

cos (2π − θ ) = cos (θ )

tan (π + θ ) = tan (θ )

tan (2π − θ ) = −tan (θ )

tan (θ )

tan (π − θ ) =

−tan

(θ )

(θ )

Worked example 3

eBook plus

Without using a calculator, find the value of: a sin (150°) Think a

b

Tutorial

 5π  b cos  .  4 


WriTe a sin (150°) = sin (180 − 30)°

1

Find the equivalent first quadrant angle.

2

As 150° is in the 2nd quadrant, sine is positive.

= sin (30°)

3

Write the exact value.

=

1

Find the equivalent first quadrant angle using 5π 4π π = + . 4 4 4

2

Decide on the sign required. As it is in the 3rd quadrant, cosine is negative.

π = − cos    4

3

Write the exact value.

=

b cos

1 2

5π π  = cos  π +   4 4

− 2 2

Angles are not restricted to values between 0 and 2π; that is, the domain is not restricted to [0, 2π]. If an angle is greater than 2π radians, it is necessary to subtract multiples of 2π so that the angle is within one turn of the unit circle. Each 2π radians is a complete turn of the circle. Worked example 4

If sin (x) = 0.6, cos (x) = 0.8, and x is in the first quadrant, find: a sin (3π − x) b cos (4π + x). Think a

1

WriTe

Write 3π as 2π + π — that is, one complete cycle and then the angle π − x.

a sin (3π − x) = sin (2π + π − x)

= sin (π − x)

x

2

y

x

= sin (x)

As it is in the 2nd quadrant, sine is positive.

Chapter 6


265

b

= 0.6

3

Substitute the given value.

1

4π = 2 × 2π — that is, two complete cycles and then the angle x.

b cos (4π + x)

y x

2

As it is in the 1st quadrant, cosine is positive.

= cos (x)

3

Substitute the given value.

= 0.8

Worked example 5

eBook plus

π < θ < π calculate cos (θ ) and hence find tan (θ ). 2

If sin (θ ) = 12 and 13 Think

x

Tutorial


WriTe

Method 1 1

2 3 4 5

In the given triangle, find the length of the third 132 = 52 + 122 side by using Pythagoras’ theorem. Let θ′ be the first quadrant angle corresponding to θ. adjacent 5 cos (θ ′) = cos (θ ′) = 13 hypotenuse

θ is in the second quadrant, therefore cos (θ ) is negative. opposite tan (θ ) = adjacent θ is in the second quadrant, therefore tan (θ ) is negative.

cos (θ ) = tan (θ ′) = tan (θ ) =

13 θ'

−5 13 12 5 − 12 5

Method 2 1

Use the rule sin2 (θ ) + cos2 (θ ) = 1.

2

Rearrange. 12 13

cos2 (θ ) = 1 − sin2 (θ ) for sin (θ ).

3

Substitute

4

Evaluate.

5

Take the square root of both sides, ing that the answer could be positive or negative.

6

cos is negative in the 2nd quadrant.

7

Use tan (θ ) = tan (θ ).

sin2 (θ ) + cos2 (θ ) = 1 144

= 1 − 169 25

= 169

sin (θ ) to find the value of cos (θ )

cos (θ ) = ± 13 5

Take cos (θ ) = tan (θ ) = tan (θ ) =

266


−5 13 12 13 5 13 − 12 5

as

π <θ <π 2

12 5

Method 3 1

On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: 12 π solve (sin (θ ) = , θ ) | < θ < π 13 2 Then press ENTER ·.

2

Complete the entry lines as:  −  12   cos  θ = π − sin 1     13     −  12   tan  θ = π − sin 1     13    Press ENTER · after each entry. Write the answers.

Given sin (θ ) = 12 , 13 cos (θ ) = tan (θ ) =

π < θ < π, 2

−5 13 − 12 5

Negative angles For negative angles, move in a clockwise direction. In the diagram, RQ = −PR, and OR = OR, so T1S = −TS An alternative way to find tan (−θ ) is: tan

(− θ ) (−θ )

(− θ )

− sin

sin (θ ) − = = = tan (θ ) cos (− θ ) cos (θ )

y P

sin (θ ) θ O –θ

R S sin (– θ) Q

−sin

sin = (θ ) cos (−θ ) = cos (θ ) tan (−θ ) = −tan (θ )

The diagram shows 0 < θ <

T

x

T1

Negative angles

π ; however, these relationships are true for all values of θ. 2

sin (−150°) = −sin (150°) cos (−190°) = cos (190°) tan (−280°) = −tan (280°)


267

Worked Example 6

Find the exact value of: a sin (−135°) b cos (−240°) c tan (−330°). Think a

b

c

Write

1

sin (−135°) = −sin (135°)

2

135° is in the 2nd quadrant.

= −sin (180° − 45°)

3

Sine is positive in the 2nd quadrant.

= −sin (45°)

4

Give the exact value.

=

1

cos (−240°) = cos (240°)

2

240° is in the 3rd quadrant.

= cos (180° + 60°)

3

Cosine is negative in the 3rd quadrant.

= −cos (60°)

4


=

1

tan (−330°) = tan (30°)

2


2

3

268

x

−1 2

c tan (−330°) = tan (30°)

=

1

=

3 3

 − 4π 

a sin  =  3 

4π is in the 3rd quadrant. 3 Sine is negative in the 3rd quadrant.


y

3

T −330°

Write

sin (−θ ) = −sin (θ )

y

–240°

Method 1: Technology-free 1

x

–135°

b cos (−240°) = cos (240°)

Find the exact value of:  − 4π   − 5π  a sin  b tan  .   3   6 

a

135°

− 2 2

Worked Example 7

Think

y

a sin (−135°) = −sin (135°)

− sin

 4π   3 

π  = − sin  π +   3  π = −  − sin     3  π = sin    3

S x

b

4


1

tan (−θ ) = −tan (θ )

2

5π is in the 2nd quadrant. 6

3

Tangent is negative in the 2nd quadrant.

4


=  − 5π 

b tan  6  =

3 2

5π   6 

− tan 

π  = − tan  π −   6  π = −  − tan     6  π = tan    6 =

1

=

3 3

 − 4π  sin  =  3 

3 2

 − 5π  tan  =  6 

3 3

3

Method 2: Technology-enabled 1

On a Calculator page, complete the entry lines as:  − 4π  a sin   3  b

 − 5π  tan   6  Press ENTER · after each entry.

2

Write the answers.

Complementary angles

π Complementary angles add to 90° or radians. Therefore, 30° and 60° are complementary angles. 2 π π π  In other words and are complementary angles, and θ and  − θ  are also complementary   2 6 3 angles.

The sine of an angle is equal to the cosine of its complement. Therefore, sin (60°) = cos (30°). We say that sine and cosine are complementary functions. The complement of the tangent of an angle is the cotangent or cot — that is, tangent and cotangent are complementary functions (as well as reciprocal functions). 1 cot (θ ) = tan (θ )


269

1st quadrant

2nd quadrant

3rd quadrant

π π   sin  − θ  = cos (θ ) sin  + θ  = cos (θ )   2 2

4th quadrant

 3π  3π   sin  − θ  = − cos (θ ) sin  + θ  = − cos (θ )  2   2 

 3π  3π π π     cos  − θ  = sin (θ ) cos  + θ  = − sin (θ ) cos  − θ  = − sin (θ ) cos  + θ  = sin (θ )  2   2    2 2  3π π π    tan  − θ  = cot (θ ) tan  + θ  = − cot (θ ) tan  − θ  = cot (θ )  2    2 2

 3π  tan  + θ  = − cot (θ )  2 

Worked example 8

eBook plus

If sin (θ ) = 0.4 and tan (α) = 0.6, find: π  3π   + α . a cos  + θ  b tan  2   2 

Tutorial


Think a

b

WriTe

π  + θ  = − sin (θ ) 2 

a cos 

1

Cosine and sine are complementary functions and cosine is negative in the second quadrant.

2

Substitute the given value for sin (θ ).

1

Tangent and cotangent are complementary functions and tangent is negative in the 4th quadrant.

= −0.4  3π



+ α = b tan   2  =

− cot (

α)

−1

tan (α ) −1

2

Substitute the given value for tan (α).

=

3

Calculate.

= −1.667, correct to 3 decimal places

0.6

1. Exact values can be determined by using an equilateral triangle and a right isosceles triangle as shown.

30° 3

45° 2 60° 1

270


2

1

1

45°

0°

30° π 6

45° π 4

60° π 3

90° π 2

180°

π

270° 3π 2

360°

θ

0

sin (θ )

0

1 2

2 2

3 2

1

0

−1

0

cos (θ )

1

3 2

2 2

1 2

0

−1

0

1

tan (θ )

0

3 3

1

Undef.

0

Undef.

0

3

2π

2. All angles in the unit circle are positive in the 1st quadrant, sine is positive in the 2nd quadrant, tangent is positive in the 3rd quadrant and cosine is positive in the 4th quadrant. Symmetry properties: sin (π − θ ) = sin (θ )

sin (π + θ ) = −sin (θ )

sin (2π − θ ) = −sin (θ )

cos (π − θ ) = −cos (θ )

cos (π + θ ) = −cos (θ )

cos (2π − θ ) = cos (θ )

tan (π + θ ) = tan (θ )

tan (2π − θ ) = −tan (θ )

tan (π − θ ) =

−tan

(θ )

3. Negative angles sin (−θ ) = −sin (θ ), cos (−θ ) = cos (θ ), tan (−θ ) = −tan (θ ) 4. Complementary angles π (a) Complementary angles add to 90o or radians. 2 (b) The sine of an angle is equal to the cosine of its complement. (c) The complement of the tangent of an angle is the cotangent.

exerCiSe

6B

π  sin  − θ  2  = cos (θ )

π  sin  + θ  2  = cos (θ )

 3π  sin  − θ  2  − = cos (θ )

 3π  sin  + θ  2  − = cos (θ )

π  cos  − θ  2  = sin (θ )

π  cos  + θ  2  = −sin (θ )

 3π  cos  − θ  2  = −sin (θ )

 3π  cos  + θ  2  = sin (θ )

π  tan  − θ  2  = cot (θ )

π  tan  + θ  2  = −cot (θ )

 3π  tan  − θ  2  = cot (θ )

 3π  tan  + θ  2  = −cot (θ )

Symmetry and exact values Note: Give answers as surds with rational denominators.

eBook plus

1 We3a

Without using a calculator, find the exact values of the following.

a sin (120°)

b cos (135°)

c tan (330°)

SkillSHEET 6.3

d cos (225°)

e sin (210°)

f tan (150°)

Rationalising the denominator

g sin (315°)

h cos (300°)

i tan (225°)

j cos (390°)

k sin (405°)

l tan (420°)

Digital doc

Chapter 6


271

2 We3b eBook plus Digital doc

Spreadsheet 142 Unit circle

Find the exact values of the following.

 3π  a sin    4

 5π  b cos    6

 2π  c tan    3

 4π  3 

 5π  e sin    4

 7π  f tan    6 

g sin  

 11π  6 

 5π  h cos    3

 7π  i tan    4 

 9π  j cos    4 

 13π  k sin   6 

 7π  l tan    6 

d cos  

3 We4 If sin (x) = 0.3, cos (a) = 0.5, tan (b) = 2.4 and x, a and b are in the first quadrant, find the value of the following. b cos (π − a) c tan (2π − b) d cos (π − x) a sin (π − x) e sin (π − a) f tan (π + b) g sin (2π − x) h cos (2π − a) i tan (π − b) j cos (2π + x) k sin (2π + a) l tan (2π + b) n cos (3π + a) o tan (3π − b) m sin (3π − x) If sin (α ) =

4

7 25

and cos(α ) =

24 , 25

find tan (α) and show that sin2 (α) + cos2 (α) = 1.

π < x < π , find cos (x) and hence find tan (x). 2 −1 3π , find sin (x) and tan (x). b If cos ( x ) = , and π < x < 2 2 If sin ( x ) = 12, and

5 a We5

c If sin ( x ) =

−

3 2

, and

3π < x < 2π , find cos (x) and tan (x). 2

3π , find sin (x) and cos (x). 2 Find the exact value of the following.

d If tan ( x ) = 3 , and π < x < 6 We6 a e i m

sin (−30°) sin (−120°) tan (−240°) sin (−420°)

b f j n

cos (−45°) tan (−135°) cos (−330°) cos (−390°)

c g k o

tan (−60°) sin (−225°) sin (−315°) tan (−405°)

d cos (−150°) h cos (−210°) l tan (−300°)

Evaluate (sin α + cos α)2 + (sin α − cos α)2.

7 8 We7

Find the exact values of:   −π  a sin   b cos    3   6  −π 

9

 − 7π   6 

 − 3π 

d cos   4 

 − 2π  e sin   3 

 − 5π  f tan   6 

g sin 

 − 5π  h cos   4 

 − 4π  i tan   3 

 − 5π  j cos   3 

 −13π  k sin   6 

 − 9π  l tan   4 

 −π   −π  Show that cos2   + sin 2   = 1.  4   4 

10 We8

If sin (θ ) = 0.3, cos (x) = 0.7 and tan (α) = 0.4, find the value of the following.

π  a sin  − x  2 

272

 −π  c tan    4 

 3π  + θ b cos   2 


π  c tan  − α  2 

π  + θ 2 

d cos 

 3π π   − x e sin  f tan  + α   2   2  3π  3π   − α + θ i tan  j cos   2   2  π 11 mC If x = , 3 sin (2x) is equal to: 12 3 3 3 3 2 c a b 2 2 2

π

 3π  − θ h cos   2   3π  + α l tan   2 



g sin  + x   2

 3π  − x k sin   2 

D

1 2

E

π 2

π  12 mC The expression 1 − sin 2  − θ  is equal to:  2

13

π  a cos  − θ   2

π  b sin  − θ   2

π  D cos2  − θ  2 

E 0

π  c sin 2  − θ  2 

A weight on a spring moves so that its speed, v cm per second, is given by the formula  πt  v = 10 + 2 sin    6 a Find the initial speed. b Find the speed of the weight after 5 seconds. c What is the greatest speed that the weight can reach?


SkillSHEET 6.4 Problem solving using trigonometry

14 The height, H, in metres, that sea water reaches up a particular tree trunk at a Caribbean resort is governed  πt  by the equation H = 0.4 cos   + 0.5,  12  where t is the number of hours past midnight. Find the height of water up the trunk at: a midnight b 8 am c 8 pm.

6C

Trigonometric equations From the general equation sin (x) = a, we can find an infinite number of solutions. An example of this general equation is: 3 sin ( x ) = . 2 3 π π One of the solutions is x = , because sin   = .  3 2 3 π 3  However, we also know that sin  π −  = because sine  3 2 is positive in the second quadrant.

y

sin (π – –3π )

–π 3

–π 3

sin ( –3π ) x

π 2π and . 3 3 (There are no solutions in the third and fourth quadrants because here, sine is negative.) For this equation there are two solutions between 0 and 2π. They are

Chapter 6


273

To find a greater number of solutions we can go around the unit circle as many times as we wish, finding new solutions each time. Since

π  sin  2π +  =  3

3 2

π  and sin  3π −  =  3

3 2

π 2π 7π 8π , , . and 3 3 3 3 We can also go in a negative direction. In the domain [−π, π] there are only 2π π 2 solutions: and . 3 3 in the domain [0, 4π] there are 4 solutions:

Worked Example 9

Find all solutions to the equation cos ( x ) =

− 2 2

in the domain [0, 2π].

Think

Write

Method 1: Technology-free − 2 2

1

Write the question.

cos ( x ) =

2

Find the equivalent angle in the first quadrant, ignoring the sign.

Basic angle is

3

In the 2nd and 3rd quadrants cos (x) is negative. So write the appropriate values of x in these quadrants.

2nd quadrant: π x=π − 4 3rd quadrant: π x=π + 4 3π 5π x= , 4 4

4

Simplify.


On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: −

2

, x ) | 0 ≤ x ≤ 2π 2 Then press ENTER ·. solve(cos( x ) =

2

274

Write the answer.

x=


3π 5π , 4 4

π . 4

y

3π — 4

x 5π — 4

Worked Example 10

Find all solutions to the equation sin (α) = 0.7 in the domain [0, 4π]. Give your answers correct to 4 decimal places. Think

Write


Find the equivalent angle in the 1st quadrant, ignoring the sign.

The basic angle is 0.7754.

2

Find in which quadrants sin (α) is positive.

sin (α) is positive in the 1st and 2nd quadrants.

3

Find the values of α in the 1st and 2nd quadrants.

α = 0.7754, π − 0.7754 = 0.7754, 2.3662

4

Since the domain is [0, 4π], we need to go around the circle twice, so add 2π to each of the first two solutions.

α = 0.7754, 2.3662, 0.7754 + 2π, 2.3662 + 2π

5

Simplify giving answers correct to 4 decimal places.

α = 0.7754, 2.3662, 7.0586, 8.6494, correct to 4 decimal places.


On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(sin (a) = 0.7,a) | 0 ≤ a ≤ 4p Then press ENTER ·.

2

Write the answer.

If sin (a) = 0.7, 0 ≤ a ≤ 4p, then a = 0.7754 or 2.3662 or 7.0586 or 8.6494, correct to 4 decimal places.

Worked Example 11

Calculate all solutions to the equation 2 sin ( 2θ ) = Think

−

3 in the domain 0° ≤ x ≤ 360°. Write

1

Since the equation contains 2θ rather than θ, multiply both end points of the domain by 2.

0° ≤ θ ≤ 360° 0° ≤ 2θ ≤ 720°

2

Simplify the trigonometric equation.

2 sin (2θ ) = − 3 sin (2θ ) =

− 3 2


275

3

Find the ‘first quadrant (basic) angle’, ignoring the negative sign.

Basic angle = 60°

4

Sine is negative in the 3rd and 4th quadrants. Find the angles in these quadrants equivalent to the basic angle.

2θ = 180° + 60°, 360° − 60° 2θ = 240°, 300°

5

As the domain has been extended to 720°, we need to go around the circle two times, so add 360° to each of the two initial values.

2θ = 240°, 300°, 240° + 360°, 300° + 360° 2θ = 240°, 300°, 600°, 660°

6

Simplify.

θ = 120°, 150°, 300°, 330°

Worked Example 12

Calculate the sum of the solutions between 0 and 2π for the equation sin (3x) = cos (3x). Think

Write

1

Adjust the domain as shown.

0 ≤ x ≤ 2π 0 ≤ 3x ≤ 6π

2

Divide both sides by cos (3x).

3

Find the basic angle.

4

Solve for x between 0 and 6π. Note that tan is positive in the 1st and 3rd quadrants.

5

Cancel where possible.

sin (3x) = cos (3x) tan (3x) = 1 π Basic angle = 4 π π π π π π 3 x = , π + , 2π + , 3π + , 4π + , 5π + 4 4 4 4 4 4 π 5π 9π 13π 17π 21π = , , , , , 4 4 4 4 4 4 π 5π 9π 13π 17π 21π x= , , , , , 12 12 12 12 12 12 π 5π 3π 13π 17π 7π x= , , , , , 12 12 4 12 12 4

6

Check that all answers are between 0 and 24π 8π . They are. 2π (2π = = ) 12 4 Calculate the sum of the solutions.

7

Sum of the solutions π 5π 3π 13π 17π 7π = + + + + + 12 12 4 12 12 4 11π = 2

General solutions of trigonometric equations π 3π and x = . 4 4 However, if a domain is not specified, there are an infinite number of solutions as multiples of π 3π 2π can be added or subtracted indefinitely to and . In this situation a general solution is 4 4 obtained where the solutions are in of a parameter, n, where n is an integer (that is, n ∈ Z). The solution to the equation sin ( x ) =

276

1

2

where x ∈ [0, 2π] is x =


π π becomes x = 2 nπ + , where n ∈ Z. 4 4 3π 3π The general solution for the second quadrant solution x = becomes x = 2 nπ + , 4 4 where n ∈ Z. 3π Note that the general solution x = 2 nπ + can be expressed as 4 π π x = 2 nπ + π − = (2 n + 1)π − . 4 4 Substituting different integer values of n will give specific solutions, as shown in the table below. π 3π n x = 2 nπ + , n ∈ Z x = 2 nπ + , n ∈Z 4 4 3π − 5π π − 7π −1 x = − 2π + = x = − 2π + = 4 4 4 4 3π 3π π π 0 x=0+ = x=0+ = 4 4 4 4 π 9π 3π 11π 1 x = 2π + = x = 2π + = 4 4 4 4 π 17π 3π 19π 2 and so on x = 4π + = x = 4π + = and so on 4 4 4 4 The general solution for the first quadrant solution x =

In general: − − • If sin (x) = a, then x = 2nπ + sin 1 (a) and x = (2n + 1)π − sin 1 (a), where a ∈[−1, 1] and n ∈ Z. − • If cos (x) = a, then x = 2nπ ± cos 1 (a), where a ∈[−1, 1] and n ∈ Z. −1 • If tan (x) = a, then x = nπ + tan (a), where a ∈ R and n ∈ Z. Worked example 13

eBook plus

Find the general solution of the equation 2 cos ( x ) − 1 = 0. Hence, find all the solutions for x ∈ [−2π, 2π]. Think


WriTe

1

Write down the general solution for cos (x) = a.

2

Substitute a = −  1  cos 1    2

Tutorial

1 2

into the general equation and evaluate

recognising that it is an exact angle.

3

Write the two separate solutions and specify n ∈ Z.

4

Note the answer from step 2 could be further simplified by combining the two . A CAS calculator will give the answer in this form.

5

Substitute n = −1, n = 0 and n = 1 into each of the general solutions.

6

Write down the solutions for x ∈ [−2π, 2π].

−

x = 2nπ ± cos 1 (a) x = 2 nπ ± cos

−1

( ) 1

2

π x = 2 nπ ± 4 π π x = 2 nπ + and x = 2 nπ − , n ∈ Z. 4 4 8nπ ± π (8n ± 1)π ,n∈Z x= = 4 4 − 7π 9π and x = 4 4 − π π n = 0: x = and x = 4 4 π 7π π 9π n = 1: x = 2π − = and x = 2π + = 4 4 4 4 − 7π − π π 7π x= , , , 4 4 4 4

n = −1 : x =

Chapter 6

−


277

Worked Example 14

Find the general solution of the equation 2 sin ( 2 x ) = domain 0 ≤ x ≤ 2π. Think

278

−

3 and hence find all solutions for x in the

Write

1

On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(2 sin (2 x ) = − 3 , x ) Then press ENTER ·.

2

Write the two separate solutions and specify n ∈ Z.

Solving 2 sin (2x) = 3 gives (6 n − 1) π (3n + 2) π x= and x = , n ∈ Z. 6 3

3

Substitute n = 0 and n = 1 and n = 2 into each of the general solutions, to find the solutions in the domain 0 ≤ x ≤ 2p.

n = 0: x =

4

This answer can be verified using a CAS calculator. Complete the entry line as: − solve(2 sin (2 x ) = 3 , x ) | 0 ≤ x ≤ 2π Then press ENTER ·.

−

−π

2π 6 3 5π 5π n = 1: x = and x = 6 3 11 π 8π n = 2: x = and x = 6 3 2π 5π 5π 11π , , , For 0 ≤ x ≤ 2p, x = 3 3 6 6


and x =

Worked Example 15

Find the general solution of the equation sin (3x) = cos (3x) and hence find all solutions for x in the domain 0 ≤ x ≤ 2π. Think 1

Divide both sides by cos (3x).

Write

sin (3 x ) = cos (3 x ) sin (3 x ) =1 cos (3 x ) tan (3x) = 1 -

2

Write down the general solution for tan (3x) = a. 3x = np + tan 1 (a)

3

Substitute a = 1 into the general equation and evaluate tan 1 (1), recognising that it is an exact angle. Divide each side by 3 to solve for x.

4

To find the solutions in the domain 0 ≤ x ≤ 2p, substitute n = 0, 1, 2 . . . into each of the general solutions.

5

Write down the simplified solutions for 0 ≤ x ≤ 2p.

6

This answer can be verified using a CAS calculator. Note: 1. The general formula here is found using − 3π the basic angle of , which means 4 that values of n start at 1 rather than 0. 2. to scroll to the right to view the full set of solutions over the given domain.

-

3x = np + tan 1 (1) π 3 x = nπ + 4 x=

(4 n + 1) π , n ∈Z 12

n= 0: x =

π 12

n = 1: x =

5π 12

n = 2: x =

9π 12

n = 3: x =

13 π 12

n= 4: x =

17 π 12

n = 5: x =

21π 12

x=

π 5π 3π 13π 17π 7π , , , , , 12 12 4 12 12 4


279

Worked Example 16

Find the solutions of the equation cos ( x ) + cos ( 3 x ) = 12 , x ∈[ − 2π , 2π ] . Think

Write

1

On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: 1 solve(cos ( x ) + cos (3 x ) = , x ) − 2π ≤ x ≤ 2π 2 Then press ENTER ·. An exact solution is not available.

2

A numerical solution will be appropriate and can be found by pressing Ctrl / ENTER ·.

3

Write the solutions.

Solving cos( x ) + cos(3 x ) =

1 2

for

x ∈ [-2p, 2p] gives x = - 5.654 87, -2.0944, -0.628 319, 0.628 319, 2.0944, 5.654 87

1. Given a general equation such as sin (x) = a, there can be an infinite number of solutions. The domain is usually restricted and it is important to find all values for x within this domain. 2. If the domain is given in radians, then the solution(s) to x should be in radians. If the domain is given in degrees, then the solution(s) to x should be in degrees. 3. Adjust the domain to match what has been done to the angle in the question.

280


4. Sine is positive in the 1st and 2nd quadrants, cosine is positive in the 1st and 4th quadrants and tangent is positive in the 1st and 3rd quadrants. 5. If sin (x) = a, then the general solution is x = sin−1 (a) + 2nπ, n ∈ Z. 6. If cos (x) = a, then the general solution is x = cos−1 (a) + 2nπ, n ∈ Z. 7. If tan (x) = a, then the general solution is x = tan−1 (a) + nπ, n ∈ Z. 8. Find all the solutions within the specified domain, by substituting integer values for n into the general solution. 9. If given cos2 (x) = a, then you will need to find values in all 4 quadrants (as x will have both positive and negative values). 10. If the equation is of the form sin (ax) = k cos (ax), divide both sides by cos (ax) to change the equation to tan (ax) = k.

exerCiSe

6C

Trigonometric equations Find all solutions to the equations below in the domain [0, 2π].

1 We9

2


Spreadsheet 135 Trig equations

a cos (θ ) = 0

b sin (θ ) =

d sin (θ ) = −1

e cos (θ ) =

−1 2

c

cos (θ ) =

c

sin (θ ) =

1 2

− 3 2

Find all the values of θ between 0° and 360° for which: a sin (θ ) = 1

b cos(θ ) =

1 2

d cos (θ ) = −1

e sin (θ ) =

1

3 2

2

3 We10 For each equation below, find all the values of x between 0 and 4π. Give answers correct to 4 decimal places. a cos (x) = −0.6591 b sin (x) = 0.9104 c cos (x) = 0.48 d sin (x) = −0.371 4 Find all the values of x between 0° and 360° for which: a sin (x) = 0.2686 b cos (x) = −0.7421 − c sin (x) = 0.5432 d cos (x) = 0.1937 Give answers correct to 2 decimal places. 5

Find the solutions to the following equations in the domain 0 ≤ x ≤ 2π. a 2 sin (x) = 1 c 2 sin ( x ) =

−

b 3 cos (x) = 0 d

3

2 cos( x ) = 1

6 We11 Find the solutions to the following equations in the domain 0° ≤ x ≤ 360°. Give exact answers where possible, otherwise give answers correct to 2 decimal places. a cos (2x) = 1 b 2 sin (2x) = −1 c 2 cos (3 x ) = e sin (3x) = g 4 sin

−

2

−0.1254

( x ) = 0.913 1 2

d 2 sin (3 x ) = 3 f 3 cos (2x) = 0.5787 h

2 cos( x ) = − 0.2751

Chapter 6


281

7 Find all the solutions between 0 and 2π to the following equations. Give exact answers where possible, otherwise give answers correct to 4 decimal places.  x   2 

a 4 sin (x) + 2 = 6

b 3 cos (x) − 3 = 0

c cos  

 x d sin   + 5 = 5.32  3

e 2 sin (3x) − 5 = −4

f

g 2 cos (2 x ) + 3 = 0

h

1 3

1

+ 4 = 4.21

2 cos (3 x ) + 2 = 3



sin  2 x − 1 = − 0.8039

8 We12 Calculate the sum of the solutions between 0 and 2π for each of the following equations. Give exact answers for questions a to d. Otherwise give answers correct to 4 decimal places. a sin (x) = cos (x) b sin (2x) = cos (2x) c sin (2 x ) = 3 cos (2 x )


WorkSHEET 6.1

f sin (x) + 3 cos (x) = 0 e sin (3x) + 2 cos (3x) = 0 d 3 sin (3 x ) = cos(3 x ) 9 A particle moves in a straight line so that its distance, x metres, from a point O is given by the equation x = 3 + 4 sin (2t), where t is the time in seconds after the particle begins to move. a Find the distance from O when the particle begins to move. b Find the time when the particle first reaches O. Give your answer correct to 2 decimal places. 10 We13 Find the general solution of the following equations. Hence, find all solutions for −2π ≤ x ≤ 2π for each equation. a 2 cos ( x ) − 3 = 0

b tan ( x ) =

1

2 sin ( x ) − 1 = 0

c

3

11 We14 Find the general solution of the equation 2 sin (2x) − 1 = 0. Hence, find all solutions for −π ≤ x ≤ π. 12 Find the general solution of the equation 2 cos (3x) − 1 = 0. Hence, find all solutions for −π ≤ x ≤ π. 13 We15 Find the general solutions for each of the following equations. Hence, find all solutions for x ∈ [0, 2π]. a

3 sin ( x ) = cos ( x )

b sin (2x) = cos (2x)

c

14 We16 Find the solutions of the equation sin (2 x ) + sin (3 x ) =

6d

3 ,x 2

−

3 sin (3 x ) = cos (3 x )

∈[− π , π ] .

Trigonometric graphs Graphs of the sine and cosine functions The graph of the function f (x) = sin (x) is drawn below. It is drawn over the domain [−2π, 2π]. The graph repeats itself every 2π radians. We say that it has a period of 2π. Half the distance between the maximum and minimum values is 1 so we say y that the amplitude is 1. 1 It is possible to take any value of x for the function 0.5 f (x) = sin (x), so the domain of the whole function is R. The range is [−1, 1]. The graph is shown at right. 3π 3π −π −–π 0 –π π — −2π − — 2π x 2

2

−0.5

2

−1 f(x) = sin (x),

282


−2π

≤ x ≤ 2π

2

The graph of the function f (x) = cos (x) is the same shape as the sine graph, but the graph has been translated to a different position. The period is also 2π. The amplitude is 1, the domain is R and the range is [−1, 1]. The graph is shown at right.

y 1 0.5 3π − π–2 0 −2π − — 2 −π −0.5

–π 2

π

3π — 2

2π x

−1 f(x) = cos (x), -2p ≤ x ≤ 2p

Dilation If we change the amplitude, the distance between the maximum value and the minimum value also changes. y The graph of f (x) = 2 sin (x) is shown at right. The amplitude is 2 2. The period is still 2π. The domain is R and the range is [−2, 2]. 1 This graph is a dilation of the basic graph of f (x) = sin (x) by a factor of 2 from the x-axis. It has been stretched vertically. 0 –π 3π — π 2π x 2 2 Generally, if f (x) = a sin (x) or f (x) = a cos (x), a is the dilation −1 factor in the direction of the y-axis. The amplitude of the −2 graph is a . f(x) = 2 sin (x), 0 ≤ x ≤ 2p If we change the coefficient of x, the period of the graph changes. The graph of f (x) = cos (2x) from 0 to 2π is shown at right. The amplitude is 1 and the period is π. This can be y found by dividing 2π by the coefficient of x. 1 2π In this case = π . The domain is R. The range is [−1, 1]. 0.5 2 The x-intercepts can be found by solving the equation 0 –π –π — 3π π — 5π — 3π — 7π 2π x 4 2 2 4 4 4 cos (2x) = 0. −0.5 π 3π 5π 7π 2 x = , , , −1 2 2 2 2 f(x) = cos (2x), 0 ≤ x ≤ 2p π 3π 5π 7π so x = , , , 4 4 4 4 1 This graph is a dilation of the basic graph of f (x) = cos (x) by a factor of 2 from the y-axis. The period has been halved or the graph has been ‘squashed up’. The decimal approximation for these solutions can be found using a graphics calculator. The graphics calculator can also be used to check the number and approximate value of the solutions when solving trigonometric equations. 1 Generally, if f (x) = sin (nx) or f (x) = cos (nx), the graph is dilated by a factor of from the n 2π y y-axis. The period of the graph is . n 4 The graph of f (x) = 4 sin (3x) is shown at right. 3 2

It is drawn from 0 to 2π. The amplitude is 4 and the 1 2π 4π 5π –π — — — 3 3 3 3 2π − 3π 0 period is . The domain is R. The range is [ 4, 4]. π –π — 2π x 2 2 −1 3 −2 The x-intercepts are found by solving 4 sin (3x) = 0. −3 So sin (3x) = 0 −4 3x = 0, π, 2π, 3π, 4π, 5π, 6π f (x) = 4 sin (3x), 0 ≤ x ≤ 2p 4π 5π π 2π x = 0, , , π, , , 2π 3 3 3 3 This is an example of a sine graph dilated in both x and y directions. It has a dilation factor of 1 4 from the x-axis, and of 3 from the y-axis.


283

Worked Example 17

State the period and the amplitude of the graphs of each of the following functions. a y = 2 sin

( x) 1 4

b y =

−1 cos 3

(2 x ) .

Think a

b

Write

1

Write the equation.

2

Find the period of the graph, using the formula.

3

State the amplitude — it is the coefficient in front of the sine function.

1

Write the equation.

2

Find the period of the graph.

a y = 2 sin

( x) 1 4

2π 1 ,n = 4 n 2π 4 So period = 1 = 2π × = 8π 1 4 Period =

Amplitude = 2 b y=

−1 cos (2 x ) 3

Period = So period =

3

Amplitude =

State the amplitude. that the amplitude is always positive, so we need only the magnitude of the coefficient.

2π , n=2 n 2π =π 2 1 3

Worked Example 18 1 Sketch the graph of y = 2 sin (3θ ) for one complete cycle stating the amplitude, period and range.

Think

Write/draw

Method 1: Technology-free

284

1

Write the equation.

2

Find the period of the graph.

3

State the amplitude.

1

y = 2 sin (3θ ) 2π ,n=3 n 2π So period = 3 Period =

Amplitude =


1 2

4

5

Sketch the basic sine shape (one full cycle) 2π 1 with the period of and the amplitude of 2 . 3 1 The scale on the x-axis should be 4 of a period (to reflect the 4 quadrants), 2π 2π π so it is ÷ 4= = . 3 12 6

The minimum value of the function is

−1 2

and

y 1– 2

0

π– 6

π– 3

π– 2

−1

1

2— π 3

− 1– 2

Range : [ 2 , 2 ]

1

the maximum value is 2 , so state the range. Method 2: Technology-enabled On a Graphs page, complete the function entry line as: 1 2π f 1( x ) = sin (3 x) | 0 ≤ x ≤ 2 3 Then press ENTER ·.

Reflection If the coefficient of the function is negative, the graph is turned upside down, that is, reflected in the x-axis. This does not alter the amplitude, which is always positive. • y = -f (x) is the image of f (x) when reflected in the x-axis. y • y = f (-x) is the image of f (x) when reflected in the y-axis. 4 3 The graph of f (x) = −4 sin (3x) is shown at right. You 2 will notice that it is the graph of f (x) = 4 sin (3x) turned 1 3π –π — upside down. (The graph of f (x) = 4 sin (3x) is shown 2 2 5π 4π 2π 0 on page 283.) –π — — π — 2π x 3 3 3 3 −1 −2 −3 −4

f (x) = -4 sin (3x), 0 ≤ x ≤ 2p

2π . The domain is R. The range is [−4, 4]. 3 4π 5π π 2π The x-intercepts are 0, , , π, , , 2π . 3 3 3 3 1 This graph has a dilation factor of 4 from the x-axis and of 3 from the y-axis. If the function f (x) = 4 sin (3x) is reflected in the x-axis, the result is f (x) = −4 sin (3x). If we reflected the graph of f (x) = 4 sin (3x) in the y-axis, the result would still be f (x) = −4 sin (3x). The amplitude is 4 and the period is


285

If we reflect f (x) = 2 cos (3x) in the x-axis, the result is f (x) = −2 cos (3x), but if we reflect it in the y-axis, the graph does not change. This is because f (x) = 2 cos (3x) is symmetrical about the y-axis. Check this on a CAS calculator. y 4 3

Translation If we add a constant to the function, the graph is moved up or down and is said to be translated parallel to the y-axis. The number that is being added becomes the median value of the function. The graph of f (x) = 3 cos (2x) + 1 is shown at right. Compared to f (x) = 3 cos (2x), the graph is shifted 1 unit up. The amplitude is 3, the period is π, the domain is R and the range is [−2, 4]. The x-intercepts are found by solving 3 cos (2x) + 1 = 0. So cos 2 x =

−1 3

2 1 −1

0

–π 2

π

3π — 2

2π x

−2 f (x) = 3 cos (2x) + 1, 0 ≤ x ≤ 2p

(cosine is negative in quadrants 2 and 3).

The basic angle is inverse cosine of

1 3

= 1.231c .

2x = π − 1.231, π + 1.231, 3π − 1.231, 3π + 1.231 2x = 1.911, 4.373, 8.194, 10.656 ⇒ x = 0.955, 2.186, 4.097, 5.328 If we add a constant to x, the graph is translated parallel to the x-axis, that is, left or right. π  The graph of f ( x ) = sin  x −  is the graph of  4

y 1 0.5

0 –π 5π π −π π— 2π x 4 4 units to the right. −0.5 4 −1 π  The graph of f ( x ) = sin  x +  is the graph of  4 π  f ( x ) = sin  x +  , − π ≤ x ≤ 2π π  4 f (x) = sin (x) translated units to the left. 4 This graph is a translation of the π  basic graph of the function The graph of f ( x ) = sin  x +  is shown at right.  4 π f ( x ) = sin( x ) units parallel to It is drawn between [−π, 2π]. 4 The amplitude is 1, the period is 2π, the domain is R and the x-axis in a negative direction. the range is [−1, 1]. The y-intercept occurs when x = 0. 2 π  . So y = sin  0 +  =  4 2 π  The x-intercepts occur when sin  x +  = 0.  4 − π 3π 7π π So x + = 0, π , 2π ⇒ x = , , 4 4 4 4 π 3π Note: The graph of sin (x) is the same as the graph of cos ( x − ) or cos ( x + ). That is, the 2 2 π 3π sine graph can be turned into the cosine graph by a translation of units left or units 2 2 right. The cosine graph can be turned into the sine graph by the opposite translations.

f (x) = sin (x) translated

If we take a general trigonometric function f (x) = a sin n(x − b) + c, it has an amplitude 2π of a , a period of , a horizontal translation of b units and a vertical translation of n c units.

286


y

π  The graph of f ( x ) = 2 cos 3  x −  − 1 is shown at right for  3 0 ≤ x ≤ 2π. 2π The amplitude is 2, the period is , the domain is R and the 3 range is [−3, 1].

1 0

–π 3

−1

2π — 3

π

4π — 3

5π — 3

2π x

−2

This graph is a dilation of the basic graph of f (x) = cos (x) by a 1 factor of 2 from the x-axis and a factor of 3 from the y-axis, π along with a translation of units to the right and 1 unit down. 3

−3

Worked Example 19

π  Sketch the graph of y = 5 cos  x +  + 5 for 0 ≤ x ≤ 2π, and state the period and amplitude.  4 Think

Write/draw

π  y = 5 cos  x +  + 5  4

1

Write the equation.

2

Determine the period. (The coefficient of x is 1.)

Period =

3

State the amplitude.

Amplitude = 5

4

Using a pencil, sketch a basic cosine shape with the period and amplitude above.

2π = 2π 1

y 5

y = 5 cos (x)

0

π

–π 2

2π x

3π — 2

−5 5

Apply the horizontal translation of +

π (the + sign 4

y –π 4

means to the left). Again, use pencil.

–π 4

0 –π

−π–4

–π 2

4

Apply the vertical translation of +5 (+ means up).

y

π

3π — 4

5π — 3π — 7π — 4 2 4 2π

x

–π 4

y = 5 cos (x + –4π ) + 5

10

+5

+5

5 +5 +5

−π–4

y = 5 cos (x)

–π 4

−5 6

–π 4

–π 5 y = 5 cos (x + 4 )

0

y = 5 cos (x + –4π )

3π — 4 –π 2

π +5

3π — 7π — 2 4

2π x

−5


287

7

Erase the pencilled stages.

y 10

y = 5 cos (x + –4π ) + 5

0

2π x

3π — 4

1. Graphs of the form y = a sin n(x − b) + c and y = a cos n(x − b) + c are transformations of y = sin (x) and y = cos (x). 2. a is the amplitude and is a dilation from the x-axis. If a is negative, the amplitude is still positive but the graph is reflected in the x-axis. 3. c is the vertical translation (the translation parallel to the y-axis). If c is positive, the graph is translated c units up, and if c is negative, the graph is translated c units down. 4. The range is [c − |a|, c + |a|]. 2π . 5. The period is n 6. The factor n is the horizontal dilation where the graph has been dilated by a 1 factor of from the y-axis. n 7. The value b is the horizontal translation (the translation parallel to the x-axis). If b is positive, the graph is translated b units to the right, and if b is negative, the graph is translated b units to the left.

exerCiSe

6d

Trigonometric graphs 1 We17


SkillSHEET 6.5 Period and amplitude of sine and cosine graphs


Spreadsheet 124 Sine graphs

State the period and amplitude of the graphs of each of the following.

a y = cos (x) 1 3

d y = cos ( x ) g y = 3 sin

1 2

c y = 4 sin (x)

e y = 2 cos (3x)

f y = 3 sin (2x)

h y = 2 cos

( x) 1 3

i

y=

−1 3

cos (2 x )

j y = −4 sin (3x) 2 We18 Sketch the graphs of each of the following for one complete cycle and state the amplitude, the period and the range. 2

a y = 3 cos(θ )

b y = 4 sin (θ )

d y = 2 cos (3θ )

e y = 2 cos (3θ )

g y = 4 sin

288

( x)

b y = sin (x)

( θ) 1 2

1

h y = 3 cos


( θ) 1 3

c y = 3 sin (2θ ) f y=

−1 3

sin (2θ )

eBook plus

3

Sketch the graph of the function f:R → R where f (x) = 4 cos (3θ ) for 0 ≤ θ ≤ 2π. State the amplitude, period and range.

4

Sketch the graph of the function f:R → R where y = 2 sin (2x) for −π ≤ x ≤ π. State the amplitude, period and range.

5

From the basic graphs of y = sin (x) and y = cos (x), state the horizontal translation and the vertical translation for each of the following. π π   a y = sin  x +  + 3 b y = cos  x −  + 1   3 2

Digital doc

Spreadsheet 012 Cosine graphs

π  c y = 3 cos  x −  − 2  4

π  d y = 2 sin  x +  − 1  3

6 Sketch the graphs of the following for one complete cycle stating the amplitude, the period and the range. a y = sin (x) + 1 b y = cos (x) − 1 c y = 2 cos (x) − 2 d y = 2 sin (x) + 3 e y = sin (3x) − 1 f y = cos (2x) + 1 1

h y = 2 sin (2 x ) + 3

g y = 3 cos (3x) − 2 j y = 2 cos

( x) − 1

y = 3 sin

i

( x) + 4 1 2

1 3

7 We19 Sketch the graphs of the following for 0 ≤ θ ≤ 2π. State the period and amplitude. π π π    a y = sin  θ −  b y = cos  θ +  c y = 3 cos  θ −     3 4 2

π  d y = 2 sin  θ −   4 π  g y = cos 3  θ −  + 1  6

π  e y = 2 sin 2  θ +   2 π  h y = 2 sin 2  θ −  − 2  2

π  f y = 3 cos 3  θ +   3 π  i y = 2 sin  θ −  − 1  4

1

j y = cos 2 (θ − π ) + 1 8

Write down the amplitude, period and range of the following graphs. a

b

y 4

y 4

2 3π −π 0 –π – −π – −π −— 4 2 4 4 −2

2 –π 2

3π π — 4

x

0 −2

−4

c

2π

3π

4π x

−4

d

y 2

y 1

1 0 −1

π

0.5 –π 2

π

3π — 2

2π

5π — 2

3π x

−0.5

−2

0 π 2π 3π 4π 5π 6π 7π 8π 9π x

−1

Chapter 6


289

y

e

y

f

4 3 2 1 −π

g

π x

0

−π–2 −1 0 −2

π

–π 2

3π — 2

x −7

h

y 1.5 1 0.5 −0.5 −1 −1.5

9

3

y 1 0.5

0

–π 2

π

3π — 2

0

2π x

−0.5

–π 2

π

2π

3π — 2

5π — 2

3π x

−1

State the maximum and minimum values for each of the following. a y = cos (x) b y = sin (x) c y = 3 sin (x) d y = 2 cos (x) e y = 2 cos (3x) f y = 3 sin (2x) g y = 4 sin (2x) + 1 h y = cos (4x) − 2 i y = 2 cos (x − π) − 3 j y = 4 sin 3 (x + π) + 1

1

k y = 2 sin 3 ( x + 2π ) + 2

1

l y = 3 cos 2 ( x − 2π ) − 4

10 Sketch the graphs of the following over the domain [0, 2π] and state the period, amplitude and range. a y = −cos (x) b y = −sin (x) c y = −2 sin (x) d y = −3 cos (x) e y = −3 cos (2x) f y = −sin (3x) g y = 1 − 4 sin (x) j y = 2 − sin (x − 2π)

h y = −2 cos (2x) − 2

i y =

−1 2

cos 3 ( x + π ) + 1

π to the left, what is the new equation? 3 π 12 If the graph of y = 2 cos (3x) − 2 is translated to the right and 3 units up, what is the new 4 equation? π 13 If the graph of y = 3 sin (x − π) + 1 is translated to the left and 3 units down, what is the new 3 equation? 11 If the graph of y = sin (x) + 1 is translated

Water level

14 The level of the water in the Banksia River was measured at hourly intervals from midnight and the results recorded. The graph below shows the results. Find: y a the amplitude 3 b the period 2 c the maximum height of the river d the minimum height of the river 1 e at what times the river has maximum height 0 2 4 6 8 10 12 14 16 18 20 22 24 x f at what times the river has a minimum Hours height g the equation of the curve which is of the form y = A sin a(x) + B.

290


6E

Graphs of the tangent function The graph of the function f (x) = tan (x) is shown at right. − x = —2π It is drawn over the domain [−2π, 2π]. The graph repeats y − x = 3—2π itself every π radians, so its period is π. There are vertical x = –2π x = 3—2π 3 asymptotes through half the period, 2 3π π that is, at x = ± and ± , so the function is not 2 2 1 defined at these points. Generally, the vertical 0 –π π 3—π −2π − 3—2π −π − –2π 2π x π 2 2 asymptotes are given by the equation x = (2k + 1) , −1 2 −2 where k ∈ Z (that is, k = 0, ±1, ±2 …). Hence, the domain π −3 of the tangent function is R \ {x : x = (2k + 1) , k ∈ Z}. 2 Unlike sine and cosine functions, the tangent graph does not have an amplitude; it extends infinitely up and down,so its range is R.

Dilation

y = 2 tan (x) y = tan (x)

y

Compared to f (x) = tan (x), the graph of f (x) = a tan (x) is dilated by a factor of a from the x-axis. Vertical dilation does not affect the period, domain or range, or the position of the asymptotes or the x-intercepts. In fact, its effect can be seen only when two graphs are sketched on the same set of axes. The diagram at right shows the graphs of f (x) = tan (x) and f (x) = 2 tan (x) over the domain [0, 2π]. The graph of f (x) = 2 tan (x) is the dilation of the basic tangent graph by a factor of 2 from the x-axis. It has been stretched vertically. Compared to f (x) = tan (x), the graph of f (x) = tan (nx) is dilated 1 by a factor of from the y-axis. Horizontal dilation y n x = π–4 3 affects the period and domain of the graph, as well as the position of the asymptotes and x-intercepts. 2 π The period of f (x) = tan (nx) is and the general equation 1 n π of the asymptotes is x = (2k + 1) ,where k ∈ Z. 0 π– π– 2n −1 4 2 π The domain changes to R \ {x : x = (2k + 1) , k ∈ Z} −2 2n accordingly. (The range is not affected by the value of n.) −3 At right is the graph of f (x) = tan (2x). This graph is a dilation of the basic f (x) = tan (x) graph by a factor of 12 π from the y-axis; its period is . As the period has been 2 halved, the resultant graph is ‘compressed’ horizontally.

3 2 1 0 −1

π 3—2π

–π 2

2π

x

−2 −3

x = –2π x = 3—2π

x = 3—4π x = 5—4π x = 7—4π

π

3— π 4

5— π 4

y

3— π 7— π 2 4

2π x

x = –2π x = 3—2π

3 2 1

Reflection If the coefficient a in f (x) = a tan (x) is negative, the graph is reflected in the x-axis. Reflecting the graph does not affect its period, domain or range, or the position of the asymptotes. At right is the graph of f (x) = −2 tan (x). This graph is a reflection in the x-axis of the graph f (x) = 2 tan (x), shown previously.

0 −1

–π 2

π

3— π 2

2π

x

−2 −3


291

Translation y If a constant is added to the function, the graph is translated x = –2π x = 3—2π vertically, that is, up or down, parallel to the y-axis. Thus, the 3 graph of f (x) = tan (x) + c represents a translation of the graph 2 f (x) = tan (x) by c units in the y direction. If c > 0, the graph is 1 shifted up, and if c < 0, it is shifted down. Vertical translation does not affect the period, domain or range, or the position of 0 –π π 3—π 2π x 2 2 the asymptotes. The axial intercepts, however, will change. −1 The diagram at right shows the graph of y = tan (x) + 2. It is −2 translated 2 units up, compared to the basic graph of y = tan (x). −3 If a constant is added to x, the graph is translated horizontally, that is, right or left, parallel to the x-axis. Thus, the graph of f (x) = tan (x − b), y x = –4π x = 5—4π represents a translation of the graph f (x) = tan (x) 3 by b units in the x direction. If b > 0, the graph is 2 shifted to the right, and if b < 0, it is shifted to the left. Horizontal translation has no effect on the 1 period or range, but it does affect the domain and 3— π π 5— π π 7— π the position of the asymptotes and the axial 2π x –π –π 3— − –π 0 4 4 2 4 4 2 4 −1 intercepts. π −2 The diagram at right shows the graph of y = tan ( x + ). 4 π translated units to the left, compared to the basic 4 graph of y = tan (x). Having considered each transformation individually, we can now summarise them as follows. Compared to the basic graph of y = tan (x), the graph of y = a tan [n(x − b)] + c is: – dilated by a factor of a from the x-axis 1 π – dilated by a factor of from the y-axis (and hence has a period of ) n n – reflected in the x-axis if a < 0 – translated c units vertically (up if c > 0 and down if c < 0) – translated b units horizontally (to the right if b > 0 and to the left if b < 0).

Worked Example 20

 x State the period and sketch the graph of y = 2 tan   , showing one full cycle.  4 Think

Write/draw

1

Write the equation.

 x y = 2 tan    4

2

Find the period.

Period =

π ; n= n

So, period =

292


π 1 4

1 4

=π ×

4 = 4π 1

3

Sketch the tangent shape with a period of 4π. The graph is dilated by a factor of 2 parallel to the y-axis, so in comparison to the graph of y = tan (x), each y-coordinate will be doubled. The vertical asymptote goes through half of the period, so it is at x = 2π.

y

x = 2π

3 2 1 0 −1

π

2π

3π

4π

x

−2 −3

Worked example 21

eBook plus

π State the period and sketch the graph of y = − tan 2 ( x − ) + 1 4 for 0 ≤ x ≤ π. Think

Tutorial


WriTe/draW

1

Write the equation.

π  y = − tan 2  x −  + 1  4

2

Find the period.

Period =

3

Using pencil, sketch the basic tangent shape with π a period of . 2 (Since 0 ≤ x ≤ π, we need to show two cycles.) that the asymptotes are at the middle of each cycle (that is, halfway through the period).

y

π π ; n = 2. so period = n 2 x = π–4 x = 3—4π

3 2 1 0 −1

–π 4

–π 2

3— π 4

π x

−2 −3 4

Since we need a negative tangent graph, reflect the graph in the x-axis.

y

x = π–4 x = 3—4π

3 2 1 0 −1

–π 4

–π 2

3— π 4

π x

−2 −3

Chapter 6


293

5

Translate the graph

π units to the right. 4

y 3

x = π–4 x = π–2

2

x = 3—4π

1 0 −1

–π 4

–π 2

3— π 4

π

x

−2 −3 6

y

Translate the graph 1 unit up.

x = π–2 x = π

3 2 1 0 −1

–π 4

–π 2

3— π 4

π

x

−2 −3

7

Erase the pencilled stages to see the final graph.

y

x = π–2 x = π

3 2 1 0 −1

–π 4

–π 2

3— π 4

π

x

−2 −3

1. The graph of y = tan (x) has a period of π and does − − y x = 3—2π x = —2π x = –2π x = 3—2π not have an amplitude. 3 2. The equations of the vertical asymptotes of 2 π y = tan (x) are given by x = (2k + 1) , where 1 2 k ∈ Z, that is, k = 0, ±1, ±2 . . . 0 –π π 3—π −2π − 3—2π −π − –2π 2π x 2 2 3. The domain of the tangent function is −1 π R \ {x : x = (2k + 1) , k ∈ Z} and its range is R. −2 2

294


4. Compared to the basic graph of y = tan (x), the graph of y = a tan [n(x − b)] + c is: – dilated by a factor of a from the x-axis 1 π – dilated by a factor of from the y-axis (and hence has period of ) n n – reflected in the x-axis if a < 0 – translated c units vertically (up if c > 0 and down if c < 0) – translated b units horizontally (to the right if b > 0 and to the left if b < 0). exerCiSe

6e

Graphs of the tangent function 1

We20 State the period and sketch the graphs of each of the following, showing one full cycle. a y = 4 tan (x) b y = tan (2x) c y = −tan (3x)  x 1 f y = − 2 tan   d y = 2 tan (4x) e y = − tan 4 x  3  x − h y = −5 tan (2x) g y = 3 tan   i y = 21 tan (4 x )  2

( )

j

y = 13 tan

( x) 1 2

2 For each of the following, state: i the period ii the transformations, compared to the basic graph y = tan (x). a y = 2 tan (3x) d y = − tan

( x) − 2 1 2

π  1 g y = 4 tan 3  x +   6

b y = −tan (4x) + 1 π  e y = − 5 tan  x +   2

c y = 3 tan (2x) − 4 π  f y = tan 2  x −   4

 x 1 h y = 6 − 2 tan    3

π 1  i y = tan 4  x −  − 3  12 

π  j y = 2 tan  x +  + 5  3 3

We21 State the period and sketch the graphs of each of the following for 0 ≤ x ≤ π.

a y = tan (x) + 2

π  d y = 2 tan  x −   2  x 1 g y = 2 tan   − 1  2 j y=

6F

1 3

b y = tan (2x) − 3

π  c y = − tan  x +   4

1 e y = − tan  4 ( x − π )      π h y = − 2 tan  2  x +    8 

 x 1 f y = − 3 tan   + 1 2  6 −3   π  tan  4  x +   + 1 i y=   4 12 

1  π tan  2  x −   − 2  2 

Finding equations of trigonometric graphs

eBook plus Interactivity

Sometimes it is necessary to be able to find the equation of a trigonometric function from a graph. The following worked examples illustrate how this can be done.

Chapter 6

int-0251 Finding equations of trigonometric graphs


295

Worked Example 22

The equation of the following graph is in the form y = a sin (nx). Find the values of a and n. Hence, find the equation of the function.

y 2 1 0 −1

π

2π

3π

4π

x

−2

Think

Write

1

State the amplitude of the graph.

Amplitude = 2

2

The coefficient a represents the amplitude. Since the graph is not reflected in the x-axis, a is positive.

The graph is not reflected, so a = 2.

3

State the period of the graph (it is the length of one full curve).

Period = 4π.

4

Use the formula for the period to find the value of n.

Period = n=

5

Substitute the values of a and n into y = a sin (nx) to find the equation of the function.

2π 2π , so = 4π ; n n

2π = 4π

1 2

y = a sin (nx); a = 2, n = ∴ y = 2 sin

1 2

( x) 1 2

Worked Example 23

The graph shown is a trigonometric function of the form y = a sin (nx) + c. Find the values of a, n and c. Hence, find the equation of the function.

y 2 1 −1 0 –4π −2 −3 −4

Think 1

2

3

296

The amplitude (a) is half the distance between the maximum and minimum values. The period is the interval from one point on the graph to the next point where the graph begins to 2π repeat itself. The period is . n The line through the centre of the graph is y = −1, so the graph has been translated down 1 unit.

3π — 4

π

5π — 4

Write

y = a sin (nx) + c 1

Amplitude a = 2 (2 + 4) = 3 2π = π , so n = 2. The period is n

c = −1 So the equation is y = 3 sin (2x) − 1.


7π — 4

2π x

Worked Example 24

This graph is a trigonometric function of the form y = c + a cos (nx). Find the values of a, n and c. Hence, write the equation of the function. y 3 2 1 0 −1

1

2

3

Think

The amplitude (a) is half the distance between the maximum and minimum values.

2

We know that the graph is a cosine graph so it must be inverted; that is, a is negative. The period is the amount of time taken to complete the pattern once.

4 5

5

6 x

Write

1

3

4

c represents the vertical translation. This graph has been translated up one unit. The equation of this trigonometric function is in the form y = c + a cos (nx).

y = c + a cos (nx) 1 Amplitude a = 2 (3 + 1) =2 a = −2 Period =

2π =6 n

2π =n 6 π so n = 3 c=1

y = c + a cos ( nx )

π 3 π  y = 1 − 2 cos  x  3  c = 1, a = − 2, n =

1. Trigonometric functions can be expressed in the form of y = a sin [n(x − b)] + c or y = a cos [n(x − b)] + c. 2. The amplitude a can be found by halving the distance between the maximum and minimum values. 3. The period is the interval from one point on the graph to the next point where the 2π graph begins to repeat itself. The period is . n 4. The vertical shift is c. 5. The horizontal shift is b. Exercise

6F

Finding equations of trigonometric graphs 1 WE22 The equations of the following graphs are of the form y = a sin (nx). Find the values of a and n. Hence, find the equation of each function.


297

a


Spreadsheet 124 Sine graphs

Digital doc

y 2 1

0

−1 −2 −3

eBook plus

b

y 3 2 1 –π 4

–π 2

0 1 2 3 4 5 6 7 8 9 10 11 12 x −1 −2

π x

3π — 4

Spreadsheet 012 Cosine graphs

2 The equations of the following graphs are of the form y = a cos (nx). Find the values of a and n. Hence, find the equation of each function. b a y y 2

2

1

1

0 −1

–π 8

–π 4

3π — 8

–π 2

x

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 x −1 −2

−2

3

We23 The equations of the following graphs are of the form y = a sin (nx) + c. Find the values of a, n, and c and hence write the equation of the function. b a y y

1

1.5

0 1 2 3 4 5 6 7 8x −1 −2

1.0 0.5 0

–π 2

π

−3 −4 −5

2π x

3π — 2

4 The equations of the following graphs are of the form y = a cos [n(x − ε)]. Find the values of a, n and ε. Hence, write the equation of the function. b a y y 5

4 2 0 −2

–π 4

–π 2

3π — 4

π x

0

−4

–π 4

–π 2

3π — 4

π x

−5

5 The equations of the following graphs are of the form y = a sin [n(x + ε)] + c. Find the values of a, n, ε and c. Hence, write the equation of the function. b a y y 1 0 −1

–π 6

–π –π 3 2

2π — 5π — 3 6

π

7π — 4π — 3π — 5π 11 — —π 6 3 2 3 6

2π x

−2 −3

298


5 4 3 2 1

0 −1

–π 2

π

3π — 2

2π

x

6

We24 The equations of the following graphs are of the form y = c + a cos (nx). Find the values of a, n and c. Hence, write the equation of the function.

a

b

y 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12 x −1

7

8

y 4 3 2 1 0 1 2 3 4 5 6 7 8 9 10 11 12x –1

mC If the amplitude is 2, the period is 6 and there is a vertical translation of −2, then the

equation of the form y = a sin (nx) + b is: a y = 2 sin (6x) − 2

b y = 6 sin (2x) − 2

π D y = 2 sin  3

π E y = 2 sin  6

 x − 2 

π  c y = 2 − 2 sin  x  3 

 x − 2 

mC If the period is π, the range is [−2, 4], and the horizontal

eBook plus

π Digital doc , the equation for the trigonometric function of the WorkSHEET 6.2 4 form y = a cos [n(x + ε)] + b is:       π π π a y = 3 − 2 cos π  x +   + 1 b y = 3 cos  2  x +   − 1 c y = 2 cos 3  x +   + 1      4 4 4      translation is

  π D y = 2 cos 3  x +   + 1 4  

6G

  π E y = 3 cos  2  x −   + 1 4  

Trigonometric modelling In real life there are many examples of periodic behaviour. Sine and cosine functions such as y = a sin [n(x − b)] + c and y = a cos [n(x − b)] + c are often used to model this behaviour.

eBook plus eLesson

eles-0092 Trigonometric modelling

Worked example 25

While out in his trawler John North, a fisherman, notes that the height of the tide in the harbour can be found by using the equation: π  h = 5 + 2 cos  t , 6  where h metres is the height of the tide and t is the number of hours after midnight. a What is the height of the high tide and when does it occur in the first 24 hours? b What is the difference in height between high and low tides? c Sketch the graph of h for 0 ≤ t ≤ 24. d John North knows that his trawler needs a depth of at least 6 metres to enter the harbour. Between what hours is he able to bring his boat back into the harbour?

Chapter 6


299

Think

Write


1

Write the given equation.

2

For high tide, find the maximum value of h.

π  t  6 

a h = 5 + 2 cos 

For maximum h, π  cos  t  = 1 6  So h = 5 + 2 × 1 = 7 Alternatively, maximum value = median + amplitude so h = 5 + 2 = 7

b

3

Find when high tide occurs.

1

Find the minimum value of h.

π  cos  t  = 1, so 6  π t = 0, 2π , 4π , . . . 6 t = 0, 12, 24, . . . A high tide of height 7 m occurs at midnight, noon the next day, and midnight the next night. b For minimum h,

π  cos  t  = −1 6 

2

c

So h = 5 + 2 × −1 = 3 Alternatively, min. value = median − amplitude so h = 5 − 2 = 3. The difference between high and low tides is 7 − 3 = 4 metres.

Find the difference between high and low tides.

Use the information from the previous page to sketch the graph.

c

h 6 4 2 0 2 4 6 8 1012 14 16 18 202224 t

d

1

2

300

Find t using the equation when h = 6.

Write the answer in words.

d When h = 6,

π  5 + 2 cos  t  = 6 6  π  2 cos  t  = 1 6  π  1 cos  t  = 2 6  π π 5π 7π 11π t= , , , .... 6 3 3 3 3 t = 2, 10, 14, 22, . . . From the graph we can see that John North can bring his boat back into harbour before 2 am, between 10 am and 2 pm and between 10 pm and 2 am the next morning.



On a Graphs page, complete the function entry lines as: π  f 1( x ) = 5 + 2 cos  x  6  f 2( x ) = 6 Press ENTER · after each entry. Select an appropriate window.

2

To find the points of intersection, press: • MENU b • 7:Points & Lines 7 • 3:Intersection Point(s) 3 Use the NavPad to move the cursor to the first point of intersection and press Click x twice and the points of intersection will appear.

t = 2, 10, 14, 22, . . .

1. Read the question carefully to find out what is being asked. 2. Answer the questions in words where appropriate. 3. Ensure that graphs are sketched over the given domain. exerCiSe

6G eBook plus Digital doc


Trigonometric modelling 1

We25 Competition is severe, so Fred Greenseas decides that he will catch more fish in an inlet several kilometres east of the place where John North fishes. There is a sandbar at the entrance to the inlet and the depth of water in metres on the sandbar is modelled by the  πt  function d (t ) = 6 + 2.5 sin   where t is the number of hours after 12 noon.  6 a What is the greatest depth of the water on the sandbar and when does it first occur? b How many hours before there is once again the maximum depth of water on the sandbar? c What is the least amount of water on the sandbar? d Sketch the graph of d for 0 ≤ t ≤ 24. e Fred Greenseas needs a depth of 7.25 metres to cross the sandbar. Between what hours is he able to enter and leave the inlet?

Chapter 6


301

2 A student wanting to catch fish to sell at a local market on Sunday has discovered that more fish are in the water at the end of the pier when the depth of water is greater than 8.5 metres. π  The depth of the water (in metres) is given by d = 7 + 3 sin  t  , where t hours is the 6  number of hours after midnight on Friday. a b c d e

What is the maximum and minimum depth of the water at the end of the pier? Sketch a graph of d against t from midnight on Friday until midday on Sunday. When does the water first reach maximum depth? Between what hours should the student be on the pier in order to catch the most fish? If the student can fish for only two hours at a time, when should she fish in order to sell the freshest fish at the market from 10.00 am on Sunday morning?

3 The mean daily maximum temperature in Tarabon, an experimental town in a glass dome, is π  modelled by the function T (m) = 18 + 7 cos  m , where T is in degrees Celsius and m is the 6  number of months after 1 January 2007. a What was the mean daily maximum temperature in March 2007, and in August 2007? b What is the highest mean daily maximum temperature in Tarabon? In which months does it occur? c What would the mean daily maximum temperature be in February 2008? d If the pattern continued, how many months would before the mean daily maximum temperature would be the same again as it was in February 2008? 4 The height above the ground of the middle of a skipping rope as it is being turned in a child’s game is found by using the equation h = a sin (nt) + c, where t is the number of seconds after the rope has begun to turn. During the game, the maximum height the rope reaches is 1.8 metres and it takes 2 seconds for the rope to complete a full turn. a Find the values of a, n and c and hence write the equation of h in of t. b Sketch the graph of h against t for 0 ≤ t ≤ 5. c After how much time from the beginning of the turn will the rope be 25 cm above the ground? Give your answer correct to the nearest tenth of a second. y

5 The graph at right shows the path of a small lizard as it runs over a hill. a Calculate the height of the hill, in metres, and hence find the amplitude of the trigonometric function. b If the ground was flat, how far would the lizard run to reach the same spot on the other side of the hill? Hence, find the period of the function. 6 When Sloane and Michael were riding on a Ferris wheel, they realised that as the wheel turned clockwise, the height of their seat in metres after t minutes could be modelled by the function h(t) = a − b cos (nt). The graph of h against t is shown for the whole ride. a State the values of a, b and n. b Write down the rule for h(t). c How many times do Sloane and Michael reach the highest point during their ride and how far above the ground is it? d How high are they when the ride begins to move? e How many minutes before they are at this point again?

302


h

h = 34

30 25 20 15 10 5 0

0

3

6

h=2 9 12 15 t

y = 1 − cos ( –4π x)

x

f How far above the ground are they after 1 minute? g If the ride began when the boys were at the height found in part f , what would the function become? h Draw a graph of the new function for the first 6 minutes of the ride. 7 On a summer day the hourly temperature, which can be approximated to a cosine curve, was recorded. The maximum temperature was 30 °C and occurred at 3.00 pm. The minimum temperature was 10 °C and occurred at 3.00 am. The temperature was first recorded at 12 midnight, then every hour for 24 hours. a What is the amplitude of the function? b What is the period of the function? c What is the middle value of the function? d How far has this middle value been translated upwards from the x-axis? e When do the maximum and minimum temperatures occur? f Using the above, write an equation that will model this function. g Check the accuracy of your work by using your equation to find the temperature after 3, 9, 15 and 21 hours. h What is the temperature at midnight? Give your answer correct to the nearest degree. Check that your answer makes sense.

6H

Further graphs Earlier in the book we discussed in detail how to graph the sum and difference of functions. In this section we will further apply these techniques to trigonometric functions.

Graphing the sum and difference of functions The sum (and difference) function is defined over the intersection of the domains of the individual functions. That is, dom (f (x) ± g(x)) = dom f (x) ∩ dom g(x). To sketch the graph of the sum function, the method of addition of ordinates can be used. This method involves sketching the graphs of the individual functions on the same set of axes and then adding their y-coordinates. This technique is illustrated in the worked example below.

Worked Example 26

Using addition of ordinates, sketch the graph of y = sin (x) + cos (x) for the domain [0, 2π]. Think

Draw


Sketch the graphs of y = sin (x) and y = cos (x) on the same set of axes over the required domain. (Note that both y = sin (x) and y = cos (x) are defined over [0, 2π], so the sum function is defined over the same domain.)

y 2 1 0 −1

–π 4

–π 2

3— π 4

π

5— π 4

3— π 2

7— π 4

2π x

−2


303

2

Moving from left to right, add the y-coordinates of the two graphs and plot the resultant points. The π π 3π 5π 3π 7π π, , , ‘key’points are at x = 0, , , 4 2 4 4 2 4 and 2π.

y 2 1 0 −1

–π 4

–π 2

3— π 4

π

5— π 4

3— π 2

7— π 4

–π 4

–π 2

3— π 4

π

5— π 4

3— π 2

7— π 4

–π 4

–π 2

3— π 4

π

5— π 4

3— π 2

7— π 4

2π x

−2 3

the points with a smooth curve.

y 2 1 0 −1

2π

x

−2 4

Erase y = sin (x) and (y) = cos (x) to see the final graph.

y 2 1 0 −1

2π x

−2

Method 2: Technology-enabled On a Graphs page, complete the function entry lines as: f 1(x) = sin (x) f 2(x) = cos (x) f 3(x) = f 1(x) + f 2(x) Press ENTER · after each entry. Select an appropriate window.

Note that a difference function can be treated as a sum function where the second additive is negative. For instance, the function y = sin (x) − cos (x) can be viewed as y = sin (x) + (−cos (x)). So to obtain the graph of y = sin (x) − cos (x), we can sketch the graphs of y = sin (x) and y = −cos (x) on the same set of axes and add their respective y-coordinates.

Graphing modulus (absolute value) functions As was discussed previously, to sketch the graph of the modulus function y = | f (x)|, sketch the graph of y = f (x) and then reflect all sections of the graph that are below the x-axis in the x-axis. The technique is shown in the worked example below.

304


Worked Example 27

Sketch the graph of y = |3 cos (2x)| over the domain [0, 2π]. Think

Draw


First sketch the graph of y = 3 cos (2x). The amplitude of this graph is 3 and its period is 2π = π . Since the domain is [0, 2π], we need 2 to sketch two full cycles.

y 3 2 1 0 −1

–π 4

–π 2

3— π 4

π

5— π 4

3— π 2

7— π 4

–π 4

–π 2

3— π 4

π

5— π 4

3— π 2

7— π 4

–π 4

–π 2

3— π 4

π

5— π 4

3— π 2

7— π 4

2π x

−2 −3 2

Reflect all sections of the graph that are below the x-axis in the x-axis.

y 3 2 1 0 −1

2π x

−2 −3 3

Erase the sections below the x-axis to see the final graph.

y 3 2 1 0 −1

2π x

−2 −3

Method 2: Technology-enabled On a Graphs page, complete the function entry line as: f 1(x) = |3 cos (2x)| Then press ENTER ·. Select an appropriate window.


305

Graphing product functions The product function is defined over the intersection of the domains of the individual functions. That is, dom (f (x) g(x)) = dom f (x) ∩ dom g(x). Some features of the product function can be established by using the following rules: 1. The x-intercepts of f (x) g(x) occur where either f (x) or g(x) have their x-intercepts. 2. f (x) g(x) is above the x-axis where f (x) and g(x) are either both positive, or both negative. 3. f (x) g(x) is below the x-axis where one of the functions f (x) or g(x) is positive and the other is negative. The general shape of the graph and the coordinates of the turning points can be found using a CAS calculator. Worked Example 28

Find the domain and sketch the graph of the product function y = x sin (x). Use a CAS calculator for assistance. Think

306

1

The required function can be viewed as a product of two functions: f (x) = x and g(x) = sin (x). The domain of the product function is equal to the intersection of the domains of the two individual functions.

2

On a Graphs page, complete the function entry line as: f1(x) = x × sin (x) Then press ENTER ·. Select an appropriate window.

3

To locate the maximums and minimums, press: • MENU b • 5:Trace 5 • 1:Graph Trace 1 Trace along the graph to find a local maximum. Press ENTER · to lock in the point. This method can also be used to locate minimums and intercepts.

write/Draw

Let f (x) = x, dom f (x) = R Let g(x) = sin (x), dom g(x) = R dom (f (x) g(x)) = dom f (x) ∩ dom g(x) =R So the domain of y = x sin (x) is R.


Of course the graphs of product functions are not limited to those involving trigonometric functions.

Worked Example 29

If f(x) = 2x and g ( x ) = x + 1, sketch the graph of f ( x) g ( x ) = 2 x x + 1. Think 1

write/Draw

Sketch the graphs of f (x) and g(x).

y

y = 2x y= x+1

(0, 1) x

(−1, 0) 0

2

Find the domain of f(x) and the domain of g(x).

Dom f = R and dom g = [−1, ∞)

3

Find the domain of f(x)g(x).

Dom fg = [−1, ∞)

4

Find the x-intercepts of both f and g and hence find the x-intercepts of the product fg.

x-intercept for f (x) is when x = 0 and f (x) = 0 x-intercept for g(x) is when x = −1 and g(x) = 0 Hence, the x-intercepts for the product are when x = 0 and x = −1.

5

Find the values of x for which the product is negative.

f (x) is negative and g(x) is positive for x ∈ (−1, 0), so fg is negative for x ∈ (−1, 0).

6

Find the values of x for which the product is positive.

f(x) and g(x) are both positive for x ∈ (0, ∞), so fg is positive for x ∈ (0, ∞).

7

Find the turning point using a graphics calculator. Round the answer to 2 decimal places as appropriate.

The turning point is (

8

Sketch the graph of the product.

y

3

, − 0.77).

y = 2x x + 1

(−1, 0) (0, 0) −3 (— , 3

−2

x

−0.77)

Graphing composite functions A composite function is formed from two functions in the following way. If f (x) = x + 5 and g(x) = 2x are two functions, then we combine the two functions to form the composite function g(f (x)) = 2f (x) = 2(x + 5). That is, f (x) replaces x in the function g(x).


307

The composite function reads g of f and can be written g f. Another composite function is f (g(x)) = g(x) + 5 = 2x + 5. In this case, g(x) replaces x in f (x). This composite function reads f of g and can be written f g. For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). Composite functions can be rather complex to graph by hand, so a CAS calculator can be used for assistance when sketching. Worked example 30

eBook plus

For the pair of functions f(x) = cos (x) and g( x ) = x : a show that f(g(x)) is defined b find f(g(x)) and state its domain c sketch the graph of f(g(x)), using a CAS calculator for assistance. Think a

b

c

Tutorial


WriTe/draW

For f(g(x)) to exist, the range of g must be a subset of the domain of f. So find both the range of g and the domain of f to show that this condition is observed. 1

Form the composite function f (g(x)) by substituting g(x) into f (x).

2

The domain of f (g(x)) must be the same as the domain of g(x). Since the domain of g(x) is R+ ∪ {0}, so is the domain of f (g(x)).

On a Graphs page, complete the function entry line as: f 1( x ) = cos( x ) Then press ENTER ·. Select an appropriate window.

a f (x) = cos (x); domain of f (x) = R

g( x ) = x : range of g(x) = R+ ∪ {0} Range of g(x) ⊆ domain of f(x) ∴ f (g(x)) is defined

b f ( g( x )) = cos ( x )

Domain of f (g(x)) = R+ ∪ {0}

c

The graphs of composite functions are not limited to those involving trigonometric functions. This is demonstrated in the following example. Worked example 31

For f(x) = x2 − 2 and g(x) = x4: a show that both f(g(x)) and g(f(x)) are defined b find both f(g(x)) and g(f(x)), stating the domain and range of each one c on separate axes, sketch the graphs of f(g(x)) and g(f(x)).

308


Think

Write/draw

a

Show that ran f ⊆ dom g and that ran g ⊆ dom f.

a Dom f = R; ran f = [−2, ∞).

b

1

Write f(x).

b f (x) = x2 − 2

2

Form the composite function f (g(x)) by substituting g(x) into f (x). Simplify.

f (g(x)) = (x4)2 − 2 = x8 − 2

3

State the domain and range of f (g(x)).

Domain = R, range = [−2, ∞).

4

Form the composite function g(f (x)) by substituting f (x) into g(x).

g(f(x)) = (x2 − 2)4

5

State the domain and range of g(f (x)).

1

Sketch f (g(x)), rounding x-intercepts to 2 decimal places.

c

Dom g = R; ran g = [0, ∞]. Range of f ⊆ domain of g, so g(f (x)) exists. Range of g ⊆ domain of f, so f (g(x)) exists.

Domain = R, range = [0, ∞)

c The y-intercept is −2, and the x-intercepts

are −1.09 and 1.09. y

y = x8 − 2 (−1.09, 0) 0

2

Sketch g(f (x)).

(1.09, 0) x (0, −2)

The y-intercept is 16, and the turning point is (2, 0). y (0, 16) y = (x2 − 2)4 0

(2, 0) x

1. For the sum/difference function, dom (f (x) ± g(x)) = dom f (x) ∩ dom g(x). The graph of the sum/difference function can be obtained by using the addition of ordinates method. 2. The graph of the modulus function, y = | f (x)|, can be obtained by sketching the graph of y = f (x) and then reflecting all of the sections of the graph that are below the x-axis in the x-axis. 3. For the product function, dom (f (x) g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: (i) the x-intercepts of f(x) g(x) occur where either f (x) or g(x) have their x-intercepts


309

(ii) f (x) g(x) is above the x-axis where f (x) and g(x) are either both positive or both negative (iii) f (x) g(x) is below the x-axis where one of the functions f (x) or g(x) is positive and the other is negative. 4. For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). exerCiSe

6h eBook plus

Further graphs 1

Digital doc

SkillSHEET 6.6 Addition of ordinates

We26 Using addition of ordinates, sketch the following graphs for the domain [0, 2π]. a y = sin (x) + cos (2x) b y = cos (x) + sin (2x) c y = 2 sin (x) + cos (x) d y = 2 cos (x) + sin (x) e y = 2 sin (x) + cos (2x) f y = 2 cos (x) + sin (2x) g y = 2 sin (2x) + cos (x) h y = 2 cos (2x) + sin (x)

2 Apply the addition of ordinates method to sketch each of the following graphs over the domain [0, 2π]. Use a CAS calculator to check your answers. a y = sin (x) + x b y = cos (x) − x c y = 3 sin (x) − 2x 1

3

1

d y = cos (2 x ) + 4 x 2

e y = 2 sin (4 x ) − 8 x 3

g y = 4 sin (x) − 5 loge (x + 1)

h y = 3 cos (2 x ) −

f y = tan (x) − 2x2

20 ( x + 2)2

We27 Sketch each of the following graphs over the domain [0, 2π].

a y = |sin (2x)|

b y = |2 sin (4x)|

1  c y = cos  x  2 

d y = |3 cos (3x)|

e y = | 2 tan (x)|

f y = | 4 sin (x) + 2|

g y = |1 − 2 cos (2x)|

h y = |tan (2x) + 3|

4 Sketch each of the following graphs over the domain [0, 2π]. to observe the appropriate order of transformations.  x a y = 3|sin (x)| b y = |2 cos (2x)| + 1 c y = − 2 cos   + 3  2 5

6

We28 Find the domain and sketch the graph of each of the following product functions. Use a CAS calculator for assistance.

a y = 0.5x sin (x)

b y = (x − 1) cos (x)

c y = 3 sin (x) loge (x)

d y = 2 cos ( x ) x

e y = 8 cos (x) sin (x)

f y = (4 − 2x) sin (2x)

g y = (1.2)x cos (x)

h y = sin

d f (x) = | x |, g(x)

= x2

−1

g f ( x ) = x , g( x ) = 1 − x

310

2− x

We29 For each of the following functions f (x) and g(x), sketch the graph of f (x)g(x).

a f ( x ) = x, g ( x ) = x + 2

7

 x    2 

b f (x) = x − 2, g(x) = ex

c f (x) = x − 1, g(x) = loge (x)

e f ( x ) = x , g( x ) =

f f (x) = 1 − x2, g(x) = ex

3

−

x+2

h f (x) = x2, g(x) = loge (x)

We30 For each of the following pairs of functions:

i show that f (g(x)) is defined ii find f (g(x)) and state its domain iii sketch the graph of f (g(x)), using a CAS calculator for assistance.


a f (x) = cos (x) and g(x) = loge (x) x2 c f ( x ) = 2 sin ( x ) and g( x ) = 4 e f (x) = x2 and g(x) = sin (x) g f ( x ) = cos

 x    4 

b f ( x ) = sin (2 x ) and g( x ) = x d f ( x ) = x + 2 and g( x ) = 2 cos ( x ) f f (x) = 2x and g(x) = cos (2x)

1

and g( x ) = 2 x 2

h f ( x ) = 2 sin ( x ) + 1 and g( x ) = x − 3

8 WE31 For each of the following pairs of functions f (x) and g(x): i state whether f (g(x)) and g(f (x)) are defined ii for the composite functions that are defined, find f (g(x)) and g(f (x)), stating the domain and range of each one iii on separate axes, sketch the graphs of f (g(x)) and g(f (x)) that are defined.

6I

a f (x) = x − 2, g(x) = ex

b f (x) = | x| g(x) = x2 − 1

c f (x) = 1 − x2, g(x) = ex

d f ( x ) = x , g( x ) = sin ( x )

e f (x) = ex, g(x) = cos (x)

f f (x) = loge x, g(x) = sin (x)

Trigonometric functions with an increasing trend Consider a trigonometric function where there is an increasing trend, for example, economic growth cycles, tidal heights due to global warming or increasing seasonal populations. These situations can be modelled by a function of the form y = ax + b + m sin (nx) where ax + b represents the increasing trend line and m sin (nx) represents the seasonal variation.

Worked Example 32

Consider a remote island where global warming has caused the temperature to increase by 0.1 degree each month. The mean daily temperature is modelled by the function π  T ( m) = 16 + 0.1 m + 6 cos  m , where T is the temperature in degrees Celsius and m is the number 6  of months after January 2008. a Sketch a graph of the function for a five year period from January 2008, using a CAS calculator for assistance. b Find the mean daily temperature for March 2009 c When will the mean daily temperature first reach 23 degrees? Think

Write

a On a Graphs page, complete the function entry

line as: π  f 1( x ) = 16 + 0.1x + 6 cos  x  6  Then press ENTER ·. Select an appropriate window: XMin: 0 XMax: 60 YMin: 10 YMax: 30


311

b

March 2009 occurs when m = 14. On a Calculator page, complete the entry line as: f1(14) Then press ENTER ·.

The mean daily temperature in March 2009 would be 20.4 degrees Celsius. c

1

2

π  To solve 23 = 16 + 0.1m + 6 cos  m , 6  for m, return to the Graphs page. Complete the function entry line as: f 2(x) = 23 Then press ENTER ·.

Press: • MENU b • 7:Points & Lines 7 • 3:Intersection Point(s) 3 Select each function and then press ENTER ·. The points of intersection will appear.

m = 11.5631 Hence, the first time the temperature reaches 23 degrees Celsius will be during the 12th month after January 2008. That is, during January 2009.

Trigonometric functions with an increasing trend can be modelled by a function of the form y = ax + b + m sin (nx) where ax + b represents the increasing trend line and m sin (nx) represents the seasonal variation.

312


Exercise

6I

Trigonometric functions with an increasing trend 1 WE32 A fisherman finds himself stranded on an island, where the mean daily temperature is increasing as a result of global warming. He finds that the temperature can be modelled by the π  function T (m) = 12 + 0.2m + 5 cos  m , where T is the temperature in degrees Celsius and 6  m is the number of months after January 2008. a Sketch a graph of the function for a 2-year period beginning with January 2008. b Find the mean daily temperature for December 2009. c When will the temperature first reach 18 degrees? 2 In a region of country Victoria, a study shows that increased wheat production causes the π  mouse population to increase according to the function M = 15 000 + 100t − 4000 cos  t  , 6  where M is the number of mice and t is the number of months after July 2008. a How many mice are being added to the average population per month? b Draw a graph of this situation for a 5-year period beginning with July 2008. c How many mice would you expect to be in the region in December 2008? d When would the mice population first reach 20 000? 3 The value of a particular stock on the market follows a trigonometric model, and inflation causes the stock’s value to have an overall upward trend. The value of the stock can be π  represented by the equation V (t ) = 20 + 0.02t + 5 sin  t  , where V is the value of the stock 6  in dollars and t is the number of months after January 2006. a What is the inflation rate per month? b What was the initial value of the stock? c What will be the stock’s value after 6 months? d When will its value first reach $25.50?


313

Summary Revision of radians and the unit circle

• 1c = the size of the angle formed where the length of an arc is equal to the radius of the circle. • πc = 180° • Angles are in radians unless a degree symbol is shown. 180 • To change radians to degrees, multiply by . π π • To change degrees to radians, multiply by . 180 • Identities 1. sin2 (θ ) + cos2 (θ ) = 1 sin (θ ) 2. tan (θ ) = cos (θ )

Symmetry and exact values

• Exact values can be determined by using the equilateral triangle and the right isosceles triangle shown at right.

θ

π

3 30°

2

45°

1

45°

60° 1 π

π

π

2

1  3π  2 

0° (0)

30°  6 

45°  4 

60°  3 

90°  2 

180° (π)

sin (θ )

0

1 2

2 2

3 2

1

0

−1

0

cos (θ )

1

3 2

2 2

1 2

0

−1

0

1

tan (θ )

0

3 3

1

Undef.

0

Undef.

0

3

360° (2π)

270° 

• The unit circle is symmetrical so that the magnitude of sine, cosine and tangent are the same in each quadrant but the sign varies. All functions (sine, cosine and tangent) are positive in the 1st quadrant, sine is positive in the 2nd quadrant, tangent is positive in the 3rd quadrant and cosine is positive in the 4th quadrant. sin (π − θ ) = sin (θ )

sin (π + θ ) = −sin (θ )

sin (2π − θ ) = −sin (θ )

cos (π − θ ) = −cos (θ )

cos (π + θ ) = −cos (θ )

cos (2π − θ ) = cos (θ )

tan (π + θ ) = tan (θ )

tan (2π − θ ) = −tan (θ )

tan (π − θ ) =

−tan

(θ )

• Negative angles sin (−θ ) = −sin (θ ), cos (−θ ) = cos (θ ), tan (−θ ) = −tan (θ ) π • Complementary angles add to 90o or radians. 2 • The sine of an angle is equal to the cosine of its complement. • The complement of the tangent of an angle is the cotangent.

( − θ ) = cos (θ ) cos ( − θ ) = sin (θ ) tan ( − θ ) = cot (θ ) sin

π 2

π 2

π 2

314

( + θ ) = cos (θ ) cos ( + θ ) = sin (θ ) tan ( + θ ) = cot (θ ) sin

π 2

π 2

−

π 2

−

( cos ( tan ( sin


3π 2 3π 2 3π 2

) − θ ) = sin (θ ) − θ ) = cot (θ )

− θ = − cos (θ ) −

sin

(

3π 2

( tan ( cos

)

+ θ = − cos (θ )

3π 2

3π 2

)

+ θ = sin (θ )

)

+ θ = − cot (θ )

Trigonometric equations

• Given a general equation such as sin x = a, there can be an infinite number of solutions. The domain is usually restricted and it is important to find all values for x within this domain. • If the domain is given in radians, then the solution(s) to x should be in radians. If the domain is given in degrees, then the solution(s) to x should be in degrees. • Adjust the domain to match what has been done to the angle in the question. • Sine is positive in the 1st and 2nd quadrants, cosine is positive in the 1st and 4th quadrants and tangent is positive in the 1st and 3rd quadrants. • If sin (x) = a then the general solution is x = sin 1 (a) + 2np, n ∈ Z. • If cos (x) = a then the general solution is x = cos 1 (a) + 2np, n ∈ Z. -1 • If tan (x) = a then the general solution is x = tan (a) + np, n ∈ Z. • Find all the solutions within the specified domain by substituting integer values for n into the general solution. • If the equation is of the form sin (ax) = k cos (ax), divide both sides by cos (ax) to change the equation to tan (ax) = k. Trigonometric graphs

Sine and cosine graphs • Graphs of the form y = a sin [n(x − b)] + c and y = a cos [n(x − b)] + c are transformations of y = sin (x) and y = cos (x). • The amplitude a is a dilation from the x-axis. If a is negative, the amplitude is still positive but the graph is a reflection in the x-axis. • The vertical translation c is a translation parallel to the y-axis. If c is positive, the graph is translated c units up, and if c is negative, the graph is translated c units down. • c represents the median value of the function. • The range is [c − | a |, c + | a |]. 2π • The period is . n 1 • The factor n is the horizontal dilation where the graph has been dilated by a factor of from the y-axis. n • The value b is the horizontal translation or a translation parallel to the x-axis. If b is positive, the graph is translated b units to the right and if b is negative, the graph is translated b units to the left. Tangent graphs − − y x = 3—2π x = —2π x = –2π x = 3—2π The graph of y = tan (x) has the following properties. 3 • It has no amplitude. 2 • The period is π. − − 1 • There are x-intercepts at x = . . ., 2π, π, 0, π, 2π, . . . − 3π − π π 3π , , , ,... • There are vertical asymptotes at x = . . ., 0 –π π 3—π −2π − 3—2π −π − –2π 2π x 2 2 2 2 2 2 −1 • The range is R. −2 The graph of y = tan (nx) has the following properties. • It has no amplitude. −3 π • The period is . n kπ • There are x-intercepts at x = ± where k = 0, 1, 2, . . . n (2k + 1)π • There are vertical asymptotes at x = ± where k = 0, 1, 2, . . . 2n • The range is R. • Compared to the basic graph of y = tan (x), the graph of y = a tan [n(x − b)] + c is: – dilated by the factor of a from the x-axis 1 π – dilated by the factor of from the y-axis (and hence has period of ) n n


315

– reflected in the x-axis if a < 0 – translated b units horizontally (to the right if b > 0 and to the left if b < 0) – translated c units vertically (up if c > 0 and down if c < 0). Further graphs

• For the graph of the sum/difference function, dom(f (x) ± g(x)) = dom f (x) ∩ dom g(x). The graph of the sum/ difference function can be obtained by using the addition of ordinates method. • The graph of the modulus function y = | f (x)| can be obtained by sketching the graph of y = f (x) and then reflecting all of the sections of the graph that are below the x-axis in the x-axis. • For the product function, dom(f (x)g(x)) = dom f (x) ∩ dom g(x). Some features of the graph of the product function are as follows: – the x-intercepts of f (x)g(x) occur where either f (x) or g(x) have their x-intercepts – f (x)g(x) is above the x-axis where f (x) and g(x) are either both positive or both negative – f (x)g(x) is below the x-axis where one of the functions f (x) or g(x) is positive and the other is negative. • For the composite function f (g(x)) to be defined, the range of g must be a subset of the domain of f. Furthermore, if f (g(x)) is defined, the domain of f (g(x)) equals the domain of g(x). Trigonometric functions with an increasing trend

• These situations can be modelled by a function of the form y = ax + b + m sin (nx), where ax + b represents the increasing trend line and m sin (nx) is the seasonal variation.

316


chapter review Short answer

1 Convert the following into radians. a 60° b 30° d 90° e 360° g 150° h 300° j 120° k 210°

c f i l

45° 270° 225° 315°

2 Find the value of:  2π  a cos    3  5π  d sin   4  g cos (135°) j tan (225°)

π b sin    4

 7π  c tan   6 

e tan (2π)

f sin (120°)

h tan (30°)

i cos (315°)

9 The depth, d metres, of water in a shallow bay at  πt  t hours after 9 am is given by d = 5 + 3 sin   ,  6 for 0 ≤ t ≤ 24 hours. a Sketch the graph of d for 0 ≤ t ≤ 24 hours. b At what times will the depth of water in the bay be 6.5 metres? c Particular water sports cannot run in the bay when the depth of water is less than 6.5 metres. At what times of the day (not night) will the water sports be able to run?

3 If sin (x) = 0.85 and x is in the first quadrant, find: b sin (π + x) a sin (π − x) c sin (2π + x) d sin (4π − x) 4 Find the general solution to the equation 2 cos ( x ) − 2 = 0. 5 Find the general solution for 3 sin (2 x ) = cos(2 x ) and hence all solutions for x between 0 and 2π. π  6 Consider the graph of y = 3 sin  x −  + 2.  4 a State the translations required to form this graph from y = sin (x). b State the amplitude and period of the transformed trigonometric function. c Sketch the graph of the transformed function over the domain 0 ≤ x ≤ 2π. 7 State the period and sketch the graph of y = tan (2x) + 2. 8 The equation of the following graph is of the form y = a sin (nx) + b. Find the values of a, n and b, and hence find the equation of the function. y 3 2 1

−1 −2 −3 −4

2

3

4

x

10 Sketch the graph of y = | 2 sin (2x) | over the domain [0, 2π]. 11 On the same set of axes and over the domain [0, 2π], sketch the graphs of: a y = 2x b y = sin (x) c y = sin (x) + 2x. 12 On the same set of axes and over the domain [0, 2π], sketch the graphs of: a y = x b y = sin (x) c y = x sin (x). 13 For the pair of functions f (x) = sin (x) and g(x) = x2 + 2: a show that f (g(x)) is defined b find f (g(x)) c state the domain and range of f (g(x)). 14 For the function π f : [− π , π ] → R, f ( x ) = 5 cos (2( x + )): 3 a Write down the amplitude and period of the function.


317

b Sketch the graph of the function f. Label intercepts with their coordinates. Label end points of the graph with their coordinates. Exam tip Make sure that you use the period and

amplitude found in part a to help sketch the graph. The graph should be a smooth symmetrical curve with all intercepts (x and y) clearly labelled with coordinates. [Assessment report 1 2006]

[© VCAA 2006]

Multiple choice

1 What is 320° expressed in radians? A 2π B 5π C 8π D 4π 9 18 9 9 c 13π 2 What is expressed in degrees? 6 A 390° B 420° C 150° D 120°

E 16π 9

E 330°

 5π  3 The expression sin   equals:  3 A

1 2

B

−1 2

C

3 2

D

− 3 2

E 1

4 A trigonometric function is given by f : R → R, f (x) = 3 cos (2x + π) − 1 The amplitude, period, and range of f are respectively: C 3, π, [−4, 2] B 2, π, R A 3, π , [− 4, 2] 2 π − E 3, π D 2, , [ 1, 3] 2

8 The domain and range of the function   π f : [0, π] → R, f ( x ) = 4 sin 3  x −   + 2 is:  3 

B [0, π], [−2, 6] C R, R A [0, π], R −π D [ , 2π ] [− 4, 4] E [0, π], [2, 4] 3 9 Using addition of ordinates for the graph of the difference function y = 2 cos (x) − 3 sin (x), the value π of y at x = is: 3 −1 1+ 3 3 2+3 3 A B C 3 2 2 D

1− 3 3 2

1

6 The function f : [−π, π] → R, f (x) = sin (2x) + cos (2x) has the following x-intercepts: − π − 5π 3π 7π − π 3π − π 3π , , , A , B , C 4 4 4 4 2 4 8 8 − π − 5π 3π 7π −π π , , , , D E 8 8 8 8 4 4 7 The solution of the equation 4 sin ( x ) = − 2 3 3π between π and is: 2 A π B π C 4π 6 3 3 7 π 2 π D E 3 3

318

2−3 3 2

 x 10 The period of the graph of y = tan   is:  3 π B π C 2π D 3π A 3

E 6π

11 The graph of the function f : [−1.5, 1.5] → R, f (x) = a + b cos (π nx) is shown below. The values of a, b, and n are respectively: y

max = 3.5

3 2 1 −1.5

−1.0

−0.5

0 −1

0.5

1.0 1.5 min = −1.5

x

−2

5 When f (x) = 0 the following function f : [0, 2π] → R, f (x) = 2 sin  2 ( x + π )  + 1 has:   A 0 solutions B 1 solution C 2 solutions D 3 solutions E 4 solutions

E

A 1, D 1,

−5 2 −5 3

2

2

5

,3

B −1, 3 , 2

,3

E

−1,

−5 3

2

C 1, 5, 3

,2

12 If 0 ≤ c ≤ π, the x-intercepts of the function f : [0, 2π] → R, f (x) = b sin (x − c) are: B π, 2π C c, π − c A π − c, 2π − c D π + c, 2π − c E c, π + c 13 The position of a particle from a fixed point O is given by the equation x = 2 − 2 sin (π t). If 0 ≤ t ≤ 2, the particle is at the point O when t equals: E 6π A 0 B 2 C 0.5 E 6π 14 The smallest positive value of x for which tan (2x) = 1 is: π π π A 0 B C D 8 4 2


E p

[© VCAA 2006]

Extended response

1 At the South Pole in midsummer on the planet Marus, the red sun of its solar system does not set. It dips towards the horizon until its lower rim just touches it, then rises until its lowest point is at an angle of D(t)° to the horizontal before sinking again. It continues in this pattern. The angle above the horizontal can be modelled by the following relation: D(t) = a − b sin n(t + c) where t is the time in hours after midnight and a, b, c and n are positive constants. The graph of D(t) for 24 hours is shown on the axes in the figure below. D(t) 10 8 6 4 2 0

2 4 6 8 10 12 14 16 18 20 22 24

t

a State the values of a, b, c and n and hence write the rule for D(t). b What would be the angle above the horizon at 6.00 am and at 9.00 pm? Give an exact answer where possible, otherwise give your answer correct to 2 decimal places. c Use your graph to find at what times the angle to the horizontal is 8°. When does the rim of the sun reach this angle again? d By using an appropriate equation, check your answer and for any difference in your two solutions. e If a spot on the surface of the sun is 5° above the horizon at midnight, what would be the relation G(t) that models its path? 2 Nathan and Rachel are competing in the National Ballroom Dancing Championships. The judges are evenly spaced around the circular dance floor, standing just outside the edge. As Nathan and Rachel waltz around the circular floor, their distance (in metres) from judge Maya can be described by the function π  d = 10.5 − 9 cos  t  , where t is time (in seconds) from the beginning of the dance.  30  a How far is the couple from judge Maya when they start dancing? b What is the couple’s maximum distance from the judge? c Assuming that, while dancing, Rachel and Nathan trace a perfect circle, what is its diameter? d How long does it take for the couple to complete one full circle around the dance floor? e What is the couple’s average speed (in m/s)? Give your answer i in exact form and ii correct to 2 decimal places. π  f If the duration of the waltz is 2.5 minutes, draw the graph of d = 10.5 − 9 cos  t  over the domain,  30  showing the full length of this dance. g Judge Joseph is positioned further down the dance floor, so that Nathan and Rachel are closest to him 6 seconds after the waltz begins. Write the equation describing the couple’s distance from judge Joseph at any time, t, from the beginning of the dance. h How far is the couple from judge Joseph when they finish the waltz? 3 Tasmania Jones is attempting to recover the lost Zambeji diamond. The diamond is buried at a point 4 km into Death Gorge, which is infested with savage insects. In order to recover the diamond, Tasmania will need to run into the gorge, dig up the diamond and return the same way that he came. The concentration of insects in the gorge is a continuous function of time. The concentration, C, of insects per square metre is given by   π (t − 8)  + 2)2 − 1000, 8 ≤ t ≤ 16 1000 (cos  C (t ) =   2  m 0 ≤ t < 8 or 16 < t ≤ 24  where t is the number of hours after midnight and m is a real constant.


319

a What is the value of m? b Sketch the graph of C for 0 ≤ t ≤ 24. c What is the minimum concentration of insects and at what value(s) of t does that occur? The insects infecting the gorge are known to be deadly if their concentration is more than 1250 insects per square metre. d At what time after midnight does the concentration of insects first stop being deadly? e During a 24-hour period, what is the total length of time for which the concentration of insects is less than 1250 insects per square metre?


Test Yourself

exam Tip

• Be aware of the word ‘continuous’ when working out what value(s) to substitute to find m. • When drawing the graph, clearly show minimum turning points and be aware of the domain stated in the question.

Chapter 6


[© Vcaa 2007]

320


eBook plus

aCTiviTieS

chapter opener Digital doc

• 10 Quick Questions: Warm up with ten quick questions on circular functions. (page 258) 6a

Revision of radians and the unit circle

Digital docs

• Spreadsheet 142: Investigate the unit circle. (page 263) • SkillSHEET 6.1: Practise changing degrees to radians. (page 263) • SkillSHEET 6.2 : Practise working with tangent ratios. (page 263) 6b

Symmetry and exact values

Tutorials

• We 3 int-0546: Watch a worked example on exact values using degrees and radians. (page 265) • We 5 int-0547: Watch a worked example on determining trigonometric ratios. (page 266) • We 8 int-0548: Watch a worked example on complementary angle formulas. (page 270) Digital docs

• SkillSHEET 6.3: Practise rationalising the denominator. (page 271) • Spreadsheet 142: Investigate the unit circle. (page 272) • SkillSHEET 6.4: Practise problem solving using trigonometry. (page 273) 6c

Trigonometric equations

Tutorial

• We 13 int-0549: Watch a worked example on determining the general solution to a trigonometric equation. (page 277) Digital docs

• Spreadsheet 135: Practise solving trigonometric equations. (page 281) • WorkSHEET 6.1: Use exact values, solve trigonometric equations and application questions. (page 282) 6D

Trigonometric graphs

Digital docs

• SkillSHEET 6.5: Practise identifying the period and amplitude of sine and cosine graphs. (page 288) • Spreadsheet 124: Investigate sine graphs. (page 288) • Spreadsheet 012: Investigate cosine graphs. (page 289) 6E Tutorial

Graphs of the tangent function

6F

Finding equations of trigonometric graphs

Interactivity int-0251

• Finding equations of trigonometric graphs: Use the interactivity to consolidate your understanding of trigonometric graphs. (page 295) Digital docs

• Spreadsheet 124: Investigate sine graphs. (page 298) • Spreadsheet 012: Investigate cosine graphs. (page 298) • WorkSHEET 6.2: Sketch trigonometric graphs over given domains, including composite and absolute value graphs, and determine the equations of trigonometric graphs. (page 299) 6G

Trigonometric modelling

eLesson eles-0092

• Trigonometric modelling: Learn how trigonometry can be used to model sinusoidal patterns. (page 299) Digital doc

• Spreadsheet 041: Investigate graphs of functions. (page 301) 6H

Further graphs

Tutorial

• We 30 int-0551: Watch how to sketch a composite function using a CAS calculator. (page 308) Digital doc

• SkillSHEET 6.6: Practise using addition of ordinates to sketch graphs. (page 310) chapter review Digital doc


• We 21 int-0550: Watch a worked example on calculating the period of trigonometric functions. (page 293)

Chapter 6


321

EXAM PRACTICE 2 Short answer

30 minutes

1 Solve the equation log2 (2x - 3) - log2 (3) = 1 for x. 1 mark 2 Solve the equation 32x + 3 33 = 9 for x.

1 mark

[-2,0]

3 The graph of the function f : → R where f (x) = e x is reflected in the y-axis, translated 2 units to the left then translated 3 units up. (Note the change to the domain.) a Specify the rule of the transformed graph. b Determine its exact range. 2 marks 4 The diagram below shows the graph of y = sin (x) and a second graph formed by transforming sin (x) by carrying out dilation(s) followed by translation(s). y 1

0 π/6

2π

y = sin (x)

x

−1

π

a Write down the transformations (type, direction and quantity) necessary to create the second graph. b Write down the equation of the second graph. 4 marks

5 Find the exact solutions of the equation sin (3x) - cos (3x) = 0 for 0 ≤ x ≤ π

2 marks

6 Specify the range of the function f :[

−π

π π ) → R, f ( x ) = 2 tan(2 x − ) + 1. 4 8 4 ,

2 marks

7 Solve the equation

e2x - e x

= 2 exactly for x. 2 marks

−1,

8 For f : [ 3] → R, where f (x) = x sin (x): a sketch f b determine the minimum value of f (x) c estimate the maximum value of f (x) to 1 decimal place.

322

Chapters 1 TO 6

Multiple choice

12 minutes

Each question is worth 1 mark. 1 If y = 2abx - 1 + 5, then x is equal to: y−5 y log a ( − 5) + 1 a 2 +1 2 B A b b y−5 +1 y−5 2 +b C D 2b b y−5 log a ( ) +1 2 E b 2 The solution set of the equation e2x - 3e x = 4 over R is: A {−1, 4} B {4} C {loge (4)} D {loge (-1), loge (4)} E {loge (1), loge (4)} 3 The graph of the function f :[2, 5] → R, f (x) = loge (x) is reflected in the y-axis, translated 2 units to the left then translated 3 units up. The domain of the new graph is: A [−7, −1] B [−4, −1] −7, −4] C [ D [loge (-7), loge (- 4)] E not defined 4 The function f : [1, ∞) → R, where f (x) = x2 + 1 has as its inverse function: A f -1: R → R, where f −1( x ) = ± x − 1 B f -1: [2, ∞] → R, where f −1( x ) = x − 1 1 C f -1: [-1, ∞] → R, where f −1 ( x ) = 2 x +1 D f –1: R+ ∪ {0} → R, where f −1( x ) = x − 1 1 1 E f −1: [ , ∞) → R, where f −1( x ) = 2 2 x +1 5 The equation a sin (3θ ) - b = 0 has θ = 2 as one solution. Which of the following could also be a solution? A 3π + 6 B π + 2 2π C -2 D +2 3 2π E +6 3

4 marks


7 The graph of y = 3 cos (2x) − 1 is shown below. If a and b are two adjacent x-intercepts, b − a is equal to:

6 The graph shown could be the graph of: y 3a

0 −a

y

a

θ

T

b x

0

 2π  cos  θ +a  T 

 T  θ +a a 2a cos   2π 

b

 2π  c a (cos  θ + 1)  T 

D 2a sin (

− 2a

2π π (θ − )) + a T 2

π   T E a (sin  (θ − ) + 1)  2π 2 

π 2 c 3a E π − 2a

b π

a

D 2π − a

exTended reSponSe

20 minutes

Bird population

An extensive study has been made of the population of silver-banded lorikeets in an area proposed for a wind farm. The variation in estimated numbers over a number of months is recorded in the graph below. It is suggested that the relationship could be modelled by a sine function of the form P(t) = a sin (b(t + c)) + d, where P is the bird population and t is the time in months since estimates commenced. y 500 400 300 200 100 0

a b c d e f g

1

2

3

4 5 6 Months

7

8

9 x

What is the amplitude of the sine model (to the nearest 10 birds)? 1 mark What is the period (to the nearest month)? 1 mark What is the mean population (to the nearest 10 birds)? 1 mark After how many months is the population a minimum? 1 mark Determine the values of a, b, c and d in the model P(t) = a sin b(t + c) + d. 4 marks What was the initial bird population when the observations commenced? 1 mark A second researcher conducts observations over a 10-year period at the same time each year and records the following estimates for the population. Time (months)

Population

0

500

60

370

120

274

She believes that the population can be modelled by an exponential function of the form Q(t) = Ae−kt, where t is the time in months since the first estimate was calculated. Determine the values of A and k. 2 marks h Explain how both researchers’ results are consistent with the model eBook plus P( x )Q( x ) Population = . 3 marks Digital doc 500 Solutions Exam practice 2

exam practice 2

323

7

7a Review — gradient and rates of change 7b Limits and differentiation from first principles 7c The derivative of xn 7d The chain rule 7e The derivative of e x 7F The derivative of loge (x (x) 7G The derivatives of sin ((x), cos ((x) and tan ((x) 7H The product rule 7I The quotient rule 7J Mixed problems on differentiation

Differentiation AREAS OF STUDY

• Graphical treatment of limits, continuity and differentiation of functions of a single real variable • Deducing the graph of a derivative function from the graph of a function • Derivatives of xn, n ∈ Q, ex, ln (x), sin (x), cos (x), tan (x) • Properties of derivatives (af af (x) ± bg(x))′ = af ′(x) ± bg′(x), a, b ∈ R

f ( x) , and g( x ) f (g(x)) where f and g are polynomial functions, exponential, circular, logarithmic or power functions (or combinations of these functions)

• Derivatives of f (x) ± g(x), f (x) × g(x),

such as x 5 + 1 − x 2 , x sin (2 x ), e cos ( x ) 2 and ln ( x + 4) x

eBook eBoo k plus

7A

Review — gradient and rates of change

Digital doc

10 Quick Questions

Gradient is a measure of how one quantity changes with respect to another quantity; in other words, the rate of change of one quantity with respect to another quantity. The gradient (or slope) is the measure of how the vertical distance (the rise) changes with respect to the horizontal distance (the run). Speed is the measure of how distance travelled changes with respect to the time taken; another way of saying this is that speed is the rate of change of distance with respect to time.

Constant rates When the rate of change of one quantity with respect to another quantity does not change, the rate is constant. If the rate of pay for babysitting is fixed at $15 an hour, this is a constant rate. This means that a babysitter would be paid $15 for 1 hour and $150 for 10 hours. A constant rate can be represented graphically by a straight line.

324


y (1, 15)

(0, 0) x

y

Average rates An average rate of change is the rate of change over a period of time. A car travelling at 60 km/h is probably not actually ing that speed at every moment of the distance travelled; more likely, its speed varies above and below 60 km/h. The average rate of change between two points can be represented by the gradient of a straight line ing the two points.

(0, 0)

Instantaneous rates

y

(1, 0) x (1_2, −33_4)

(−2, 3)

If the police want to find how fast a car is travelling, they measure the speed at a particular instant. This is known as instantaneous (–1, 0) (1, 0) rate of change. Graphically, this is found by drawing a tangent x 0 to the curve at a particular point and finding the gradient of the (−3, 0) tangent. If a section of the graph is increasing, the gradient of the tangent is positive, and if a section of the graph is decreasing, the (0.15, −3.08) gradient of the tangent is negative. For example, for the curve y = (x − 1)(x + 1)(x + 3), the gradient is positive when x < −2 and when x > 0.15; the gradient is negative when −2 < x < 0.15; and the gradient is zero when x = −2 and x = 0.15.

Graphing the gradient function from the graph of a function Differentiation is the process of calculating the gradient function from a given function. It can be used to find the gradient of a curve at a particular point and to find the maximum and minimum values of a function. y f(x) Derivatives can be found at a given point if: There is a break in the 1. the function is continuous at that point graph at x = −1(discontinuous), 2. the function is smooth at that point. so gradient does not exist at x = −1. That is, gradient A function is continuous if there are no exists for R\{−1}. ‘breaks’, ‘jumps’ or ‘asymptotes’ on its graph. x 0 −1 The gradient does not exist where the function is not continuous. An example is shown at right. A function is smooth if there are no ‘sharp’ points on its graph. The gradient does not exist where the function is y not smooth. An example is shown at right. f (x) is not smooth at x = 2 so gradient does not exist at x = 2. That is, gradient exists for R\{2}.

0

x

2 f(x)

The gradient of a function exists wherever the function is smooth and continuous. That is, the gradient of a function exists at a point providing only one tangent can be drawn at that point. An example is shown at right. A function is not differentiable where there is a gap, hole or asymptote, a sharp corner or an end point.

y f(x)

2 −2

0

The gradient of f(x) exists for x ∈ R as f(x) is smooth and continuous. x

Chapter 7 Differentiation

325

Gradient function of straight lines In general, for straight lines in the form f (x) = mx + c (or y = mx + c), the rule for the gradient is f ′(x) = m, that is, a horizontal straight line through y = m. WORKED EXAMPLE 1

For the straight line function shown, sketch the graph of its gradient function.

y

f(x)

0

x

1

−2 THINK

WRITE/DRAW

1

Find the gradient of f (x) using m =

2

Sketch the graph of y = gradient.

1 2

rise . run

m=

to represent the

1 2 y 1– 2

f'(x) x

0

Note: The domain of f ′(x) is R as f (x) is smooth and continuous.

Gradient function of quadratic functions The gradient function of a polynomial function is also a polynomial function but the degree is reduced by 1. That is, the gradient function of f (x) = ax2 + bx + c is of the form y = mx + c. So the gradient of a quadratic function is a linear function.

WORKED EXAMPLE 2

Sketch the graph of the gradient function for the quadratic function shown and state its domain. y

0 THINK

326

f(x)

1

x

WRITE/DRAW

1

Find when f ′(x) = 0.

f ′(x) = 0 when x = 1.

2

Find when f ′(x) > 0.

f ′(x) > 0 when x > 1.

3

Find when f ′(x) < 0.

f ′(x) < 0 when x < 1.


4

Sketch f ′(x).

y

f'(x)

0

5

Find where f (x) is smooth and continuous and hence find the domain of f ′(x).

x

1

The domain is R.

Gradient function of cubic functions The gradient function of a cubic is a quadratic function. WORKED EXAMPLE 3

eBook plus

For the cubic function shown, sketch the gradient function and state its domain.

y

Tutorial

f(x)


−3 THINK

x

0 1

WRITE/DRAW

1

Find when f ′(x) = 0.

f ′(x) = 0 when x = −3 and x = 1.

2

Find when f ′(x) > 0.

f ′(x) > 0 when x < −3 and x > 1.

3

Find when f ′(x) < 0.

f ′(x) < 0 when −3 < x < 1.

4

Sketch the graph of the gradient function.

y

−3

5

Find the domain by determining where f (x) is smooth and continuous.

0

f'(x)

x

1

The domain is R.

WORKED EXAMPLE 4

For the function f (x) shown, state the domain of the gradient function f ′(x).

y

−1 0 2

x f(x)

Chapter 7

Differentiation

327

THINK

WRITE

The domain = R \{−1, 2}.

The function is smooth and continuous everywhere except at x = −1 (discontinuous) and x = 2 (not smooth).

1. 2. 3. 4. 5.

6. 7. 8. 9. 10.

EXERCISE

7A eBook plus Digital doc

A function is smooth if there are no sharp points on its graph. A function is continuous if the graph can be drawn without lifting pen from paper. The gradient of a function exists where the function is smooth and continuous. The graph of the gradient function is a graph of the gradients of all the points of the original function. The gradient function of a polynomial function is also a polynomial function but the degree is reduced by 1. For example, the gradient function of f (x) = ax2 + bx + c is of the form y = mx + c. Wherever the gradient of a function is zero, its gradient function will have an x-intercept. Wherever the gradient of a function is positive (sloping /), the gradient function graph is above the x-axis. Wherever the gradient of a function is negative (sloping \), the gradient function graph is below the x-axis. The gradient of a horizontal line is 0. The gradient of a vertical line is undefined.

Review — gradient and rates of change 1 WE 1 For each straight line function shown below, sketch the graph of its gradient function. b c a y y y f(x)

SkillSHEET 7.1

3

1

Gradient — positive, negative and zero

x

0

−1

f(x) −2

0

x

1 f(x)


SkillSHEET 7.2

d

y 0

Gradient function

e

f(x) 2

y f(x)

x 3 0

−5

328


x

0 −1

x

2 MC a The gradient of the line in the graph at right is: a 1

b 2

d

c 0

1 2

y f(x)

−1

e

2

x

0 1

b The graph of the gradient function in the graph above is represented by which of the diagrams below? b c a y y y f'(x)

2

f'(x)

d

0

x

0 1

e

y 1– 2

x

2

x

0

y x

0

f'(x)

f'(x)

2

1

f'(x) x

0

3 WE 2 a

−2

Sketch the graph of the gradient function for each quadratic function shown below. b c y y y g(x)

g(x)

x

0

x

0

d

g(x)

x

0

−2

y

e

g(x)

y −2

0

3

x

0 −1

x

g(x)

4 MC a The gradient of the function shown in the graph at right is: a always increasing b always decreasing c decreasing then increasing d increasing then decreasing e constant

y

0

x f(x)

Chapter 7

Differentiation

329

b The gradient function for the graph in question 4a is shown by which of the graphs shown below? a

y

b

y

x

0

c

f'(x)

y

x

0

f'(x)

x

0

f'(x)

d

e

y

y

f'(x)

x

0

x

0 f'(x)

5 WE 3 a

For each cubic function f (x) graphed below, sketch the gradient function. y b y c y f(x)

−3

0

2

x

0

1

4

x

0

f(x)

x

5

f (x)

d

e

y Gradient = 0

f(x)

0

x

f

y

y

Gradient = 0

f(x) 0

x −3

0

x f(x)

g

f(x)

y Gradient = 0 2 0

x

6 MC a The figure at right has a positive gradient where: A −1 < x < 2 B x < −1 only C x > 2 only D x < −1 and x > 2 E x > 0 b The figure above has a negative gradient where: A x > −1 B x < 2 C −1 < x < 2 D x < −1 and x > 2 E x < 0 330


y f(x)

−1

0

2

x

c The graph of the gradient function for the figure on the previous page is: b c a y y y f '(x)

f '(x) −1 0

2

x

d

2x

−1 0

f '(x)

e

y

y

−1 0

2

x

f '(x)

f '(x) −1 0

2

x

−1 0

2

x

7 For the functions graphed below, state the domain (where applicable) where the gradient: i is equal to zero ii is positive iii is negative iv does not exist. a

b

y

f(x)

1

d

x

y

0

e

g(x)

2

x

−1 0

f(x)

f

g(x) y 0

0

x

4

y

1 0

i

y

3

y

−1 0

2

f(x)

x 0

j

x f(x)

g(x)

Gradient = 0 x

2

x

y

−2 0

h

f(x)

g(x)

x Gradient = 0

−4

g

y 5

6

0

−1

c

y

x

y 4 2 −2 0

4

x f(x)

8 Sketch the gradient function for each function in question 7.

Chapter 7 Differentiation

331

9 WE 4 For each function f (x) graphed below, state the domain of the gradient function f ′(x). (Do not sketch the graph of f ′(x).) b c a y y y f(x)

f(x) 2 0

0

8 x

2

2

x

−4

−3

d

e

y

x

0 f(x)

f

y

y

f(x)

f(x) −1

0

x

2

0

x

0

x

3

f(x)

g

h

y

i

y

y

f(x)

f(x)

f(x) −2

j

x

0

0

2

x

0

x

5

y

f(x) −5

7B

0

x

4

Limits and differentiation from first principles The limit of a function is the value that the function approaches as x approaches a given value. If the function is continuous at the point in question then the limit exists and can be found by direct substitution. For example, the limit of f (x) = 3x + 1 as x approaches 1 is denoted as lim (3 x + 1)

eBook plus eLesson

eles-0093 Limits and differentiation from first principles

y

f(x)

4

Continuous

x →1

In this example f (x) is a continuous function at x = 1. (In fact, it is continuous for all values of x.) Therefore, the limit is found by direct substitution, that is, lim (3 x + 1) = 3(1) + 1 = 4

x →1

332


1 0

1

x

If the function is discontinuous at the point in question, then the limit exists if the function is approaching the same value from the left as it is from the right. For example: For example: y

Discontinuous 3

y

Discontinuous at x = −2

f(x)

f(x) 5 3

–2 0

2

x

0

x

The limit exists at x = 2, as the function is approaching 3 from the left and from the right. Therefore: lim f ( x ) = 3 x→2

The limit does not exist at x = −2, as the function is approaching 3 from the left and 5 from the right. That is, if the left-hand limit is not equal to the right-hand limit, then the limit does not exist.

WORKED EXAMPLE 5

Evaluate the following limits. a lim ( x 2 − 3 x ) x→ 5

eBook plus

x2 + 5 x + 6 x→ 0 x+3

Tutorial

b lim

int-0053

THINK a

b

WRITE

1

Decide whether f (x) is continuous at x = 5. If so, substitute x = 5 into f (x).

2

Evaluate.

1 2

Worked example 5

a

lim ( x 2 − 3 x ) = 52 − 3(5)

x→5

= 10

Decide whether f (x) is continuous at x = 0. If so, substitute x = 0 into f (x).

b

lim

+ 5 x + 6 0 2 + 5(0) + 6 = x+3 0+3 6 = 3 =2

x2

x→0

Evaluate.

If direct substitution makes the denominator zero, the limit of a rational expression can be evaluated by first simplifying the expressions and then using direct substitution. WORKED EXAMPLE 6

Evaluate

lim

x→ − 3

x2 + 5 x + 6 . x+3

THINK

WRITE

x 2 + 5x + 6 ( x + 3) ( x + 2) = lim 3 x → 3 x+3 x+3

1

If the limit can’t be found because the denominator becomes zero, factorise the numerator.

2

Simplify by cancelling.

− = lim ( x + 2), x ≠ 3

3

Substitute x = −3.

= −3 + 2

4

Evaluate.

= −1

lim

x→

−

−

x → −3

Chapter 7

Differentiation

333

Differentiation using first principles The gradient function is the rule for the instantaneous rate of change of a given function at any point. It also gives the gradient of the tangent drawn at any point of the given function. Consider the chord (straight line) PQ to the curve below. f(x) Q

y f(x + h)

f(x + h) − f(x) P

f(x)

h

0

x+h

x

x

rise run f ( x + h) − f ( x ) = h As Q moves along the curve towards P, the value of h gets smaller and smaller. Or as Q gets as close as possible to P, h → 0, and PQ becomes a tangent at P. The gradient of the curve at a point P is the gradient of the tangent at that point. f ( x + h) − f ( x ) That is, gradient at point P is f ′( x ) = lim , h ≠ 0. h→0 h Finding the gradient this way is known as differentiation from first principles. Differentiating f (x) gives f ′(x) or f ′(x) is the derivative of f (x). dy dy Differentiating y gives , or is the derivative of y with respect to x. dx dx The gradient of PQ =

WORKED EXAMPLE 7

Find the gradient of the chord PQ drawn to the curve f (x) = x2 + 2 in the diagram. y Q [1 + h, f(1 + h)]

P [1, f(1)] x

0 THINK

Find the gradient and simplify. rise Gr ient = Grad . run

WRITE

f ( x + h) − f ( x ) ,h≠0 h f (1 + h) − f (1) = h Gr ient = Grad

(1 + h)2 + 2 − (12 + 2) h 3 + 2h + h 2 − 3 = h =

334


2h + h 2 h h(2 + h) = h =2+h The gradient is 2 + h. =

WORKED EXAMPLE 8

Find the gradient of f (x) = x2 − 1 at the point where x = 2 by: a sketching a graph and finding the gradient of the tangent at x = 2 b differentiating using first principles. THINK a

1 2

b

WRITE

Sketch the graph of f (x) over a domain which includes the given value of x. Construct a tangent at the given point on the curve. (It is difficult to be accurate.)

3

Find the gradient of this tangent by evaluating rise . run

1

Find f ′(x) using first principles.

a

y 8 7 (3, 6) 6 5 4 3 2 11 1 0 12345 −2 −3 −4 −5 (0, −5) 3

x

Gradient of tangent at x = 2 is 11 approximately = 3 23 . 3 f ( x + h) − f ( x ) b f ′( x ) = lim ,h≠0 h→0 h ( x + h)2 − 1 − ( x 2 − 1) h→0 h

= lim

x 2 + 2 xxhh + h 2 − 1 − x 2 + 1 h→0 h

= lim

2 xh + h 2 h→0 h h(2 x + h) = lim h→0 h = lim (2 x + h) = lim

h→0

= 2x 2

Evaluate f ′(2) to find the gradient at x = 2.

f ′(2) = 2(2) = 4

Note: Answer a (3 23 ) is very close to answer b (4).

Chapter 7

Differentiation

335

WORKED EXAMPLE 9

Use first principles to differentiate g(x) = x2 − x. THINK

WRITE

1

Find g(x + h) and simplify.

2

Find g ′(x) using first principles.

g(x + h) = (x + h)2 − (x + h) = x2 + 2xh + h2 − x − h g( x + h) − g( x ) g ′( x ) = lim , h ≠ 0, h→0 h x 2 + 2 xxhh + h 2 − x − h − ( x 2 − x ) h→0 h 2 2 x + 2 xxhh + h − x − h − x 2 + x = lim h→0 h 2 2 xh + h − h = lim h→0 h h(2 x + h − 1) = lim h→0 h = lim (2 x + h − 1) = lim

h→0

= 2x − 1

1. The limit of a function is the value that y approaches as x approaches a given value. 2. If a function is continuous at a point then a limit exists at that point and can be found by direct substitution. 3. If a function is discontinuous at a point then the limit exists only if the function is approaching the same value from both left and right. If there is a break in the curve and the point lies within the break then the limit does not exist. 4. The limit of expressions with a denominator can be found. (a) If direct substitution makes the denominator equal to zero then factorise the numerator, cancel and then use direct substitution. (b) If direct substitution does not make the denominator equal to zero then use direct substitution. f ( x + h) − f ( x ) 5. The gradient of a chord or secant is found using . h f ( x + h) − f ( x ) , h ≠ 0. 6. The gradient at a point P on a curve is f ′( x ) = lim h→0 h 7. The gradient at a point P on a curve is the gradient of the tangent to the curve at that point. EXERCISE

7B

Limits and differentiation from first principles 1 WE 5 Evaluate the following limits. a lim ( x + 4) b lim (2 p − 3) x→2

d lim ( x 2 − 5)

e

g

h lim

x→3

336

p→ −2

lim (10 − x + x 2 − x 3 )

x→ 2 −

c

lim ( x 2 + 4 x − 3)

f

x 2 + 5x + 6 x→0 x+2

i

x → −1


lim (8 − 3h)

h→0

lim ( x 3 − 5 x + 2)

x→3

x2 − 2x − 3 x →1 x−3 lim

2 WE 6 By first simplifying the rational expression, evaluate the following limits. 2x2 + 2x x 2 + 3x 3x 2 − 3x a lim c lim b lim x→ 1 x +1 x→0 x →1 x − 1 x 2 −4 2 + 3x + 2 x x x 2 − 5x − 6 d lim e lim f lim x→2 x − 2 x→ 1 x→6 x +1 x−6 h3 − 8 x 3 + 27 x2 + 4x − 5 g lim h lim i lim h→2 h − 2 x→ 3 x + 3 x→ 5 x+5 −

−

−

3 Evaluate the following. a lim (3 x − 4) x→3

x3 + 4 x →1 x + 2 x 2 + 3x g lim x→3 x − 1 d lim

b

−

x2 − 9 3 x+3

lim

x→ −

c

h3 − 64 h→4 h − 4 x3 + 1 h lim x→ 1 x −1 e lim

f

−

i

x2 + x − 6 x→2 x−2 lim

lim ( x 3 + x 2 − 6)

x → −2

x 2 + 7 x + 12 4 x+4

lim

x→

−

Questions 4 and 5 refer to the following diagram. Consider the chord PQ drawn to the curve f (x) as shown. 4 WE 7 Find the gradient of the chord PQ drawn to the curve f (x) = x2 + 1 in the diagram at right.

y f(x) = x2+1 Q (2+h, f(2+h))

5 MC The gradient of the tangent at P in the diagram is: a h b 4+h c 4h e 4 d 4 + h2

P (2, f(2))

1 0

x

2

6 a Find the gradient of the chord PQ to the function f (x) = x(x + 2) if the x-coordinates of P and Q are −1 and −1 + h respectively. b Hence, find the gradient of the function at P.

7 If the gradient of the chord ing two points on the curve f (x) = x2 + 2x + 3 is find the gradient of the curve at the point where x = 1.

f (1 + h) − f (1) , h

8 WE 8 Find the gradient of g (x) = 4 − x2 at the point where x = 2 by: a sketching a graph and finding the gradient of the tangent at x = 2 b differentiating using first principles. 9 Find the gradient of h (x) = 2x2 − 6x at x = −1 by using: a a sketch graph to find the gradient of the tangent at x = −1 b differentiation from first principles. 10 MC The gradient of a function f (x) at the point where x = 3 is: f ( x + h) − f ( x ) f (3 + h) − f ( x ) f (3 + h) − f (3) a lim b lim c lim h→0 h→0 h→0 h h h f (3 + h) − f (3) f ( x + h) − f ( x ) d e h h 11 WE9 Use first principles to differentiate f (x) if: a f (x) = 3x + 5 b f (x) = x2 − 3 d f (x) = (x − 4)(x + 2) e f (x) = 8 − 3x2 dy 12 Use first principles to find if: dx a y = 9 − 4x b y = x2 + 3x 3 d y = x − 4x e y = 5x − 2x3

c f

f (x) = x2 + 6x f (x) = x3 + 2

c f

y = 3x2 + 8x − 5 y = −x2 − 2x

Chapter 7

Differentiation

337

The derivative of xn

7C

Instead of using the procedure of differentiating from first principles, rules can be applied to find derivatives. These rules can be derived from first principles and have been looked at in detail in Maths Quest 11 Mathematical Methods CAS. If f (x) = axn then f ′ (x) = naxn − 1, where a and n are constants and n is rational. If f (x) = c then f ′ (x) = 0, where c is a constant. (This is because c = x0 and, using the rule, the derivative of x0 is 0 × x −1 or 0.) dy For example, if y = x7 then = 7 x 6 . If f (x) = 5x4 then f ′(x) = 20x3. dx If f (x) = g(x) + h(x) then f ′(x) = g′(x) + h′(x), that is, differentiate each term of a function separately. If f (x) = a × g(x), where a is a constant, then f ′(x) = a × g′(x). WORKED EXAMPLE 10

Differentiate y = x 4 −

3 2 x + 7. 2

THINK

WRITE

y = x 4 − 32 x 2 + 7

1

Write the equation.

2

Differentiate each of the three separately.

dy = 4 x 4 − 1 − 32 (2) x 2 − 1 + 0 dx = 4x3 − 3x

WORKED EXAMPLE 11

Find the derivative of: 1 1 a f ( x) = + x x

b f ( x) =

x+ x . x2

THINK a

WRITE

1

Write the equation.

2

Rewrite

3

Differentiate each term.

a

1 1 and using negative indices. x x

f ( x) =

f ( x ) = x −1 + x

b

Write the function in the form originally given.

1

Write the equation.

−1 −1 −1 − 1 − 1 2 x 2 −3 2 x − −x 2 −

= b

f ( x) =

2

−

1 1 − x 2 2 x3

x+ x x2 1

2

338

Rewrite x using indices.


−1 2

f ′( x ) = − x =

4

1 1 + x x

x + x2 = x2

1

3

Separate the function into two expressed in index form.

x x2 = 2+ 2 x x

4

Simplify each term.

=x 1+x

5


−

f ′( x ) = −1x = −x

6

′( x ) . Simplify f ′(

= =

−

−

−3 2

−1 − 1

−2

−

1 − x2

−

3 − 32 − 1 x 2

3 − 25 x 2 3 5

2x 2 3

1 − x2 2 x5

WORKED EXAMPLE 12

If f ( x ) = x 3 − 2 x 2 + a f '(x)

8 , use a CAS calculator to find: x b f '(2).

THINK 1

2

WRITE

On a Calculator page, press: • MENU b • 4:Calculus 4 • 1:Derivative 1 Complete the entry lines as: 8 d  3 x − 2x2 +   x dx  d  3 8 x − 2x2 +  x = 2  x dx  Press ENTER · after each entry. Alternatively, press / r to obtain the expression template and choose the derivative template. Write the answers using the correct notation for the derivative.

a f '( x ) = 3 x 2 − 4 x − b f '(2) = 2

8 x2

1. Rules for differentiating axn (a) If f (x) = axn then f '(x) = naxn − 1, where a and n are constants. (b) If f (x) = c then f '(x) = 0, where c is a constant. (c) If f (x) = ag(x), where a is a constant then f '(x) = ag'(x). 2. If f (x) = g(x) + h(x) then f '(x) = g'(x) + h'( x). Differentiate each term of a function separately. 3. To evaluate f (a) where a is a constant, replace every x in the f (x) equation with the value of a. For example, if f (x) = 2x2 − 3x + 1, f (2) = 2 × 22 − 3 × 2 + 1 = 3.

Chapter 7

Differentiation

339

EXERCISE

7C

The derivative of xn 1

Find the derivative of each of the following. a y = x6 b y = 3x2 c y = 5x4 f y = −5x g y = 12 x 3 j y = 8x5 Differentiate each of the following.

e y = −4x3 i y = 10 2 WE 10


SkillSHEET 7.3 Index laws

d y = x20 x4 h y= 3

a f (x) = 4x3 + 5x

b g(x) = −5x2 + 6x + 1

c h( x ) = 9 +

d h(x) = 4 − 3x + 6x2 + x3

e g(x) = 7x11 + 6x5 − 8

f

g f (x) = −6x + 3x2 − 4x3

h g( x ) = 7 x 2 − 4 x + 23

x3 5

2x 5 x3 + + 10 5 3 i h (x) = (x + 4)(x − 1) f ( x) =

j f (x) = (x2 + 2x)(3x − 6) 3 WE 11 a

Find the derivative of each of the following. 1

2 x3

5

d 4x 4 g x

c x3

b 3 x

−1 2

j

3 4x

m

1 3

x − 2x2

e

f

x+3 x 2 k 5x 2

2

+ x3

h

i l

1 + x2 x x 2 + x3 x 2 − + 3x 2 x

1

x3

− 4x +

− x 3

n

x + x4 x

4 WE 12 If f (x) = 2x5 − 10x + 5 find: a f ′(x) b f ′(2). 5 MC If f (x) = x2 − 6x then f ′(4) is equal to: a 8 b −12 c 12

d 2

e

−16

3

6 MC The value of f ′(9) if f ( x ) = x 2 + x 2 − 10 x is: b 18 d 8 e 0 c 12 12 1 7 Find g′ (−2) if g( x ) = 2 + 3 x − 8. x 5 8 Find the gradient of the curve y = 4 at the point where a x = 2 and b x = 0. x 1 9 Find the gradient of f ( x ) = 2 x 3 − x 2 + at the point where x equals: x a 1 b 4 c 9. a

1 2

10 If g (x) =

3

x + 4 x , find:

a g′ (x)

b g′(1)

c g′ (8)

d g′ (−8).

11 Show that the derivative of y = k, where k is a constant, is zero. 12 For each of the following: i expand the brackets ii differentiate the expanded expression iii factorise. a (x + 1)2 b (x + 1)3 c (2x + 1)2 d (2x + 1)3 e (3x + 1)2 f (3x + 1)3 13 Using the results of question 12 give the derivative of (ax + b)n in factorised form. (a, b, n are constants.)

340


The chain rule

7D

A function which can be expressed as a composition of two simpler functions is called a composite function. For example, y = (x + 3)2 can be expressed as y = u2 where u = x + 3. That is, to obtain y from x, the first function to be performed is to add 3 to x (u = x + 3), then this function has to be ‘squared’ ( y = u2). Composite functions can be differentiated using the chain rule. For example, using the previous function, y = (x + 3)2: Let u = x + 3, so y = u2. dy du = 2u. Then = 1 and du dx dy dy dy du = × . This is known as the chain rule. It is known as the But we require and dx dx du dx chain rule because u provides the ‘link’ between y and x. dy Now = 2u × 1 dx = 2(x + 3) × 1 (replacing u with x + 3) = 2(x + 3) The chain rule is used when it is necessary to differentiate a ‘function of a function’ as above. WORKED EXAMPLE 13

eBook plus

If y = (3x − is expressed as y = dy dy ddu u dy dy a b and hence c . du du dx dx dx dx 2)3

un,

find:

Tutorial


THINK a

b

c

WRITE

1

Write the equation.

a y = (3x − 2)3

2

Express y as a function of u.

3

Differentiate y with respect to u.

1

Express u as a function of x.

2

Differentiate u with respect to x.

Let y = u3 where u = 3x − 2 dy = 3u 2 du b u = 3x − 2 du =3 dx

1

Find

dy using the chain rule. dx

c

dy dy du = × dx du dx = 3u2 × 3 = 9u2

2

= 9(3x − 2)2

Replace u as a function of x.

WORKED EXAMPLE 14

If f ( x ) =

1 2 x2

− 3x

, find f ′ (x). Check your answer using a CAS calculator.

THINK 1

Write the equation.

WRITE

f ( x) =

1 2x2

− 3x

Chapter 7

Differentiation

341

2

Express f (x) in index form.

3


4


5

Express u as a function of x.

6


7

Find f ′ (x) using the chain rule.

8

Replace u as a function of x and simplify.

y = (2 x 2 − 3 x ) −1

Let y = u 2 where u = 2x2 − 3x − dy − 1 32 = 2u du u = 2x2 − 3x du = 4x − 3 dx −3 dy − = f ′( x ) = 12 u 2 × (4 x − 3) dx = = =

9

−1 2

−1 (4 x − 3) (2 x 2 2 − (4 x − 3)

− 3x)

−3 2

3

2(2 x 2 − 3 x ) 2 3 − 4x 2 ( 2 x 2 − 3 x )3

On a Calculator page, press: • MENU b • 4:Calculus 4 • 1:Derivative 1 Complete the entry line as:  d  1  2 dx  2 x − 3 x  Press ENTER ·.

10

Note the answer from the CAS is not expressed in the same format obtained previously.

∴ f '( x) =

− (4 x − 3)

2 x (2 x − 3) x (2 x − 3)

A quicker way to apply the chain rule when a function can be expressed in index form is as follows. If f (x) = [g(x)]n then f ′(x) = n[g(x)]n − 1 × g ′(x). That is, differentiate the bracket and then what is inside the bracket; ‘outside then inside’. WORKED EXAMPLE 15

Find the derivative of f (x) = (x2 − 2 x)3. THINK

342

WRITE

1

Write the equation.

f (x) = (x2 − 2x)3

2

Let g(x) equal what is inside the bracket.

g(x) = x2 − 2x

3

Find g ′(x).

g′(x) = 2x − 2


4

Use the rule f ′ (x) = n[g(x)]n − 1 × g′(x) to differentiate f (x).

f ′ (x) = 3(x2 − 2x)3 − 1 × (2x − 2) = 3(x2 − 2x)2 × [2(x − 1)] = 6(x − 1)(x2 − 2x)2

5

Simplify f ′(x) as far as possible.

= 6(x − 1)[x(x − 2)x(x − 2)] = 6x2(x − 1)(x − 2)2

1. A composite function is a function composed of two (or more) functions. 2. Composite functions can be differentiated using the chain rule, dy dy du = × . dx du dx 3. A short way of applying the chain rule is: If f (x) = [g(x)]n then f ′(x) = n[g(x)]n − 1 × g′(x). EXERCISE

7D

The chain rule 1 If each of the following functions are expressed in the form y = un, state i u and ii n. 1 a y = (5x − 4)3 b y = 3x + 1 c y= (2 x + 3)4 4 1 − d y = 7 − 4x e y = (5x + 3) 6 f y = (4 − 3 x ) 3 2 MC If y = (x + 3)5 is expressed as y = u5 then: a u = (x + 3)5 b u=x+3 d u=3 e u = x5

c u=x

3 WE 13 If each of the following composite functions are expressed as y = un, find: du dy dy i ii and hence iii . dx du dx 1 b y = (7 − x)3 c y= a y = (3x + 2)2 2x − 5 3 1 d y= e y = 5x + 2 f y= 4 (4 − 2 x ) 3x − 2 −2

g y = 3(2x2 + 5x)5

h y = (4x − 3x2)

i

−4

1  y = x+   x

6

j y = 4(5 − 6x)

For questions 4, 5 and 6 below, y = x 2 − 3 x + 2 is expressed as y = un. 4 MC a

dy is equal to: du

1 u 2

b u

−1 2

c

1 2 u

d

3 1 2 u 2

d

1 2

e

1 1 2 u 2

du 5 MC is equal to: dx a 2x − 3

b x2 − 3x + 2

c x2 − 3x

3

x2 − 2 x + 1

Chapter 7

e x−3

Differentiation

343

6 MC Using the chain rule, dy is equal to: dx 2x − 3 2x − 3 b a 2 x 2 − 3x + 2 u d 7

1 (2 x 2

− 3)

e

1 (2 x 2

c

x 2 − 3x + 2 2 u

1 − 3)( x 2 − 3 x + 2) 2

Use the chain rule to find the derivative of the following. a y = (8x + 3)4

b y = (2x − 5)3

c f (x) = (4 − 3x)5

d y = 3x 2 − 4

e

1

−2

f g(x) = (2x3 + x)

f ( x) = ( x 2 − 4 x) 3

1  g g( x ) =  x −   x

6

−1

h y = (x2 − 3x) 1

8 WE 14 If f ( x ) =

4x + 7

, find f ′(x).

9 Use the chain rule to find the derivative of the following. (Hint: Simplify first using index notation and the laws of indices.) a y=

6x − 5 6x − 5

b f ( x) =

( x 2 + 2)2 x2 + 2

10 WE 15 Find the derivative of: a f (x) = (x2 + 5x)8 b y = (x3 − 2x)2 c

1

f ( x ) = ( x 3 + 2 x 2 − 7) 5

3

d y = (2 x 4 − 3 x 2 + 1) 2

11 If f (x) = (2x − 1)6, find f ′(3). −

12 If g (x) = (x2 − 3x) 2, find g′(−2). 13 If f ( x ) = x 2 − 2 x + 1, find: a f (3) c f ′ (3) eBook plus Digital doc

WorkSHEET 7.1

b f ′(x) d f ′(x) when x = 2.

14 Find the gradient of the function h( x ) = 3 x 2 + 2 x at the point where x = 2. 3 15 Find the value of f ′ (−1) if f ( x ) = . 5 − 4x 16 If f (x) = (2x − 1)5 and f ′ (x) = 10 (2x − 1)n, find the value of n. 17 If f (x) = (3x + 2)7 and f ′ (x) = a (3x + 2)6 × 3, find the value of a. 18 If f (x) = (5x − 3)10 and f ′ (x) = 10m (5x − 3)9, find the value of m.

7E

The derivative of e x If f (x) = ex then using first principles f ( x + h) − f ( x ) f ′( x ) = lim ,h≠0 h→0 h

344

ex + h − ex h→0 h

= lim

= lim

e x eh − e x h→0 h


e x (e h − 1) h→0 h h e −1 = e x lim h→0 h h e −1 Note that lim can be deduced by using a calculator and substituting values of h close to h→0 h zero.

= lim

h

eh − 1 h

0.01

1.0050

0.0001

1.000 05

0.000 001

1.000 000

eh − 1 = 1. h→0 h Therefore, f ′ (x) = ex × 1 = ex That is, lim

If f (x) = ex then f ′ (x) = ex.

WORKED EXAMPLE 16

Differentiate y = e

−5x

.

THINK

WRITE

1

Write the equation.

2

Express u as a function of x and find

3 4 5

du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx Replace u as a function of x.

−5x

y=e

Let u = −5x so

du − = 5 dx

dy = eu du

y = eu so

dy − u = 5e du −5x

= −5e

This example shows that if f (x) = ekx then f ' (x) = kekx. WORKED EXAMPLE 17

Find the derivative of y = e2x + 1. THINK 1

Write the equation.

2


3 4

5

du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx Replace u as a function of x.

WRITE

y = e2x + 1 Let u = 2x + 1 so y = eu so

du =2 dx

dy = eu du

dy = eu × 2 dx = 2eu = 2e2x + 1

Chapter 7

Differentiation

345

WORKED EXAMPLE 18

a f (x) = ex(ex − 2)

Differentiate:

b f ( x) =

e2 x − 2 e− x . ex

THINK a

b

WRITE

Write the equation.

2

Expand.

3

Differentiate.

4

Factorise in order to leave the answer in the form it was given.

1

Write the equation.

2

Write each term in the numerator over each term in the denominator. Divide the numerator of each term by its denominator using the laws of indices.

3

f (x) = ex(ex − 2)

a

1

= e2x − 2ex f ′ (x) = 2e2x − 2ex = 2ex(ex − 1) b

f ( x) =

e 2 x − 2e − x ex

=

e 2 x 2e − x − x ex e −x − x

= e2x − x − 2e − = ex − 2e 2x −

4


5

Write your answer in the form it was given.

f ′ (x) = ex + 4e 2x 4 = ex + 2x e

WORKED EXAMPLE 19 3 − x

Find the derivative of y = ex

.

THINK

WRITE

1

Write the equation.

2


3 4

5

3

y = ex

− x

du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx

y = eu so

Replace u as a function of x.

Let u = x3 − x so

du = 3x 2 − 1 dx

dy = eu du

dy = eu × (3 x 2 − 1) dx = (3x2 − 1)eu 3 − x

= (3x2 − 1)ex

This example shows that if f (x) = eg(x) then f ′ (x) = g′ (x) e g (x).

1. 2. 3. 4.

346

If f (x) = ex, f ′ (x) = ex. If f (x) = ekx, f ′ (x) = ke kx. If f (x) = aekx + c, f ′ (x) = ake kx + c. If f (x) = ae g (x), f ′ (x) = g ′(x) × ae g(x).


EXERCISE

7E

The derivative of ex 1 WE 16

Differentiate each of the following. 1x

x

−

a y = e10x b y = e3 c y = e4 −5x −2x − f y = 4e g y = 6e h y = 5e0.2x 2 WE 17 Find the derivative of each of the following. a y = e6x − 2 b y = e8 − 6x 7 − 2x d y = 4e e y = −3e8x + 1 6 − 9x g y = 10e h y = −5e3x + 4 j

x +1

y = 2e 2

k y = 3e

c y = 2e5x + 3 f y = −2e6 − 5x − i y = 6e 7x

2− x 3

l

3 MC The derivative of y = e3x + 2 is equal to: 3x + 2 a 3e b (3x + 2)e3x + 2 c 3e3x d 3xe3x + 2 4 WE 18 Differentiate each of the following: a f (x) = 2(ex + 1) b f (x) = 3e2x(ex + 1) −x

d f (x) = (ex + 2)(e + 3)

e

f ( x) =

3e3 x

e2

4e5x

2

2x2

d f (x) = e2 − 5x

e f (x) = e6 − 3x + x

e

+ 2x) − 2e − x

−2x

−3

2

f g(x) = ex + 3x − 2

2

i y = e(2x + 1)

−2

4

f ( x) =

4e7 x

c y = ex − 2x

h y = −5e1 − 2x − 3x

j f (x) = e(4 − x)

f

−4x

2

b y = ex − 3x + 1 2

e 3xe3x

−e

2

a y = ex + 3x g h(x) = 3e4x − 7x

x +5

y = − 4e 4

c f (x) = 5(e

− e 6x

+ ex

g f (x) = + h f (x) = + 5 WE 19 Find the derivative of each of the following: ex

e y = 2e3x

d y=e x − i y = −2e 11x

k g(x) = e(x + 2)

3

3

l

y=e

3x + 4

2

1 3

n h(x) = e(x2 + 3x) m f ( x ) = e( x +1) 6 MC The derivative of 6ex3 − 5x is equal to: a 3x2 − 5 b 6(3x2 − 5)ex3 − 5x 3 − 5x 3 x d 6(x − 5x)e e 6(3x2 − 5)e3x2 − 5 9 − 4x 7 If f (x) = 5e , find the exact value of f ′ (2).

c (3x2 − 5)ex2 − 5x

8 If g (x) = 2ex2 − 3x + 2, give the exact value of g′(0). 9 Find the exact value of h′(−1) if h(x) = −5ex3 + 2x.

7F

The derivative of loge (x) −

The inverse of the function f (x) = ex is f 1(x) = loge (x). If y = loge (x) then ey = x, as shown in chapter 4, Exponential and logarithmic equations. Let x = ey dx = ey dy dy 1 But = dx dx and ey = x dy dy 1 Therefore, = . dx e y 1 = x

Chapter 7

Differentiation

347

That is, if f (x) = loge (x) then f'( x ) =

1 . x

WORKED EXAMPLE 20

Differentiate y = loge (7x). THINK 1 2 3 4

WRITE

Write the equation. du Express u as a function of x and find . dx dy Express y as a function of u and find . du dy Find . dx

y = loge (7x) du u = 7x, so =7 dx dy 1 y = loge (u), so = du u dy 1 = ×7 dx 7 x 1 = x

If f (x) = loge (kx), where k is a constant then f ′( x ) =

1 . x

WORKED EXAMPLE 21

Find the derivative of y = 2 loge (3x − 4). THINK 1

WRITE

2

Write the equation. Express u as a function of x.

3


4


5


6

Find

7

Replace u with 3x − 4.


y = 2 loge (3x − 4) Let u = 3x − 4. du =3 dx y = 2 loge (u) dy 1 = 2× du u 2 = u dy 2 = ×3 dx u 6 = u 6 = 3x − 4

WORKED EXAMPLE 22

Differentiate y = loge (x2 + 4x − 1). THINK

348

WRITE

1

Write the equation.

y = loge (x2 + 4x − 1)

2

Let u equal the section in brackets.

Let u = x2 + 4x − 1


3


4


5


6

Find

7

Replace u with what is in the brackets.

du = 2x + 4 dx y = loge (u) dy 1 = du u dy 1 = × (2 x + 4) dx u 2x + 4 = 2 x + 4x − 1


This example shows that if f (x) = loge [g(x)] then f ′( x ) =

g ′( x ) . g( x )

WORKED EXAMPLE 23

Differentiate y = loge (x2 + 5x − 2)2. THINK

WRITE

1

Write the equation.

y = loge (x2 + 5x − 2)2

2

Simplify the right side by using log laws. du Express u as a function of x and find . dx dy Express y as a function of u and find . du dy Use the chain rule to find . dx

y = 2 loge (x2 + 5x − 2) du u = x2 + 5x − 2, so = 2x + 5 dx dy 2 y = 2 loge (u), so = du u dy dy du dy 2(2 x + 5) = × so = dx du dx dx u 2(2 x + 5) = 2 x + 5x − 2

3 4 5

1 1. If f (x) = loge (x), then f ′( x ) = . x 1 2. If f (x) = loge (kx), then f ′( x ) = . x g ′( x ) . 3. If f (x) = loge [g(x)], then f ′( x ) = g( x ) EXERCISE

7F

The derivative of loge (x) 1 If y = loge (4x) is expressed as y = loge (u), then find: du a u b dx c

dy du

d

dy using the chain rule. dx Chapter 7

Differentiation

349

2 WE 20

Differentiate each of the following.

a y = loge (10x)

b y = loge (5x)

c y = loge (−x)

d y = loge (−6x)

e y = 3 loge (4x)  x h y = log e    3

f y = −6 loge (9x)  x i y = 4 log e    5

 x g y = log e    2 j

 − 2x  y = − 5 log e   3 

3 MC The derivative of loge (8x) is:

8 1 d x x 4 MC To differentiate y = loge (3x + 7) using the chain rule: a ‘u’ would be used to represent d loge (3x) a 3x + 7 b 3x c loge (x) b

a 8

b

1 8

x

c

e loge (8)

e x

dy du and are respectively du dx

1 1 1 1 and 3x c 3 and b and 3x + 7 d and 3 3x u u u dy c Hence is equal to dx 3 1 1 d c a 3 b 3x + 7 3x + 7 x 5 WE 21 Find the derivative of each of the following. a y = loge (2x + 5) b y = loge (6x + 1) c y = loge (3x − 4) a

e

d y = loge (8x − 1)

e y = loge (3 − 5x)

f y = loge (2 − x)

g y = loge (4 − 7x)

h y = 6 loge (5x + 2)

i y = 8 loge (4x − 2)

j y = −4 loge (12x + 5)

k y = −7 loge (8 − 9x)

6 WE 22

3 x

Differentiate the following.

a y = loge (3x4)

b y = loge (x2 + 3)

d y = loge (x2 − 2x3 + x4) g y = log e (5 x +

1 2) 3

j f (x) = loge (3x − 2)4

e y = log e

(

h f ( x ) = log log e

2x + 1

c y = loge (x3 + 2x2 − 7x)

)

1 (2 − 3 x ) 5 (2 −2

k f (x) = loge (5x + 8)

(

3 − 4x

a y = loge (x2 + 1)2

b y = loge (3 − x2)2

c y = loge (x2 − 2x + 3)3

d y = loge (x2 + 4x + 4)3

y = log e

i

 1  f ( x ) = log e   x + 3 

l

 2  f ( x ) = log e   4 + 3 x 

8 MC Using the chain rule the derivative of f (x) = loge (x2 − 5x + 2) would be: − 1 5 c 2x − 5 a 2 b 2 x − 5x + 2 x − 5x + 2 1 2x − 5 d e 2 x (2 x − 5) x − 5x + 2 9 Find the gradient of the function f (x) = 6 loge (4 − 3x) when x = −1. Maths Quest 12 Mathematical Methods CAS for the TI-Nspire

)

f

7 WE 23 Differentiate each of the following.

350

e 1 and 3

10 If g(x) = 3 loge (3x + 5) find the value of g′(0). 11 Find the exact value of f ′(2) if f (x) = 3x2 + 4 loge (x2 + x). 12 If y = eloge (x), find: dy a dx b the exact gradient when i x = 1 result?

ii x = 2

iii x = 4 iv x = 10. Can you explain this

13 If f (x) = eloge (x2), find: a f ' (x) b the exact value of i f ' (1) ii f ' (5) iii f ' (−2).

7G

The derivatives of sin (x), cos (x) and tan (x) The derivatives of sin (x), cos (x) and tan (x) can be found by differentiation from first principles but are beyond the requirements of this course. The derivatives of sin (x) and cos (x) can be shown by drawing the graphs of the gradient functions. The domain of each of these functions is R, but we will use only part of that domain. Consider the graph of f (x) = sin (x), domain [0, 2π] shown below. y π (— , 1) 2

0

π

2π

x

π ( 3— , −1) 2

π 3π , 2 2 π 3π f '(x) < 0 when < x < 2 2 3π π f '(x) > 0 when 0 < x < and < x < 2π 2 2 By sketching the graph of the gradient function, we can see that it is y = cos (x). f '(x) = 0 when x =

y 1 0 −1

π– 2

π

3π π — 2

2π x

Similarly, by sketching the graph of the gradient function of y = cos (x), we can see that the derivative of y = cos (x) is y = −sin (x). The derivative of tan (x) can be found using the quotient rule y (which appears later in this chapter). If f (x) = sin (x) then f ′ (x) = cos (x). y = cos (x) 1 If f (x) = cos (x) then f ′ (x) = −sin (x). y = −sin (x) 1 0 — π 3π 2π x π — If f (x) = tan (x) then f '( x ) = , which can be 2 2 2 ( x) –1 2 cos s written as sec (x).

Chapter 7

Differentiation

351

WORKED EXAMPLE 24

Find the derivative of y = sin (5x). THINK

WRITE

y = sin (5x)

1

Write the equation.

2


du . dx

Let u = 5x so

du =5 dx

3

Express y as a function of u and find

dy . du

y = sin (u) so

dy = cos (u) du

4

Find

5

Replace u with 5x.

6

It is important to note what happens when a CAS calculator is used to answer this question. On a calculator page, complete the entry line as: d (sin (5x)) dx Press ENTER ·. Note: The calculator gives an incorrect answer as it is set in Degree mode.


dy = 5cos (u) dx = 5 cos (5x)

In differentiating circular functions, the calculator needs to be set in Radian mode as the angles are given in radians. Write the answer, using the correct notation for the derivative.

∴

dy = 5 cos (5 x ) dx

This example shows that if f (x) = sin (ax), then f ′ (x) = a cos (ax). Similarly, if f (x) = cos (ax), then f ′ (x) = −a sin (ax). 352


WORKED EXAMPLE 25

Find the derivative of y = tan (3x). THINK

WRITE

1

Write the equation.

2


3 4

y = tan (3x)

du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx

du =3 dx dy y = tan (u) so = secc 2 (u) du dy 3 = 3 secc 2 (u) or dx coss2 (u) 3 = 3 sec 2 (3 x ) or 2 cos (3 x ) Let u = 3x so

This example shows that if f (x) = tan (ax), then f ' ( x ) =

a or a sec2 (ax). coss 2 ( ax )

WORKED EXAMPLE 26

eBook plus

Differentiate y = cos (x2 + 2x − 3).

Tutorial

THINK 1

Write the equation.

2


3 4 5

int-0555

WRITE

Worked example 26

y = cos (x2 + 2x − 3)

du . dx dy Express y as a function of u and find . du dy Find using the chain rule. dx Replace u with the part in brackets in the rule and simplify.

Let u = x2 + 2x − 3 so y = cos (u) so

du = 2x + 2 dx

dy − = sin(u) du

dy − = sin u × (2 x + 2) dx = −2(x + 1) sin (x2 + 2x − 3)

This example shows that the chain rule can be applied as follows. If f (x) = sin [g(x)] then f ′(x) = g′(x) cos [g(x)]. If f (x) = cos [g(x)] then f ′(x) = −g′(x) sin [g(x)]. g'( x ) If f (x) = tan [g(x)] then f '( x ) = cos 2 [ g( x ))]] = g′(x) sec2 [ g(x)]

1. If f (x) = sin (x) 2. If f (x) = cos (x) 3. If f (x) = tan (x) 4. If f (x) = sin (ax) 5. If f (x) = cos (ax)

then f ′(x) = cos (x). then f ′(x) = −sin (x). 1 then f ′( x ) = = secc 2 ( x ). cos2 ( x ) then f ′(x) = a cos (ax). then f ′(x) = −a sin (ax).

Chapter 7

Differentiation

353

a

6. If f (x) = tan (ax)

then f ′( x ) =

7. If f (x) = sin [g(x)] 8. If f (x) = cos [g(x)]

then f ′(x) = g′ (x) cos [g(x)]. then f ′(x) = −g′ (x) sin [g(x)]. g ′( x ) = g ′( x ) secc 2 [ g( x ))]]. then f ′( x ) = cos2 [ g( x)] x

9. If f (x) = tan [g(x)]

= a sec 2 (aax ).

cos2 (ax )

EXERCISE

7G

The derivatives of sin (x), cos (x) and tan (x) 1 WE 24

Find the derivative of each of the following.

a y = sin (8x)

b y = sin (−6x)

c y = sin (x)

 x d y = sin  3 

 − x e y = sin    2

f

2

 2x  y = sin    3

Differentiate each of the following. a y = cos (3x)

b y = cos (−2x)

 x c y = cos    3

 π x d y = cos    4

 − x e y = cos    8 

f

3 WE 25

 2x  y = cos    5

Differentiate each of the following.

a y = tan (2x)

b y = tan (−4x)

 x ta   c y = tan  5

 − 3x  d y = tan ta   4 

4 MC a The derivative of sin (6x) is: a 6 cos (6x)

b 6 cos (x)

d −6 cos (6x)

e

1 6

cos (6 x )

b The derivative of cos (4x) is: a 4 sin (4x) b 4 sin (x) c The derivative of sin (−4x) is: a 4 cos (−4x) b 4 cos (−x) d The derivative of cos (−8x) is: a 8 cos (−8x) b 8 sin (−8x) e The derivative of tan (7x) is:  x ta   a 7 tan  7

b

1 (7 x )

coss2

c 6 sin (x)

c −4 sin (x)

d 4 cos (4x)

e

−4

sin (4x)

c −4 cos (4x)

d −4 cos (−4x)

e

−4

sin (4x)

c −8 sin (−8x)

d −8 sin (−x)

e 8 sin (−x)

c 7 sec2 (7x)

d sec2 (7x)

e

5 WE26 If y = sin (4x + 3) is expressed as y = sin (u), find: a

dy du

b

du dx

c


6 If y = cos (3x + 1) is expressed as y = cos (u), find: a

354

dy du

b

du dx


c


1 7

sec ( x )

7

Differentiate each of the following. a y = sin (2x + 3) b y = sin (6 − 7x)

c y = sin (5x − 4)

 3x + 2  d y = sin  4 

f

e y = 5π sin (2π x)

8 Differentiate each of the following. a y = cos (3x − 2) b y = cos (4x + 7)  2x + 3 d y = cos  3 

 3x  y = − 4 sin    8

c y = cos (6 − 5x) f y = −6 cos (−2x)

e y = 4π cos (10π x)

9 Differentiate each of the following. a y = tan (2x + 1) b y = tan (8 − x) d y = tan 2(x + 1) e y = −3 tan (−x)

c y = tan (5x − 2)

10 Find the derivative of each of the following. a cos (x2 − 4x + 3)

b sin (10 − 5x + x2)

c sin (ex)

d cos (x2 + 7x)

e tan (4x − x2)

f tan (x2 + 3x)

g cos [loge (x)]

h sin (e4x)

i

 1 cos    x

j sin [loge (2x − 1)]

k cos (2e3x)

l 3 cos [loge (10x)]

m 4 tan (x3 + 2x2)

n

o

− 2 cos

 x  4 

− 8 tan ta

 − 3x   5 

p cos (x2 + 2x) + sin (3x − 9)

11 If f (x) = 3 sin (x2 + x), find f ′ (1) (answer correct to 3 decimal places). 12 Find the gradient of the curve g (x) = 2 cos (x3 − 3x) at the point where x = 0. 13 For each of the following functions find: i f ′ (x) and π ii the exact value of f ′   .  6 a f (x) = esin (x)

7H

b f (x) = ecos (x)

c f (x) = loge [sin (x)]

d f (x) = loge [cos (x)]

The product rule Any function which is a product of two simpler functions, for example, f (x) = (x + 2)(x − 5) or f (x) = (x2 − 5x + 6) sin (3x + 5) can be differentiated using the product rule of differentiation. Although the first example can be expanded and then differentiated, the second example cannot and therefore can be differentiated only using the product rule.

Product rule

dy dy dv dv ddu u = u× +v× . ddxx dx dx dx dx Or if f (x) = u(x) × v(x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x). If y = uv then

Chapter 7

Differentiation

355

WORKED EXAMPLE 27

If y = (3x − 1)(x2 + 4x + 3) is expressed as y = uv, find: du du dv dv dy dy dy dy dv dv du du =u +v . a u and v b and c using dx dx dx dx dx dx dx dx dx dx dx dx THINK a

b

c

WRITE a y = (3x − 1)(x2 + 4x + 3)

1

Write the equation.

2

Identify u and v, two functions of x which are multiplied together.

1


2

Differentiate v with respect to x.

1

Apply the product rule to find

Let u = 3x − 1

and

v = x2 + 4x + 3.

du =3 dx dv = 2x + 4 dx c dy = u × dv + v × du dx dx dx

b

dy . dx

= (3x − 1)(2x + 4) + (x2 + 4x + 3) × 3 2

= 6x2 + 10x − 4 + 3x2 + 12x + 9 = 9x2 + 22x + 5

Expand and simplify where possible.

WORKED EXAMPLE 28

Find the derivative of y = loge (4x) × sin (3x − 2). THINK

WRITE

1

Write the equation.

y = loge (4x) × sin (3x − 2)

2

Identify u and v. du dv Find and . dx dx dy Find using the product rule. dx

Let u = loge (4x) and let v = sin (3x − 2). du 1 dv = = 3 cos (3 x − 2) dx x dx dy 1 = log e (4 x ) × 3 cos (3 x − 2) + sin (3 x − 2) × dx x 1 = 3 log e (4 x ) × cos (3 x − 2) + sin (3 x − 2) x

3 4 5

Simplify wherever possible.

dy dv du =u× +v× . dx dx dx 2. If f (x) = u(x) × v(x) then f ′ (x) = u(x) × v′(x) + v(x) × u′(x). 1. If y = u × v then

EXERCISE

7H

The product rule 1 WE27 If y = (x + 3)(2x2 − 5x) is expressed as y = u × v, find: du dv dy dy dv du a u and v b and c using the product rule, =u× +v × . dx dx dx dx dx dx

356


2 WE28 a y=

Find the derivative of: 4x3

× loge (6x)

b g (x) = (3x − 2) loge (2x).

3 MC The derivative of f (x) = a f ′ (x) = 2x cos (2x) c f ′ (x) = 2x sin (2x) + x2 cos (2x) e f ′ (x) = 2x sin (x) + 2x2 cos (x)

x2

sin (2x) is: b f ′ (x) = 4x cos (2x) d f ′ (x) = 2x sin (2x) + 2x2 cos (2x)

4 Use the product rule to differentiate each of the following. a y = x cos (x) b y = 3x sin (x) c y = (5x − 2) ex d y = e3x (2 − 11x) e y = x5 cos (3x + 1) f y = 2x3 loge (7x) −2x g y = e loge (2x − 5) h y = 8 tan (5x) loge (5x)  4x  i y = 5 cos (2x) sin (x) j y = sinn   cos ( x )  3 − 4x − 3 k f (x) = e loge (6x) l f (x) = 4e 5x sin (2 − x) − 1 n f ( x) = x e 3 x cos (6 x ) m f ( x) = x −

−

o f (x) = 2x 3 sin (2x + 3) − q f (x) = (x2 + e3x)(4 − e 3x)

p f (x) = e 2x loge (3x2 + 5) r f (x) = (x2 − 6)(2 + 3x − x2)

5 If f (x) = (2x + 1) loge (x + 3), find the exact value of f ′(1). 6 Find g′(0) if g(x) = 5e2x cos (4x). eBook plus

7 Find the value of f ′(−2) if f (x) = (x2 + 2) sin (4 − 3x) (answer correct to 3 decimal places).

Digital doc

8 If g (x) = (6x + x2) ex − 3, find the exact value of g′(2).

WorkSHEET 7.2

9 Find the exact value of f ' (π) if f (x) = (3 − x) tan (2x).

7I

The quotient rule The quotient rule is used to differentiate functions which are rational expressions (that is, one function divided by another). For example, e3 x + 8 x2 − 6x + 3 or f ( x ) = f ( x) = coss ((6 − x ) 5x + 2

Quotient rule

du du dv dv −u× u dy v × dx dy dx dx . dx If y = then = v ddxx v2 u( x ) v ( x )u ′( x ) − u( x )v ′( x ) Or if f ( x ) = then f ′( x ) = . v( x) [v ( x)] x 2 WORKED EXAMPLE 29

3− x u is expressed as y = , find: v x2 + 4 x du du dv dv a u and v b and dx dx dx dx

If y =

c

dy dy . ddxx

Chapter 7

Differentiation

357

THINK a

b

c

WRITE

Write the equation.

2

Identify u and v.

1


2

Differentiate v with respect to x.

1

Apply the quotient rule to obtain

2

3− x x2 + 4x Let u = 3 − x and v = x2 + 4x. du − b = 1 dx dv = 2x + 4 dx du dv v× −u × dy dx dx c = dx v2 ( x 2 + 4 x ) × −1 − (3 − x )( )(2 x + 4) = 2 2 ( x + 4 x) a y=

1

dy . dx

dy where possible, factorising the dx final answer where appropriate. Simplify

= =

− x2

− 4 x − (12 + 2 x − 2 x 2 ) ( x 2 + 4 x )2

− x2

− 4 x − 12 − 2 x + 2 x 2 ( x 2 + 4 x )2

=

x 2 − 6 x − 12 ( x 2 + 4 x )2

=

x 2 − 6 x − 12 x 2 ( x + 4) 2

WORKED EXAMPLE 30

Find the derivative of f ( x ) =

eBook plus 3x

2e . cos ( 2 x − 3)

THINK

Tutorial

int-0556

WRITE

2e 3 x coss ((2 (2x 2xx − 3)

1

Write the equation.

f ( x) =

2

Identify u(x) and v(x).

Let u(x) = 2e3x. Let v(x) = cos (2x − 3).

3

Differentiate u(x) and v(x) with respect to x.

4

Apply the quotient rule to obtain f ′(x).

u′(x) = 6e3x v′(x) = −2 sin (2x − 3) v ( x )u ′( x ) − u( x )v ′( x ) f ′( x ) = [v ( x)] x 2 =

5

358

Simplify where possible.

Worked example 30

coss ((22x − 3) × 6e3 x − 2e3 x [− 2 ssin in (2 x − 3))]] [cos cos (2 (2 x − 3))]]2

6e3 x cos (2x − 3) + 4 e3 x sin (2 x − 3) [cos cos ((22xx − 3))]]2 3 x 2e [3 cos (2 x − 3) + 2 sin (2 x − 3)] = cos2 (2 x − 3)

=


du dv v× −u × u dy dx dx . 1. If y = then = 2 v dx v u( x ) v ( x )u ′( x ) − u( x )v ′( x ) 2. If f ( x ) = then f ′( x ) = . v( x) [v ( x)] x 2

EXERCISE

7I

The quotient rule 1 WE 29 If y =

x+3 u is expressed as y = , find: x+7 v

a u and v du dv b and dx dx du dv v× −u × dy dy dx dx . c using the quotient rule, = dx dx v2 x2 + 2x u( x ) 2 If f ( x ) = is expressed as f ( x ) = , find: 5−x v( x) a u(x) and v(x) b u′ (x) and v′ (x) c f ′(x) using the quotient rule. sin ( x ) 3 f (x) = tan (x) can be written as f ( x ) = . If u(x) = sin (x) and v(x) = cos (x), use the cos ( x ) 1 quotient rule to show that the derivative of tan (x) is . coss2 ( x ) 4 WE 30 Find the derivative of each of the following. a d g j

x2

2x − 4x

4x − 7 10 − x sin (2 x ) cos (2 x ) x2

4x + 3x − 2

− 3x

b

x2 + 7x + 6 3x + 2

c

e

e2 x x

f

h

log e ( x + 1) x2 + 2

i

k

2x3 + 7x e5x

l

m

e 3x + 8

n

4 log e (8x 8x) 2 x − 2x

o

p

2 cos (3 − 2 x ) x2

q

3e 2 − 7 x x+3

r

5 MC If h( x ) = 9x2 − 8 x2 − 3x 2 − 8 d x2 a

8 − 3x 2 then h′(x) equals: x 8 − 9x 9 2 b x2 − 3x 2 + 8 e x

c

cos ( x ) ex 3x 2 logg e (4 x ) e3 x + 2 cos (2 x ) x2 − 5 x sin ( x ) x 2 x e 2x − 3

− 3x 2

+8

x2

Chapter 7

Differentiation

359

sinn (4 x ) is: 4x + 1 4(4 x + 1) cos ( x ) − 4 sin (4 x ) a f ′( x ) = (4 x + 1)2 4(4 x + 1) cos (4x ) − 4 sin (4 x ) c f ′( x ) = 4x + 1 4 ssin (4x ) − 4(4 x + 1) cos (4x ) e f ′( x ) = (4 x + 1)2 cos (33 x − 2) 7 MC If g( x ) = then g′(x) is equal to: ex 3e x sin (3 x − 2) − e x cos (3 x − 2) a e2 x e x cos (3 x − 2) + 3e x cos (3 x − 2) c e2 x − 3e x sin (3 x − 2) − e x cos (3 x − 2) e e2 x 6 MC The derivative of f ( x ) =

8 If y =

cos (2 x ) dy find when x = 0. dx e3 x

9 Find the gradient of the function f ( x ) = 10 Find the exact value of g′(5) if g( x ) =

7J

(4 x + 1) cos (4 x ) − 4 ssin in (4 x ) (4 x + 1)2 4(4 x + 1) cos (4 x ) − 4 sin (4 x ) d f ′( x ) = (4 x + 1)2 b f ′( x ) =

−ex

sin (3 x − 2) − e x cos (3 x − 2) e2 x − 3e x sin (3 x − 2) − e x cos ((33x − 2) d ex b

2 x − 3x 2 at the point where x = −1. log e (3 x + 4)

4 log e (2 x ) . 3x


Mixed problems on differentiation

int-0252 Mixed problems on differentiation

Problems on differentiation may involve any combination of chain, product and quotient rules. WORKED EXAMPLE 31

For each of the following decide which rule of differentiation, that is, chain, product or quotient rule, would be useful to find the derivative. logg e (x ( ) 2 a b (x2 − 5x)6 c (x2 + 2x − 3) cos (2x) d ex + 3x sin ( x ) THINK a

b

c

360

WRITE

1

Write the equation.

2

It is of the form

1

u , that is, a rational function. v Write the equation.

2

It is a composite function of the form u6, where u = x2 − 5x.

1

Write the equation.

2

It is of the form u × v, that is, the product of two functions.


a y=

log e ( x ) sin ( x )

Quotient rule b y = (x2 − 5x)6

Chain rule c y = (x2 + 2x − 3) cos (2x)

Product rule

d

d y = ex2 + 3x

1

Write the equation.

2

It is a composite function of the form eu, where u = x2 + 3x.

Chain rule

WORKED EXAMPLE 32

Find the derivative of y = 2x3 cos (x2 + x). THINK

WRITE

1

Write the equation.

y = 2x3 cos (x2 + x)

2

Decide which rule to use and identify u and v to apply the rule.

Use the product rule. u = 2x3 and v = cos (x2 + x).

3


du = 6x2 dx

4

v is a composite function, so differentiate v with respect to x using the chain rule.

5

Apply the product rule to find

v = cos (x2 + x) Let w = x2 + x. dw = 2x + 1 dx v = cos (w) dv − = sin ( w) dw dv dv dw = × dx dw dx dv − So = ( ssin in ( w)))) (2 x + 1) dx = −(2x + 1) sin (x2 + x) dy dv du =u× +v× dx dx dx

dy . dx

= 2x3[−(2x + 1)] sin (x2 + x) + cos (x2 + x) × 6x2 6

= −2x3(2x + 1) sin (x2 + x) + 6x2 cos (x2 + x) = 2x2[3 cos (x2 + x) − x(2x + 1) sin (x2 + x)]


Derivatives involving the absolute value function  x if x ≥ 0 Consider the absolute value function, f ( x ) = x =  − .  x if x < 0 As discussed earlier, the graph of f (x) = |x| is continuous for all x and has a cusp (sharp point) at x = 0.

y

1 if x > 0 ′( x ) =  − The derivative of f (x) =  x, f ′( . x  1 if x < 0 In order for the derivative to be defined at x = 0 for a function f (x), the limit as x approaches 0 from the left, written as lim f '( x ), and the limit as x approaches 0 from the right, written as x →0−

lim f '( x ), must be equal.

x →0+

Chapter 7

Differentiation

361

y

This is not the case, as lim f ' ( x ) = − 1 and lim f ' ( x ) = 1. Thus, x → 0−

x →0+

1 the function f '(x) =  x is not differentiable at x = 0 and the graph of y = f '(x) has open circles at x = 0 as shown right. x −1 The chain rule is used to differentiate the composite function f (x) = h(g(x)), to give f '(x) = g′(x) × h′(g(x)). 1 if g( x ) > 0 Hence, for f (x) = h(g(x)), where h (x) =  x, then f '( x ) = g'( x ) ×  − .  1 if g( x ) < 0 Note the derivative is a hybrid function and the domain is obtained by examining the graph of the function y = g(x). WORKED EXAMPLE 33

eBook plus

=  x2

For the function: f (x) − 4x: a find the derivative b sketch the graphs of y = f (x) and y = f ′(x) on the same set of axes. THINK a

1

2


WRITE

As f (x) =  x2 − 4x is a composite function, apply the chain rule to find the derivative, f ′(x) where g(x) = x2 − 4x and h(x) =  x.

To determine the domain of the derivative, consider the graph of the function g(x) = x2 − 4x. From the graph, x2 − 4x > 0 if x < 0 or x > 4 and x2 − 4x < 0 if 0 < x < 4.

f (x) = h(g(x)) f ′(x) = g ′(x) × h′(g(x)) 1 f '( x) = 2 x − 4 ×  −  1  2 x − 4 if f '( x) =  −  2 x + 4 if

x 2 − 4 x > 0. x2 − 4x < 0

1 2 3 4 5x

 2 x − 4 if x < 0 or x > 4 ∴ f ' ( x ) =  − 2 x + 4 if 0 < x < 4 

3

Write the derivative with the correct domain.

4

To differentiate f (x) =  x2 − 4x with a CAS calculator, on a Calculator page, press: • MENU b • 4:Calculus 4 • 1:Derivative 1 Alternatively press /r to obtain the expression template and choose the derivative template. d Complete the entry line as: x2 − 4x dx Press ENTER ·.

(

if x 2 − 4 x > 0 if x 2 − 4 x < 0

y 4 3 2 1 0 −1 −1 −2 −3 −4

)

Note: The answer from the CAS is not expressed as a hybrid function with the correct domain and hence cannot be used.

362

Tutorial

 x.sign

(x − 4) + sign (x). x − 4 is not an acceptable answer.


b

1

Sketch the graph of f (x) =  x2 − 4x.

2

Sketch the graph of the derivative,  2 x − 4 if x < 0 or x > 4 f '( x) =  −  2 x + 4 if 0 < x < 4 For the graph of y = f ′(x), put open circles at x = 0 and x = 4 as the derivative is not defined at these points.

3

y

f'(x) = 2x − 4, x > 4

10 8 6 4 2 f(x) =x − 4x 2 x −3 −2 −1 0 1 2 3 4 5 6 7 −2 −4 f '(x) = −2x + 4, 0 < x < 4 −6 −8 f '(x) = 2x − 4, x < 0

To graph f (x) = x2 − 4x and y = f ′(x) with a CAS calculator, on a Graphs page, complete the function entry lines as: f1(x) = x2 − 4x d f 2(x) = x2 − 4x dx and press ENTER · after each entry. Note: The open circles at x = 0 and x = 4 are not displayed.

(

)

WORKED EXAMPLE 34

For the function f (x) =  sin (x) for x ∈[0, 2π]: a find the derivative b sketch the graphs of y = f (x) and y = f ′(x) on the same set of axes. THINK a

1

WRITE

As f (x) = |sin (x)| is a composite function, apply the chain rule to find the derivative, f ′(x) where g(x) = sin (x) and h(x) = |x|.

f (x) = h(g(x)) f ′(x) = g ′(x) × h′(g(x)) sinn ( x ) > 0 1 if si f ' ( x ) = cos cos ( x ) ×  − sin ( x ) < 0  1 if sin  cos ( x ) if sin ( x ) > 0 f '( x) =  − in ( x ) < 0  cos ( x ) if ssin

2

3

To determine the domain of the derivative, consider the graph of the function g(x) = sin (x) for x ∈ [0, 2π]. From the graph, sin (x) > 0 if 0 < x < π and sin (x) < 0 if π < x < 2π. Write the derivative with the correct domain.

y 1 0.5 0 −0.5 −1

π 2

π

3π 2

2π

x

 cos ( x ) if 0 < x < π ∴ f '( x) =  −  cos ( x ) if π < x < 2π

Chapter 7

Differentiation

363

b

1

Sketch the graph of f (x) = |sin(x)| for x ∈ [0, 2π].

y 1

2

Sketch the graph of the derivative,  cos ( x ) if 0 < x < π f '( x) =  −  cos ( x ) if π < x < 2π For the graph of y = f ′(x), put open circles at x = 0, x = π and x = 2π as the derivative is not defined at these points.

f(x) =sin (x)

0.5 0

π 2

−0.5 −1

π

2π

3π 2

x

f '(x) = cos (x), 0 < x < π f '(x) = −cos (x), π < x < 2π

1. Chain rule: dy dy du (a) = × dx du dx (b) A short way to apply the chain rule is: If f (x) = [g(x)]n then f ′(x) = n[g(x)n − 1 × g′(x). 2. Product rule: dy dv du (a) If y = uv then =u× +v× . dx dx dx (b) If f (x) = u(x) × v(x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x). 3. Quotient rule: du dv v× −u × u dy dx dx. (a) If y = then = v dx v2 u( x ) v ( x ) u ′( x ) − u( x ) v ′( x ) (b) If f ( x ) = then f ′( x ) = . v( x) [v ( x)] x 2 EXERCISE

7J

Mixed problems on differentiation 1 WE 31 For each function given below, state which rule of differentiation would be used to find the derivative, that is, chain (C), product (P) or quotient (Q). 3x + 7 a f (x) = loge (8x) b f (x) = 3x sin (x) c g( x ) = 4x2 d h( x ) =

4x cos ( x )

e g(x) = e5x sin (x) −

g h(x) = cos (x2 − 4x)

h f (x) = e x loge (5x)

j f (x) = sin2 (x)

k h( x ) =

m g(x) = ecos (x)

x−2 ex n f (x) = tan (x)

f

g( x ) =

x2 + 9x − 8 log e ( x )

i g(x) = loge [sin (x)] l

f ( x ) = log e ( x )

2 Using the appropriate rule find the derivative of each function in question 1. 3 WE 32 Find the derivative of each of the following. (Note that more than one rule will need to be applied in some cases.) x−2 − a y = e 5x cos (4x − 7) b y= 3x + 1 c y = loge (x + 1)3

364


d y = cos (x2 − 6x)

e f (x) = ex cos (2x) 1 sin ( x )

g f ( x) =

j f (x) = (x − 1)(x2 + 5x + 3)

k g(x) = ex(x2 + 3) −4x

l

cos (4x − 3)

s y=

p f ( x) =

cos (2 x ) sin (2 x )

log e

g( x ) =

(2 x + 3)5 x3 − 5

n y = cos2 (3x)

o y = loge [cos (3x)] q f ( x) =

sin (2 x ) cos (2 x )

h y = loge [sin (3x)]

i y = 4e3x2 − 5x + 2

m f (x) = e

f ( x) =

f

sin ( x 4 ) x2

r f (x) = [loge (5x − 1)]4

( x)

t

x

 x + 3 y = sin   x − 2  3

log e ( x 2 ) v g( x ) = x2 − x f (x) = 3 cos2 (x) + e 7x − x3

u f (x) = 3x5 cos (2x + 1) w y = ex sin (x) y f (x) = 3 sin (6x) + loge

(5x2)

−

x 2 4e

z h (x) = cos3 (x)

4 WE 33 For the following functions: a f (x) = x2 − 1 b f (x) = x2 + 2x  i find the derivative ii sketch the graphs of y = f (x) and y = f ′(x) on the same set of axes. 5 WE 34 For the following functions: a f (x) = sin (2x) for x ∈ [0, π] b f (x) = cos (x) for x ∈ [0, 2π] i find the derivative ii sketch the graphs of y = f (x) and y = f ′(x) on the same set of axes.

Chapter 7

Differentiation

365

SUMMARY Review — gradient and rates of change

• The gradient of a function exists wherever the graph of the function is smooth and continuous. • If the gradient of a function, f (x), is zero at x = a, then the graph of its gradient function, f ′(x), will have an x-intercept at x = a. • When the gradient of a function is positive, the graph of the gradient function is above the x-axis and when the gradient of a function is negative, the graph of the gradient function is below the x-axis. • A polynomial function has a gradient function which is also a polynomial function, but its degree is reduced by one. Limits and differentiation from first principles

• The gradient of a chord (secant) or the average rate of change is given by: f ( x + h) − f ( x ) . h • A limit is the value that y approaches as x approaches a given value. • A limit exists if the function is approaching the same value from both left and right. • The gradient of the tangent to a curve at a point P is the gradient of the curve at P and is f ( x + h) − f ( x ) given by lim . h→0 h dy • For a function y = f (x), its derivative is expressed as either or f ′ (x). dx The derivative of xn

• • • •

If f (x) = axn then f ′(x) = naxn − 1, where a and n are constants. If f (x) = c then f ′(x) = 0, where c is a constant. If f (x) = ag(x) where a is a constant then f ′(x) = ag′(x). If f (x) = g(x) + h(x) then f ′(x) = g ′(x) + h ′(x). Differentiate each term separately. The chain rule

• The chain rule of differentiation is: dy dy du = × dx du dx The derivative of ex

• • • •

If f (x) = ex then f ′(x) = ex. If f (x) = ekx then f ′(x) = kekx. If f (x) = aekx + c then f ′(x) = akekx + c. If f (x) = aeg(x) then f ′(x) = g ′(x) × ae g(x). The derivative of loge (x )

1 . x 1 • If f (x) = loge (kx) then f ′(x) = . x g ′( x ) • If f (x) = loge [g(x)] then f ′(x) = . g( x )

• If f (x) = loge (x) then f ′(x) =

The derivatives of sin (x ), cos (x ) and tan (x )

• If f (x) = sin (x) then f ′(x) = cos (x). • If f (x) = cos (x) then f ′(x) = −sin (x).

366


1 = sec2 (x). cos2 ( x ) If f (x) = sin (ax) then f ′(x) = a cos (ax). If f (x) = cos (ax) then f ′(x) = −a sin (ax). a If f (x) = tan (ax) then f ′(x) = = a sec2 (ax). 2 cos (ax ) If f (x) = sin [g(x)] then f ′(x) = g ′(x) cos [g(x)]. If f (x) = cos [g(x)] then f ′(x) = −g ′(x) sin [g(x)]. g ′( x ) If f (x) = tan [g(x)] then f ′(x) = = g ′(x) sec2 [g(x)]. cos2 [ g( x )]

• If f (x) = tan (x) then f ′(x) = • • • • • •

The product rule

• The product rule of differentiation states: dy dv du 1. if y = u × v, then =u× +v× dx dx dx 2. if f (x) = u(x) × v (x) then f ′(x) = u(x) × v′(x) + v(x) × u′(x). The quotient rule

• The quotient rule of differentiation states: du dv v× −u × u dy dx dx . 1. if y = then = v dx v2 u( x ) 2. if f ( x ) = then v( x) v ( x )u ′( x ) − u( x )v ′( x ) f ′( x ) = [v ( x)] x 2 • Summary of derivatives: f (x)

f ′ (x)

c

0

axn

naxn − 1

[g(x)]n

ng′(x)[g(x)]n − 1

ex

ex

ekx

kekx

eg(x)

g′(x)eg(x)

loge (x)

1 x 1 x g ′( x ) g( x ) a cos (ax) −a sin (ax) a cos2 (ax )

loge (kx) loge [g(x)] sin (ax) cos (ax) tan (ax)

(= a sec2 (ax))

Chapter 7

Differentiation

367

CHAPTER REVIEW y

SHORT ANSWER

1 The graph of a cubic function is shown below. y

−4

−1 0

1 2 3

3

4

5

Sketch the graph of its gradient function. h 3 + 2h 2 + 4 h Find lim . h→0 h a Find the derivative of f (x) = x3 + 2x using first principles. b Hence find the gradient at the point where x = 1. x3 a Find the gradient function if g( x ) = − 4x . 3 b Find the gradient of g (x) when x = 3. 3x 4 x3 If h( x ) = + − 3 x, find: 2 4 a h′ (x) b i h′(−1) ii h′(2). 3 (4 x + 1) 2 ,

6 If y = find the gradient when a x = 2 and b x = 1. 7 Find the derivative of x 2 + 4 . 8 Differentiate f (x) = e2x − 1. −x2

9 a Find f ′ (x) if f (x) = e . b What is the value of x when f ′ (x) = 0? dy if y = loge (2x3 − 4). 10 Find dx 11 The tangent to the curve f (x) = loge (ax − 1) when x = 2, has a gradient of 1. Find the value of a.

a On the same set of axes, sketch the graph of the derivative function. b Write down the domain of the derivative function. EXAM TIP

a Common errors were drawing the cusp point at x = 0 as a closed circle as well as the point at x = 3. b In this question students often gave responses that were inconsistent with the graph they had drawn in a. Union and intersection were sometimes confused as was the use of round, square or curly brackets. [Assessment report 1 2007]

[© Vcaa 2007]

19 Find the derivative of y = f (x) where f (x) = x2 − 2x . 20 Find the derivative of y = f (x) where f (x) = sin (x) − π π  for x ∈  , .  2 2  MULTIPLE CHOICE

1 The graph of f (x) is shown below. y

0

13 Find f '(x) if f (x) = tan (5x). 14 Find f '(x) if f (x) = tan (2x2 − 3). dy 15 If y = 3x2 loge (6x), find . dx cos ( x 2 ) 16 Find f '(x) if f ( x ) = . x2 17 Differentiate esin (2x).

f(x)

4

12 Find f '(x) if f (x) = 3 sin (2x) and hence find f '(2π).

3

The graph of its gradient function is: y b a

18 The diagram (top, right) shows the graph of a function with domain R.

368

x

3

x

f(x)

2

0

−3


0

3

x

x

y 0

3

x

c

y

0

e

y

d

3

c

x

0

x

3

y 4

e

4

2 4x − 9 −6x

9 If y = 5e

then

−

0

a −30e 6x − c 5e 6x − 1 − e 5e 7x

x

2 For the function g(x) graphed below, the gradient function g′(x) is defined over the domain: y a R b R \{1} c R \{4} d R \{1, 4} x e [1, 4] 0 1 4 x2 − 2x − 8 is: x→4 x−4 a undefined b 0 c 1

d 4

− 5x is: x +1 a 22 b −22 c 6 d undefined 5 The derivative of f (x) = 4x3 − x2 + 3x is: a 12x2 − 2x + 3 b 4x2 − 2x + 3 c 12x2 − 2x d 12x2 − x + 3 e 4x2 − 2x 1 6 The derivative of g( x ) = 2 − 2 x is: x −1 −2 2 2 a − b 3 − x x x x −2 −2 1 2 c d − − x x3 x x −2 1 − 3 e x3 x2 4 lim−

e 6

x3

7 The derivative of (2x + 5)6 is: a 6(2x + 5)5 b 12x(2x + 5)5 5 d 12(2x + 5) e 12(2x + 5)4 1 8 The derivative of is: 4x − 9 b

e

−6

c 6x(2x + 5)5

−2

(4 x − 9)3

dy is equal to: dx − b −6e 6x − d −30e 6x − 1 dy is: dx

a 4e4x + 6 c e4x + 7 e 4e3x + 7

b e4x + 6 d 4e4x + 7

11 The derivative of loge (3x − 2) is: 1 1 a b 3x − 2 3x 1 3(3 x − 2)

e

c

1 x

3 3x − 2

12 The derivative of 2 loge (x2 + x) is: 2(2 x + 1) c 2x + 1 x x2 + x 4x e 2 x +x dy 13 If y = cos (8x) then is: dx a 8 sin (8x) b sin (8x) 2(2 x + 1) x2 + x 2x d 2 x +x a

x→ 3

a 2 4x − 9

10 If y = e4x + 7 then

d

3 The value of lim

d 4 4x − 9

3

(4 x − 9) 2

b

c −8 sin (8x) e

−sin

d −8x sin (8x)

(8x)

14 If y = 2 sin (2x + 3) then a 4x cos (2x + 3) c 4 cos (2x + 3) e 2 cos (2x + 3)

dy is equal to: dx b −4 cos (2x + 3) d 4 cos (2x)

15 If f (x) = tan (6 − 5x), then f ′(x) is equal to: −5 5 a b 2 (6 − 5x 2 cos s 5 ) coss (6 − 5 x ) 2 2 c sec (6 − 5x) d 5 cos (6 − 5x) e −5 cos2 (6 − 5x) 16 If f (x) = x2 e2x then f ′ (x) is equal to: a 2xe2x + 2x2e2x c 4xe2x e 2xe2x + x2e2x

b 2xe2x d 2xe2x − 2x2e2x

Chapter 7

Differentiation

369

17 If g (x) = 2x loge (3x) then g′(x) must be: 2

a 2 log e (3x 3 )+ 3

b 2 loge (3x) + 2

c 2 loge (3x) + 6x

3 )− 3 d 2 log e (3x

e 2 loge (3x) + 6x loge (3x) 2x + 1 18 The derivative of is: x−2 4x − 5 a ( x − 2)2 4x − 3 c ( x − 2)2 −5 e ( x − 2)2 19 The derivative of

2

b

−3

( x − 2)2

d 4x − 5

e4 x is: x2

( x − 2)e 4 x x3 2(2 x − 1)e 4 x c x3 a

b

2(1 − 2 x ) x3

d

x 2 e 4 x − 2e 4 x x4

2e 4 x x3 20 If g(x) = (x2 + 3x − 7)5 then g′(x) is equal to: a 5(x2 + 3x − 7)4 b (2x + 3)(x2 + 3x − 7)4 4 c 5(2x + 3) d 5(2x + 3)(x2 + 3x − 7)4 2 4 e (x + 3x − 7) e

21 The derivative of sin (x) cos (x) is: a 2 sin (x) cos (x) b sin2 (x) − cos2 (x) 2 2 c sin (x) + cos (x) d cos2 (x) − sin2 (x) − 2 2 e sin (x) − cos (x) 22 Which one of the following is not true about the function f: R → R, f (x) = |2x + 4|? a The graph of f is continuous everywhere. b The graph of f ′ is continuous everywhere. c f (x) ≥ 0 for all values of x. d f ′(x) = 2 for all x > 0 e f ′(x) = −2 for all x < −2 [© Vcaa 2007] 23 If y = |cos (x)|, the rate of change of y with respect to x at x = k, π < k < 3π , is: 2 2 a −sin (k) b sin (k) c −cos (k) d −sin (1) e sin (1) [© Vcaa 2005] 24 Let f: R → R be a differentiable function. For all real values of x, the derivative of f (e2x) with respect to x will be equal to: a 2e2x f ′(x) b e2x f ′(x) c 2e2x f ′(e2x) d 2f ′(e2x) e f ′(e2x) [© Vcaa 2005]

EXTENDED RESPONSE 1

1 A section of a roller-coaster ride follows part of the curve with the equation y = 200 ( x 3 + 30 x 2 ) as shown below. a For what values of x (domain) is the gradient: y i zero? ii positive? iii negative? b Sketch the gradient function. c Use the graph of the gradient function to find the value of x where the 0 −28 −20 gradient is steepest over the domain [−25, 10]. dy d Find . dx e Find the gradient where x equals: i −25 ii −10 iii 10. f Does this your answer to part c? Briefly explain. g What is the highest point reached by the roller-coaster? (Give your answer in metres.) 2 Consider the functions f ( x ) = 2 x and g(x) = x2 + 1. a State the domain and range for each function.

370


12

x

b Find the composite functions: i f (g(x)) ii g(f (x)). c State the domain and range for f (g(x)) and g(f (x)). d Find: d i ( f ( g( g( x )))) dx d ii ( g( f ( x )))) dx e Evaluate: i f ′(g(2)) ii g ′(f (2)) 4 − x 2 , x ≤ 2  f ( x ) = 3 Consider the function 2 x − 4, 2 < x < 5 .  x − 1, x ≥ 5 a Sketch the graph of f (x). b For what values of x is f (x) discontinuous? c For what values of x is f (x) not differentiable? d Find f ′(x). e Sketch the graph of f ′(x).



Chapter 7

Differentiation

371

eBook plus

ACTIVITIES


• 10 Quick Questions: Warm up with ten quick questions on differentiation. (page 324) 7a

Review — gradient and rates of change

Tutorial

• WE 3 int-0552: Watch how to sketch a gradient function. (page 327) Digital docs

• SkillSHEET 7.1: Practise identifying positive negative and zero gradients. (page 328) • SkillSHEET 7.2: Practise sketching the gradient function given the original function. (page 328) 7b

Limits and differentiation from first principles

eLesson eles-0093

• Limits and differentiation from first principles. Watch an eLesson on related rates of change (page 332) Tutorial

• WE 5 int-0553: Watch how to evaluate limits. (page 333) 7c

The derivative of x n

Digital doc

• SkillSHEET 7.3: Practise using index laws. (page 340) 7d

The chain rule

Tutorial

• WE 13 int-0554: Watch a worked example on using the chain rule. (page 341) Digital doc

• WorkSHEET 7.1: Sketch gradient functions, identify where the derivative exists, evaluate limits, apply first principles and differentiation rules to determine derivatives. (page 344) 7G

The derivatives of sin (x), cos (x) and tan (x)

7H

Digital doc

• WorkSHEET 7.2: Differentiation of mixed expression involving the product, quotient and chain rules. (page 357) 7I

The quotient rule

Tutorial

• WE 30 int-0556: Watch a worked example on how to use the quotient rule. (page 358) 7J

Mixed problems on differentiation


• Differentiation: Consolidate your understanding of differentiation. (page 360) Tutorial

• WE 33 int-0557: Watch a worked example on how derivatives of an absolute value function. (page 362) chapter review Digital doc


Tutorial

• WE 26 int-0555: Watch a worked example on using the chain rule to differentiate trigonometric functions. (page 353)

372

The product rule


8

8A Equations of tangents and normals 8B Sketching curves 8c Maximum and minimum problems when the function is known 8D Maximum and minimum problems when the function is unknown 8e Rates of change 8F Related rates 8G Linear approximation

Applications of differentiation AreAS oF STudy

• Application of differentiation to curve sketching and identification of: – key features of curves – intervals over which a function is constant, stationary, strictly increasing or strictly decreasing – the maximum rate of increase or decrease in a given application context (consideration of second derivative is not required) – tangents and normals to curves

• Identification of: – local maximum/minimum values over an interval and application to solving problems – interval end point maximum and minimum values • Average and instantaneous rates of change, including formulation of expressions and solution and interpretation of problems involving rates of change and simple cases of related rates of change • The relationship f ( x + h) ≈ f ( x ) + hf hf '( x ) for a small value of h and its geometric interpretation eBook plus

8A

Digital doc

equations of tangents and normals As we have seen, a tangent to a curve is a straight line that touches the curve at a given point and whose gradient represents the gradient of the curve at that point. A normal to a curve is a straight line ing through the point where the tangent touches the curve and is perpendicular (at right angles) to the tangent at that point.

10 Quick Questions

y

0

f(x)

Tangent

Point of tangency Normal x

− 1 If the gradient of the tangent to a curve is m, then the gradient of the normal is (as the m − product of the gradients of two perpendicular lines equals 1).

The equation of a straight line ing through the point (x1, y1) and having a gradient of m is: y − y1 = m(x − x1) The gradient of the tangent at x = a is f ′(a). − 1 Therefore the gradient of the normal is . f ′(a)

Chapter 8

Applications of differentiation

373

y

The equation of the tangent at x = a is y − f (a) = f ′(a) (x − a) and the equation of the normal is  −1  y − f (a) =  ( x − a).  f ′(a) 

f(x)

f(a) 0

(a, f(a)) a

Worked exAmple 1

Find the equation of the tangent to y = x3 − 2x + 3 at the point (1, 2). Think

WriTe

1

Write the equation.

2

Find

dy . dx

dy where x = 1 to find the gradient of dx the tangent where x = 1.

y = x3 − 2x + 3 dy = 3x2 − 2 dx

Evaluate

At x = 1, dy =3−2 dx =1 So gradient of tangent is 1.

4

Substitute (1, 2) for (x1, y1) and m = 1 into the rule for the equation of a straight line: y − y1 = m(x − x1).

Equation of tangent at the point (1, 2) is y − 2 = 1(x − 1) y−2=x−1

5

Rearrange the rule to a simple form.

3

y=x+1

Worked exAmple 2

Find the equation of: a the tangent b the normal to the curve with the equation y = 3 loge (2x) at x = 1 c the tangent and the normal to the curve y = 3 loge (2x) at x = 1 using a CAS calculator. Think

Write the equation.

y = 3 loge (2x)

2

Evaluate y when x = 1. dy Find . dx

At x = 1, y = 3 loge (2) dy 3(2) = dx 2x 3 = x dy 3 a At x = 1, = dx 1 =3 So gradient of tangent is 3.

a

374

Tutorial


WriTe

1

3

eBook plus

dy when x = 1 to obtain the dx gradient of the tangent at x = 1.

1

Evaluate

2

Determine the equation of the tangent at (1, 3 loge (2)).

Equation of tangent is y − 3 loge (2) = 3(x − 1) = 3x − 3 y = 3x − 3 + 3 loge (2)

maths Quest 12 mathematical methods CAS for the Ti-nspire

x

b

1

Evaluate the gradient of the normal which is

c

−1

 dy   dx 

−1 . 3

b Gradient of normal is

.

2

Determine the equation of the normal at (1, 3 loge (2)).

1

On a Calculator page, press: • MENU b • 4:Calculus 4 Select either 9:Tangent Line 9 or A:Normal Line A. Complete the entry lines as: tangentLine(3ln (2x),x,1) normalLine(3ln (2x),x,1) Press ENTER · after each entry.

2

Note the answers from the CAS are not written as equations.

Equation of normal is 1 y − 3 loge (2) = 3 (x − 1) 3y − 9 loge (2) = −1(x − 1) = −x + 1 x + 3y = 1 + 9 loge (2) −

c

∴ y = 3 x + 3 (log e (2 − 1)) is the equation of the tangent. −x 9 log e (2 + 1) ∴y= is the equation + 3 3 of the normal.

− 1 1. If m is the gradient of the tangent, then is the gradient of the normal. That is, − m m1 × m2 = 1. 2. The equation of a straight line ing through the point (x1, y1) with gradient m is y − y1 = m(x − x1)

3. Parallel lines have the same gradient. exerCiSe

8A

equations of tangents and normals 1 We1 Find the equation of the tangent to the curve y = x2 + x at the point (2, 6). 2 3

Find the equations of the tangent to the curve y = x2 + 5x − 6 at the points where it crosses the x-axis.


Spreadsheet 129 Tangent and normal

Find the equation of the normal to the curve y = 3x2 − 5x + 4 at the point where x = 1.

Chapter 8


375

4

Find the equation of the normal to the curve y = 12 x2 + 3x − 7 at the point where it crosses the y-axis.

5 We2 For each of the following functions, find the equation of: i the tangent ii the normal at the given value of x. a y = x2 + 1, x = 1 b y = x3 − 6x, x = −2 1 c y = , x = 2 d y = (x − 1)(x2 + 2), x = −1 x e g i k

y = x , x = 4 y = x(x + 2)(x − 1), x = −1 y = 2x3 + x2 − 6x + 2, x = 1 y = e3x + 2, x = −1

y = 2 x + 3, x = 3 y = x3 − 3x2 + 4x, x = 0 y = e2x, x = 0 y = loge (x), x = 2 π n y = sin (2x), x = 3 π  p y = sin  2 x +  , x = 0  4 f h j l

m y = loge (2x + 3), x = 0  x o y = 3 cos   , x = π  2

6 mC If y = (2x + 3)4 then at the point (−1, 1) a the equation of the normal is: A y + 4x − 3 = 0 B 8y + x − 7 = 0 D y + 2x + 8 = 0 e 2y − x = 0

c y − 4x + 5 = 0

b the value of x where the gradient of the tangent is parallel to the x-axis is: A

−2 3

B

1 3

c

−3 2

D

−1 3

e

2 3

7 Find the equation of the tangent to f (x) = x2 + 4x + 1 which is parallel to the line y = 2x + 3. x2 + 1 at x = 0. x2 − 1 π 9 Find the equation of the normal to y = x sin (x) at x = . 2 10 Find the equation of the normal to y = loge (x + 2) which is parallel to the line with equation y + 3x − 5 = 0. 8 Find the equation of the tangent to y =

11 Find the equations of the tangent and normal for each of the following curves. a f (x) = x2 + 1 at x = a b f (x) = x at x = a


WorkSHEET 8.1

c f (x) = e x at x = 2a. 2

12 Find the equation of the tangent to the curve f (x) = e4x that is perpendicular to the line x + 8y = 16. 13 The graph of y = x has a normal with equation y = −8x + b, where b is a real constant. Find the value of b.

8B

Sketching curves When the graphs of polynomial functions are being sketched, four main characteristics should be featured: 1. the basic shape (whenever possible) 2. the y-intercept 3. the x-intercept(s) 4. the stationary points.

376


Stationary points A stationary point is a point on a graph where the function momentarily stops rising or falling; that is, it is a point where the gradient is zero. y

y

or x Gradient = 0 where function stops rising momentarily, then continues to rise again after this point

0 0

x Function stops falling and rises after this point

The stationary point (or turning point) of a quadratic function can be found by completing a perfect square in the form y = (x + h)2 + k. In this case the stationary point is (−h, k). For cubics, quartics or higher-degree polynomials there is no similar procedure. Differentiation enables stationary points to be found for any polynomial function where the rule is known. The gradient function of a function f (x) is f ′(x). Stationary points occur wherever the gradient is zero. f (x) has stationary points when f ′(x) = 0 or y has stationary points when

dy = 0. dx

The solution of f ′(x) = 0 gives the x-value or values where stationary points occur. If f ′(a) = 0, a stationary point occurs when x = a and y = f (a). So the coordinate of the stationary point is (a, f (a)).

Types of stationary points There are four types of stationary points. 1. A local minimum turning point at x = a. If x < a, then f ′(x) < 0 (immediately to the left of x = a, the gradient is negative). If x = a, then f ′(x) = 0 (at x = a the gradient is zero). If x > a, then f ′(x) > 0 (immediately to the right of x = a, the gradient is positive).

y

f(x)

f '(x) > 0 f '(x) < 0 0

2. A local maximum turning point at x = a. If x < a, then f ′(x) > 0. If x = a, then f ′(x) = 0. If x > a, then f ′(x) < 0. The two cases (1 and 2) can be called ‘turning points’ because the gradients each side of the stationary point are opposite in sign (that is, the graph turns). The term ‘local turning point at x = a’ implies ‘in the vicinity of x = a’, as polynomials can have more than one stationary point.

f '(x) = 0 x

a

y 0

a

x

f(x)

Chapter 8 Applications of differentiation

377

y

3. A positive stationary point of inflection at x = a. If x < a, then f ′(x) > 0. If x = a, then f ′(x) = 0. If x > a, then f ′(x) > 0. That is, the gradient is positive either side of the stationary point.

Gradient = 0

0

x

a

f'(x)

4. A negative stationary point of inflection at x = a. If x < a, then f ′(x) < 0. If x = a, then f ′(x) = 0. If x > a, then f ′(x) < 0. In cases 3 and 4 above the word ‘stationary’ implies that the gradient is zero. Not all points of inflection are stationary points. y

y

f(x)

Gradient = 0

x

0

a

y

or 0

x

0

x

Gradient ≠ 0

When determining the nature of stationary points it is helpful to complete a ‘gradient table’, which shows the sign of the gradient either side of any stationary points. This is known as the first derivative test. Gradient tables are demonstrated in the examples that follow.

Worked exAmple 3

eBook plus

a Find the stationary points and their nature for the function b c d e

f (x) = x3 + 5x2 − 8x − 12. Show that the curve es through (−1, 0). Find the coordinates of all other intercepts. Hence, sketch the graph of f (x). Use a CAS calculator to find the stationary points.

Think a

378

Tutorial


WriTe/drAW a f (x) = x3 + 5x2 − 8x − 12

1

Write the rule for f (x).

2

Differentiate f (x) to find f ′(x).

f ′(x) = 3x2 + 10x − 8

3

Find all values of x where f ′(x) = 0.

For stationary points 3x2 + 10x − 8 = 0 (3x − 2)(x + 4) = 0 x = 23 or x = −4


4

2

2

2

2

2

When x = 3 , f ( 3 ) = ( 3 )3 + 5( 3 )2 − 8( 3 ) − 12

Find the value of f (x) for each value of x where f ′(x) = 0.

22

= −14 27 so ( 23 , −14 22 ) is one stationary point. 27 When x = −4, f (−4) = (−4)3 + 5(−4)2 − 8(−4) − 12 = 36 so (-4, 36) is another stationary point.

5

6

b

c

d

Complete a gradient table to determine the nature of the stationary points.

To find the x-intercepts, factorise f (x) by long division, or by another appropriate method, knowing that (x + 1) is a factor of f (x).

x

−5

−4

0

2 3

1

f ′(x)

+

0

−

0

+

Slope

/

—

\

—

/

Therefore (−4, 36) is a local maximum stationary point and ( 23 , −14 22 ) is a local minimum stationary 27 point.

State each stationary point and its nature.

Show that f (−1) = 0.

1

Gradient table:

b f (−1) = (−1)3 + 5(−1)2 − 8(−1) − 12

= −1 + 5 + 8 − 12 =0 Therefore f (x) es through (−1, 0).

c As f (−1) = 0 then (x + 1) is a factor of f (x)

and f (x) = (x + 1)(x2 + 4x − 12) = (x + 1)(x + 6)(x − 2)

2

Solve f (x) = 0.

x-intercepts: (x + 1)(x + 6)(x − 2) = 0 x = −1 or −6 or 2

3

State the coordinates of the x-intercepts.

The x-intercepts are (−1, 0), (−6, 0) and (2, 0).

4

Evaluate f (0) to determine the y-intercept.

5

State the coordinate of the y-intercept.

The y-intercept is (0, −12).

Sketch the graph of f (x) showing all intercepts and stationary points.

f (0) = (0)3 + 5(0)2 − 8(0) − 12 = −12

d (−4, 36)

f(x)

y

(−6, 0) (−1, 0)

0

(2, 0) x

(0, −12) —) ( 2–3 , −14 22 27


379

e

1

To find the stationary points with a CAS calculator, on a Calculator page: • define the function f (x) • find f ′(x) • solve f ′(x) = 0 • find the y-values of the stationary points. To do this, complete the entry lines as: Define f (x) = x3 + 5x2 − 8x − 12

e

d ( f ( x)) dx solve(3x2 + 10x − 8 = 0,x)  2  f   − 4,  3   Press ENTER · after each entry. 2

 ∴ The stationary points are (−4, 36) and  23 , 

Write down the coordinates of the stationary points.

− 400  27

 .

Worked Example 4

Sketch the graph of g(x) = x2(4 − x2), clearly indicating all stationary points and intercepts. Think

Write/draw

1

Write the rule for g(x).

g(x) = x2(4 − x2)

2

Expand g(x) to make it easier to differentiate.

g(x) = 4x2 − x4

3

Differentiate g(x).

g′(x) = 8x − 4x3

4

Solve g′(x) = 0.

For stationary points, g′(x) = 0 8x − 4x3 = 0 4x(2 − x2) = 0 x = 0 or x2 = 2 x = 0 or x = − 2 or 2

5

Find g(x) for each value of x where g′(x) = 0.

When x = 0, g(0) = 0 When x = − 2 , g(− 2 ) = 4(− 2 )2 − (− 2 )4 =4 When x = 2, g( 2) = 4( 2)2 − ( 2)4 =4 Therefore the stationary points are (− 2 , 4), (0, 0) and ( 2, 4).

6

380


Gradient table: x

−2

g′(x)

+

Slope

/


−

−1

0

1

0

−

0

+

0

−

—

\

—

/

—

\

2

2

2

7

State the stationary points and their nature.

Therefore (− 2 , 4) is a local maximum stationary point. (0, 0) is a local minimum stationary point. ( 2, 4) is a local maximum stationary point.

8

Solve g(x) = 0 to determine the x-intercepts.

x-intercepts: When g(x) = 0, x2(4 − x2) = 0 x2 = 0 or x2 = 4 x = 0 or x = −2 or 2 The x-intercepts are (−2, 0), (0, 0) and (2, 0).

9

Find g(0) to determine the y-intercept.

y-intercept: When x = 0, g(0) = 02(4 − 02) =0 The y-intercept is (0, 0).

10

Sketch the graph of g(x).

(− 2, 4)

y

( 2, 4)

(−2, 0) (0, 0) 0

(2, 0)

x g(x)

Worked Example 5

If f (x) = x3 + 4x2 − 3x − 7: a sketch the graph of f ′(x) b state the values of x where f (x) is i increasing and i i decreasing. Think a

Write/draw a f (x) = x3 + 4x2 − 3x − 7

1

Write the rule for f (x).

2

Differentiate f (x) to find f ′(x).

f ′(x) = 3x2 + 8x − 3

3

Solve f ′(x) = 0 to find the x-intercepts of f ′(x).

x-intercepts: When f ′(x) = 0, 3x2 + 8x − 3 = 0 (3x − 1)(x + 3) = 0 x = 13 or −3 The x-intercepts of f ′(x) are ( 13 , 0) and (−3, 0).

4

Evaluate f ′(0) to find the y-intercept of f ′(x).

5

Sketch the graph of f ′(x) (an upright parabola).

y-intercept: When x = 0, f ′(0) = −3 so the y-intercept of f ′(x) is (0, −3). y

−3

0

f '(x)

1– 3

x

−3


381

b

i

f (x) increases where f ′(x) > 0. By inspecting the graph of f ′(x) deduce where f ′(x) is positive (that is, above the x-axis).

b

i f ′(x) > 0 where x < −3 and x >

so f (x) is increasing where x < −3 and x > 13 .

i i f ′(x) < 0 where −3 < x <

ii f (x) decreases where f ′(x) < 0.

By inspecting the graph of f ′(x), deduce where f ′(x) is negative (that is, below the x-axis).

1 3

1 3

so f (x) is decreasing where −3 < x < 13 .

Worked Example 6

Consider the function f ( x ) = ( x + 3)( x − a)2 where a is a positive real constant. a Find f ′(x). b Find the coordinates of the stationary points. c Determine the nature of the stationary points. d Find the value of a in exact form if the straight line with equation y = 5x + 15 intersects y = f (x) at the maximum turning point. Think a

b

382

Write

1

Store the function to f (x). On a Calculator page, press: • MENU b • 1:Actions 1 • 1:Define 1 Complete the entry line as: Define f (x) = (x + 3)(x − a)2 Press ENTER ·.

2

On the same Calculator page, select the derivative template: Complete the entry line as: d ( f ( x)) dx Press ENTER ·.

3

Write the derivative with the correct notation.

1

Solve f ′(x) = 0 to find the x-values of each stationary point. On the same Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve((x + 3)(x − a)2 = 0, x) Press ENTER ·.

a

∴ f ′(x) = (x − a)(3x − a + 6) b


2

 a − 6 4(a + 3)3  ∴ and (a, 0) are the , 27   3

Substitute each value of x into f (x) to find the y-coordinates of the stationary points. On the same Calculator page, complete the entry line as:

coordinates of the stationary points.

  a − 6  f  , a  .  3  Press ENTER ·. Write the stationary points in coordinate form. c

1

To determine the nature of the stationary points, let a = 1 (as a is given as a positive real constant).

c  1 − 6 4(1 + 3)3 

 3 ,

 − 5 256  ∴ , and (1, 0) are the stationary points.  3 27 

Substitute a = 1 into the answer found previously to obtain the stationary points. 2

3

4

d


State the stationary points and their nature.

The maximum turning point lies on the straight line y = 5x + 15. Substitute the coordinates of the turning point into y = 5x + 15 and solve for a.

2

On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as:

x

−2

f ′(x)

+

Slope

/

−5

0

1

2

0

−

0

+

—

\

—

/

3

(1, 0) is a local minimum turning point.  − 5 256   3 , 27  is a local maximum turning point. ∴ (a, 0) is a local minimum turning point and

State the stationary points and their nature in of a.

1

 and (1, 0)

27

 a − 6 4(a + 3)3  is a local maximum turning point.  3 , 27  d

 4 ( a + 3)3 5(a − 6)  solve = + 15, a 27 3   Press ENTER ·. 3

Select the appropriate value of a given that it is positive.

∴a=

3

(

5−2

)

2


383

1. A stationary point (SP) occurs when f ′(x) = 0. 2. There are four types of stationary points: (a) local maximum, where the gradient is positive on the left of the SP and negative on the right (b) local minimum, where the gradient is negative on the left of the SP and positive on the right (c) positive point of inflection, where the gradient is positive on both sides of the SP (d) negative point of inflection, where the gradient is negative on both sides of the SP. Exercise

8B

Sketching curves 1 WE3 Find the stationary points and their nature for each of the following functions. a y = 8 − x2 b f (x) = x3 − 3x c g(x) = 2x2 − 8x 2 3 3 4 d f (x) = 4x − 2x − x e g(x) = 4x − 3x f y = x2(x + 3) 2 3 g y = 5 − 6x + x h f (x) = x + 8 i y = −x2 − x + 6 4 3 2 2 j y = 3x − 8x + 6x + 5 k g(x) = x(x − 27) l y = x3 + 4x2 − 3x − 2 3 3 m h(x) = 12 − x n g(x) = x (x − 4) 2 WE3 Sketch the graph of each function in question 1 , clearly indicating all stationary points. 3 a WE3 Find the stationary points of the function f (x) = x3 − 2x2 − 7x − 4 and state their nature. b Show that the graph es through (4, 0). c Give the coordinates of all other intercepts and hence sketch the graph of f (x). 4 a Find the stationary points, and their nature, for the curve y = x3 − x2 − 16x + 16. b Show that the graph es through (1, 0) and give the coordinates of all other intercepts. c Sketch the graph. 5 a WE3 Find the stationary points of the function g(x) = x4 − 4x2 and state their nature. b Find the coordinates of all the intercepts. c Sketch the graph of g(x). 6 a b c d

If y = x4 − 6x2 + 8x − 3, find each stationary point and its nature. Show that the point (1, 0) lies on the curve. Find all other intercepts. Sketch the graph.

7 a If y = x4 + x3 − 5x2 − 6, find each stationary point and its nature. b Find the y-intercept. c Sketch the graph without finding the x-intercepts. 8 WE4 Sketch the graph of each of the following functions, clearly indicating all stationary points and intercepts. a f (x) = x4 − x2 b f (x) = x3 − 3x2 3 4 c g(x) = x + 3x d g(x) = x3 − 4x2 + 4x 3 2 e h(x) = x − 4x − 11x + 30 f h(x) = x(x + 3)(x − 5) g f (x) = x4 − 2x2 + 1 h f (x) = x(x2 + 1) 3 2 i g(x) = x + 9x + 24x + 20 j h(x) = (x2 − 1)3 9 MC If f ′(x) < 0 where x > 2 and f ′(x) > 0 where x < 2, then at x = 2, f (x) has a: A local minimum B local maximum c point of inflection D discontinuous point e gradient of 2 384


10 mC The function f (x) = x3 + x2 − 8x − 3 has stationary points when x is equal to: A −2 and

4 3

c −2 only

B 3 only

D 3 and

4 3

e 0 and 2

11 mC The graph of y = + has: A a local maximum where x = 0 B a local minimum where x = 0 x4

x3

−3 4

c a local minimum where x =

−3 4

D a local maximum where x =

−4 3

e a local maximum where x =

12 mC A quadratic function has a turning point (2, 1) and a y-intercept of (0, 9). The equation must be: A y = (x − 2)2 + 9 B y = (x − 1)2 + 8 c y = (x − 2)2 + 1 2 2 D y = 2(x − 2) + 9 e y = 2(x − 2) + 1 13 The graphs of f ′(x) are shown below. Find all values of x for which f (x) has stationary points and state their nature. y y y a b c

f '(x)

f '(x)

x

0

−3

−2

0

1

−2

x

4

0

3

x

f'(x) y

d

−5

0

2

y

e

f '(x)

x

−3

f

0

x

2

y f '(x)

0 1

5

x

f'(x)

14 Show that f (x) = x2 − 4x + 3 is decreasing for x < 2 and increasing for x > 2. 15 We5 For each of the following functions i sketch f ′(x) and, hence, state the values of x where f (x) is ii increasing and iii decreasing. a f (x) = 13 x3 + 2x2 + 2 b g(x) = x3 + 2x2 − 7x − 5 c h(x) = x4 + 4x3 + 4x2 16 If y = f (x) has the following properties then sketch its graph. f ′(x) = 0 if x = −2 and x = 3 f ′(x) < 0 if −2 < x < 3 f ′(x) > 0 for all other x −

17 If f (x) = x3 + ax2 + bx has stationary points at x = 2 and x = 43 , find: a the value of a and b b the nature of the stationary points. eBook plus Digital doc

SkillSHEET 8.1 Review of derivatives other than polynomials

18 If f (x) = x4 + ax2 + b has a stationary point (1, 4), find: a a and b b the other stationary points c the nature of each stationary point. 19 We6 Consider the function f ( x ) = (b − x )( x + 2)2 where b is a positive real constant. a Find f ′(x). b Find the coordinates of the stationary points. c State the nature of the stationary points. d Find the equation of the tangent at x = 2. e If the maximum turning point occurs at x = 4, find the value of b.

Chapter 8


385

20 WE6 Consider the function f ( x ) = ( x + b)2 (2 x − 1) where b is a positive real constant. a Find f ′(x). b Find the coordinates of the stationary points. c Find the equation of the tangent to the curve when x = 2. d Find the value of b in exact form if the straight line with equation y = 4x − 2 intersects y = f (x) at the minimum turning point.

8c

Maximum and minimum problems when the function is known In many practical situations the maximum local or minimum value of a quantity is desired. y maximum maximum For example it is important for manufacturers or business operators to minimise the costs involved in running their businesses. Equally, it is just as important to maximise their profits. local A graph is always useful and helps us to find minimum x 0 approximately where a maximum or minimum minimum occurs. As we have seen, local maximum and/or minimum stationary points occur where the derivative is zero. dP If P = f (x) then a local maximum and/or minimum may exist where = 0. dx To decide whether a solution is a maximum or a minimum the first derivative test must be applied (by setting up a gradient table). A graph of the function could be sketched if it is not difficult, to ensure that the maximum or minimum value is applicable. y f(x)

0

1 2 3 4 5

x

The function f (x) has a limited domain of [1, 5]. The maximum value occurs at the point where x = 4. The minimum value, however, is not at x = 2 but at the end point x = 5. This example shows that the derivative test on its own is not always reliable for finding maximum or minimum values and a graph is sometimes necessary.

Solving maximum and minimum problems When finding the maximum/minimum value of f (x) the following steps are taken. 1. Find f ′(x) (to obtain the gradient function). 2. Solve for x where f ′(x) = 0 (to find the values of x where the maximum or minimum occur). 3. Apply the first derivative test as a check. Also, sketch the graph of f (x). 4. Check end points if domain is restricted. 5. Substitute the appropriate value of x into f (x) to obtain the maximum or minimum.

386


Worked exAmple 7

The population of a colony of birds at any time, t months, after observation began can be modelled by the function: −t

P(t) = 400 te 5 + 600, where P is the number of birds. Find: a the initial population b when the largest number of birds is reached c the maximum number of birds. Think a

b

WriTe

1

Write the rule for P(t).

2

The initial value occurs when t = 0 and is P(0).

1

Find P′(t) using the product rule.

a P(t) = 400te

P(0) = + 600 = 0 + 600 = 600 So the initial population of birds is 600. b P′(t) = 400te

= Solve P′(t) = 0.

+ 400t

( )

−t −t 400 e 5 − 80te 5 −t 80 e 5 (5 − t)

−t 5

−1 5

e

−t 5

+0

(5 − t) = 0

t = 5 (as 80 e Check that it is a maximum using the first derivative test.

−t 5

For maximum and/or minimum, P′(t) = 0 80 e

3

+ 600

400(0)e0

= 2

−t 5

−t 5

cannot equal 0)

Gradient table: t

0

5

10

P′(t)

+

0

−

Slope

/

—

\

Therefore the maximum population occurs after 5 months. c

Evaluate P(5) to find the maximum number of birds.

c P(5) = 400(5)e

−5 5

+ 600 − = 2000e 1 + 600 = 735.8 + 600 = 1335.8 Therefore the maximum number of birds is approximately 1335 (as it does not reach 1336).

Worked exAmple 8

eBook plus

The displacement of a particle moving in a straight line from the origin at any time, t, is given by x(t) = 13 t3 − 4t2 + 12 t + 1, 0 ≤ t ≤ 7. Find the maximum and minimum displacement.

Chapter 8

Tutorial



387

Think

Write

t3 − 4t2 + 12t + 1 3

1

Write the rule.

x(t) =

2

Find x′(t).

x′(t) = t2 − 8t + 12

3

Solve x′(t) = 0.

For maximum and/or minimum, t2 − 8t + 12 = 0 (t − 2)(t − 6) = 0 t = 2 or t = 6

4

Test for maximum and minimum using the first derivative test.

Gradient table: x

0

2

4

6

8

x′(t)

+

0

−

0

+

Slope

/

—

\

—

/

When t = 2 a local maximum occurs and when t = 6 a local minimum occurs. 5

Evaluate x(2) to find the local maximum.

x(2) = 13 (2)3 − 4(2)2 + 12(2) + 1

= 83 − 16 + 24 + 1

= 11 3

2 2

so a local maximum occurs at (2, 11 3 ). 6

Evaluate x(6) to find the local minimum.

x(6) = 13 (6)3 − 4(6)2 + 12(6) + 1 = 72 − 144 + 72 + 1 =1 so a local minimum occurs at (6, 1).

7

8

Sketch the graph of x(t) over the domain [0, 7] to test that the maximum and minimum have been found.

Evaluate x(0) and x(7) to see if the end points give a smaller or larger value.

This graph indicates that we have to find the value of x at the end points in case they produce larger or smaller values than the stationary points do.

x

0

x(t)

1 2 3 4 5 6 7 t

x(0) = 1 which is equal to the local minimum above x(7) = 13 (7)3 − 4(7)2 + 12(7) + 1

1

= 114 3 − 196 + 84 + 1 1

= 3 3 which is between the local maximum and minimum above. Therefore the maximum displacement is 2 11 3 units from the origin and the minimum displacement is 1 unit from the origin.

388


1. To find a maximum and/or minimum, let f ′(x) = 0. 2. If the function has a restricted domain, f ′(x) may not be useful for finding maximum and/or minimum values. 3. You can prove your maximum and/or minimum value using a slope or gradient diagram. Exercise

8c

Maximum and minimum problems when the function is known 1

For each of the following, write the expression for the derivative indicated. a C = 5x − 2x2 + 10,

dC dx

dV dh dM e M = k loge 3k, dk 1

2

c V = 4 h4 − 3 h3,

b P = n3 + 2n2 − 5 n, d h = t3 + 2e−t, f L = t2 +

dP dn

dh dt

4 dL − 2 cos (πt), t dt

Questions 2 and 3 refer to the following information. The function Q = n3 − 3n2 + 5, 0 ≤ n ≤ 5, is graphed at right.

Q

2 MC The minimum value occurs where n equals: A −2 B 0 c 1 d 2 e 5 3 MC The maximum value occurs where n is equal to: A 0 b 2 c 5 d 1 e −1

0

2

5

n

4 WE7 The profit, $P per week, of a small manufacturing company is related to the number of workers, n, by: P = − 12 n3 + 96n + 600 Find: a the number of workers needed for maximum profit per week b he maximum profit per week. 5

The cost, $C, of producing x-metre lengths of a certain tow rope is: C = 15 x2 − 8x + 100 per rope, x > 0 What is the length of the cheapest tow rope that can be produced?

6 WE8 The number of people, P, visiting a certain beach on a particular day in January depends on the number of hours, x, that the temperature is below 30 °C according to the rule P = x3 − 12x2 + 21x + 105 where x ≥ 0. Find the value of x for the maximum and minimum number of people who visit the beach. 7 The number of rabbits, N, which feed on a particular farmland on any night can be modelled − as N = 13 x3 + 5x2 + 75x + 500 where x is the average overnight temperature in °C and x ≥ −5. Find: a the temperature for the minimum and maximum number of rabbits b the minimum and maximum number of rabbits that will be found on the farmland.


389

8 The velocity, v cm/s, of an engine’s piston t seconds after the engine is started is approximated by v = 0.8 sin (2t). a Find the minimum velocity and the time it first occurs. b Find the maximum velocity and the time it first occurs. 9 The length of a snake, L cm, at time t weeks after it is born is modelled as:  πt  L = 12 + 6t + 2 sin   , 0 ≤ t ≤ 20  4 Find: a the length at i birth and ii 20 weeks b R, the rate of growth, at any time, t c the maximum and minimum growth rate. 10 The population of cheetahs, P, in a national park in Africa since 1 January 1986 can be −t

modelled as N = 100te 12 + 500 where t is the number of years. a When does this model predict that the maximum population will be reached? b What is the maximum population of cheetahs that will be reached? c How many cheetahs will there be on 1 January in i 2010 and ii 2070? 11 In the same park as in question 10 the number of elephants, N, since 1 January 1986 is − t

modelled as N = 100 + 4t + 400 e 10 . Find: a the minimum number of elephants predicted b when this minimum will occur. 12 The profit, $P, per item that a store makes by selling n items of a certain type each day is P = 40 n + 25 − 200 − 2 n. a Find the number of items that need to be sold to maximise the profit on each item. b What is the maximum profit per item? c Hence, find the total profit per day by selling this number of items. 13 The weight (in kg) of a bodybuilder t months after starting a training program is W = 5t − 20 loge (t + 1) + 90, 0 ≤ t ≤ 15 a Find the weight of the bodybuilder at the start of the program. b Find the minimum weight obtained and how many months it takes to reach it.

8d

Maximum and minimum problems when the function is unknown When solving maximum and/or minimum problems when the function is not given directly, a rule for the function must be obtained from the given information. This rule should have the quantity being maximised or minimised in of one variable only. Sometimes a diagram will assist in establishing the rule. Then solve the problem using differentiation. These steps should be followed. 1. Draw a diagram if appropriate. 2. Identify the quantity to be maximised or minimised. 3. Express this quantity in of one variable only. 4. Solve f ′(x) = 0. 5. it is a maximum or minimum using the first derivative test. 6. Sketch a graph to confirm the maximum or minimum found. 7. Answer the question that has been asked.

390


The following formulas may be useful in problem solving: Area of circle Curved surface area of a cylinder Surface area of a sphere

A = π r2 S = 2π rh S = 4π r2

Volume of a sphere

V = 3 πr3

Volume of a cylinder

V = π r2h

Volume of a cone

V = 13 π r 2 h

4

1

V = 3 Ah , where A is the area of the base

Volume of a right pyramid

d = ( x2 − x1 )2 + ( y2 − y1 )2

Distance between two points

Worked Example 9

The sum of two positive numbers is 10. Find the numbers if the sum of their squares is a minimum. Think

Write

1

Let one number be x and the other number be y. Form an equation.

Let x = 1st number and y = 2nd number. x + y = 10

2

Express y in of x.

So y = 10 − x

3

Write an expression for S(x), the sum of the squares of x and y, in of x only.

S(x) = x2 + y2 = x2 + (10 − x)2

4

Simplify the equation.

5

Find S′(x).

S′(x) = 4x − 20 Function has a stationary point when S′(x) = 0

6

Find x for minimum S by solving S′(x) = 0.

4x − 20 = 0 4x = 20 x=5

7

that it is a minimum by the first derivative test.

Gradient table:

= x2 + 100 − 20x + x2 = 2x2 − 20x + 100

x

4

5

6

P′(t)

−

0

+

Slope

\

—

/

So x = 5 gives a minimum for S. 8

Find y.

When x = 5, y = 10 − 5 =5

9

State the two numbers.

Therefore, the two numbers which give a minimum of their squares are both 5.

Note: The actual sum was not asked for in this example.


391

Worked exAmple 10

eBook plus

A cuboid container with a base length twice its width is to be made with 48 m2 of metal. 8 2x a Show that the height is given by the expression h = − , x 3 where x is the width of the base. b Express the volume, V, in of x. c Find the maximum volume. Think a

1

Tutorial


WriTe

Draw a diagram of a cuboid.

a

h

2x x

2

Let x be the width of the base and hence express length in of x.

Let x = width and h = height so length = 2x

3

Calculate the total surface area (TSA) of the cuboid in of x and h only.

TSA = 2[2x(x) + 2x(h) + x(h)] = 2(2x2 + 3xh) = 4x2 + 6xh

4

Express h in of x.

As TSA = 48 m2 4x2 + 6xh = 48 6xh = 48 − 4x2 h=

48 − 4 x 2 6x

=

48 4 x 2 − 6x 6x

h= b

1

Find the volume, V, in of x and h.

2

Express the volume in of x by substituting for h.

b V = x(2x)h

 8 2x  V(x) = 2x2 −  x 3  = 16x −

c

1

2

Solve V′(x) = 0.

that x = 2 gives a maximum.

8 2x − x 3

4 x3 3

c V′(x) = 16 − 4x2 = 0 for maximum or minimum

4x2 = 16 x2 = 4 x = 2 or −2 (reject −2 as width cannot be negative)

Gradient table: x

1

2

3

V′(x)

+

0

−

Slope

/

—

\

The maximum volume is achieved when x = 2 m. 392


3

Substitute x = 2 into the rule for V(x) to obtain the maximum volume.

V(2) = 16(2) −

4(2)3 3

32 3

= 32 −

= 32 − 10 23

= 21 13 1

Therefore the maximum volume is 21 3 m3.

Worked Example 11 y

Find the minimum distance from the straight line with equation y = x − 4 to the point (1, 1).

1

P (1, 1)

0 1 −4 Think

y=x−4 x

4 Q (x, y)

Write

1

The minimum distance between a straight line and a point is a perpendicular line from a point on the straight line, Q (x, y) to the point P (1, 1).

Let Q be the point on the line with coordinates (x, y).

2

From the given rule for the straight line, find y in of x.

As Q is on the line y = x − 4 then Q is (x, x − 4).

3

Find the distance, d(x) between P and Q in of x only using the formula for the distance between 2 points.

d(x) = ( x2 − x1 )2 + ( y2 − y1 )2

4

Differentiate d(x) using the chain rule (or differentiate the expression inside the square root only, as the smallest value of this expression will give the minimum distance when the square root is taken).

= ( x − 1)2 + ( x − 4 − 1)2

= ( x − 1)2 + ( x − 5)2

= x 2 − 2 x + 1 + x 2 − 10 x + 25

= 2 x 2 − 12 x + 26

1 d (2 x 2 − 12 x + 26) 2 dx

= 12 × (4x − 12) × =

1 1

(2 x 2 − 12 x + 26) 2

4 x − 12 2 2 x 2 − 12 x + 26

OR d (2 x 2 − 12 x + 26) dx = 4x − 12


393

5

Solve for x where the derivative equals zero.

For maximum or minimum, 4 x − 12 =0 2 2 x 2 − 12 x + 26 4x − 12 = 0 4x = 12 x=3

6

that x = 3 gives a minimum.

Gradient table: x

2

3

4

Derivative (4x − 12)

−

0

+

Slope

\

−

/

So x = 3 gives the minimum distance. 7

Evaluate d(3) to obtain the minimum distance. The exact answer is 8 and 2.828 is an approximate answer.

d(3) = 2(3)2 − 12(3) + 26

d = 18 − 36 + 26

d= 8=2 2

or

≈ 2.828

Therefore the minimum distance is 2 2 units or approximately 2.828 units.

1. Express the quantity being maximised or minimised in of one variable only. 2. Find the derivative and let it equal zero to find the maximum or minimum. 3. Test for maximum and/or minimum using the first derivative test (gradient table). 4. Use exact answers where appropriate. If using an approximate answer show by use of the ≈ symbol. 5. Ensure you have answered the question asked. Exercise

8d

Maximum and minimum problems when the function is unknown 1 WE9 The sum of two positive numbers is 10. Find the numbers if their product is a maximum. 2 The sum of two positive numbers is 8. Find the numbers if the sum of the cube of one and the square of the other is a minimum. 3 A rectangular frame is to be made from a piece of wire 120 cm long. a If the width is x cm, show that the length is 60 − x. b Find an expression for the area of the rectangle in of x. c Hence, find the dimensions for maximum area. d Find the maximum area. 4 Find the area of the largest rectangular paddock that can be enclosed with 400 metres of fencing.

394


5 Cylindrical, cardboard postal tubes are made with the restriction that the sum of the length and the circular circumference are 120 cm. What should the dimensions be for maximum volume?

Length

Circumference

6 The frame of a container in the shape of a cuboid is shown at right. If it is to be made with a total length of 18 metres of steel edging, find: a the value of L in of x 2x b the expression for the volume in of x only c the length of each edge for maximum volume d the maximum volume.

L

x

7 WE10 A cuboid with a square base is to be made with 200 cm2 of material. 50 x − , where x is the side length of the base. x 2 b Express the volume, V, in of x. c Find the maximum volume (to the nearest unit).

a Show that the height, h =

8 A window frame is in the shape of a semicircle ed to a rectangle. Find the maximum area of a window using 300 cm of framework. 9 A rectangular sheet of cardboard, 50 cm by 40 cm, is to have square corners cut out so it can be folded into a rectangular tray. Find the maximum volume possible for such a tray.

10 A bushwalker can walk at 5 km/h through clear land and 3 km/h through bushland. If she has to get from point A to point B following a route indicated at right, find the value of x so that the route is covered in a minimum time. (Note: time =

50 cm 40 cm

B Clear

Bush 3 km

distance ) speed

x

A

11 The cost of running a train at a constant speed of v km/h is C = 50 +

2 km

v2 dollars/hour. 1000

a Find the time taken for an 800 km journey in of v. b Hence, find an expression for the cost of an 800 km journey. c Find the most economical speed for this journey. 12 Find the side length of the largest cube which can fit inside a sphere of diameter 24 cm.

24 cm

x


395

13 A cylinder of cheese is to be removed from a spherical piece of cheese of radius 8 cm. What is the maximum volume of the cylinder of cheese? (Express the answer to the nearest unit.) 8 cm

y

14 We11 Find the minimum distance from the line y = 2x + 3 to the point (1, 0).

y = 2x + 3 3

Minimum distance (1, 0)

−2

0 y


WorkSHEET 8.2

x

1 y = x2

15 Find the minimum distance from the parabola y = to the point (5, 0). (Express the answer to the nearest hundredth.) x2

(5, 0) 5 x

0

8e

rates of change

eBook plus

P (x1, f (x1)) and Q (x2, f (x2)) are two points on the function with rule y = f (x) as shown in the diagram.

Interactivity

int-0253 Rates of change

y y = f(x) Q (x2, f(x2))

P (x1, f(x1)) 0

x1

x2

x

The average rate of change of y with respect to x over the interval x ∈ [ x1, x2] is equal to the gradient of the straight line (or chord) PQ. Average rate of change =

change in f ( x ) change in x

f ( x2 ) − f ( x1 ) x2 − x1 The instantaneous rate of change is the rate of change at a specific point. dy The instantaneous rate of change of y with respect to x is given by the derivative . dx dy If > 0, then y is increasing as x increases (gradient is positive). dx dy If < 0, then y is decreasing as x increases (gradient is negative). dx =

396


Note: Rates of change are often calculated with respect to time, but not always. If you are required to find the rate of change with respect to some quantity other than time then the quantity must be stated. If this quantity is not stated then the rate of change is taken as being with respect to time. Worked exAmple 12 a Find the rate of change of the surface area of a melting ice cube with respect to its side length (x). b What is the rate of change when x = 2 cm? (Assume that the ice cube remains in the shape of

a cube.)

Think a

b

WriTe

1

Express the surface area, S, of the cube in of its side length, x.

2

Find

3

Place a negative sign in front of the rate as the ice cube is melting (that is, the rate is decreasing). dS Substitute x = 2 into . dx

1

2

dS . dx

State the solution.

a S = 6x2 (total surface area of a cube)

dS = 12x dx But

dS − = 12x because the surface area is decreasing. dx

b When

x=2 dS − = 12(2) dx = −24

Therefore, the rate of change of the surface area when x = 2 cm is −24 cm2/cm (it is decreasing at a rate of 24 cm2/cm).

Worked exAmple 13

eBook plus

The number of mosquitoes, N, around a dam on a certain night can be modelled by the equation N = 100 loge (2t + 1) + 5t + 1000

Tutorial


where t equals hours after sunset. Find: the initial number of mosquitoes the average rate of change in the first 4 hours the rate of change at any time, t the rate of change when t = 4 hours.

a b c d

Think a

WriTe

1

Write the rule.

2

Find N when t = 0.

a N = 100 loge (2t + 1) + 5t + 1000

When t = 0, N = 100 loge (2 × 0 + 1) + 5 × 0 + 1000 = 100 loge 1 + 1000 = 100 × 0 + 1000 = 1000 The initial number of mosquitoes is 1000.

Chapter 8


397

b

1

2

Find N when t = 4.

b When t = 4,

N = 100 loge (2 × 4 + 1) + 5 × 4 + 1000 = 100 loge (9) + 20 + 1000 = 1020 + 100 loge (9) = 1020 + 219.7 = 1239.7 that is, 1239 mosquitoes (as the number has not reached 1240).

Calculate the average rate between t = 0 and t = 4 using

The average rate of change in the first 4 hours 1239 − 1000 = 4−0

f ( x2 ) − f ( x1 ) . x2 − x1

239 4 = 59.75 mosquitoes per hour. =

c Differentiate N with respect to t, to find

dN . dt

c

dN 100 × 2 = +5 dt 2t + 1 =

d Find

dN when t = 4. dt

200 +5 2t + 1

d When t = 4 the rate of change is:

dN 200 = +5 dt 2 × 4 × 1 =

200 9

+5

2

= 22 9 + 5 2

= 27 9 mosquitoes per hour.

1. The average rate of change is the gradient of a straight line between two points. f ( x2 ) − f ( x1 ) . x2 − x1 3. The instantaneous rate of change at any point is given by the value of the derivative at that point. 2. The average rate of change =

exerCiSe

8e

rates of change In the following exercise, use a CAS calculator to assist with any graphing. 1 Express the following in simplest mathematical notation. a The rate of change of volume, V, with respect to radius, r. b The rate of change of surface area, S, with respect to height, h. c The rate of change of area, A, with respect to time, t.

398



SkillSHEET 8.2 Review of rates of change

d The rate of change of cost, C, with respect to distance, x. e The rate of change of intensity, I, with respect to pressure p. f The rate of change of velocity. 2 WE12 a Find the rate of change of the area, A = π r2, of an increasing circular oil spill with respect to the radius, r. b What is the rate of change when r = 10 metres? 3 a Find the rate of change of the volume, V = 43 π r3, of a deflating spherical balloon with respect to the radius, r. b Hence, find the rate when r = 5 cm. 4 A sugar cube dissolves in a cup of tea. Find the rate of decrease of its surface area, S, when the side length, x, is 0.5 cm. 5 The height of a projectile t seconds after being fired is h = 10 + 20t − 5t2 metres. a Find the rate of change of height after i 1 second and ii 3 seconds. b Explain the difference between these two answers. 6 The volume of water (in litres) which has flowed through a swimming pool filter t minutes t4 1 (30t3 − ) where 0 ≤ t ≤ 90. after starting it is V = 100 4 a At what rate is water flowing through the filter at any time, t? dV over the domain [0, 90]. b Sketch a graph of dt c When is the rate of flow greatest? 7 A particle moves in a straight line so that its displacement from a point, O, at any time, t, is x = 3t 2 + 4 . Find: a the velocity as a function of time b the acceleration as a function of time c the velocity and acceleration when t = 2. 8 If a particle is moving in a straight line so that its displacement from the origin at any time, t, is x = t3 − 12t2 + 36t, find: a the velocity b the time and displacement when the velocity is zero c the acceleration when the velocity is zero. 9 WE13 The number of people with the flu virus, N, in a particular town t days after a vaccine is introduced is N = 3000 − 500 loge (8t + 1). a How many people are infected in the town before the vaccine is introduced? b Find the average rate of change over the first 5 days. c Find the rate of change of the number of people in the town infected with flu. d Find the rate of change after 5 days. 10 MC If V = t3 − 2t2 + 8 t , where V is in m3 and t is in hours a the rate of change of V with respect to t when t = 4 is: B 32 m3/h c 28 m3/h d 34 m3/h e −20 m3/h A 48 m3/h b the average rate of change from t = 1 to t = 4 is: d −10 13 m3/h e 0 m3/h A 13 23 m3/h B 10 13 m3/h c 3 m3/h 11 MC Temperature, T °C, is related to height, h metres, by T = h2 + 4eh. a The rate of change, °C/m, at h = 0 is: A 0 b 4 c 4e d −4e e 1 Chapter 8 Applications of differentiation

399

b The rate of change, °C/m, at h = 4 is: A 16 + 4e4 b 16 + e4 D 8 + e4 e 8 + 16e4

c 8 + 4e4

c The average rate of change between h = 0 and h = 4 in °C/m is: A 1 + e4 b e4 c 12 + 4e4 D 4 + e4 e 3 + e4 12 The height of water (in metres) at the entrance to a bay t hours after high tide is:  πt  H = 10 + 2 cos    12 

Find: a the rate of change of H at any time, t b the rate of change of H: i 6 hours ii 15 hours iii 20 hours after high tide c the minimum and maximum value of H and the time when they first occur after the initial high tide. 13 The value of an antique ornament t years after being purchased is A = 2000 e 0.1 t dollars. a Find the value of the antique i when purchased and ii 3 years after purchase. b Hence, find the average rate of change of value in the 3 years since purchased. c Find the rate of change of value 3 years after purchase. 20 14 The amount of chlorine in a jug of water t hours after it was filled from a tap is C = , t +1 where C is in millilitres. Find the rate of decrease of chlorine 9 hours after being poured. 15 The graph below shows the price of the shares of a particular company over a period of time. Price (¢) 180 170 160 150 140 130 120 0

2

4

6

8

10 12 14 16 t (months)

Use the graph to estimate the answers to the following questions. a Estimate the share price when t = 5. b State the time interval over which the share price was decreasing. c Find the average rate of change of the share prices over the entire period shown on the graph. d Estimate the rate of change of the share price at t = 11. e What was the maximum share price over the period shown? f When was the share price at its lowest value? 400


16 The graph below shows how the velocity of a car varies over a period of time. v (m/s) 14 12 10 8 6 4 2 0

2

4

6

8

10 12 14 16

t (s)

Use the graph to estimate the answers to the following questions. a Estimate the velocity of the car at t = 12. b Estimate the acceleration of the car at t = 5. c State the time intervals over which the car is accelerating. d What was the average rate of change of velocity of the car over the interval [4, 8]?

8F

eBook plus

related rates

eLesson

eles-0094

When two variables are both functions of the third variable, we may need to Related rates use related rates to solve the problem. For example, we may need to dx dy dx dx dy find when given . In such instances we would need to use the chain rule: = × . dt dt dt dy dt When solving problems involving related rates, the following steps may be helpful: 1. Draw a diagram where appropriate. Sometimes two or more diagrams may be necessary. 2. Identify the variables. 3. Identify which rate is given and which rate is required. 4. Use the chain rule to connect the required and given rates. 5. Find an equation, or relationship, that connects the variables if not given. 6. Differentiate the equation. 7. Substitute into the chain rule and simplify. 8. Answer the question noting correct units. Tools for finding relationships: • similar triangles • Pythagoras’ theorem • right-angled triangle trigonometry • sine and cosine rules. Worked exAmple 14

eBook plus

cm3

An ice cube is melting at a constant rate of 2 per hour. At what rate is the side length of the cube changing when the side length is 1.2 cm? Think 1

We are given the rate of change of volume with respect to time. Write it using appropriate notation. Since the volume is decreasing, the rate is negative.

Tutorial


WriTe

dV − = 2 dt

Chapter 8


401

dx when x = 1.2 dt

2

State the rate of change that needs to be found.

Need to find

3

Use the chain rule to connect the rate that we need to find with the one that is given.

dx dx dV = × dt dV dt

4

We do not know dx . To find it, we need the dV rule that connects x and V, that is, the rule that connects the side length of the cube with its volume. So write the formula for the volume of the cube.

5

Differentiate.

6

Reciprocate both sides to obtain dx . dV

7

8

9

Substitute 1 2 for dx and (−2) for dv into the dV dt 3x chain rule and simplify.

Vcube = x3, where x is the length of the side.

dV = 3x2 dx dx 1 = 2 dV 3 x dx 1 = 2 × (−2) dt 3 x

=

−2

3x 2

We now have the formula that allows us to find the rate of change of the side length for any value of x. To find the rate of change when length is 1.2 cm, simply substitute 1.2 for x into this formula and evaluate.

So when x = 1.2, −2 dx = dt 3 (1.2 )2

State the answer, adding appropriate units.

When the side length of the cube is 1.2 cm, its

=

−

25 54

− length is changing at a rate of 25 cm/h. 54

Worked Example 15

An empty inverted right circular cone has a radius of 4 cm and a height of 20 cm. Water is being poured in at a constant rate of 0.1 cm3/second. Find the rate at which the depth of the water is increasing at the instant the depth is 6 cm. Give your answer correct to 2 decimal places. Think 1

Draw a diagram showing half the cross-section at any depth, h cm.

Write 4 cm

r cm

20 cm h cm

402


2

Identify the variables.

3

Identify the given rate. Identify the required rate.

4

Use the chain rule to connect the rate that we need to find with the one that is given.

5

Find a rule that connects volume, V with the depth of water, h. Use similar triangles to connect r and h.

Depth of the water, h cm Radius, r cm, of surface of the water Time, t seconds Volume, V cm3 dV = 0.1 cm3/second dt dh Find when h = 10 cm. dt dh dh dV = × dt dV dt 1 V = π r 2h 3 r 4 = h 20 h ∴ r= 5 2

Rewrite the formula with the two variables V and h.

1  h V = π  h 3  5

π 3 h 75 dV 3π 2 = h dh 75 dV π 2 = h dh 25 dh 25 = dV π h 2 dh dh dV = × dt dV dt dh 25 = × 0.1 dt π h 2 dh 2.5 = dt π h 2 dh 2.5 = dt π × 62

∴V= 6

7

8

Differentiate V with respect to h.

dh by taking reciprocals of both sides. dV Substitute into the Chain Rule and simplify. Obtain

Substitute h = 6 cm to find the required rate of change. State the answer with appropriate units.

dh = 0.02 cm/second (correct to dt 2 decimal places) ∴ The depth of water is increasing at a rate of 0.02 cm/second when the depth is 6 cm.

dx dx dy 1. To solve a problem involving related rates, use the chain rule, for example: = × . dt dy dt 2. Show appropriate units.


403

Exercise

8F

Related rates 1 WE 14 A gym fitness ball is being inflated and its volume is increasing at a constant rate of 5 cm3 per second. At what rate is the radius of the ball changing when the radius is 34 cm? 2 A particle moves along a path that can be described using the Cartesian equation y = 3x3 − 2x + 1. If

dy − dx = 3 when x = 5, find at that moment. dt dt

3 The surface area of a cube is decreasing at a constant rate of 9 cm2/s. Find the rate at which the sides of the cube are decreasing when the sides are 1.5 cm long. 4 A spherical balloon is being deflated and its radius, r cm, is decreasing at a constant rate of 5 cm/min. At what rate is its volume, V cm3, decreasing when the radius of the balloon is 4 cm? 5 A bowl is being filled with water at a rate of 12 cm3/s. The volume, V cm3, of water in the 5

bowl is given by V = 8h 2 where the depth of water in the bowl is h cm. Find the rate at which the depth of water in the bowl is increasing when the depth is 9 cm. 6 The radius of a circular puddle of water is increasing at a rate of 2.5 cm/s. Find the exact rate at which the area is increasing at the instant the radius is 12 cm. 7 MC A cylinder with radius of 4 m is being filled with water. If the rate of change of the depth 1 of the water is m/s, the rate of change of volume, in m3/s, in the cylinder is: 8π 1 1 1 A B 2 c 2 d e 2π 2 2π 108π 8 WE15 An inverted right circular cone is filled with liquid. The cone has a radius of 3 m and height of 7 m. The liquid flows from the apex of the cone at a constant rate of 0.6 m3/min. Find the rate at which the depth of the liquid is dropping, correct to 2 decimal places, when the depth of the liquid is 2 m. 9 The upper end of an 8 m ladder rests against a vertical wall with the lower end on the horizontal ground. The lower end of the ladder on the ground slips away from the wall at a rate of 6 m/s. Find the rate at which the upper end of the ladder is moving the instant the ladder is 4 m from the wall. 10 A stainless steel cylindrical tank of radius 6 m is being filled with milk at a constant rate of 1.5 m3/min. At what rate is the level of milk rising? Give your answer in of π. 11 A sand timer consists of two cones ed at the apex. Each cone has height h, radius r and an angle at the apex of 60°. a Express the radius of the top cone in of its height. Give your answer in exact form. b Write the volume of the top cone as a function of its height. c When the timer is turned over, the sand starts pouring from the top cone into the bottom 1 one at a constant rate of 32 cm3/s. Find the rate of change of the depth of the sand in the top cone when the depth is 0.8 cm.

8G

Linear approximation It is useful to be able to find how much a small change in the independent variable affects the dependent variable. For example, how will a small increase in the radius of a circle affect the area of a circle?

404


This concept can be illustrated graphically. y

y = f(x)

B δy

δy = f(x + h) − f(x)

A

0

x

δx =h

Consider the function y = f (x) as shown with the points A (x, f (x)) and B (x + h, f (x + h)). The gradient of the chord AB =

f ( x + h) − f ( x ) . If h is very small, the gradient of the chord is h

very close to the gradient of the tangent at the point A (x, f (x)). f '( x ) ≈

f ( x + h) − f ( x ) h

h f '( x ) ≈ f ( x + h) − f ( x ) f ( x + h) ≈ f ( x ) + h f '( x ) The formula f ( x + h) ≈ f ( x ) + h f '( x ) is called the linear approximation formula. Alternatively, if a small change in y, δ y, corresponds to a small change in x, δ x, then ∴ δ y ≈ δx

dy δ y . ≈ dx δ x

dy where δ y ≈ f ( x + h) − f ( x ). dx

If δ y is the small change in a quantity, y, the percentage change (or error) in y is given by h f '( x ) δy × 100% . × 100% or f ( x) y

Worked Example 16

Use the approximation formula, f (x + h) ≈ f (x) + hf '(x), with f (x) = x2 and x = 1 to find an approximate value of (1.01)2. Think

Write

f (x) = x2, f '(x) = 2x, x = 1 and h = 0.01

1

Write f (x), f '(x), x and h.

2

Find x + h and hence f (x + h).

x + h = 1.01 and f (x + h) = (x + h)2 = 1.012

3

Write the formula and hence find the value of 1.012.

f (x + h) ≈ f (x) + hf '(x) 1.012 ≈ 12 + 0.01 × 2 1.012 ≈ 1.02


405

Worked Example 17

A circular metal disc is being cooled. If the radius is decreased from 10 cm to 9.8 cm, find correct to 2 decimal places: a the approximate change in the area b the percentage change in the area. Think a

b

Write

1

Write the relationship between area and radius.

A(r) = πr2

2

Write A′(r), r and h.

A′(r) = 2πr, r = 10 cm, h = −0.2

3

Write the approximation formula and substitute.

A(r + h) ≈ A(r ) + hA'(r ) A(9.8) ≈ A(10) − 0.2 A'(10)

4

State the answer with correct units.

1

Write the change in area as fraction of the original area, expressing as a percentage.

2

State the answer.

Change in A ≈ A(9.8) − A(10) ≈ −0.2 × 20π Area decreases by 12.57 cm2 correct to 2 decimal places.

δ A − 0.2 × 20 × π = × 100% A π × 10 2 = −4% ∴ Area is decreased by approximately 4%

1. Where h is a small change in x then f ( x + h) ≈ f ( x ) + hf '( x ). 2. Alternatively f ( x + h) − f ( x ) ≈ hf '( x ) dy or δ y ≈ δ x × . dx Exercise

8G

Linear approximation 1 WE 16 Use the approximation formula, f (x + h) ≈ f '(x)h + f (x), with each of the following. a For f (x) = x3 and x = 1, find 1.013. b For f (x) = x2 and x = 1, find 0.9992. c For f (x) = 5x2 and x = 1, find 5 × 0.992. d For f (x) = x and x = 1, find 1.001 2 If a spherical balloon has a radius of 5 cm, find the increase in volume of the balloon when the radius expands by 0.02 cm. 3 A circular disc has a radius of 10 cm. Find the percentage increase in its area if the radius is increased by 2%. 4 WE 17 An isosceles triangle has its equal sides of length 10 cm θ with an included angle of θ as shown in the diagram. 10 cm

406


10 cm

If θ changes from 60° to 61°, find, correct to 2 decimal places: a the approximate area of the triangle when θ = 61° b the approximate increase in the area, A, of the triangle. 5 MC If f (x) = x2 and x = 3, an approximate value of (2.9)2 is given by: A f (3) + 0.1 f '(3) B f (2.9) + 0.1 f '(2.9) C f (3) − f (2.9) D f (2.9) − 0.1 f '(2.9) E f (3) − 0.1 f '(3) 6 The volume, V cm3, of a gas is inversely proportional to the pressure, P atmospheres, of a gas, k that is, V = where k is a constant (that depends on the mass of the gas and the P temperature). If the pressure of the gas increases from 2 atmospheres to 2.25 atmospheres, find the approximate change in the volume of the gas in of k. Hence find the corresponding percentage change in the volume. 2 7 Find the approximate change in x as y decreases from 2 to 1.5 if f ( x ) = . x−4 8 The length of a rectangle is four times its width. If the width increases by 5%, find the corresponding percentage change in the perimeter and area of the rectangle.


407

Summary Stationary points

• A function f (x) has stationary points wherever its derivative f ′(x) = 0. • If f ′(a) = 0, then f (x) has a stationary point (a, f (a)) which is: 1. a local minimum turning point if: for x < a, f ′(x) < 0 and x > a, f ′(x) > 0. 2. a local maximum turning point if: for x < a, f ′(x) > 0 and x > a, f ′(x) < 0. 3. a positive stationary point of inflection if: for x < a, f ′(x) > 0 and x > a, f ′(x) > 0. 4. a negative stationary point of inflection if: for x < a, f ′(x) < 0 and x > a, f ′(x) < 0. Equations of tangents and normals

• The gradient of the tangent at x = a to the curve y = f (x) is f ′(a). −1 • The gradient of the normal at x = a to the curve y = f (x) is . f'(a) • The equation of a straight line with gradient m and ing through the point (x1, y1) is: y – y1 = m(x – x1) Maximum and minimum problems

• When solving maximum or minimum problems follow these steps. 1. Draw a diagram if appropriate. 2. Identify the quantity to be maximised or minimised (say, f (x)). 3. Express the quantity in of one variable only (say x). 4. Solve f ′(x) = 0. 5. it is a maximum or minimum using the first derivative test. 6. Sketch a graph to confirm the maximum or minimum found. 7. Answer the question. Rates of change

• The instantaneous rate of change of y = f (x) at x = x1 is found by evaluating f ′(x1) (or finding x = x1). • The average rate of change of y = f (x) between x = x1 and x = x2 is: f ( x2 ) − f ( x1 ) . x2 − x1

408


dy where dx

Related rates

• To solve a problem involving related rates, use the chain rule, for example

dx dx dy = × . dt dy dt

Linear approximation

• f ( x + h) ≈ f ( x ) + hf '( x ) dy • δ y ≈ δ x dx • Percentage change: δ y × 100% y


409


1 Determine the stationary points and their nature for the function f (x) = 2x3 + 3x2 − 36x + 5.

9 P is the point on the line 2x + y − 10 = 0 such that the length of OP, the line segment from the origin O to P, is a minimum. Find the coordinates of P and this minimum length.

2 Sketch the graph of y = x2(4 − x2), clearly indicating all stationary points and intercepts.8B

Exam tip Students failed to recognise this as a maximum/minimum question. A common, incorrect response was to find intercepts (0, 10) and (5, 0) on the line then use the distance between two points to

3 Find the equation of the tangent to the curve y = 6 − 3x + 2x2 − x3 at the point where x = 1. 4 Find the equation of the tangent to the curve y = 3 loge (x2) (x ≠ 0) which is parallel to the line y − 3x − 7 = 0.

find the distance to be 5 5. Other common, incorrect responses involved taking P as the midpoint of the line, guessing and checking solutions without justification, incorrectly differentiating the distance equation, or using an incorrect diagram.

5 Find the equation of the normals to the curve 1 y = x + at the point where the normals are x


[© VCAA 2007]

parallel to the line 3y + 4x − 10 = 0.C 6 The number of bees, N, in a hive can be modelled by the function N = 2t(50 − t) + 180 where t is the number of days the hive has existed. What is the maximum number of bees in the hive?

10 A normal to the graph of y = x has the equation y = −4x + a, where a is a real constant. Find the value of a.

7 The area of a certain triangular shape is: A=

Exam tip Students found this question quite challenging, and a significant number seemed to have little or no idea of what was required. Many were able to find the correct derivative but did not know how to proceed, or equated the derivative to –4 1 instead of 4 . Students who got to this point without error often did not solve correctly for x or y or both. It was not uncommon to see two values found for x and y and therefore two values for a, 18 and –14.

2x2 ,x>6 3( x − 6)

where x is the length of the base of the triangle. a Find the value of x for minimum area. b Hence, find the minimum area. 8 A wine glass is being filled with wine at a rate of 8 cm3/s. The volume, V cm3, of wine in the glass when the depth of wine in the glass is x cm is given 3 4x 2.

by V = Find the rate at which the depth of wine in the glass is increasing when the depth is 4 cm. Exam tip Some students had difficulty deciding on how to label variables; x was sometimes interchangeable with d or h and little or no consistency was evident through the question. Most students were able to correctly differentiate V with respect to x but then either incorrectly used the chain rule or could not manage the arithmetic required. Common errors were to simply substitute 4 into V or to substitute into the derivative. [Assessment report 1 2007]

[© VCAA 2007]

410


[© VCAA 2006]

Multiple choice

1 The graph of y = (x + 2)3 has: A 1 turning point B 2 turning points C 1 point of inflection D 2 points of inflection E 0 stationary points 2 The graph of x3 + 2x2 + x − 2 has: A 2 points of inflection B 1 point of inflection C 1 turning point and 1 point of inflection D 3 turning points E 2 turning points


3 The graph of 13 x3 − 4x2 − 9x + 5 has a local maximum turning point at: 2

A (−1, 9 3 ) B (9, 20) C (9, 18) d (1, −10 23 ) E (−1, 7 13 ) 4 If the graph of g(x) has the following properties: i g'(x) = 0 if x = −3, 1 and 4 ii g'(x) < 0 if x < −3 and 1 < x < 4 iii g'(x) > 0 for all other x then the graph of g(x) could be: y

A

g(x) −3

x

0 1

4

01

4 g(x)

y

B

−3

C

x

y g(x) 0

−3

1

x

4

g(x) 01

−3

4

x

y

E

g(x) −3

01

4

8 The tangent to the curve y = 2 loge (3 − 2x) at the point where x = 1 intercepts with the y-axis at the point where y equals: A 1 B 3 C 0 D −2 E 4 9 The graph of f ′(x) y shown at right f '(x) indicates that the graph of f (x) has: x A a turning point at 0 –4 2 x = 2 and x = −4 B a turning point at x = 2 and point of inflection at x = −4 C a turning point at x = −4 and point of inflection at x = 2 D 3 stationary points E 2 points of inflection at x = −4 and x = 2 Questions 10 and 11 refer to the following information. The volume of liquid in a container varies with time according to the rule V = 1 + t2e−t, where t is in hours and V is in thousands of litres, and 0 ≤ t ≤ 40. 10 The minimum value of V occurs when t equals: A 1 hour B 2 hours C 0 hours d 3 hours E 40 hours

y

D

7 The equation of the normal to the curve y = 3ex + 2 at the point where x = −2 is: A 3y + x − 7 = 0 B 3y − x + 7 = 0 C y − 3x − 9 = 0 d y − 3x − 3 = 0 E y + 3x − 9 = 0

11 The maximum volume, to the nearest tens of litres, is: A 1450 B 1540 C 1390 d 1360 E 1630 Questions 12 to 16 follow from the isosceles triangle below which has a perimeter of 40 cm.

x

5 The equation of a line with a gradient of 2 and ing through the point (1, −3) is: A y = 2x + 5 B y = 2x C y = 2x − 1 d y = 2x − 5 E y = x + 1 6 The equation of the tangent to the curve y = x (x + 2)(x − 1) at the point where x = −1 is: A y − x + 1 = 0 B y − x + 3 = 0 C y + x + 3 = 0 d y + x − 1 = 0 E y − x − 1 = 0

y

x

x

8D

12 The value of y in of x is: A 40 − 2x B 20 − x C 40 − x d 20 − 2x E x2


411

13 The height of the triangle in of x is: A 400 − 40 x

B 20 − 40 x

C

400 − 40 x + 2 x 2

E

40x

D

400 − 40 x + x 2

14 The area in of x is: A ( 20 − x ) 400 − 40 x

B x 400 − 40 x + x 2

C 2 x 400 − 40 x + x 2

D x 400 − 40 x

E 2 x 400 − 40 x 15 The maximum area of the triangle is obtained if x equals: A 6 23 cm B 10 cm C 20 cm d 5 cm

e 10 25 cm

16 Therefore, the maximum area possible is: A 50 2 cm2 d 40 3 9

B 64 cm2 400 3 E 9

C 32 5 cm2

17 The average rate of change of the function f (x) = x4 − 3x3 + 5x between x = 1 and x = 3 is: A 15 B 6 C 12 d 4 12 E 16 18 An iceberg in the shape of a cube is slowly melting. The rate of change of the surface area of the iceberg in m3/m when the side length is 40 metres is: A 480 B -240 C 720 d -480 E 240 19 The radius of a sphere is increasing at a rate of 3 cm/min. When the radius is 6 cm, the rate of increase of the volume, in cm3/min, in the sphere is: A 432π B 48π C 144π d 108π E 16π [© VCAA 2006] 20 Using the approximation formula f ( x + h) ≈ f ( x ) + hf '( x ) , with f (x) = x3 and x = 2, an approximate value of (1.8)3 is given by: A f (2) + 0.2 f '(2) B f (1.8) + 0.2 f '(1.8) C f (2) − f (1.8) d f (2) − 0.2 f '(2) E f (1.8) − 0.2 f '(1.8) [© VCAA 2004]

Extended response

1 Consider the cone with a slant side of 10 cm shown at right. 10 cm

a Show that the height, h, is 100 − r 2 . h

b Show that the volume of the cone is V = 13 πr2 100 − r 2 . c Hence, find the maximum volume.

r 2 The number of bacteria present in a lasagne, t minutes after it is placed in a − microwave to heat up, can be modelled by N = 3000e 0.5t, where t > 0. a Find the initial number of bacteria. b Find the rate of change of bacteria after being in the microwave for 10 minutes. 3 The height above the ground of a person on a ferris wheel at any time, t, seconds after the ride has  πt  started is h = 5.4 − 4 cos   metres.  15  a Find the initial height of the person above ground level. d (m) b Find the height after 5 seconds. 700 c Hence, find the average rate of change of height during the first 5 seconds. 600 d Find the rate of change of height at t = 5. 500 4 Lena is walking in the park. The graph at right shows her 400 displacement from the park entrance over a period of time. a Estimate Lena’s displacement from the park entrance at t = 20. 300 b Find Lena’s average velocity over the interval [20, 30]. 200 c What was Lena’s maximum displacement from her starting 100 point? d Estimate Lena’s velocity at t = 15. 0

412


5

10 15 20 25 30 35 t(min)

5 A manufacturing company is required to produce cylindrical cans (for tuna) of volume 50 cm3. Tin used to produce the cans costs 40 cents per 100 cm2. a Find the area of tin required, A, in of the radius, r. b Find the radius of the can (to the nearest tenth) for minimum area. c Hence, find the minimum area (to the nearest tenth). d What is the cost of tin to produce 10 000 such cans? 6 A small manufacturing company needs to order new cardboard boxes for packaging their product. Each box is to be in the shape of a prism with a square end w and is to have a volume of 27 000 cm3. To hold each box together, tape is used all around the box as shown in the diagram at right. w l a Express the length of the box (l) in of its width (w). b Write the formula for the total area of cardboard (A) in of w. c Find the dimensions of the box which uses the least amount of cardboard. d Write the formula for the total length of the tape (L) in of w. (Ignore the width of the tape and any overlaps.) e Find the dimensions of the box that uses the least amount of tape. f The cardboard used to make the boxes costs 0.01 cents per square centimetre, and the tape can be purchased at $0.50 per metre. Write the formula for the total cost (in cents) of the package (that is, the cost of the cardboard and tape) in of w. g Find the dimensions of the box with the minimum total cost. h Find the minimum total cost to the nearest cent. 7 The cross-section of a pencil head when it is first placed in a sharpening device is shown at right. The gradient of the tangent at point A is 4. The equation of the pencil head is y = 4x2. If the x- and y-axes are as indicated and all distances are in centimetres, find: a the coordinates of points A and B b the equation of the tangent to the curve y = 4x2 at point A c the coordinate of point C d the minimum distance from the pencil head to point C e the length of the pencil head if it starts at the point where the normal at point A meets the y-axis.

y

y = 4x2

Pencil head B

A

Tangent x

0 C

8 The population of rabbits on a particular island t weeks after a virus is introduced is modelled by − P = 1200e 0.1t, where P is the number of rabbits. Find: a the time taken for the population to halve (to the nearest week) b the rate of decrease of the population after: i 2 weeks and ii 10 weeks. After 15 weeks the virus has become ineffective and the population of rabbits starts to increase again according to the model P = P0 + 10(t − 15) loge (2t − 29) where t is the number of weeks since the virus was first introduced. Find: c the value of P0 d the population after 30 weeks e the rate of change of the population after i 20 weeks and ii 30 weeks f how many weeks the population takes to get back to its original number.


413

9 After washing the kitchen floor, Alex put his favourite mop flat against the wall and left it there. A few minutes later, the mop starts to slide down the wall. Let h be the height of the top end of the mop above the floor and let y be the horizontal distance of h the bottom end of the mop from the wall at any time, t. a If the mop is 1.2 m tall, express y in of x. b If the top end of the mop slides down with a constant speed of 5 cm per second, find the speed (in of h) with which the bottom end of the mop moves away y from the wall. c Find the speed with which the bottom end of the mop moves away from the wall when the top of the mop: i is 0.8 m from the floor ii has slid down by 20 cm. d Find the speed with which the bottom end of the mop moves away from the wall 6 seconds after the top of the mop started sliding down. 10 Consider the function f ( x ) = − ( x − 3)( x + 2a)2 where a is a positive real constant. a Find f ′(x). b Find the coordinates of the stationary points. c Determine the nature of the stationary points. d Find the equation of the tangent at x = −2. e Find the x-intercept of the tangent. 11 Consider the function y = (x2 − a)2 where a ∈ R. dy a Find at x = 2. dx b Find the equation of the tangent to the curve at x = 2 in of a. c Hence find the x-intercept of the tangent line in of a. d A straight line with equation y = −2x + 1 es through the x-intercept of the tangent line. Find the value of a. e Using the value of a found previously, what is the equation of the tangent line? eBook plus Digital doc


414


eBook plus

ACTiviTieS


• 10 Quick Questions: Warm up with ten quick questions on applications of differentiation. (page 373) 8A

equations of tangents and normals

Tutorial

• We 2 int-0558: Watch a worked example on finding equations of tangents and normals. (page 374) Digital docs

• Spreadsheet 129: Investigate tangents and normals. (page 375) • WorkSHEET 8.1: Find stationary points and their nature and equations of normals and tangents. (page 376) 8B

Sketching curves

Tutorial

• We 3 int-0559: Watch a worked example on sketching curves and finding stationary points. (page 378) Digital doc

• SkillSHEET 8.1: Practise differentiating expressions other than polynomials. (page 385) 8c

Maximum and minimum problems when the function is known

Tutorial

• We 8 int-0560: Watch a worked example on applications of maximum and minimum problems. (page 387) 8D

Maximum and minimum problems when the function is unknown

Tutorial

• We 10 int-0561: Watch a worked example on maximising volume. (page 392) Digital doc

• WorkSHEET 8.2: Apply differentiation skills to a variety of problems. (page 396) 8e

Tutorial

• We 13 int-0562: Watch a worked example on application of rates of change and average rates of change. (page 397) Digital doc

• SkillSHEET 8.2: Practise identifying rates of change. (page 398) 8F

Related rates

eLesson eles-0094

• Related rates: Learn ing differentiation for related rates. (page 401) Tutorial

• We 14 int-0563: Watch a worked example on using related rates to determine an instantaneous rate of change. (page 401) 8G

Linear approximation

chapter review Digital doc


Rates of change


• Rates of change: Consolidate your understanding of applying differentiation techniques by riding and rolling. (page 396)

Chapter 8


415

9

9A 9B 9C 9D 9E 9F 9G 9H 9I 9J

Antidifferentiation Integration of ex, sin ((x) and cos ((x) Integration by recognition Approximating areas enclosed by functions The fundamental theorem of integral calculus Signed areas Further areas Areas between two curves Average value of a function Further applications of integration

Integration areaS of STudy

• Informal treatment of the fundamental theorem

• Antiderivatives of polynomial functions and of f (ax + b), where f is xn for n ∈ Q, ex, sin (x), cos (x) or linear combinations of these • Definition of the definite integral as the limiting b

a

n

∑ f ( xi* ) δ xi, δ x→0

value of a sum ∫ f ( x ) dx = lim

i =1

where the interval [a, b] is partitioned into n subintervals, with the ith subinterval of length δx δ i and containing xi*, and δx = max{δxi: i = 1, 2, … n} and evaluation of numerical approximations based on this definition • Informal approximation using areas under curves by left and right rectangles • Examples of the definite integral as a limiting value of a sum involving quantities such as area under a curve, distance travelled in a straight line and cumulative effects of growth such as inflation • Antidifferentiation by recognition that F ′( x ) = f ( x ) implies ∫ f ( x ) dx = F ( x ) + c

9a

b

of calculus, ∫ f ( x ) dx = F (b) − F (a) a

• Properties of antiderivatives and definite integrals: ( x ) ± bg( x )] dx = a ∫ f ( x ) ddxx ± b ∫ g( x ) dx ∫ [af (x b

∫a

b

∫a

a

∫a

c

c

f ( x ) dx +

∫b f ( x) dx = ∫a f ( x) dx

f ( x ) dx =

−

a

∫b

f ( x ) dx

f ( x ) dx = 0

• Application of integration to problems involving calculation of the area of a region under a curve and simple cases of areas between curves, such as distance travelled in a straight line; average value of a function; other situations modelled by the use of the definite integral as a limiting value of a sum over an interval; and finding a function from a known rate of change

antidifferentiation


10 Quick Questions

As we have seen, the process of differentiation enables us to find the gradient of a function. The reverse process, antidifferentiation (or integration), will find the function for a particular gradient. Integration has wider applications including calculation of areas, volumes, energy, probability and many more quantities in science and business. Note that d f ( x ) means differentiate f (x) with respect to x; that is, d f ( x ) = f ′( x ). dx dx

416

maths Quest 12 mathematical methods CaS for the TI-nspire

So f (x) is the antiderivative of f '(x), denoted as f ( x ) = ∫ f '( x ) dx where ∫ means antidifferentiate, or integrate, or find an indefinite integral and dx indicates that the integration of the function is with respect to x. Since

d (ax + c) = a, where a and c are constants dx

then

∫ a dx = ax + c

Since then

d  ax n + 1  = ax n dx  n + 1 

∫ ax n dx =

ax n + 1 + c, n ≠ − 1 n +1

In the expression above, the term c is used to denote a constant. In the antiderivative of a function, there are an infinite number of possibilities for c. However, when we are finding an antiderivative, we set c to zero. That is, finding an antiderivative means ‘let c = 0’, or ‘do not add on the c’. For example, the antiderivative of 3x2 + 4x + 5 is x3 + 2x2 + 5x + c. An antiderivative of 2 3x + 4x + 5 is x3 + 2x2 + 5x.

Properties of integrals Since d is a linear operator, so too is ∫ ..Therefore, dx ∫ [ f ( x) ± g( x)] dx = ∫ f ( x) dx ± ∫ g( x) dx That is, each term can be integrated separately, and

∫ k f ( x) dx = k ∫ f ( x) dx That is, a ‘constant’ factor of the function can be taken to the front of the integral. So ∫ [af ( x) ± bg( x)]dx = a ∫ f ( x )dx ± b ∫ g( x )dx Worked Example 1

Antidifferentiate each of the following, expressing answers with positive indices. − 3 a 2x7 b 4x 3 c x Think a

b

c

Write

1

Integrate by rule; that is, add 1 to the index and divide by the new index.

2

Simplify.

1

Integrate by rule.

2

Simplify.

3

Express the answer with a positive index.

1

When a square root is involved, replace it with a fractional index.

a

∫ 2 x 7 dx =

2 x8 +c 8

x8 +c 4 − 4x 2 dx = − + c 2 =

b

∫ 4x

−3

−

c

∫

= −2x 2 + c − 2 = 2 x 3 3 dx = ∫ 1 dx x x2

Chapter 9 Integration

417

2

= 3∫ x

Bring the x to the numerator and change the sign of the index.

−1 2 dx

1

3x 2

+c

3

Integrate by rule.

=

4

Simplify.

= 6x 2 + c

5

Write the answer in the form it was given.

=6 x +c

1 2 1

Worked Example 2

Find the following indefinite integral. ∫(x. - 1)(3x + 5) dx Think

Write

∫ ( x − 1)(3x + 5) dx = ∫ (3x 2 − 3x + 5x − 5) dx = ∫ (3 x 2 + 2 x − 5) dx

1

Expand the expression.

2

Collect like .

3

Integrate each term separately.

=

4

Simplify each term.

= x3 + x 2 − 5x + c

3x3 2 x 2 + − 5x + c 3 2

Integration of (ax + b)n where n ≠ − 1 By applying the chain rule for differentiation: d (ax + b) n + 1 = a( n + 1)(ax + b) n dx

so

∫ a(n + 1)(ax + b)n dx = (ax + b)n + 1 + c

or

a( n + 1) ∫ (ax + b) n dx = (ax + b) n + 1 + c

or

∫ (ax + b)n dx =

(ax + b) n + 1 +c a( n + 1)

Worked Example 3 n+1

Antidifferentiate 4(5x − 2)3 by using ( ax + b) n dx = ( ax + b) ∫ a( n + 1) Think

418

+ c.

Write

∫ 4(5x − 2)3 dx = 4 ∫ (5x − 2)3 dx

1

Express as an integral and take 4 out as a factor.

2

Apply the rule where a = 5 and n = 3.

=

3

Simplify the antiderivative by cancelling the fraction.

= 5(5x − 2)4 + c


4(5 x − 2)4 +c 5(4) 1

Integration of x1 Since then

1 d log e ( x ) = dx x 1 ∫ x dx = loge ( x) + c, where x > 0

or

∫x

−1 dx

= log e x + c.

Worked Example 4

Antidifferentiate 4 . 7x Think 1

Write

4

4

= 2

1

∫ 7 x dx = ∫  7 × x  dx

Take 47 out as a factor.

4 7

1

∫ x dx

= 47 log e x + c

Integrate by rule.

- 1

Integration of (ax + b)

This can be done by applying the chain rule for differentiation: d a log e (ax + b) = , where a and b are constants. dx ax + b Multiplying both sides by 1 gives a 1 d a 1 × log e (ax + b) = × a dx ax + b a 1 ax + b 1 1 d log e ax + b dx dx = × ax + b a dx =

So

∫

so

∫ ( ax + b)

−1

dx =

1 log e ax + b + c a

1 Note that the a in the fraction a is the derivative of the linear function ax + b. Worked Example 5

Antidifferentiate

5 . 2x + 3

Think

Write


Express as an integral and take 5 out as a factor.

2

Integrate by rule where a = 2.

5

1

∫ 2 x + 3 dx = 5 ∫ 2 x + 3 dx =

5 log e 2 x + 3 + c 2


419


On a Calculator page, press: • MENU b • 4:Calculus 4 • 3:Integral 3 Alternatively, select the indefinite integral template from the Maths expression template and fill in the required fields. Complete the entry line as:  5  dx ∫  2 x + 3  Press ENTER ·.

2

Write the answer and put in the constant of integration, as the CAS calculator omits it. Also note that the CAS uses ln instead of loge.

5  5  ∴ ∫ dx = log e 2 x + 3 + c 2  2 x + 3 

Worked Example 6

Find ∫

6x + 5 dx . x2

Think

Write

∫

6x + 5  6x 5  dx = ∫  2 + 2  dx 2 x x x 

1

Express as separate fractions.

2

Simplify each fraction.

= ∫ (6 x

3

Integrate each term separately by rule.

= 6 log e x +

4

Simplify leaving the answer with positive indices.

= 6 log e x − 5 x + c 5 = 6 log e x − + c x

−1

−

+ 5 x 2 ) dx 5x

−1

−1 −1

+c

Worked Example 7

Find the equation of the curve g(x) given that g′(x) = 3 x + 2 and the curve es through (1, 2). Think

420

Write

1

Write the rule for g′(x).

g′(x) = 3 x + 2

2

Rewrite g′(x) in index form.

g′(x) = 3 x 2 + 2

3

Express g(x) in integral notation.

g(x) =

4

Antidifferentiate to obtain a general rule for g(x).


1



1



∫  3x 2 + 2 dx 3

= 3x 2 ÷ 3 + 2 x + c 2

5

Simplify.

3 2

= 3x × 2 + 2 x + c 3 1

3

g(x) = 2 x 2 + 2 x + c

6

Substitute coordinates of the given point into g(x).

7

Find the constant of antidifferentiation, c.

8

State the rule for g(x) in the form that it is given.

3

As g(1) = 2, 2(1) 2 + 2(1) + c = 2 2+2+c=2 so c = −2 3

g(x) = 2 x 2 + 2 x − 2

= 2 x3 + 2x − 2

Worked Example 8

If a curve has a stationary point (2, 3), and a gradient of 2x − k, where k is a constant, find: a the value of k b y when x = 1. Think a

1

2

Write

The gradient is dy so write the rule for the dx gradient.

a

Let dy = 0 (as stationary points occur when dx the derivative is zero) and substitute the value of x into this equation.

dy = 2x − k dx For stationary points, dy = 0, so 2x − k = 0 dx 2(2) − k = 0 as x = 2

b

3

Solve for k.

1

Rewrite the rule for the gradient function, using the value of k found in a above.

2

Integrate to obtain the general rule for y.

4 − k = 0 so k = 4 b

dy = 2x − 4 dx y = ∫ (2 x − 4)dx = x2 − 4x + c

3

Substitute the coordinates of the given point on the curve to find the value of c.

Since curve es through (2, 3), 3 = 22 − 4(2) + c 3=4−8+c c=7

4

State the rule for y.

So y = x2 − 4x + 7

5

Substitute the given value of x and calculate y.

When x = 1, y = (1)2 − 4(1) + 7 =4

The relationship between the graph of an antiderivative function and the graph of the original function f (x) is the antiderivative of f ′(x) and is written as f ( x ) = ∫ f ′( x )dx. f ′(x) is the gradient function of the antiderivative function f (x).


421

The graph of the antiderivative function f (x) can be derived from the graph of f ′(x) and results in a family of curves. For example if f ′(x) = 1 then the antiderivative function is f (x) = x + c, where c can take any real value.

Sketching the antiderivative function from the graph of the original function 1. The general shape of the graph of the antiderivative function can be determined from the graph of a polynomial function by increasing the degree by one. For example, if f ′(x) is a quadratic function, then f (x) is a cubic function. 2. The x-intercepts of f ′(x) become the turning points on the graph of f (x). 3. When f ′(x) is above the x-axis, the gradient of f (x) is positive. 4. When f ′(x) is below the x-axis, the gradient of f (x) is negative. Worked Example 9

Sketch the graph of the antiderivative function from the graph of the gradient function f ′(x) shown. f ′(x)

(1, 0)

(−2, 0) 0

x

(0, −1) Think

Write

1

State the shape of the antiderivative function.

The antiderivative function will be a positive cubic function.

2

Find the x-intercepts of the gradient function, f ′(x), and hence find the x-coordinates of the turning points.

There are x-intercepts when x = −2 and when x = 1, so f (x) has turning points when x = −2 and x = 1.

3

Find when the given graph, f ′(x), is above the x-axis and hence find when f (x) has a positive gradient.

f ′(x) is above the x-axis when x < −2 and when x > 1, so f (x) has a positive gradient when x < −2 and when x > 1.

4

Find when the given graph, f ′(x), is below the x-axis and hence find when f (x) has a negative gradient.

f ′(x) is below the x-axis when −2 < x < 1, so f (x) has a negative gradient when −2 < x < 1.

5

Sketch the curve.

The graph could be any of the family of graphs formed by vertical translations of the graph shown. f(x)

−2

422


0

1

x

1.

d f ( x ) = f ′( x ) dx

2. f ( x ) = ∫ f ′( x ) dx 3. ∫ a dx = ax + c 4. ∫ ax n dx =

ax n + 1 + c, n ≠ − 1 n +1

5. ∫ [ f ( x ) ± g( x )] dx =

∫ f ( x) dx ± ∫ g( x) dx

6. ∫ k f ( x ) dx = k ∫ f ( x ) dx 7. ∫ (ax + b) n dx =

(ax + b) n + 1 + c, n ≠ − 1 a( n + 1)

1 dx = log e x + c, x ≠ 0 or ∫ x −1 dx = log e x + c x

8. ∫

1 1 1 dx = log e ax + b + c or ∫ (ax + b)−1 dx = log e ax + b + c ax + b a a 10. Write the answer in the same form as the question unless otherwise stated. 11. To draw the graph of the antiderivative function f (x) from the graph of the gradient function f ′(x), use the x-intercepts as the x-coordinates of the turning points, and make the gradient positive when f ′(x) is above the x-axis and negative when f ′(x) is below the x-axis. 12. For a polynomial function, the graph of f (x) is one degree greater than the graph of f ′(x). 9. ∫

Exercise

9a

Antidifferentiation 1 WE1 a x

Antidifferentiate each of the following, giving answers with positive indices. c x7 d 3x5 b x4

−

e 5x 2 x4 i 5

f

x3 j 2

2

3

m x 3 q

n 4 x 4

9 x2

2 WE2

−

−2x4

r

− 10

x6

g −6x 4 − x 4 k 3

h 2 x

o x

p

s

l

−3 7

8 x

t

x 5 x3 −6

(x x )

Find the following indefinite integrals.

a ∫ (2 x + 5) dx

b ∫ (3 x 2 + 4 x − 10) dx

c

∫ (10 x 4 + 6 x3 + 2) dx

d ∫ ( − 4 x 5 + x 3 − 6 x 2 + 2 x ) dx

e

∫ ( x3 + 12 − x 2 ) dx

f

∫ ( x + 3)( x − 7) dx

g ∫ 5( x 2 + 2 x − 1) dx

h ∫ ( x 2 + 4)( x − 7) dx

i

∫ x( x − 1)( x + 4) dx Chapter 9 Integration

423

3 MC

∫ ( x 2 + x + 2) dx is equal to:

A ∫ x 2 dx + x + 2

B ∫ x 2 dx + ∫ x dx + ∫ 2 dx

C ∫ (x 2 + x ) dx + 2

D x 2 + ∫ ( x + 2) dx

E x 2 + x + ∫ 2 dx 4 MC

∫ x( x + 3) dx is equal to:

A ∫ x dx ∫ ( x + 3) dx

B x ∫ ( x + 3) dx

C ( x + 1) ∫ x dx

D ∫ x dx + ∫ ( x + 3) dx

E

∫ ( x 2 + 3x) dx

5 WE3 Antidifferentiate each of the following by using ∫ (ax + b) n dx = a (x + 3)2 c 2(2x + 1)4 e (6x + 5)4 g (4 − x)3 i 4(8 − 3x)4 − k (2x + 3) 2 − m 6(4x − 7) 4 − o (6 − 5x) 3 6 MC

b d f h j l n p

(ax + b) n + 1 + c. a( n + 1)

(x − 5)3 −2(3x − 4)5 3(4x − 1)2 (7 − x)4 −3(8 − 9x)10 − (6x + 5) 3 − (3x − 8) 6 −10(7 − 5x)−4

∫ 3( x + 2)4 dx is equal to:

A 3 + ∫ ( x + 2)4 dx

B ∫ 3 dx + ∫ ( x + 2)4 dx

C 3 ∫ ( x + 2)4 dx

D 3 ∫ dx × ∫ ( x + 2)4 dx

E ( x + 2)4 ∫ 3 dx 7 WE4, 5 Antidifferentiate the following.

424

a

∫ x dx

3

b

∫ x dx

8

c

∫ 5x dx

d

∫ 3x dx

7

e

∫ 7x dx

4

f

∫ x + 3 dx

g

∫ x + 3 dx

3

h

∫ x + 4 dx

i

∫ x + 5 dx

j

∫ 3x + 2 dx

4

k

∫ 5x + 6 dx

8

l

∫ 2 x − 5 dx

o

∫ 5 − x dx

−5

m

∫ 3 + 2x

p

3 ∫ 6 − 11x dx

dx

n q

−2

−2

∫ 6 + 7x dx −2

∫ 4 − 3x

dx


r

6

1

−6

3

1

−8

∫ 5 − 2x dx

6

∫ x + 5 dx is equal to:

8 mC A 6∫

1 dx x+5

D

∫ 6dx

B E

∫ ( x + 5) dx

9 We6 Find ∫

1

∫ 6dx ∫ x + 5 dx

∫ 3(4 x + 1)

j

∫ 

m eBook plus Digital doc

SkillSHEET 9.1 Substitution and evaluation



∫

x+

∫ 6dx + ∫ x + 5 dx

c

∫

f

∫ 4(2 x − 5)5 dx

i

∫

( x − 5)( x + 3) dx x3

l

∫

x2 + x4 dx x

6

∫ ( x + 5) dx

(2 x + 7) dx. x

10 For the following mixed sets, find: 1  a ∫  x4 + 2x +  b ∫ (3 x + 1)5 dx  x −5 3 d ∫ e ∫ dx dx 2x + 1 6 − 10x g

1

C

−3

( x + 4) 2 dx 2x −1  3 k ∫  5 x 2 − 2 x + 3 x 3  dx  

h

dx

2  dx 3 − x 

x2 + 2x − 1 x

n

dx

∫

3x 2 + 2 x − 1 dx x2

10 − x + 2 x 4 dx ∫ x3

11 We7 Find the equation of the curve f (x) given that: a f ′(x) = 4x + 1 and the curve es through (0, 2) b f ′(x) = 5 − 2x and the curve es through (1, −1) − c f ′(x) = x 2 + 3 and the curve es through (1, 4) d f ′(x) = x + x and f (4) = 10 1

e f ′(x) = x 3 − 3 x 2 + 50 and f (8) = −100 1 f f ′(x) = − 2 x and f (1) = −5 x g f ′(x) = (x + 4)3 and the curve es through (−2, 5) − h f ′(x) = 8(1 − 2x) 5 and f (1) = 3 −1 i f ′(x) = (x + 5) and the curve es through (−4, 2) 8 j f ′′(x) and f (3) = 7 ( x) = 7 − 2x 12 We8 If a curve has a stationary point (1, 5), and a gradient of 8x + k, where k is a constant, find: a the value of k b y when x = −2. 13 A curve g(x) has g ′( x ) = kx +2 x , where k is a constant, and a stationary point (1, 2). Find: x a the value of k b g(x) c g(4). 14 We9 Sketch the graph of the antiderivative functions from each of the following graphs. a b c f ′(x) f′(x) f ′(x)

(−1, 0)

(1, 0) x

0

(−2, 0) 0

x

0

(0, 2)

(0, −3)

x

(0, −1)

Chapter 9

Integration

425

f ′(x)

d

e

f

f ′(x)

f ′(x)

(0, 1) (0, 0)

( 1–2 , 0) 0 (1, 0)

0

x

0

x

x

(0, −1)

g

f ′(x)

(−2, 0) (0, 0)

9B

(1, 0)

x

Integration of e x, sin (x) and cos (x) Integration of the exponential function e x Since

d x (e ) = e x dx

then

∫ e x ddxx = e x + c

and or

Therefore,

d kx (e ) = kke kx , where k is a constant dx

∫ kekx dx = ekx + c k ∫ e kx dx = e kx + c 1 ∫ ekx ddx = k ekx + c.

Worked examPle 10

Antidifferentiate each of the following. − e 5x 4x a 3e b c (e x − 1)2 4 ThInk a

b

426

WrITe

1

Integrate by rule where k = 4.

2

Simplify.

1

Rewrite the function to be integrated so that the coefficient of the e term is clear.

2

Integrate by rule where k = −5.


a

b

3e 4 x +c 4 3 = 4 e4 x + c

∫ 3e4 x ddxx = ∫

e

−5 x

4

dx = =

1

−

∫ 4 e 5 x ddx 1 4

e −

−5 x

5

+c

3

1 = 4e

Simplify the antiderivative.

−5x 5x

1 = − 20 e

c

1

Expand the function to be integrated.

2

Integrate each term by the rule.

c

×

−5 x

1 −5

+c

+c

∫ (e x − 1)2 dx = ∫ (e2 x − 2e x + 1) dx = 12 e 2 x − 2e x + x + c

Integration of trigonometric functions Since

d [sin (ax)] = a cos (ax) dx

and

d [cos (ax )] = − a sin (ax ) dx

it follows that

∫ sin ( ax) dx =

−

1 cos ( ax ) + c a

1

∫ cos ( ax) dx = a sin ( ax) + c Worked examPle 11

Antidifferentiate the following. a sin (6x)

b 8 cos (4x)

c 3 sin

 − x  2 

ThInk

WrITe

a

Integrate by rule.

a

b

1

Integrate by rule.

b

2

Simplify the result.

1

Integrate by rule.

2

Simplify the result.

c

−

1

∫ sin (6 x)dx = 6 cos (6 x) + c 8 ∫ 8 cos (4 x)dx = 4 sin (4 x) + c = 2 sin (4x) + c

c

 − x dx = 2 

∫ 3 sin 

−

3 1 2

−

 − x cos   + c  2

 − x = 6 cos   + c  2

Worked examPle 12

Find ∫ [ 2 e 4 x − 5 sin ( 2 x ) + 4 x ] dx. ThInk

WrITe


Integrate each term separately.

∫ [2e4 x − 5 sin (2 x) + 4 x]dx = 42 e 4 x −

2

Simplify each term where appropriate.

−5 2

cos (2 x ) + 42 x 2 + c

= 12 e 4 x + 25 cos (2 x ) + 2 x 2 + c

Chapter 9

Integration

427


On a Calculator page, press: • MENU b • 4:Calculus 4 • 2:Integral 2 Complete the entry line as:

∫ (2e4 x − 5 sin (2 x) + 4 x) dx Press ENTER ·. Note: When integrating trigonometric functions, ensure the CAS is set to radian mode.

2

∴∫ (2e 4 x − 5 sin (2 x ) + 4 x ) dx

Write the answer and put in the constant of integration.

e 4 x 5 cos (2 x ) + + 2x2 + c 2 2

=

Exercise

9b

1

1.

∫ e x dx = e x + c and ∫ ekx dx = k ekx + c

2.

∫ sin (ax) dx =

−1

a

cos (ax ) + c and ∫ cos (ax ) dx =

1 sin (ax ) + c a

Integration of e x, sin (x) and cos (x) 1 WE10 a e2x − d e 3x e6 x g 2 j

−8e−2x x

m 3e 2 e x − e− x p 2

Antidifferentiate each of the following. b e4x e 5e5x 2e 3 x h 3 x

−x 3

i

−3e6x x

l 0.1e 4

k e 3 n 3e

c e−x f 7e4x

o e x + e−x

2 Find an antiderivative of (1 + ex)2. 3 Find an antiderivative of (ex − 1)3. 4 Find an antiderivative of x3 − 3x2 + 6e3x. 5 MC If f ′(x) = e2x + k and f (x) has a stationary point (0, 2), where k is a constant, then: a k is equal to: A e B e2 C 1 D −1 E −e b f (1) is equal to: A e2 − 1 B e 2 + 12 C 12 e 2 + 12 D e4 E 1 12

428


6 WE11 Antidifferentiate the following. a sin (3x) b sin (4x) cos ( 2 x ) d e sin (−2x) 3 4 sin (6 x ) g h 8 cos (4x) 3  x j −2 cos (−x) k sin    3  − x  x m 3 sin   n − 2 sin    5  4  − x p − 6 cos    2 s

− 2 sin

 5x    2

t

− 3 cos

−6

sin (3x)  x l cos    2  x o 4 cos    4

 3x  r 6 cos    4

 7x    4

u 5 sin (πx)  − 4x  s x − sin   π 

b [sin (2x ) − cos ( x )] dx ∫

∫ [cos (4 x) + sin (2 x)] dx e ∫ [4 cos (4 x ) − 13 sin (2 x)] dx c



i

 π x w − 2 cos    3

7 WE12 Find: a [sin ( x ) + cos ( x )] dx ∫

∫ 3 sin ( π2x

f cos (−3x)

 2x  q 4 sin    3

 π x v 3 cos    2

g

c cos (7x)

) + 2 cos ( π3x ) dx

) − cos (2x) dx

d

∫ sin ( 2x

f

∫ [5x + 2 sin ( x)] dx



h ∫ [3e6 x − 4 sin (8 x ) + 7] dx

8 Find the antiderivative of e4x + sin (2x) + x3. 9 Find an antiderivative of 3x2 − 2 cos (2x) + 6e3x. 10 Antidifferentiate each of the following. 1 a x 3 − b x2 + 4 cos (2x) − e−x + e2 x 2x + 3 x  x c sin   + e 2 − (3 x − 1)4  3

d

x  x  x e 3 sin   − 2 cos   − e 5  3  2 −

f

1  x + e 4 x + cos    5 3x − 2  π x x + 2 x − 2 sin   + 5  3

11 In each of the following, find f (x) if: π a f ′(x) = cos (x) and f   = 5 b f ′(x) = 4 sin (2x) and f (0) = −1  2  x  x  x c f ′( x ) = 3 cos   and f (π ) = 9 2 d f ′( x ) = cos   − sin   and f (2π) = −2.  2  4  4 dy  π x 12 If = sin   + k, where k is a constant, and y has a stationary point (3, 4), find:  6 dx a the value of k b the equation of the curve c y when x = 6. 13 A curve has a gradient function f ′(x) = 4 cos (2x) + ke x, where k is a constant, and a stationary point (0, −1). Find: a the value of k b the equation of the curve f (x) π c f   correct to 2 decimal places.  6


429

9C

Integration by recognition As we have seen, if d [ f ( x )] = g( x ) dx then

∫ g( x) dx = f ( x) + c, where g( x) = f '( x).

This result can be used to determine integrals of functions that are too difficult to antidifferentiate, via differentiation of a related function.

Worked examPle 13

eBook plus

a Find the derivative of the function y = (5x + 1)3. b Use this result to deduce the antiderivative of 3(5x + 1)2.

Tutorial

int-0564

ThInk a

b

WrITe

1

Write the function and recognise that the chain rule can be used.

2

Let u equal the function inside the brackets. du Find . dx

3 4

Express y in of u.

5

Find

6

Write the chain rule.

7

Find

8

Replace u with the expression inside the brackets and simplify where applicable.

1

2

3

Worked example 13

a y = (5x + 1)3

Let u = 5x + 1 du =5 dx y = u3

dy . du

dy = 3u 2 du dy dy du = × dx du dx dy = 3u 2 × 5 dx


du dx = y + c1, express the relationship dx in integral notation. dy Remove a factor from so that it resembles dx the integral required. Since ∫

Divide both sides by the factor in order to obtain the integral required.

= 15(5x + 1)2 b

∫ 15(5x + 1)2 ddxx = (5x + 1)3 + c1 5 ∫ 3(5 x + 1)2 dx = (5 x + 1)3 + c1

∫ 3(5x + 1)2 dx = 15 (5x + 1)3 + c, where c =

c1 5

Therefore, the antiderivative of 3(5x + 1)2 is 15 (5x ( 5x ++ 11))33 + cc..

Note that the shorter form of the chain rule below can be used to differentiate. If f ( x ) = [ g( x ))]]n then f ′ ( x ) = ng ′ ( x )[ g( x ))]]n − 1 .

430


Worked examPle 14 a Differentiate e x . 3

b Hence, antidifferentiate 6x2 e x . 3

ThInk a

1

Write the equation and apply the chain rule to differentiate y.

2

Let u equal the index of e.

3

Find

4

Express y in of u.

5

Find

6

b

WrITe a

y = ex

3

Let u = x3

du . dx

du = 3x 2 dx y = eu

dy . du dy Find using the chain rule and dx replace u.

dy = eu du dy = 3 x 2 eu dx = 3x 2 e x

dy in integral notation. dx

b

1

Express

2

Multiply both sides by a constant to obtain the integral required.

3

∫ 3x 2 e x dx = e x 3

3

+ c1

2∫ 3 x 2 e x dx = 2e 2e x + 2c1 3

3

2e x ∫ 6 x 2 e x dx = 2e 3

3

+ c, c where c = 2c1.

Therefore, the antiderivative of 6 x 2 e x is 2e x + c. c 3

3

Note that the shorter form of the chain rule below can be used to differentiate. If y = e f ( x ) then

dy = f ′( x)e f ( x ) . dx

Worked examPle 15 a Find the derivative of sin (2 x + 1) and use this result to deduce the antiderivative of 8 cos (2 x + 1). b Differentiate loge(5x2 − 2) and hence antidifferentiate ThInk a

b

x . 5 x2 − 2 WrITe a f (x) = sin (2x + 1)

1

Define f (x).

2

Differentiate using f ′(x) = g′(x) cos [g(x)] where f (x) = sin [g(x)].

3

Express f (x) using integral notation.

4

Multiply both sides by whatever is necessary for it to resemble the integral required.

4 ∫ 2 cos (2 x + 1) 1)dx = 4 sin (2 x + 1) + c

5

Write the integral in the form in which the question is asked.

∫ 8 cos (2 x + 1)dx = 4 ssinin (2 x + 1) + c

1

Define f (x).

f ′(x) = 2 cos (2x + 1)

∫ 2 cos (2 x + 1)dx = sin (2 x + 1) + c1

The antiderivative of 8 cos (2x + 1) is 4 sin (2x + 1) + c. b f (x) = loge (5x2 − 2)

Chapter 9

Integration

431

2

Differentiate using f ′( x ) = where f (x) = loge [g(x)].

g ′( x ) g( x )

3

Express f (x) using integral notation.

4

Take out a factor so that f ′(x) resembles the integral required.

5

Divide both sides by the factor to obtain the required integral.

10 x 5x 2 − 2

f ′( x ) =

10 10x

∫ 5x 2 − 2 dx = loge 10 ∫

5 x 2 − 2 + c1

x dx = log e 5 x 2 − 2 + c1 −2

5x 2

x

∫ 5x 2 − 2 dx = 101 loge

5x 2 − 2 + c

The antiderivative of x is 1 log e 5 x 2 − 2 + c. 5 x 2 − 2 10

Worked examPle 16

Differentiate x cos (x) and hence find an antiderivative of x sin (x). ThInk

WrITe

Let y = x cos (x)

1

Write the rule.

2

Apply the product rule to differentiate x cos (x).

3

Express the result in integral notation. (Do not add c, as an antiderivative is required.)

∴ ∫ [cos ( x ) − x sin sin ( x )] ddxx = x cos ( x )

4

Express the integral as two separate integrals.

5

Simplify by integrating. (Do not add c.)

6

Make the expression to be integrated the subject of the equation.

cos ( x ) dx − ∫ x ssin in ( x ) ddxx = x cos ( x ) ∫ cos sin ( x ) − ∫ x ssin in ( x ) ddxx = x cos ( x ) − x sin ( x ) dx = x cos ( x ) − sin ( x ) ∫

7

Simplify.

dy = x [−sin (x)] + [cos (x)](1) dx = −x sin (x) + cos (x) = cos (x) − x sin (x)

x ) dx = ssin in ( x ) − x cos ( x ) ∫ x sin ( x) Therefore, an antiderivative of x sin (x) is sin (x) − x cos (x).

Worked examPle 17 a Show that

5x + 1 4 = 5− . x+1 x+1

b Hence, find

ThInk a

432

Use algebraic long division to divide the numerator into the denominator.

∫

5x + 1 dx. x+1 WrITe

)

5

a x + 1 5x + 1


5x + 5 −4

So

5x + 1 4 = 5− x +1 x +1

b

b

∫

5x + 1 4   dx = ∫  5 −  x +1 x + 1

1

Write the expression using integral notation.

2

Express as two separate integrals.

= ∫ 5 dx − ∫

3

Antidifferentiate each part.

= 5 x − 4 log e x + 1 + c

4 dx x +1

1. To differentiate using the chain rule, use one of the following rules. (a) If f (x) = [g(x)]n then f ′(x) = ng′(x)[g(x)]n − 1 (b) If y = e f (x),

dy = f ′( x )e f ( x ) dx

dy dy du = × dx du dx (d) f ′(x) = g′(x) cos [g(x)] where f (x) = sin [g(x)] f ′(x) = −g′(x) sin [g(x)] where f (x) = cos [g(x)] (c)

(e) f ′( x ) =

g ′( x) where f ( x ) = logg e g( x ) . g( x )

2. To antidifferentiate, use ∫ g( x ) dx = f ( x ) + c where g(x) = f ′(x).

exerCISe

9C

Integration by recognition 1 We13 For each of the following find the derivative of the function in i and use this result to deduce the antiderivative of the function in ii. a i (3x − 2)8 ii 12(3x − 2)7 2 5 b i (x + 1) ii 5x(x2 + 1)4 1 c i 2x − 5 ii 2x − 5 d i

4x + 3

+ 3x − 1 f i 2 x −1 e i

(x2

ii 7)4

2 4x + 3

ii (2x + 3)(x2 + 3x − 7)3 4x ii 2 ( x − 1)2

2 mC The derivative of (x + 7)4 is 4(x + 7)3. a Therefore, the antiderivative of 4(x + 7)3 is: A (x + 7)4 + c B 14 (x + 7)4 + c D 3(x + 7)4 + c E 12(x + 7)4 + c 3 b The antiderivative of (x + 7) is: A (x + 7)4 + c B 14 (x + 7)4 + c D 3(x + 7)4 + c

C 4(x + 7)4 + c

C 4(x + 7)4 + c

E 12(x + 7)4 + c

Chapter 9

Integration

433

3 mC If the derivative of (2x − 3)6 is 12(2x − 3)5, then ∫ 6(2 x − 3)5 dx is: A 2(2x − 3)6 + c B 4(2x − 3)6 + c C (2x − 3)6 + c 1 6 6 D 6(2x − 3) + c E 2 (2x − 3) + c 4 We14

For each of the following differentiate i and hence antidifferentiate ii.

e4x − 5

a i c i ex

ii 2e4x − 5 ii x e x

2

b i e6 − 5x d i ex − x

2

ii 10e6 − 5x ii (1 − 2 x )e x − x

2

2

5 We15 For each of the following find the derivative of the function in i and use this result to deduce the antiderivative of the function in ii. a i sin (x − 5) ii cos (x − 5) b i sin (3x + 2) ii 6 cos (3x + 2) c i cos (4x − 7) ii sin (4x − 7) d i cos (6x − 3) ii 3 sin (6x − 3) e i sin (2 − 5x) ii 10 cos (2 − 5x) f i cos (3 − 4x) ii −2 sin (3 − 4x) 12 x 20 g i loge (5x + 2) ii h i loge (x2 + 3) ii 2 5x + 2 x +3 i i loge (x2 − 4x) ii

x−2 x2 − 4x

6 We16

Differentiate i and hence find an antiderivative of ii. 2[ x cos cos ( x ) − sin ( x )] sin ( x ) a i x cos (x) + 2 sin (x) ii x sin (x) b i ii x x2 c i e x sin (x) ii 3e x [sin (x) + cos (x)] d i x sin (x) ii x cos (x) e i x ex ii x ex

7 For each of the following differentiate i and use this result to antidifferentiate ii. a i (2x − 3x2)6 ii 6x5(1 − 3x)(2 − 3x)5 x3 + 2x

b i 8 We17

a Show that

9 a Show that 10

x3 + 2x

3x − 2 1 =3+ . x −1 x −1

8x − 7 5 =4+ . 2x − 3 2x − 3

6x − 5 − 4 = 3+ . 3 − 2x 3 − 2x 2

b Hence, find ∫

3x − 2 dx. x −1

b Hence, find ∫

5x + 8 dx. x+2

b Hence, find ∫

8x − 7 dx. 2x − 3

b Hence, find ∫

6x − 5 dx. 3 − 2x

If y = loge[cos (x)]: a find

dy . dx

13 Differentiate 14

3x 2 + 2

5x + 8 2 =5− . x+2 x+2

a Show that

11 a Show that 12

ii

b Hence, find ∫ ta tan ( x ) dx. cos ( x ) 1 and hence find an antiderivative of 2 . sin ( x ) sinn ( x )

Differentiate loge (3x2 − 4) and hence find an antiderivative of

x . −4

3x 2

15 Differentiate sin (ax + b) and hence find an antiderivative of cos (ax + b). (Here, a and b are constants.) 16 Differentiate cos (ax + b) and hence find an antiderivative of sin (ax + b). (Here, a and b are constants.)

434


17 Differentiate eax + b and hence find an antiderivative of eax + b. (Here, a and b are constants.) 18 Antidifferentiate each of the following. a sin (3π x + 1) b cos (1 − 4π x) π x  πx  d sinn  2 + e 3 cos  + 5    2  3

9d

eBook plus

c eπ x + 3

Digital doc

f cos (x)esin (x)

WorkSHEET 9.1

approximating areas enclosed by functions


int-0254 Approximating areas enclosed by functions

There are several ways of finding an approximation to the area between a graph and the x-axis. We will look at two methods: 1. the left rectangle method 2. the right rectangle method

The left rectangle method Consider the area between the curve f (x) shown at right, the x-axis y f(x) and the lines x = 1 and x = 5. If the area is approximated by ‘left’ rectangles of width 1 unit then the top left corner of each rectangle touches the curve at one point. So, the height of rectangle R1 is f (1) units and the area of R1 = 1 × f (1) square units (area of a R1 R2 R3 R4 rectangle = height × width). Similarly, the area of R2 = 1 × f (2) square units, 0 1 2 3 4 5 x the area of R3 = 1 × f (3) square units, the area of R4 = 1 × f (4) square units. Therefore, the approximate area under the graph between the curve f (x), the x-axis and the lines x = 1 to x = 5 is 1[ f (1) + f (2) + f (3) + f (4)] square units, (the sum of the area of the four rectangles). If the same area was approximated using rectangle widths of 0.5 there would be 8 rectangles and the sum of their areas would be: 0.5[ f (1) + f (1.5) + f (2) + f (2.5) + f (3) + f (3.5) + f (4) + f (4.5)] square units. From the diagram it can be seen that the left rectangle approximation is less than the actual area under the curve. Worked examPle 18

Find an approximation for the area between the curve f (x) shown and the x-axis from x = 1 to x = 3 using left rectangles of width 0.5 units. The graph shown has the equation f (x) = 0.2x2 + 3.

eBook plus

y

f(x)

Tutorial


3

0 0.5 1 1.5 2 2.5 3 x ThInk

WrITe


Write the number of rectangles and their width.

There are 4 rectangles of width 0.5 units.

Chapter 9

Integration

435

2

Find the height of each rectangle (left) by substituting the appropriate x-value into the f (x) equation.

h1 = f (1) = 0.2(1)2 + 3 = 3.2 h2 = f (1.5) = 0.2(1.5)2 + 3 = 3.45 h3 = f (2) = 0.2(2)2 + 3 = 3.8 h4 = f (2.5) = 0.2(2.5)2 + 3 = 4.25

3

Area equals the width multiplied by the sum of the heights.

Area = width × (sum of heights of 4 rectangles) = 0.5(3.2 + 3.45 + 3.8 + 4.25) = 0.5(14.7)

4

Calculate this area.

5

State the solution.

The approximate area is 7.35 square units.

= 7.35


Store the function to f (x). On a Calculator page, press: • MENU b • 1:Actions 1 • 1:Define 1 Complete the entry line as: Define f (x) = 0.2x2 + 3 Press ENTER ·.

2

To find the approximate area, complete the entry line as: 0.5( f (1) + f (1.5) + f (2) + f (2.5)) Press ENTER ·.

3

Write the answer with the correct units for area.

∴ The approximate area is 7.35 square units.

The right rectangle method y f(x) Consider the area between the curve f (x) shown at right, the x-axis and the lines x = 1 and x = 5. If the area is approximated by ‘right rectangles’ of width 1 unit then the top right corner of each rectangle touches the curve at one point. So, the height of R1 is f (2) units and the area of R1 is 1 × f (2) square units. R1 R2 R3 R4 Similarly, the area of R2 = 1 × f (3) square units, 0 1 2 3 4 5 x the area of R3 = 1 × f (4) square units the area of R4 = 1 × f (5) square units. Therefore, the approximate area between the curve f (x), the x-axis and the lines x = 1 to x = 5 is (R1 + R2 + R3 + R4) = 1[ f (2) + f (3) + f (4) + f (5)] square units. If the same area was approximated with upper rectangle widths of 0.5 units, the sum of their areas would equal: 0.5[ f (1.5) + f (2) + f (2.5) + f (3) + f (3.5) + f (4) + f (4.5) + f (5)] square units. From the diagram it can be seen that the right rectangle approximation is greater than the actual area under the curve.

For an increasing function, left rectangle approximation ≤ actual area ≤ right rectangle approximation. For a decreasing function, left rectangle approximation ≥ actual area ≥ right rectangle approximation.

436


Worked Example 19

Find an approximation for the area in the diagram in worked example 18 using right rectangles of width 0.5 units. f (x) = 0.2x2 + 3 Think

Write

1

Find the number of rectangles and the height of each one (from left to right).

There are 4 rectangles: h1 = f (1.5) = 0.2(1.5)2 + 3 = 3.45 h2 = f (2) = 0.2(2)2 + 3 = 3.8 h3 = f (2.5) = 0.2(2.5)2 + 3 = 4.25 h4 = f (3) = 0.2(3)2 + 3 = 4.8

2

Area is the width of the interval multiplied by the sum of the heights.

Area = 0.5(3.45 + 3.8 + 4.25 + 4.8)

3

Calculate the area.

4

State the solution.

The approximate area is 8.15 square units.

= 0.5(16.3) = 8.15

It can be seen that the left rectangle approximation (7.35 units) is less than the right rectangle approximation (8.15 units). If the area is divided into narrower strips, the estimate of the area would be closer to the true value. Worked Example 20

With width intervals of 1 unit, calculate an approximation for the area between the graph of f (x) = x2 + 2 and the x-axis from x = −2 to x = 3 using: a left rectangles b right rectangles c averaging of the left and right rectangle areas. Think

Write

1

Sketch the graph of f (x) over a domain which exceeds the width of the required area

2

Draw the left and right rectangles.

y

y = x2 + 2

2 −2 −1

0 1 2 3 x = Left rectangles = Right rectangles

a

1

2

Calculate the height of the left rectangles by substituting the appropriate values of x into the equation for f (x). Note the two rectangles to the right and left of the origin have the same height and are equal in area. Find the area by multiplying the width by the sum of the heights.

a Left rectangle heights:

f (−2) = (−2)2 + 2 = 6 f (−1) = (−1)2 + 2 = 3 f (0) = 02 + 2 = 2 f (1) = 12 + 2 = 3 f (2) = 22 + 2 = 6

Area = 1(6 + 3 + 2 + 3 + 6) = 20 Using left rectangles, the approximate area is 20 square units.


437

b

1

2

Calculate the height of the right rectangles by substituting the appropriate values of x into the equation for f (x).

b Right rectangle heights:

Find the area by multiplying the width by the sum of the heights.

Area = 1(3 + 2 + 3 + 6 + 11) = 25 Using right rectangles, the approximate area is 25 square units. 20 + 25 c Average of the areas = 2 = 22.5 The approximate area is 22.5 square units when averaging the left and right rectangle areas and using widths of 1 unit.

c Find the average by adding the area of the

left rectangles and right rectangles and dividing by 2.

f (−1) = 3 (from above) f (0) = 2 f (1) = 3 f (2) = 6 f (3) = 32 + 2 = 11

Note that this average is between the area of the left rectangles and the area of the right rectangles and is closer to the actual area.

1. An approximation to the area between a curve and the x-axis can be found by dividing the area into a series of rectangles that are all the same width. The approximation is found by finding the sum of all the areas of the rectangles. 2. For an increasing function, left rectangle approximation ≤ actual area ≤ right rectangle approximation. 3. For a decreasing function, left rectangle approximation ≥ actual area ≥ right rectangle approximation. exerCISe

9d

approximating areas enclosed by functions 1 We18 Find an approximation for the area between the curve f (x) at right and the x-axis from x = 1 to x = 5 using left rectangles of width 2 units. 2 Find an approximation for the area between the curves below and the x-axis, from x = 1 to x = 5, by calculating the area of the shaded rectangles. y a b y f(x) (5, 4)

4 2

19

(1, 2)

12 0

438

1

5

x

f(x)

(4, 19) (3, 12)

7 (2, 7) 4 (1, 4) 3 0 1 2 3 4 5x


y

f(x)

(3, 3) 3 (1, 2) 2 0 1

3

5

x

3 MC Consider the graph of y = x2 from x = 0 to x = 4 at right. a The width of each rectangle is: A 1 unit B 2 units C 3 units D 4 units E varying b The height of the right-most rectangle is: A 9 units B 4 units C 16 units D 12 units E 1 unit c The area between the curve y = x2 and the x-axis from x = 0 to x = 4 can be approximated by the area of the left rectangles as: A 20 sq. units B 14 sq. units C 18 sq. units D 15 sq. units E 30 sq. units 4 WE 19 a Find an approximation for the area in the diagram at right using right rectangles of width 1 unit. b A better approximation for the area under this curve can be found by averaging the right and left rectangle areas. State this approximate value. 5 Find an approximation for the area between the curves below and the x-axis, from x = 1 to x = 5, by calculating the area of the shaded rectangles. b y c a y (1, 8)

8

8 7

6

0

(1, 8) (3, 8)

5 f(x)

1

x

(5, 5)

5

6 4

0 1

g

5

3

1 2 3 4

y 11 10

1 2 3 4

(2, 11)

(3, 10) (4, 7)

x

1 2 3 4 5

f y 4

f(x)

(1, 4)

f(x)

(2, 6)

(3, 2)

2

(1, 4)

x 0

x

f(x)

(4, 10) (3, 9)

x

y = x2

y

x 5 f(x)

y 10 9

f(x)

(3, 3)

3

0

8 7

0 1 3

e

y

y = x2

0

0

d

y

1

2 3 4 5

x

0

3

1

x

5

y 7

(1, 7) (2, 5)

5 4

0

(4, 7)

1

2

3

4

5

f(x) x

6 WE20 With width intervals of 1 unit calculate an approximation for the area between the graph of f (x) = x2 + 4 and the x-axis from x = 1 to x = 4 using: a left rectangles b right rectangles c averaging of the left and right rectangle areas.

y y = x2 + 4

0

1

2 3 4

x


439

7 Find the approximate area between the curves below and the x-axis, over the interval indicated, by calculating the area of the shaded rectangles. Give exact answers. y a y b y = ex y = −x2 + 3x + 8

0

1 2 3 4 x = 1 to x = 4

c y

x −1 0 1 2 x = −1 to x = 2

x

y = loge(x)

d y

0

0 1 2 3 4 5 x x = 1 to x = 5

e y

f(x) =

1– 3 3x

0 1 2 3 4 5 x = 1 to x = 5

g

y

− 3x2 + 8x

f

x

y = (x − 4)2

1 2 3 4 5 6 x = 2 to x = 6

f(x) = −x2 − 4x

−3 −2.5 −2 −1.5 −1

x

y

0

x

x = −3 to x = −1

x = 2 to x = 6 2 3 4 5 6 x

0

y = x3 − 6x2

8 Calculate an approximation for the area between the graph of y = x(4 − x), the x-axis and the lines x = 1 and x = 4, using interval widths of 1 unit and: a left rectangles b right rectangles c averaging the left and right rectangle areas. 9 Calculate an approximation for the area under the graph of y = x2 − 4x + 5 to the x-axis between x = 0 and x = 3, using interval widths of 0.5 units and: a left rectangles b right rectangles c averaging the left and right rectangle areas. 10 Find an approximation for the area under the graph of y = 2x between x = 0 and x = 3, using interval widths of 1 unit, by averaging the left and right rectangle areas.

440


9e

The fundamental theorem of integral calculus Consider the region under the curve f (x) between x = a and x = b, where f (x) ≥ 0 and is continuous for all x ∈ [a, b]. Let F (x) be the function that is the measure of the area under the curve between a and x. F (x + h) is the area under the curve between a and x + h and F (x + h) − F (x) is the area of the strip indicated on the graph. The area of the strip is between the areas of the left and right rectangles; that is, f (x)h < F (x + h) − F (x) < F (x + h)h or f ( x ) <

F ( x + h) − F F(( x ) < f ( x + h), h ≠ 0 (dividing by h). h

y

0

y = f(x)

a x x+h b F(x) F(x + h) − F(x)

x

As h → 0, f (x + h) → f (x) F ( x + h) − F ( x ) = f ( x) h that is, F ′(x) = f (x) (differentiation from first principles). lim

or

h→0

F ( x ) = ∫ f ( x ) dx

Therefore, that is, F (x) is an antiderivative of f (x)

∫ f ( x) dx = F ( x) + c

or but when x = a,

∫ f ( x) dx = F (a) + c = 0 (as the area defined is zero at x = a) c = −F F((a).

or Therefore,

∫ f ( x) dx = F ( x) − F (a)

and when x = b,

∫ f ( x) dx = F (b) − F (a).

That is, the area under the graph of f (x) between x = a and x = b is F (b) − F (a).

∫ f ( x) dx is the indefinite integral, which represents the general antiderivative of the function

being integrated. This is the fundamental theorem of integral calculus and it enables areas under graphs to be calculated exactly. It applies only to functions that are smooth and continuous over the interval [a, b]. b

It can be stated as area = ∫a f ( x ) dx = [ F ( x )]ba [do not add c as F (x) is an antiderivative of f (x)] = F (b) − F (a) a and b are called the terminals of this definite integral and indicate the domain over which the integral is taken.

Chapter 9

Integration

441

b

∫a

f ( x ) dx is called the definite integral because it can be expressed in of its terminals a and b, which are usually real numbers. In this case the value of the definite integral is a real number and not a function. The function being integrated, f (x), is called the integrand.

Properties of definite integrals

eBook plus

Definite integrals have the following five properties. 1.

a

∫a

b

Digital doc

f ( x ) dx = 0

Investigation Definite integrals c

b

a

c

2.

∫a

f ( x ) dx = ∫ f ( x ) dx + ∫ f ( x ) dx, a < c < b

3.

∫a k f ( x) dx = k ∫a

4.

∫a [ f ( x) + g( x)] dx = ∫a

5.

∫a

b

b

f ( x ) dx

b

b

b

f ( x ) dx =

−

a

∫b

b

f ( x ) dx + ∫ g( x ) dx a

f ( x ) dx

Worked examPle 21

eBook plus

Evaluate the following definite integrals. 3

a

∫0

b

∫1 (2 x + 1)3 dx

Tutorial

int-0566

( 3 x 2 + 4 x − 1) dx

2

Worked example 21

4

ThInk

WrITe


b

1

Antidifferentiate each term of the integrand and write in the form [ F ( x )]ba.

a

3

∫0 (3x 2 + 4 x − 1) dx = [ x 3 + 2 x 2 − x ]30

2

Substitute values of a and b into F (b) − F (a).

= [33 + 2(3)2 − 3] − [0 3 + 2(0)2 − 0] =

3

Evaluate the integral.

= 42 − 0 = 42

1

Express the integrand with a negative power.

b

2

4



2

−

∫1  (2 x + 1)3  dx = ∫1 4(2 x + 1) 3 dx 2

2

Antidifferentiate by rule.

 4(2 x + 1) − 2  =   2 × − 2  1

2

− =  − (2 x + 1) 2   1

3

442

Express the integral with a positive power.


2

− 1  =  2  (2 x + 1)  1

2

 −1  −1 =  2 − 2  5  3  −1 1 = + 25 9 16 = 225

Evaluate the definite integral.

5

 −1  = 2  (2 x + 1)  1

Substitute the values of a and b into F (b) − F (a) where a = 1 and b = 2.

4


On a Calculator page, press: • MENU b • 4:Calculus 4 • 3:Integral 3 Alternatively, select the definite integral template from the expression template and fill in the required fields. Complete the entry lines as: 3

∫0 (3x 2 + 4 x − 1) dx 2

4



∫1  (2 x + 1)3  dx Press ENTER · after each entry. 2

Write the answers.

3

a

∫0 (3x 2 + 4 x − 1) dx = 42

b

∫1  (2 x + 1)3  dx = 225

2

4



16

Worked Example 22

Find the exact value of each of the following definite integrals. a

2π

∫π

 x sin   dx  6

b

∫0 ( e 3 x − e 1

−3x

) dx

Think a

Write

1

Antidifferentiate the integrand, writing it in the form [ F ( x )]ba.

2

Substitute values of a and b into F (b) − F (a).

a

2π

∫π

2π

  x  x  sin   dx =  − 6 cos     6  6  π 

−  2π    − π  =  6 cos  6   −  6 cos  6       − π   − π  =  6 cos    −  6 cos     3    6  


443

3

 −  1   −  3   =  6  2   −  6  2        


= [− 3] − [− 3 3 ] = −3+ 3 3 b

1

Antidifferentiate the integrand, using [ F ( x )]ba.

2

Substitute values of a and b into F (a) − F (b).

3

Evaluate.

b

1

∫0 (e3 x − e

−3 x

) dx 1

−   =  13 e3 x + 13 e 3 x  0  −   =  13 e3 + 13 e 3  −  13 e 0 + 13 e 0   

= 13 e3 + 13 e = 13 ( e3 + e

−3

−3

− 23

− 2)

Worked Example 23

If

k

∫0 8 x dx = 36, find k.

Think

Write


2

Antidifferentiate the integrand, using [ F ( x )]ba.

3

Substitute the values of a and b into F (a) − F (b). Simplify the integral.

4

Solve the equation.

k

∫0 8x dx = [4 x 2 ]0k So [4 x 2 ]k0 = 36 [4k2] − [4(0)2] = 36 4k2 − 0 = 36 4k2 = 36 k2 = 9 k=± 9 k = 3 or −3

Method 2: Technology-enabled 1 On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 • MENU b • 4:Calculus 4 • 3:Integral 3 Complete the entry line as: k solve ∫ (8 x ) dx = 36, k   0 

Press ENTER ·. 2

444

Write the solutions.

k

Solving ∫ (8 x ) dx = 36 for k implies k = −3 or k = 3.


0

Sigma notation

y

An alternative notation for the definite integral of f (x) for x ∈[a, b] is the sigma notation (meaning ‘the sum of’). y = f(x)

(xi, f(xi))

a

x

b

x

Divide the interval [a, b] into n equal subintervals with the ith subinterval of width δxi and height f (xi), i ∈[1, n]. The area of the rectangle formed by the ith subinterval is δAi = f (xi) × δxi. n

b

As n → ∞, δxi → 0 and ∑ δ Ai → ∫ f ( x ) dx . a

1

So

n

b

∑ f ( xi )δ xi δ x→0

∫a

f ( x ) dx = lim

i =1

Worked Example 24

The interval [1, 3] is divided into n equal subintervals by the points x0, x1, . . . xn − 1, xn, where 1 = x0 < x1 < . . . < xn − 1 < xn = 3. Let δx = xi − xi − 1 for i = 1, 2, . . . n. n

∑ ( 6 x i 2δ x ) . δ x→ 0

Rewrite as a definite integral and thus evaluate lim Think 1

2

i=1

Write n

Recognise the alternate notation for the definite integral.

∑ (6 xi 2δ x) = ∫ 1 6 x 2 dx δ x→0

Evaluate.

∫ 1 6 x 2 dx =  2 x3  1

3

lim

i =1

3

3

= 2 × 27 − 2 × 1 = 54 − 2 = 52

1. The fundamental theorem of calculus is where F (x) is an antiderivative of f (x). 2. The expression

b

∫a

b

∫a

f ( x ) dx = [ F ( x )]ba = F (b) − F (a).

f ( x ) dx is called the definite integral where a and b are the

terminals and represent the upper and lower values of x. 3. Properties of definite integrals a

(a) ∫ f ( x ) dx = 0 a

b

c

b

a

a

c

(b) ∫ f ( x ) dx = ∫ f ( x )dx + ∫ f ( x ) dx, a < c < b


445

4.

exerCISe

9e

b

b

(c)

∫a k f ( x) dx = k ∫a

(d)

∫a [ f ( x) + g( x)] dx = ∫a

(e)

∫a

b

∫a

b

b

b

Digital doc

SkillSHEET 9.2 Subtracting function values

a

∫b

b

f ( x )dx + ∫ g( x ) dx a

f ( x ) dx

n

f ( x ) dx = lim

δ xi → 0

∑ f ( xi ) δ xi i =1

The fundamental theorem of integral calculus Evaluate the following definite integrals.

1

a

∫0 x 2 dx

d

∫2 x 2 dx

b

1

6

1

e

g

∫ −1

j

∫1 (4 x −2 + 2 x − 6) dx

(6 − 2 x + x 2 ) dx

2

0

m

∫−2

p

∫1

s

∫−2

− 4 ( 2 − 3 x )3

3 2x3

6

0

2

dx

+ 5x 2 dx x

8 − 3x

2 We22

446

−

f ( x ) dx =

1 We21 eBook plus

f ( x ) dx

∫0

d

∫1 (3e6 x + 5x) dx

g

∫32 sin ( x) dx

j

∫π

2

π

− 2 sin

( 3x ) dx

 x 3 cos   dx  6

m

∫π

p

∫0 π  2 + sin

π

∫0

( x 3 + 3 x 2 − 2 x ) dx

−2

h

∫− 4 ( x3 + x − 4) dx

k

∫0 2( x + 4)4 dx

3

3

5 dx (2 x − 7)3

n

∫0

q

∫1 5x dx

3

5

∫3 ( x 2 − 2 x) dx

f

∫1 (3x 2 + 2 x 2 ) dx

i

∫4 3

l

∫ 1 3(5x − 2)4 dx

o

∫1  2 x 2 − 3x

r

∫3

1

4

9

x dx

2

−

4

3

7

1 2x − 5

−1 

 dx

dx

Find the exact value of each of the following definite integrals.

e 4 x dx

2π

2

4

c

dx

a

2π

3

∫0 x3 dx



( 4x ) dx

0

x

b

∫− 2

e

∫1  x + e 2  dx

h

∫π2 3 sin (4 x) dx

k

∫− π cos (2 x) dx

4

e 3 dx 5

x



π

0

0

 x

n

∫− 2π − 7 cos  2  dx

q

∫1 [ x 2 + 3 − 6 sin (3x)] dx

3


1

−

c

∫−1 − 4e 2 x dx

f

∫−3 (e2 x − e 2 x ) dx

i

∫0 5 sin

l

∫−2π 8 cos (4 x) dx

−1

π

−

( 4x ) dx

π

2

−π 2 −4 −π

o

∫

r

∫1  x + 3 cos

π

1

cos (3 x ) dx

( 2x ) dx

3 If

5

∫1

f ( x ) dx = 6, find the value of

4 MC Given that a

5

∫ 1 f ( x) dx = 6,

5

∫ 1 [ f ( x) + 1] dx is equal to: A 16

b

5

∫1 3 f ( x) dx.

B 10

C 11

D 19

E 22

C −5

D 6

E 0

1

∫5 f ( x) dx is equal to: A −6

B 5

5 Evaluate the following. 3

a

∫0

d

∫5π

g

∫1 t

(t 2 − 4t ) dt

10π

 − sin  x   dx  5   

4 −3

t

dt

π 2 0

2 cos (3t ) dt

π



b

∫

e

∫0  e 4 − cos (2 x)  dx

h

∫−1 [3 sin (2 x) − e



x

1

−3x

3

7

c

∫4

f

∫1 4m − 3 dm

2

3 ( p − 3) 2

dp

8

] dx

k

6 WE23 If ∫ (2 x + 3) dx = 4, find k. 0

k

7 If ∫ 3 x 2 dx = 8, find k. 0

8 If ∫

k 1

2 dx = log e (9), find k. x x

a

9 If ∫ e 2 dx = 4, find the value of a. 0

π

10 If ∫ cos (2 x) dx = − k

π 3 , find the value of k given that 0 < k < . 2 4

6

11 Given that ∫ f ( x ) dx = 8, evaluate the following. 2

a

6

∫ 2 2 f ( x) dx

b

2

∫6

f ( x ) dx c

6

∫2

[3 f ( x ) − 3] dx d

6

∫2 [ f ( x) + 2 x] dx

12 WE24 MC The interval [1, 3] is divided into n equal subintervals by the points x0, x1, . . . xn − 1, xn, where 1 = x0 < x1 < . . . < xn − 1 < xn = 3. Let δx = xi − xi − 1 for i = 1, 2, . . . n. n

Then lim

δ x→0

A

1

∫3

∑ ( xi3 δ x) is equal to: i =1

x 3 dx b ∫

3 1

3 x4 dx c  x 3  D 20 E 26 1 4

13 MC The interval [0, 3] is divided into n equal subintervals by the points x0, x1, . . . xn − 1, xn, where 0 = x0 < x1 < . . . < xn − 1 < xn = 3. Let δx = xi − xi − 1 for i = 1, 2, . . . n. n

∑ ( xi 2δ x) is equal to: δ x→0

Then lim

i =1

A 9 b

3

∫0

3 x3 dx c  x 3  0 D 3

0

∫ 3 x 2 dx

E 27


447

9f

Signed areas When calculating areas between the graph of a function f (x) and the x-axis using the definite integral

b

∫a

f ( x ) dx, the area is signed; that is, it is positive or negative. If f (x) > 0, the region is

above the x-axis; if f (x) < 0 it is below the axis. We shall now examine these two situations and look at how we calculate the area of regions that include both.

Region above axis

y

If f (x) > 0, that is, the region is above the x-axis, then b

∫a

y = f(x)

f ( x ) dx > 0, so the value of the definite integral is positive. For example, if f (x) > 0, then the area =

b

∫a

f ( x ) dx. 0

Region below axis

a

b

x

If f (x) < 0, that is, the region is below the x-axis, then b

∫a

f ( x ) dx < 0, so the value of the definite integral is negative. b

For example, if f (x) < 0, then the area = − ∫ f ( x ) dx

y

a

or

b

∫a

f ( x ) dx , as the region is below the x-axis or

y = f(x) a

b x

0

a

area = ∫ f ( x ) dx (reversing the terminals changes the sign). b

Therefore, for areas below the x-axis, ensure that the area has a positive value. (Areas cannot be negative.)

Combining regions For regions that are combinations of areas above and below the x-axis, each area has to be calculated by separate integrals — one for each area above and one for each area below the x-axis. For example, from the diagram, Area = A1 + A2 b

However, ∫ f ( x ) dx = A1 − A2 , because the areas are signed.

y

y = f(x)

A1 a

0 A 2

c

b

a

To overcome this difficulty we find the correct area as: b

c

c

a

Area = ∫ f ( x ) dx − ∫ f ( x ) dx (= A1 − − A2 = A1 + A2 ) b

or = ∫ f ( x ) dx + c

c

∫a f ( x) dx

b

a

c

c

or = ∫ f ( x ) dx + ∫ f ( x ) dx Note: When calculating the area between a curve and the x-axis it is essential that the x-intercepts are determined and a graph of the curve is sketched over the interval required. The term |x| means that we should make the value of x positive even if it is negative. 448


x

Worked Example 25 a Express the shaded area as a definite integral. b Evaluate the definite integral to find the shaded area, giving your

y

y = –x1

answer as an exact value.

0 Think

1

4

x

Write 4

∫1

1 dx x

a

Express the area in definite integral notation where a = 1 and b = 4.

a Area =

b

1

Antidifferentiate the integrand.

b Area =  log e x  1

2

Evaluate.

3

State the solution as an exact answer.

4

= [loge (4)] – [loge (1)] = loge (4) − 0 = loge (4) The area is loge (4) square units.

Worked Example 26

Calculate the shaded area.

y

y = x2 − 4x

0

Think 1

Express the area in definite integral notation, with a negative sign in front of the integral as the region is below the x-axis.

2

Antidifferentiate the integrand.

3

Evaluate.

4

State the solution.

1

3 4

x

Write 3

Area = − ∫ ( x 2 − 4 x ) dx 1

= − [ 1 x 3 − 2 x 2 ]13 3

= − [( 3 (3)3 − 2(3)2 ) − ( 3 (1)3 − 2(1)2 )]

1 − = [(9 − 18) − ( 3 − 2)]

− − − 2 = [ 9 − ( 1 3 )]

= − [− 9 + 1 23 ]

= − (− 7 13 )

= 7 13

1

1

1

The area is 7 3 square units.


449

Worked examPle 27 y

a Express the shaded area as a definite integral. b Calculate the area. c Calculate the area using a CAS calculator.

0

−2 ThInk

y = x3 − 4x 2

WrITe 0

2

∫−2 ( x3 − 4 x ) dx − ∫0 ( x3 − 4 x ) dx

a

Express the area above the x-axis as an integral and the area below the x-axis as an integral. For the area below the x-axis, take the negative of the integral from 0 to 2.

a Area =

b

1

Antidifferentiate the integrands.

b =  14 x 4 − 2 x 2  − −  14 x 4 − 2 x 2  2 0

2

Evaluate.

= 

3

Simplify.

4

State the solution. On a Calculator page, press: • MENU b • 4:Calculus 4 • 3:Integral 3 Complete the entry line as:

= [0 − (4 − 8)] − [(4 − 8) − 0] = 4 − (−4) =8 The area is 8 square units.

c

1

x

0

0

2

( 14 (0)4 − 2(0)2 ) − ( 14 (−2)4 − 2(−2)2 ) − ( 14 (2)4 − 2(2)2 ) − ( 14 (0)4 − 2(0)2 ) 

c

2

∫ − 2 ( x3 − 4 x) dx − ∫ 0 ( x3 − 4 x) dx Press ENTER ·. Alternatively, take the absolute value for the area below the x-axis as shown.

2

Write the area with the correct units.

∴ The area is 8 square units.

Worked examPle 28

eBook plus

a Sketch the graph of y = ex − 2 showing all intercepts and using exact values

for all key features. b Find the area between the curve and the x-axis from x = 0 to x = 2. ThInk a

450

1

WrITe

Find the x-intercept by letting y = 0 and solving for x.

a When y = 0, ex – 2 = 0


ex = 2

Tutorial


loge (ex) = loge (2) x = loge (2) (or approximately 0.69) so the x-intercept is loge (2). When x = 0, y = e0 − 2 =1−2 = −1 so the y-intercept is −1.


b

2

Find the y-intercept by letting x = 0.

3

Note the vertical translation and hence sketch the graph showing the appropriate horizontal asymptote and intercept. Shade the region required.

1

Express the area above the x-axis as an integral and the area below the x-axis as an integral. Subtract the area below the x-axis from the area above the x-axis. Antidifferentiate the integrands.

3

Evaluate. (: eloge (a) = a)

4

Simplify.

5

State the solution.

0 −1

−2

4

2

y y = ex − 2

−2

b Area =

loge2

x 2 y=−2

2

log e ( 2)

∫loge (2) (e x − 2) dx - ∫0

= [e x − 2 x ]2log

(e x − 2) dx

log e ( 2)

e ( 2)

− [e x − 2 x ]0

= [e2 − 2(2)] − [eloge (2) − 2 loge (2)] − [[eloge (2) − 2 (loge (2)] − [e0 − 2(0)]] = [e2 − 4] − [2 − 2 loge (2)] − [[2 − 2 loge (2)] − [1 − 0]] = e2 − 4 − 2 + 2 loge (2) − 2 + 2 loge (2) + 1 = e2 − 7 + 4 loge (2) The area is [e2 − 7 + 4 loge (2)] or approximately 3.162 square units.

1. If f (x) > 0, area =

b

∫a

f ( x ) dx .

b

2. If f (x) < 0, area = −∫ f ( x ) dx , as the region is below the x-axis or a

b

∫a

=

= ∫ f ( x ) dx , (reversing the terminals changes the sign).

f ( x ) dx , or

a

b

y

3. Check if the required area lies above and below the x-axis. 4. Area =

b

∫c

f ( x ) dx − ∫ f ( x ) dx (= A1 − −A2 = A1 + A2)

b

or = ∫ f ( x ) dx + c

5. e

log e ( a )

y = f(x)

c

a

A1

c

∫a f ( x) dx

a

0 A 2

c

b

x

=a


451

Exercise

9f

Signed areas y

1 Find the area of the triangle at right. a geometrically b using integration

y=x

0

x

4

y

2 Find the area of the triangle at right. a geometrically b using integration

3 0

3 y=3−x

x

3 WE25a Express the following shaded areas as definite integrals. a y b y c y = 2x

y

y = x2

4

01 y

d

−3

y = 3x2

y

e

x

−1 0 y

g

0

x

3

0

y = ex

j

x y=4–x

y = x3 − 9x2 + 20x

1

0

3

x

1 2 y

f

0

−2

x

h y

x

y = −x3 − 4x2 − 4x

i

y 2

y = 2 sin (2x)

−2x

y=e

1 0

−1 y

4

x

1

0

0

1

4

π – 2

x

x

y = cos (–3x)

1 0

x

3— π 2

4 WE25b Evaluate each of the definite integrals in question 3 to find the shaded area. Give your answer as an exact value. 5 For each of the following, sketch a graph to illustrate the region for which the definite integral gives the area.

452

3

a

∫0 4 x dx

d

∫−1 (4 − x 2 ) dx

g

∫2 loge

1

4

( 2x ) dx

2

b

∫1 (6 − x) dx

e

∫1

h

4

x dx

π 2 3 sin (2 x ) dx 0

∫


3

c

∫1 x 2 dx

f

∫−3 2e x dx

0

6 WE26 Calculate each of the shaded areas below. y y a b − y=x−2

0

y y = x2 − 4

c

x

−1 0

−2

x

2

y = 4 − 2x

0

2

−2 y

d −2

−1

x

0

y

e

y = x3

y −1

y = x3 + 2x2 − x −2

0

h

x

1

0 1 x

−1

y = 1− x2

g

y

f

x

0

−2

x

y

1– 2

0

1

i

y y = sin (x)

x 0

y=

y = −ex

π

−e−2x

3π — 2

2π

x

y

j

−2π

x

−π 0

y = 2 cos (–2x )

7 MC a The area between the graph, the x-axis and the lines x = −2 and x = −1 is equal to: 2

A

∫1

C

∫− 1 f ( x) dx

f ( x ) dx

0

y

−1

B

∫− 2 f ( x) dx

D

∫2 f ( x) dx

1

0

−2

−1

3 y = f(x)

x

E − ∫ − f ( x ) dx 2 b The area between the graph, the x-axis and the lines x = −2 and x = 3 is equal to: 3

−2

1

3

0

3

A

∫0 f ( x) dx + ∫0

C

∫− 2 f ( x) dx + ∫1

E

∫− 2 f ( x) dx − ∫0 f ( x) dx

f ( x) dx

B

f ( x) dx

D

3

∫− 2 f ( x) dx −2

∫3

f ( x ) dx

8 WE27a Express the following shaded areas as definite integrals which give the correct area. b c a y y y g(x) f(x) 0

2

5

x

−3

0 1

3

x

−3

−1

0 2

x h(x)


453

y

d

−5 −4 −2

y

e

0

f(x)

x

g(x)

0

−3 −2

2 3

x

9 mC Examine the graph. y

0

−2

1

3

x

y = x3 − 2x2 − 5x + 6

a The area between the curve and the x-axis from x = −2 and x = 1 is equal to: 1

B 154 sq. units

3

E 10 sq. units

A 1712 sq. units D −154 sq. units

3

1

C −1712 sq. units

b The area between the curve and the x-axis from x = 1 and x = 3 is equal to: A −6 2 sq. units B 2 sq. units C −513 sq. units 3

D 513 sq. units

E 6 2 sq. units 3

c The area between the curve and the x-axis from x = −2 and x = 3 is equal to: 5

3

A 1012 sq. units

B 114 sq. units

D 12 sq. units

E 2112 sq. units

5

C 2212 sq. units

1

10 We28 Sketch the graph of the curve y = x2 − 4, showing all intercepts and using exact values for all key features. Find the area between the curve and the x-axis: a from x = 0 to x = 2 b from x = 2 to x = 4 c from x = 0 to x = 4. 11 Sketch the graph of the curve y = x3 + x2 − 2x, showing all intercepts. Find the area between the curve and the x-axis between the lines: a x = −2 and x = 0 b x = 0 and x = 1 c x = −2 and x = 1. 12 Sketch the graph of the curve y = 1 + 3 cos (2x) over [0, π]. Find the exact area between the curve and the x-axis from: π a x = 0 to x = 4 3π b x= to x = π. 4 eBook plus Digital doc

WorkSHEET 9.2

454

13 Sketch the graph of f (x) = x − 1 and find the area between the curve and the x-axis and the lines x = 2 and x = 3. Give both an exact answer and an approximation to 3 decimal places. 14 Find the exact area between the curve y = 1 , the x-axis and the lines x = 12 and x = 2. x 15 Find the exact area bounded by the curve g(x) = ex + 2, the x-axis and the lines x = −1 and x = 3.


9g

Further areas Areas bound by a curve and the x-axis For graphs with two or more x-intercepts, there is an enclosed region (or regions) between the graph and the x-axis. The area bound by the graph of f (x) and the x-axis is: −

b

∫a

or

y

f ( x ) dx (negative because the area is below the x-axis) b

∫a

y = f(x)

0

a

x

b

f ( x ) dx . y

The area bound by the graph of g(x) and the x-axis is: b

c

∫c g( x) dx − ∫a g( x) dx or

c

b

∫a g( x) dx + ∫c g( x) dx

a

c

0

b

x

That is, if the graph has two x-intercepts then one integrand is y = g(x) required. If the graph has three x-intercepts then two integrands are required, and so on. Note: Wherever possible it is good practice to use sketch graphs to assist in any problems involving the calculation of areas under curves. Worked Example 29 a Sketch the graph of the function g(x) = (3 − x)(2 + x). b Find the area bound by the x-axis and the graph of the function. Think a

b

Write

1

Determine the type of graph by looking at the number of brackets and the sign of the x-.

2

Solve g(x) = 0 to find the x-intercepts.

3

Sketch the graph.

1

Shade the region bound by g(x) and the x-axis.

a g(x) = (3 − x)(2 + x) is an inverted parabola.

For x-intercepts, g(x) = 0 (3 − x)(2 + x) = 0 x = 3 and x = −2. The x-intercepts are −2 and 3. b

y

x 3 y = g(x)

−2 0

3

∫− 2 (6 + x − x 2 ) dx

2

Express the area as an integral.

Area =

3

Evaluate.

= [6 x + 12 x 2 − 13 x 3 ]3− 2

= [6(3) + 2 (3)2 − 3 (3)3]

1

1

1

1

−[6(−2) + 2 (−2)2 − 3 (−2)3]


455

4

9

8

= (18 + 2 − 9) − (−12 + 2 + 3 )

= 13 2 − (−7 3 )

= 13 2 + 7 3

= 20 6

1 1

1

1

5 5

The area bound by g(x) and the x-axis is 20 6 square units.

State the solution.

Finding areas without sketch graphs When finding areas under curves that involve functions whose graphs are not easily sketched, the area can be calculated providing the x-intercepts can be determined. Note: The symbol  f (x)is known as the absolute value of f (x), which means that we must make the f (x) function or value positive whenever it is negative. For example,  −5= 5

 3= 3

−2 3

2

=3

As areas cannot be negative, taking the absolute value of the integrands involved in a problem will ensure that all areas are made positive. y For example, The shaded area at right =

or

=

b

∫c

b

∫c

f ( x ) dx + f ( x ) dx +

c

∫a f ( x) dx

y = f(x) a

c 0

b

c

∫a f ( x) dx

Worked Example 30 a Find the x-intercepts of y = sin (2x) over the domain [0, 2π]. b Calculate the area between the curve, the x-axis and x = 0 and x = π. Think a

Write

1

To find the x-intercepts, let y = 0.

2

Solve for x over the given domain.

a For x-intercepts, y = 0

sin (2x) = 0 2x = 0, π, 2π, 3π, 4π, etc.

b

456

1

Pick the x-intercepts that are between the given end points of the area.

2

State the regions for which it is necessary to calculate the area.

b x=

x = 0,

3π π , π , , 2π 2 2

π is the only x-intercept between 0 and π. 2 π

Area =


π

∫02 sin (2x) dx + ∫π sin (2x) dx 2

x

π

3

Evaluate the absolute value of the integral for each region.

π

= [− 12 cos (2x )] 2 + [− 12 cos (2x )]π

0

=

[− 12

cos

2

(π ) − (− 12

cos (0))]

+ [− 12 cos (2π ) − (− 12 cos (π ))]

= [ 12 − (− 12 )] + [− 12 − ( 12 )]

= 1 + −1 =1+1 =2

4

Add the result to give the total area.

5

State the solution.

The area is 2 square units.

Worked Example 31 a Differentiate loge (x2 − 1).

x . −1 x c Find the area between the graph of 2 , the x-axis, x = 2 and x = 3, giving your answer correct x −1 to 2 decimal places. b Hence, find an antiderivative of

x2

Think a

Write a Let y = loge (x2 − 1)

1

Let y equal the expression to be differentiated.

2

Express u as a function of x in order to apply the chain rule for differentiation. (Let u equal the function inside the brackets.)

3

Find

4

Write y in of u.

5

Find

dy . du

dy 1 = du u

6

Find


So

dy 1 = × 2x dx u

Let u = x2 − 1

du . dx

du = 2x dx

y = loge (u)

=

b

dy

∫ dx dx = y + c, express the

1

Since

2

relationship in integral notation. dy Remove a factor from so that it dx resembles the integral required.

b

2x

2x −1

x2

∫ x 2 − 1 dx = loge 2∫

x2

x2 − 1 + c

x dx = log e x 2 − 1 + c −1


457

3

∫

Divide both sides by the factor in order to obtain the required integral.

x dx = 12 log e x 2 − 1 + c x2 − 1

An antiderivative of c

1

Find the x-intercepts. x (For 2 = 0, the numerator = 0.) x −1

2

If the x-intercepts are not between the terminals of the area, find the area by evaluating the integrand.

c For x-intercepts,

Area =

State the solution.

x 1 is 2 log x2 − 1. −1

x =0 x2 − 1 x=0

x

∫2 x 2 − 1 dx 3

1 2 =  2 log e x − 1  2

=  1 log e (32 − 1)  −  1 log e (22 − 1)  2 2

=

=

3

3

x2

1 2

log e (8) − 12 log e (3)

() = 12 log e ( 83 ) 1 log e 83 2

The area is 12 log e square units.

( 83 ) or approximately 0.49

1. For graphs with two or more intercepts, there is an enclosed region (or regions) between the graph and the x-axis. 2. The number of regions is one less than the number of intercepts. 3. Where possible, sketch graphs to make it easier to calculate the areas under curves. 4. As areas cannot be negative, take the absolute values of the integrals. 5. When graphs are not easily drawn, areas can be calculated by finding the x-intercepts and determining whether they are within the bounds of the required area. Exercise

9G

Further areas In the following exercise give all answers correct to 2 decimal places where appropriate, unless otherwise stated. 1 WE29 i Sketch the graph of each of the following functions. ii Find the area bound by the x-axis and the graph of each function. a f (x) = x2 − 3x b g(x) = (2 − x)(4 + x) 2 Find the area bound by the x-axis and the graph of each of the following functions. a h(x) = (x + 3)(5 − x) b h(x) = x2 + 5x − 6 2 c g(x) = 8 − x d g(x) = x3 − 4x2 e f (x) = x(x − 2)(x − 3) f f (x) = x3 − 4x2 − 4x + 16 3 2 g g(x) = x + 3x − x − 3 h h(x) = (x − 1)(x + 2)(x + 5) 3 MC The area bound by the curve with equation y = x2 − 6x + 8 and the x-axis is equal to: A 1 13 square units B 6 23 square units C 12 square units D 3 square units

458

E −1 13 square units


y

4 MC The area between the curve at right, the x-axis and the lines x = −3 and x = 4 is equal to: A

4

∫− 3 f ( x) dx −3

C

∫4

E

∫− 3 f ( x) dx − ∫2

f ( x ) dx

2

2

4

∫2

D 4

4

∫− 3 f ( x) dx + ∫2

B

f ( x) dx

−3

0 2

x

4 y = f(x)

2

f ( x) dx − ∫ − f ( x ) dx 3

f ( x) dx

5 MC The area between the curve y = x2 − x − 6, the x-axis and the lines x = 2 and x = 4 is equal to: 2 3

A 2 65 square units

B

C 5 square units

D 2 43 square units

E 4 12 square units

square units

6 For each of the following: i sketch the graph of the curve over an appropriate domain, clearly labelling any x-intercepts in the interval required. ii find the area between the curve, the x-axis and the lines indicated below. 2 a y = 3 − 3x2, x = 0 and x = 2 b y = , x = 1 and x = 3 x −1 c y = , x = 1 and x = 2 d y = x3 − 4x, x = −2 and x = 1 x2

e y = e2x, x = −2 and x = 0

g y = 2 sin (x), x =

i y = sin (3x), x =

π π and x = 6 3 −π

− and x = π 2 6

f y = e−x, x = 0 and x = 2

π  x h y = cos   , x = and x = 2π 2  2 j y = x x, x ≥ 0, x = 0 and x = 4

7 WE30 For each of the following functions: i find the x-intercepts over the given domain ii calculate the area between the curve, the x-axis and the given lines. Use sketch graphs to assist your workings. − a y = x − 4x 1, x ≠ 0, x = 1 and x = 3 b y = sin (x) − cos (x), x ∈ [0, π], x = 0 and x = π c y = e x − e, x = 0 and x = 3 1 d y = x − 2 , x ≠ 0, x = 12 and x = 2 x x

e y = e 2 , x = −2 and x = 2 f y = x4 − 3x2 − 4, x = 1 and x = 4 g y = (x − 2)4, x = 1 and x = 3 8 Find the exact area of the region enclosed by the x-axis, y = e3x and the lines x = 1 and x = 2. π 9 Find the exact area of the region enclosed by the x-axis, y = −cos (x) and the lines x = and 3 5π x= . 6 10 Find the area bound by y = (x − 1)3, the x-axis and the y-axis. 1 11 a Sketch the graph of y = showing all asymptotes and intercepts. ( x − 3)2 b Find the area under the curve between x = −1 and x = 1.


459

−

12 a Give the equation of the asymptotes for the function f (x) = (x + 2) 3. b Find the area between the curve, the x-axis, x = −1 and x = 1. 13 Find the area bound by the curve y = 3 − e2x, the x-axis, x = −2 and x = 0. (Find the x-intercepts first.) −π 14 Find the area bound by the curve y = 4 − sin (2x), the x-axis, x = and x = π . (Check the 2 x-intercepts first.) 15 WE31 a Differentiate x loge (x). (x > 0) b Hence, find an antiderivative of loge (x). c Find the area bound by the graph of loge (x), the x-axis, x = 1 and x = 4 giving exact answers. x 16 a Differentiate loge (x2 + 2). b Hence, find an antiderivative of 2 . x +2 x c Find the area between , the x-axis, x = −1 and x = 1. 2 +2 x 17 a Find the area between the graph of y = x2, the x-axis, x = 0 and x = 2. y b Use this result to calculate the area between the graph, the y-axis y = e2x and the line y = 4.

0 2

x

18 Find the exact area of the shaded region on the graph y = e2x below. y

y = x2 (2, 4)

0

2

x

19 Find the shaded area below. (Hint: It is easier if you use symmetry.) y

− π–2

0

π – 2

x

y = 2 sin (x)

20

9H

a The area of the region bounded by the y-axis, the x-axis, the curve y = 2e-x and the 3 line x = k, where k is a positive real constant, is square units. Find k. 2 b The area of the region bounded by the y-axis, the x-axis, the curve y = sin (2x) and the line x = k, where k is a positive real constant, is 1 square unit. Find k.

Areas between two curves We shall now consider the area between two functions, f (x) and g(x), over an interval [a, b]. Our approach depends on whether the curves intersect or do not intersect over this interval.

If the two curves f (x) and g(x) do not intersect over the interval [a, b] Here, we may look at three circumstances: when the region is above the x-axis, when it is below the x-axis, and when it crosses the x-axis.

460


y

Region above x-axis b

∫a

The brown shaded area =

b

f ( x ) dx − ∫ g( x ) dx

g(x)

a

b

∫a [ f ( x) − g( x)] dx.

=

f(x)

0 a

x

b

Note: The lower function is subtracted from the higher function to ensure a positive answer. y

Region below x-axis Again, the lower function is subtracted from the higher function to ensure a positive answer. Brown shaded area =

0

a

b

b f(x)

∫a [ f ( x) − g( x)] dx, as f (x) is above g(x)

over the interval [a, b].

y

b

∫a [ f ( x) − g( x)] dx

0

a

g(x)

f(x) g(x)

Region crosses x-axis Shaded area =

x

x

b

Worked Example 32 y

a State the definite integral that describes the shaded area on the

y = 2x + 1

graph at right.

y=x

b Find the area.

0

Think a

x

Write a f (x) = 2x + 1 and g(x) = x

1

State the two functions f (x) and g(x).

2

Subtract the equation of the lower function from the equation of the upper function and simplify.

f (x) − g(x) = 2x + 1 − x =x+1

3

Write as a definite integral between the given values of x.

Area =

b

2

=

2

∫0 [ f ( x) − g( x)] 2

∫0 ( x + 1) dx

b Area = [ 12 x 2 + x ]20

1

Antidifferentiate.

2


3

State the area.

The area is 4 square units.

= [ 12 (2)2 + 2] − [ 12 (0)2 + 0] = (2 + 2) − (0) =4

Worked Example 33

a Find the values of x where the functions y = x and y = x2 − 2 intersect. b Sketch the graphs on the same axes.

c Hence, find the area bound by the curves. Think a

1

Write/draw

State the two functions.

a y = x and y = x2 − 2


461

2

Find where the curves intersect.

3

Solve for x.

For points of intersection: x = x2 − 2 2 x −x−2=0 (x − 2)(x + 1) = 0 x = 2 or x = −1 b For y = x,

b Find the key points of each

y when x = 0, y = 0 y = x2 − 2 when x = 2, y = 2 (2, 2) when x = −1, y = −1 y=x Line es through (0, 0), (2, 2) and (−1, −1) x 0 (−1, −1) For y = x2 − 2, when x = 0, y = −2 Hence the y-intercept is −2. Parabola also es through (2, 2) and (−1, −1).

function and sketch.

c

c Let f (x) = x and g(x) = x2 − 2

1

Define f (x) and g (x).

2

Write the area as a definite integral between the values of x at the points of intersection.

Area = ∫ − [ f ( x ) − g( x )] dx

= ∫ − ( x − x 2 + 2) dx

3

Antidifferentiate.

= [ 12 x 2 − 13 x 3 + 2 x ]2−1

4


= [ 12 (2)2 − 13 (2)3 + 2(2)] − [ 12 (− 1)2 − 13 (− 1)3 + 2(− 1)]

= (2 − 83 + 4) − ( 12 +

= (3 13 ) − (-1 16 )

= 3 13 + 1 16

= 4 12

5

2

1

State the area.

2

= ∫ − [ x − ( x 2 − 2)] dx 1

2

1

1 3

− 2)

The area is 4 12 square units.

If the two curves intersect over the interval [a, b]

y

Where c1 and c2 are the values of x where f (x) and g(x) intersect over the interval [a, b], the area is found by considering the intervals [a, c1], [c1, c2] and [c2, b] separately. For each interval care must be taken to make sure the integrand is the higher function. Subtract the lower function. So the shaded area equals: c1

c2

∫ a [ g( x) − f ( x)] dx + ∫c

1

a c1 0

g(x)

c2

b

x f(x)

b

[ f ( x ) − g( x )] dx + ∫ [ g( x ) − f ( x )] dx c2

Therefore, when finding areas between two curves over an interval, it must be determined whether the curves intersect within that interval. If they do, the area is broken into sub-intervals as shown above. As with areas under curves, sketch graphs should be used to assist in finding areas between curves.

462


If sketch graphs are not used, the absolute value of each integral, for each sub-interval, should be taken to ensure the correct value is obtained.

Worked examPle 34

eBook plus

a Find the values of x where the graph of the functions f (x) =

4 and x

Tutorial

int-0568 g(x) = x intersect. Worked example 34 b Sketch the graphs on the same axes. Shade the region between the two curves and x = 1 and x = 3. c Find the exact area between f (x) and g(x) from x = 1 to x = 3 using a CAS calculator.

ThInk a

WrITe/draW a f (x) = 4 , g (x) = x

1

State the two functions.

2

Let f (x) = g(x) to find the values of x where the graphs intersect.

For points of intersection, x =

3

Solve for x.

x2 = 4 −4=0 (x − 2)(x + 2) = 0 x = −2 and x = 2

x

4 x

x2

b Sketch f (x) and g(x) on the same axes and

b

y

shade the region between the two curves from x = 1 to x = 3.

f(x) = x4– g(x) = x 01 2 3

c

1

State the area as the sum of two integrals for the two sub-intervals.

2

On a Calculator page, press: • MENU b • 4:Calculus 4 • 3:Integral 3 Complete the entry line as: 2

∫1

c Area =

∫1 ( x − x 2

4

x

) dx + ∫ ( x − 4x ) dx 3

2

( 4x − x ) dx + ∫ ( x − 4x ) dx 3

2

Press ENTER ·.

3

Write the area with the correct units.

 4 ∴ The area is 4 log e   + 1 square units.  3

Chapter 9

Integration

463

1. If two curves f (x) and g(x) do not intersect over the interval [a, b] and f (x) > g(x), then the area enclosed by the two curves and the lines x = a and x = b is found by using the formula: b b b x ) − g( x)] x dx ∫ f ( x) dx − ∫ g( x) dx = ∫ [ f ( x) a

a

a

2. If two curves f (x) and g(x) intersect over the interval [a, b], it is necessary to find the points of intersection and hence find the area of each section, because sometimes f (x) > g(x) and sometimes g(x) > f (x). 3. If sketch graphs are not used to determine which is the upper curve, then it is necessary to take the absolute value or positive value of each integral. exerCISe

9h

areas between two curves 1 We32a State the definite integral that will find the shaded areas on each graph below. y y y a b c 2 y=x

y = 2x

y = 3x

y=x y=x+1

y

d

y = x2

e

y

2 x y = 8 − x2

0

−2

g

0

x

0 1

y

y=x 01

2

x

1

y = x3

0

h

y

f

y = 3x

x

3

x 1 y = 4 − x2

0

x

2

y

−1

y = ex

x 1 y = 9 − x2

0

y = x2 − 5

−1 0

1

x y = −4

y=−e

x

2 We32b Find each of the areas in question 1.

y

f(x)

3 mC Which one of the following does not equal the shaded area? 5

5

∫1 g( x) dx + ∫5 f ( x) dx

f ( x ) dx − ∫ g( x ) dx

D

∫1 [g( x) − f ( x)] dx

C

∫1

E

∫5 [ f ( x) − g( x)] dx

5

1

g(x)

1

B

∫1 g( x) dx − ∫1 5

5

f ( x ) dx

A

0 1

5

1

5 y

g(x)

4 mC The area bound by the curves f (x), g(x) and the lines x = −3 and x = 1 at right is equal to: A

464

−3

∫− 1

[ f ( x ) − g( x )] dx

x

B

f(x)

−1

∫− 3 [ f ( x) + g( x)] dx


−4

−3 −1

0

x

C E

−1

∫− 3

[ g( x ) − f ( x )] dx

D

−3

−1

∫− 3 [ f ( x) − g( x)] dx

∫− 1 [ f ( x) + g( x)] dx y

5 MC The shaded area at right is equal to: A B

4

g(x)

∫0 [ f ( x) − g( x)] dx 3

f(x) 4

∫0 [g( x) − f ( x)] dx + ∫3 [ f ( x) − g( x)] dx

0

4

C

∫0 [g( x) − f ( x)] dx

D

∫0 [ f ( x) − g( x)] dx

E

∫0 [ f ( x) − g( x)] dx + ∫3 [g( x) − f ( x)]

3 4

x

3

3

4

6 WE33 In each of the following: i find the values of x where the functions intersect ii sketch the graphs on the same axes iii hence, find the area bound by the curves. a y = 4x and y = x2 b y = 2x and y = 3 − x2 2 2 c y = x − 1 and y = 1 − x d y = x2 − 4 and y = 4 − x2 2 2 e y = (x + 1) and y = 1 − x f y = x and y = x2 7 WE34 i Find the values of x where the functions intersect. ii Sketch the graphs on the same axes. iii Find the area between f (x) and g(x) giving an exact answer. a y = x3 and y = x b y = 3x2 and y = x3 + 2x 8 Find the area between the pairs of curves below, over the given interval. a y = x3, y = x2, x ∈ [−1, 1] b y = sin (x), y = cos (x), x ∈ [0, π] c y = (x − 1)2, y = (x + 1)2, x ∈ [−1, 1] d y = x3 − 5x, y = 6 − 2x2, x ∈ [0, 3] 1 e y = , y = 4x, x ∈ [ 14 , 1] x f y = e x, y = e−x, x ∈ [0, 1]

π g y = 2 cos (x), y = x − π , x ∈ [0, ] 2 2 x − x − h y = e , y = e , x ∈ [ 2, 1] 9 Find the area between the curve y = ex and the lines y = x, x = 1 and x = 3. 10 Find the area between the curve y = x2 and the lines y = x + 3, x = 1 and x = 3. 2 11 Calculate the area between the curves y = sin (2x) and y = cos (x) from x = 0 to x = π . 2 12 Calculate the area between the curves y = 3 − sin (2x) and y = sin (2x) from x = 0 to x = π . 4 13 Find the exact area bound by the curves y = ex and y = 3 − 2e−x. y 1 14 The graph at right shows the cross-section of a bricked archway. (All measurements are in metres.) a Find the x-intercepts of f (x). b Find the x-intercepts of g(x). c Find the cross-sectional area of the brickwork.

f(x) = 4 − –4 x2

x 0 g(x) = 3 − 1–3 x2


465

15 The diagram below shows the outline of a window frame. If all measurements are in metres, what is the area of glass which fits into the frame? y

y = 2 1–2 − 2x2 y = 2x2

− 1–2

0

x

1– 2

y

16 The diagram at right shows the side view of a concrete bridge. (All measurements are in metres.) Find: a the x-intercepts of the curve b the length of the bridge c the area of the side of the bridge d the volume of concrete used to build the bridge if the bridge is 9 metres wide. 17 The cross-section of a road tunnel entrance is shown at right. (All measurements are in metres.) The shaded area is to be concreted. Find: a the exact area, above the entrance, which is to be concreted b the exact volume of concrete required to build this tunnel if it is 200 metres long.

9I

5 2

x y = 4 − —– 100

0

x

y

0

x f(x) = 5 sin (π–– 30 )

x

Average value of a function Consider the function y = f (x). The average value, yav, for the function y = f (x) over the interval [a, b] is given by: b 1 yav = f ( x ) dx. b − a ∫a b

This can be rearranged to give yav × (b − a) = ∫ f ( x ) dx. a

y

y = f(x)

yav a

x

b

Geometrically, the average value of the function is the height, yav, of the rectangle of width (b − a) that has the same area as the area under the graph of y = f (x) for the interval [a, b]. Worked Example 35

Find the average value of f (x) = 2x2 for the interval [1, 4]. Find the value of x that corresponds to the average value. Think 1

466

Write the relationship for the average value.

Write

yav =


b 1 f ( x ) dx b − a ∫a

4 1 2 x 2 dx ∫ 1 4 −1

2

Identify f (x) = 2x2 and substitute in the values for a and b.

yav =

3

Antidifferentiate.

1  2x3  yav =   3  3 1

4

Evaluate the definite integral and multiply by 1 . 3

4

1   2 × 43   2 × 13  yav =   −  3   3   3 

=

1  128 2  × −  3  3 3

1 126 × 3 3 = 14 ∴ The average value is 14.

5

Solve f (x) = 14 to find x.

=

2 x2 = 14 x2 = 7 x=± 7

6

∴x = 7 as x ∈ [1, 4].

Choose the value of x that satisfies the given interval.

Worked Example 36

Find the average value of f (x) = loge (2x) for the interval [2, 4]. Give your answer in exact form. Think 1

Write the relationship for the average value and substitute the values for a and b.

2

The integral of loge (2x) is not covered in this course. The average value can only be evaluated using a CAS Calculator. On a Calculator page, press: • MENU b • 4:Calculus 4 • 3:Integral 3 Complete the entry line as:

Write

yav =

4 1 log e (2 x ) dx ∫ 2 4−2

1 4 ( ln (2 x)) dx 2 ∫2 Press ENTER •. 3

Write the answer.

∴ The average value f (x) = loge (2x) for the interval [2, 4] is 4 loge (2) − 1.


467

The average value, yav, for the function y = f (x) over the interval [a, b] is given by yav =

b 1 f ( x ) dx . ∫ b−a a

exerCISe

9I

average value of a function 1

2

We35

Find the average value, in exact form, of the function for the given interval.

a y = x3 − x, x ∈ [1, 3]

 π b y = sin (x), x ∈  0,   6

c y = x, x ∈ [1, 4]

d y = e3x, x ∈ [0, 2]

We36 Find the average value of the function, in exact form, for the given interval.

a y = x loge (x), x ∈ [2, 5]

 π b y = tan (2x), x ∈  0,   6

c y = xex, x ∈ [0, 4]

d y = x x + 1, x ∈ [1, 7] −

3 For the function y = e 2x for x ∈ [−1, 0], find: a the average value of the function b the corresponding x-value in exact form. 4

mC The average value of the function f (x) = loge (2x + 1) over the interval [0, 4] is:

A 5

B loge (9)

C

1 [ 9 log e (3) − 4 ] 4

D 9 loge (3) − 4

mC The average value of the function y = sin (2x) over the interval  0,

A

9J

log e (9) 4

4 π

B 2π

C

π 8

D

1 2

E

2 π



E

−9

π is: 4 

log e (3) + 4 4

further applications of integration Differentiation can be applied to rates of change and related rates. If the rate of change of a function is known, antidifferentiation allows the original function to be found. So integration has many practical applications.

Worked examPle 37

eBook plus

The rate of change of position, velocity, of a particle travelling in a straight line is dx − given by = 40 − 10 e 0.4 t m/s, t ≥ 0, where x is measured in metres and t in dt seconds. a Find the velocity: i initially ii after 10 seconds, correct to 2 decimal places. b Find the time taken to reach a velocity of 35 m/s. dx c Sketch the graph of against t. dt d Find the total distance travelled by the particle in the first 10 seconds. 468


Tutorial


Think a

i

write 1

The initial velocity occurs when t = 0. Substitute t = 0 into

2

ii

1

dx . dt

1

Answer with the correct units. dx Substitute = 35. dt

dx − = 40 − 10 e 0.4 t dt − = 40 − 10 e 0.4 × 0 = 40 − 10 e 0 = 40 − 10 = 30 ∴ Velocity is initially 30 m/s. dx − = 40 − 10 e 0.4 ×10 dt − = 40 − 10 e 4 = 39.82

Answer with the correct units. dx Substitute t = 10 into . dt

2

b

a

∴ After 10 seconds, the velocity is 39.82 m/s. dx − b = 40 − 10 e 0.4 t dt 35 = 40 − 10 e

2

Solve for t.

−5

= − 10 e

− 0.4 t

− 0.4 t

e −0.4 t = 0.5 − 0.4t = ln (0.5) ln (0.5) − 0.4 = 1.73 s

t=

3

c

d

1

∴ Time taken is 1.73 seconds.

Answer the question correct to 2 decimal places. To graph dx = 40 − 10 e − 0.4 t , on a dt Graphs page, complete the function entry line as: − f1(x) = 40 − 10 e 0.4 x Press ENTER ·. Note the viewing window settings: Xmin:0, Xmax:15, Ymin:0 and Ymax:55

2

When drawing your graph, label the axes with the given variables. Note the horizontal asymptote dx = 40 is not displayed. The dt asymptotic behaviour of the function, however, is clearly visible.

1

Distance travelled = area under the graph State the distance as a definite integral.

c

d dx = 40 − 10 e −0.4 t

dt

x=

10

∫0

(40 − 10 e

−0.4 t )dt


469

2

Antidifferntiate.

3


4

State the distance travelled with correct units.

10

− =  40t + 25e 0.4 t   0 −

= (40 × 10 + 25e 0.4 ×10 ) − (40 × 0 + 25e − = (400 + 25e 4 ) − (25e 0 ) −4 = 400 + 25e − 25 − = 375 + 25e 4 = 375.46

− 0.4 × 0

∴ The distance travelled in the first 10 seconds is 375.46 metres.

Worked Example 38

The rate of change of pressure, P atmospheres, of a given mass of gas with respect to its volume, dP − k , k > 0. = dV V 2 a If d P = −5 when V = 10, find k. dV b Find the pressure, P, as a function of V given that when P = 10 atmospheres, V = 50 cm3. c Find the volume when the pressure is 20 atmospheres. V cm3, is given by

Think a

1

Write

dP − Substitute the conditions = 5 and V = 10 dV into the given rate.

a

dP − k = dV V 2 −5

b

2

Simplify.

3

Solve for k.

1

Write the rate with the value of k found previously.

2

Antidifferentiate.

=

−

k

(10)

2

−

k 100 ∴ k = 500 b dP = − 500 2 dV V −5

=

P=∫

−

500 V

2

dV

= − 500 ∫ V

470

Substitute the given conditions P = 10 and V = 50 to find c.

4

Write the relationship for P.


−2

500 +c V 500 P= +c V 500 10 = +c 50 c=0 500 ∴P= V =

3

)

dV

c

1

Substitute P = 20.

2

Solve for V.

3

State the answer with correct units.

Exercise

9J

c

500 V 500 20 = V 500 V= 20 ∴ The volume is 25 cm3. P=

Further applications of integration 1 If f ′(x) = (2 − x)2 and the y-intercept of f (x) is 4 , find the rule for f (x). 3 π dy 2 If = 1 − 4 cos (2 x ) and the y-intercept is 2, find the exact value of y when x = . 12 dx 3 The rate of deflection from a horizontal position of a 3-metre diving board when an 80-kg dy − person is x metres from its fixed end is given by = 0.03( x + 1)2 + 0.03, where y is the dx deflection in metres. y (Metres) 0 Board

(Metres) x Deflection

a What is the deflection when x = 0? b Determine the equation that measures the deflection. c Hence, find the maximum deflection. 4 On any day the cost per item for a machine producing n items is given by dC = 40 − 2e 0.01n, where n ∈ [0, 200] and C is the cost in dollars. dn a Use the rate to find the cost of producing the 100th item. b Express C as a function of n. c What is the total cost of producing the first 100 items? d Find the average cost of production for: i the first 100 items ii the second 100 items. 5 WE37 The rate of change of position (velocity) of a racing car travelling down a straight dx stretch of road is given by = t (16 − t ), where x is measured in metres and t in seconds. dt a Find the velocity when: i t = 0 ii t = 4. b Determine: i when the maximum velocity occurs ii the maximum velocity. dx c Sketch the graph of against t for 0 ≤ t ≤ 16. dt


471

d Find the area under the graph between t = 0 and t = 10. e What does this area represent?

6 WE38 The rate at which water is pumped out of a dam, in L/min, t minutes after the pump is dV  πt  = 5 + cos   . started is  40  dt a How much water is pumped out in the 40th minute? b Find the volume of water pumped out at any time, t, after the pump is started. c How much water is pumped out after 40 minutes? d Find the average rate at which water is pumped in the first hour. e How long would it take to fill a tank holding 1600 litres? 7 The rate of flow of water into a hot water system during a 12-hour period is thought to be dV  πt  = 10 + cos   , where V is in litres and t is the number of hours after 8 am.  2 dt dV a Sketch the graph of against t. dt b Find the length of time for which the rate is above 10.5 L/h. c Find the volume of water that has flowed into the system between: i 8 am and 2 pm ii 3 pm and 8 pm. 8 The roof of a stadium has the shape given by the function f : [−25, 25] → R, f (x) = 20 − 0.024x2. The stadium is 75 metres long and its cross-section is shown at right. a Find the volume of the stadium. b The stadium is to have several airconditioners strategically placed around it. Each can service a volume of 11 250 m3. How many airconditioners are required? 9 The cross-section of a channel is parabolic. It is 3 metres wide at the top and 2 metres deep. Find the depth of water, to the nearest cm, when the channel is half full. 10 For any point P on the curve y = x3, prove that the area under the curve is one quarter of the area of the rectangle.

472


y (metres) 20 5 −25

(metres) x 25

0

y

y = x3 P

0

x

11 The arch of a concrete bridge has the shape of a parabola. It is 6 metres high and 8 metres long. a Find the rule for the function corresponding to the arch of the bridge. b Find the area of the shaded region. c If the bridge is 10 metres wide, find the volume of concrete in the bridge. 12 In the figure at right f (x) and g(x) intersect at O and B. a Show that the coordinate of B is (loge (2), 1). b Find the exact area of the region bound by f (x) and g(x). c Show that the sum of the areas under f (x) and g(x), from x = 0 to x = loge (2), is equal to the area of the rectangle OABC.

y (metres) 7 6

−4 y

0

4 5

(metres) x

f(x) = ex − 1 B g(x) = −2e−x + 2

C

O

A

x


473

Summary Antidifferentiation rules

• The relationships between f (x) and ∫ f ( x ) dx are: f (x)

∫ f ( x) dx

a

ax + c

ax n

ax n + 1 +c n +1

(ax + b)n

(ax + b) n + 1 +c a( n + 1)

1 x

loge |x| + c

1 ax + b

1 log |ax + b| + c a

ex

ex + c

ekx

1 kx e +c k −

sin (ax) cos (ax) • ∫ [ f ( x ) ± g( x )] dx =

∫ f ( x) dx ± ∫ g( x) dx

∫ g( x) dx = f ( x) + c, where g(x) = f ′(x) dx

•

1 cos (ax) + c a 1 sin (ax) + c a

∫ kf ( x) dx = k ∫ f ( x) dx • ∫ f ( x) dx is the indefinite integral •

Definite integrals

• The fundamental theorem of integral calculus: b

∫a

b

f ( x ) dx = [ F ( x )]ba = F (b) − F (a) where F (x) is an antiderivative of f (x).

•

∫a

•

∫a f ( x) dx = ∫a f ( x) dx + ∫c

•

∫a [ f ( x) ± g( x)] dx = ∫a

f ( x ) dx is the definite integral

b

c

b

b

b

∫a

•

b

f ( x ) dx =

−

a

∫b

•

b

b

∫a kf ( x) dx = k ∫a

f ( x ) dx

f ( x ) dx, a < c < b b

f ( x ) dx ± ∫ g( x ) dx a

f ( x ) dx

Graphs of the antiderivative function

• • • •

474

For a polynomial function, the graph of f (x) is one degree higher than the graph of f ′(x). The x-intercepts of f ′(x) are the x-coordinates of the turning points of f (x). When f ′(x) is above the x-axis, the gradient of f (x) is positive. When f ′(x) is below the x-axis, the gradient of f (x) is negative.


Approximating areas under curves

• An approximation to the area between a curve and the x-axis can be found by dividing the area into a series of rectangles that are all the same width. The approximation is found by finding the sum of all the areas of the rectangles. • For an increasing function, left rectangle approximation ≤ actual area ≤ right rectangle approximation. • For a decreasing function, left rectangle approximation ≥ actual area ≥ right rectangle approximation. Area under curves b

• Area = ∫ f ( x ) dx, if f ( x ) > 0 for x ∈[a, b] a

y

•

b

Area = − ∫ f ( x ) dx, if f ( x ) < 0, or a

b

∫ a f ( x)dx , for x ∈ [a, b] y

y = f(x)

y = f(x) a

b x

0

0

• Area =

b

∫c

a

x y

c

f ( x ) dx − ∫ f ( x ) dx

y = f(x)

a

b

b

= ∫ f ( x ) dx + c

c

∫a f ( x) dx , if

f ( x ) > 0 for x ∈[c, b]

and f (x) < 0 for x ∈[a, c]

A1 0 A 2

a

c

x

b

Area between curves y

b

• Area = ∫ [ f ( x ) − g( x )] dx, if f ( x ) > g( x ) for x ∈[a, b]

f(x)

a

g(x) 0 a

c

b

a

c

• Area = ∫ [ g( x ) − f ( x )] dx + ∫ [ f ( x ) − g( x )] dx,

b

y

x

g(x) f(x)

if g(x) > f (x) for x ∈ [a, c] and f (x) > g(x) for x ∈[c, b] a 0

c

b

x

Limiting value of a sum

•

b

∫a

0

∑ f ( xi ) δ xi δ x→0

f ( x )dx = lim

i =1

Average value of a function

• yav =

1 b−a

b

∫a

f ( x )dx


475

chapter review x

Short answer

1 Find the equation of the curve f (x) if it es 3x3 − 2 x 2 through (1, −3) and f ′( x ) = . x dy πx = cos   + k, where 2 A particular curve has  4 dx k is a constant, and it has a stationary point (2, 1). Find: a the value of k b the equation of the curve c the value of y when x = 6. dy 3 If y = sin (x2 + 2x), find and hence dx antidifferentiate (x + 1) cos (x2 + 2x).

12 The graph of f: R → R, f ( x ) = e 2 + 1 is shown. The normal to the graph of f where it crosses the y-axis is also shown. y

0

a Find the equation of the normal to the graph of f where it crosses the y-axis. b Find the exact area of the shaded region.

4 A curve has a gradient function f ′(x) = 2ex + k. It has a stationary point at (0, 3). Find: a the value of k b the equation of the curve f (x).

Exam tip a Students often had difficulty correctly finding the y-intercept; (0, 1) was a common answer, leading to the normal being y = −2x + 1. Quite a few students differentiated but then did not find the value of the derivative at x = 0 and simply substituted the expression for the derivative into the equation of the normal, which led to incorrect attempts at algebraic manipulation and nonlinear normals. Some students found a tangent instead. b Most students correctly set up an area expression but some were unable to proceed due to their ‘normal’ not being a linear expression. Many had the correct terminals, although 2 instead of 1 was a popular incorrect value. Arithmetic errors appeared frequently and subtraction mistakes in the integrand were also common. Some students also lost ‘1’ from the equation of the curve. The use of ‘dx’ was surprisingly good. A few students used the area under the curve minus the triangle with some success.

5 Calculate the exact area between the curve y = ex − 4 and the lines y = x, x = 1 and x = 2. (Hint: y = ex − 4 and y = x do not intersect between x = 1 and x = 2.) 6 Evaluate each of the following definite integrals. 0 − 9 a ∫ − dx 1 (2 x + 3) 4 b

∫

2π 3 π 3

cos (2 x ) dx

7 Given that

k

∫0 (4 x − 5) dx = − 2, find two

possible values for k. 1 . x−2 b Find the exact area between the graph of f (x), the x-axis and the lines x = 3 and x = 6.

8 a Sketch the graph of the function f ( x ) =

9 Find the area bound by the x-axis and the curve g(x) = (4 − x)(6 + x). 10 Calculate the area between the curve y = 2 cos (x) π and the lines y = −x, x = 0 and x = . 2 11 Use the method of left rectangles to approximate the area under the curve y = x2 + 1, from x = 1 to x = 4, using interval widths of 1 unit. 476

x


[© VCAA 2007]

Multiple choice

1 The antiderivative of 4 x 3 − A x4 − loge (1 − x) + c B x4 + loge (1 − x) + c c 16x4 − loge (1 − x) + c D 16x4 + loge (1 − x) + c 1 +c E x 4 − (1 − x )2


1 is: 1− x

2 The indefinite integral ∫ (5 x − 4)4 dx is equal to: A 25(5x − 4)5 + c

B 5(5x − 4)5 + c

C (5x − 4)5 + c

D

E

1 (5 x − 4) 5 25

1 (5 x − 4) 5 5

+c

−

−3

−2 9

E

−8(3x

(3x + 4)

D

−3 −3x

−1 2

−3x

e

−5

(3x + 4)

+ 4)

4 The antiderivative of 6e − A −2e 3x + c − C −18e 3x + c E

−2 9

is: − B −3e 3x + c − D −2e 3x +1 + c

+c

x 5 The indefinite integral ∫  cos   − 3 sin ( 3 x )  dx    3 is equal to: x A sin + cos (3x) + c 3

() x B sin ( ) + 3 cos (3x) + c 3 x C 3 sin ( ) + cos (3x) + c 3 x D 3 sin ( ) + cos (3x) + c 3 x E 3 sin ( ) − cos (3x) + c 3 1 3

−

6 An antiderivative of x3 + sin (4x) + e4x is: A 4[x4 − cos (4x) + e4x] x 4 + cos (4 x ) + 14 e 4 x

B

1 4

C

1 4 x 4

D

1 4

[x4 − cos (x) + e4x]

E

1 4

[x4

− 4 cos (4x) + 14 e4x

A

3 3e x + 3 x

2

+3

+c

3

+3x

+ 1)e x + 3 x

is:

is 3( x 2 + 1)e x (x2 B D

3

1 3( x 2 + 1) 1 x3 +3 x e 3

,

+c

+c

E undefined

− x 2 10 If the derivative of loge (5 − x2) is then the 5 − x2 x antiderivative of is: 5 − x2 A − 12 loge (5 − x2) + c B −2 loge (5 − x2) + c

C 12 loge (5 − x2) + c D 2 loge (5 − x2) + c E undefined 11 The approximation for the y (3.5, 9) area under the graph at (3, 6) right from x = 2 to x = 4, (2.5, 4) using the ‘lower (2, 3) rectangles’ is: A 22 sq. units B 14 sq. units x 0 3 2 4 C 11 sq. units D 10 sq. units E 20 sq. units 12 The area under the graph y at right from x = −5 to (−5, 8) x = −1 can be (−4, 6) approximated by the area (−3, 5) (−2, 4) of the ‘upper rectangles’ and is equal to: A 20 sq. units x −5 −1 0 B 21 sq. units C 23 sq. units D 11 12 sq. units E 10 sq. units 13 The interval [0, 4] is divided into n equal subintervals by the points x0, x1, . . . xn − 1, xn, where 0 = x0 < x1 < . . . < xn − 1 < xn = 4. Let δx = xi − xi − 1 for i = 1, 2, . . . n. n

∑ ( xiδ x) is equal to: δ x→0

Then lim

7 If f (x) has a stationary point at (0, 3) and f ′(x) = ex + k, where k is a constant, then f (x) is: A ex − 2x + 2 B ex − x + 2 C ex − x + 2 −x x D e +x+2 E e + 2x + 1

0

8 If the derivative of (x − x2)8 is 8(1 − 2x)(x − x2)7 then an antiderivative of 24(1 − 2x)(x − x2)7 is: A 2(x − x2)8 B 3(x − x2)8 C 12 (x − x2)8 1 2 8 2 8 D 3 (x − x ) E 8(x − x )

+3x

+c

e4x]

− cos (4x) +

3

then the antiderivative of

C e3 x

+c

3 An antiderivative of 2(3x + 4) 4 is: − − A − 23 (3x + 4) 3 B − 23 (3x + 4) 3 + 5 C

9 If the derivative of e x

i =1

A

∫ 4 x dx

B

4

∫0

x2 dx 2

C 0

D 4 E 8

exam TIP The definition of the definite integral as the limiting value of a sum, and its corresponding representation in integral form, is a fundamental property of integral calculus and must be understood.


[© VCAA 2003]

Chapter 9

Integration

477

Questions 20 to 22 apply to the curve with equation f (x) = ex − 1.

4

14 The expression ∫ (3 x − x ) dx is equal to: 0

A 2 B 8 C −2 12 D 20 E 16

20 The graph of f (x) is best represented by: y y A B f(x) f(x) 2

15 The exact value of the definite integral 2

C

−4

A −e − B 2e4 − e 4 − 4 C e − 2e 4 −4 4 D e + e − E e4 − e 4 3e4

()

f(x)

1 8 The shaded area on the graph at right is: A 20 sq. units B −20 sq. units C −16 sq. units D 16 sq. units E −18 sq. units 19 The area bound by the curve on the graph at right and the x-axis is equal to: 5 20 12 1

5 1012

sq. units

5

D −10 12 sq. units 7

E 20 12 sq. units

y

D

−1

f(x) 0

x

f(x) x

−1

21 The area bound by the graph of f (x), the x-axis and the line x = 2 is equal to: A e2 − 1 B e2 − 2 C e2 + 1 D e2 + 2 E e2 − 3

y

y = (x − 2)3

0

4 x

2

22 The area bound by the graph of f (x), the y-axis and the line y = e2 − 1 is equal to: A e2 − 5 B e2 − 3 C e2 + 2 D e2 + 1 E 5 − e2 Use the graph below to answer questions 23 and 24. y

y 0

−2

2

x

y = 2x + 3 0

−1 y = −1 − 3x2

y

y = x2

x

y = x(x + 2)(x − 3)

2 3 The two graphs intersect where x is equal to: A 1 and −3 B −1 and 3 C 1 and 2 D −1 and − 2 E 1 and 3 24 The area bound by the two graphs is equal to: 2 1 A 10 3 sq. units B 7 3 sq. units 1

C −7 3 sq. units

sq. units

B 21 12 sq. units

x

0

x

−1 0

E 3 2 1 7 The shaded area on the graph at right is equal to: A 12 sq. units B 16 sq. units C 10 sq. units D 4 sq. units E 8 sq. units

−1 y

E

D 3 3

478

y

0

π x  16 The exact value of ∫  − 2 cos dx is: 0  3  A −3 B 3 C −3 3

C

x

−

∫ − 2 (4e2 x − 2e 2 x ) dx is:

A

0

1

D 11 3 sq. units

2

0

x

E 6 3 sq. units 25 The average value of the function y = cos (2x) over  π the interval  0,  is:  4 4 2 π A B c π π 8 1 d E 2π 2


Extended response

1 From past records it has been found that the cost rate of maintaining a certain car is dC = 75t 2 + 50t + 800, where C is the accumulated cost in dollars and t is the time in years since the car dt was first used. Find: a the initial maintenance cost b C as a function of t c the total maintenance cost during the first 5 years of use of the car d the total maintenance cost from 3 to 5 years e the maintenance cost for the second year. 2 Over a 24-hour period on a particular autumn day, starting at 12 midnight, the rate of change of the temperature for Melbourne was approximately dT − 5π  πt  cos   , where T is the temperature =  12  dt 12 in °C and t is the number of hours since midnight when the temperature was 10 °C. Find: a the temperature at any time, t b whether the temperature reaches 17 °C at any time during the day c the maximum temperature and the time at which it occurs d the minimum temperature and the time at which it occurs e the temperature at i 2 am i i 3 pm f the time when the temperature first reaches 14.33 °C.

x

3 The diagram at right shows part of the curve with equation y = e 2 . Find: a the coordinate of point A b the equation of the normal to the curve at point A c the coordinate of point B d the coordinate of point C e the area bound by the curve and the lines AB and BC. 4 a Find the derivative of x loge (x). b Hence, find an antiderivative of loge (x). The cross-section of a platform is shown at right. (All measurements are in metres.) c Find the height of the platform. d Find the cross-sectional area of the platform. e Find the volume of concrete required to build this platform if it is 20 metres long.

y –x

Normal

y = e2

B A

C 0

2

x

y 1 0

e−2

e x 1 f(x) = logex


479

5 A thick metal pipe is filled with boiling water and is kept boiling. The temperature, T °C, of the metal in the pipe decreases relative to its distance, x cm, from the centre of the pipe. dT − 20 It is known that and 4 ≤ x ≤ 8. = dx x a Find the rate of change of the temperature in the metal on the outside of the pipe. b Express T as a function of x. c Find the temperature of the metal, correct to 2 decimal places: i when x = 6 cm ii on the outside of the pipe. eBook plus Digital doc


480


x

eBook plus

aCTIvITIeS


• 10 Quick Questions: Warm up with ten quick questions on integration. (page 416) 9A

Antidifferentiation

Digital doc

• SkillSHEET 9.1: Practise substitution and evaluation. (page 425) 9C

Integration by recognition

Tutorial

• We 13 int-0564: Watch a worked example on performing integration by recognition. (page 430) Digital doc

• WorkSHEET 9.1: Determine functions using antidifferentiation. (page 435) 9D

Approximating areas enclosed by functions


• Approximating areas enclosed by functions: Consolidate your understanding of varying approximations to areas enclosed by functions. (page 435) Tutorial

• We 18 int-0565: Watch a worked example on the approximation of the area under a curve. (page 435) 9E

The fundamental theorem of integral calculus

Tutorial

• We 21 int-0566: Watch a worked example on evaluating definite integrals. (page 442)

9H

Areas between two curves

Tutorial

• We 34 int-0568: Watch a worked example on calculating the area between two curves using CAS. (page 463) 9J

Further applications of integration

Tutorial

• We 37 int-0569: Watch a worked example on applications of integration. (page 468) Chapter review Digital doc


Digital docs

• Definite integrals: Investigate the properties of definite integrals. (page 442) • SkillSHEET 9.2: Practise subtracting function values. (page 446) 9F

Signed areas

Tutorial

• We 28 int-0567: Watch a worked example on finding the area bound by an exponential curve above and below the x-axis. (page 450) Digital doc

• WorkSHEET 9.2: Approximate areas between curves and calculate areas between curves using integrals. (page 454)

Chapter 9

Integration

481

EXAM PRACTICE 3

Short answer

Chapters 1 TO 9

y

40 minutes

4

1 The graph of f (x): R → R, where f (x) = 10xe-x is graphed below.

3

y 4

2 1

2 −2

0

2

4

6

−2

a By selecting approximate values from the graph, determine a left-rectangle approximation to the area bounded by the graph, the x-axis and the lines x = 1 and x = 5 by dividing the area into four strips of equal width. dy b If y = (x + 1)e−x, find . dx c Hence determine an exact value for the shaded area. 3 marks

2 Write down an antiderivative of: a y = 4x3 - 2 cos (x) + e x 3 b y = x + x

3

x

4 marks

Multiple choice

8 minutes

Each question is worth 1 mark. 1 Which integral could be evaluated in order to calculate the area enclosed between the graphs of y = f (x) and y = g(x) and the vertical lines x = a and x = b? f x=b

4 The area of the region bound by the curve with  x equation y = a cos   , the x-axis and the line  3 2 marks

5 A spa is being filled with water at the rate of 30 litres per minute. The volume (V litres) of water in the spa is related to its depth (x cm) by V = 250 loge (x). At what rate (cm per min) is the depth of the spa increasing? 2 marks

6 The graph of f : R → R, f (x) = (x - 1)(5 - x) is shown. The tangent to the graph of f where it crosses the x-axis is also shown.

482

2

a Find the equation of the tangent to the graph of f where it crosses the x-axis. b Find the exact area of the shaded region.

x=a

2 marks

with equation x = π is 3. Find a.

1

y

3 A is (0, 2), a point on the y-axis. P is a point on the parabola y = x2 such that the length of AP, the line segment from the A to P is a minimum. a Find the coordinates of P. b Find the minimum length of AP. 2 marks

0

−1

8x

g x

a

A ∫ ( f ( x) − g( x )) dx b b

C ∫ ( f ( x) − g( x )) dx a

b

B ∫ ( f ( x) + g( x )) dx a a

D ∫ ( f ( x) + g( x )) dx b

a

E

∫ (g( x) − f ( x)) dx b

π 2 The average value of the function y = 3 sin (2 x + ) 3 π over the interval [0, ] is: 3 3 3 A 0 B 4 1 3 3 C 2 D 4 2 E


27 4π

2

3 Using the fact that ∫ f ( x ) dx = 5, what is the value 1

1

of ∫ (3 f ( x ) − 1) dx? 2 −16

B −14 E 16

A D 14

C −12

4 A function with the rule y = f (x) is shown in the first sketch below. y

y

x

x

Which of these rules could correspond to the second sketch? A y = f ′ (x) B y = f −1(x) C y = −f (x) D y = f (−x) 1 E y= f ( x) 5 If f (x) = (2x + 1)2 and f (0) = 1 then the antiderivative of f (x) will be equal to: (2 x + 1)3 2 + 3 3 1 C ((2 x + 1)3 + 5) 6 E 4x3 + 4x2 + x + 1

B 4x + 1

A

D 2(2x + 1) + 1

exTended reSPonSe

30 minutes

The volume of a cylindrical soft drink can is 250 mL. The volume of a cylinder can be calculated using V = π r2h, the area of a circle is given by A = π r2 and its perimeter is given by P = 2π r. Also 1 mL = 1 cm3. a Create an expression for the total surface area of the can, S, in of r and h.

r

h

1 mark

b Use the fact that the volume is 250 mL to show that S can be expressed in of r 500 only as S = 2π r 2 + . 1 mark r 500 y c The graphs at right are of y = 2π r2 and y = for 0 ≤ r ≤ 10. Use these graphs r 700 600 500 to create a sketch of S = 2π r 2 + on the same axes. 500 r 1 mark d Hence obtain an estimate, to 1 decimal place, of the radius corresponding to the minimum total surface area.

1 mark

400 300 200

100

e Using calculus determine an exact value for the radius corresponding to the 0 2 minimum surface area. f i Write down a decimal approximation to 2 decimal places for the radius corresponding to the minimum total surface area. ii Confirm using the sign of gradient test that the stationary point located is in fact a minimum. iii Write down the value of the minimum surface area to the nearest cm2.

4

6

8 10 x

1 + 1 = 2 marks

g If the cost of the material for the bottom and top of the can is twice the cost for the sides determine the radius of the can for minimum total cost. h The actual radius for a can of soft drink is 3.0 cm. Determine to 2 decimal places the ratio cost of material for top and bottom which would be consistent with this can having the minimum cost of material for the sides cost of materials.

2 marks

2 marks


Solutions Exam practice 3

exam practice 3

483

10

10A Probability revision 10B Discrete random variables 10C Measures of centre of discrete random distributions 10D Measures of variability of discrete random distributions

Discrete random variables areaS oF STuDy

• Random variables, including: – the concept of discrete and continuous random variables – calculation and interpretation of the expected value, variance and standard deviation of a random variable – calculation and interpretation of central measures (mean, median, mode) – the property that, for many random variables, approximately 95% of their probability distribution is within 2 standard deviations of the mean

• Discrete random variables, including: – specification of probability distributions for discrete random variables using graphs, tables and probability functions – interpretation and use of mean (µ), median, mode, variance (σ 2) and standard deviation of a discrete random variable – probabilities for specific values of a random variable and intervals defined in of a random variable, including conditional probability


10a

Probability revision

10 Quick Questions

To introduce this chapter we shall revise important concepts and skills that were covered in Mathematical Methods (CAS) Units 1 and 2.

Terminology The circular spinner at right is divided into 8 equal sectors. When the 2 1 spinner is spun, the possible outcomes are 1, 2, 3, 4, 5, 6, 7, 8. These outcomes may be listed as the elements of a set. The set of all possible 3 8 outcomes of an experiment is called the sample space (or the universal set) and is denoted by ε, and each possible outcome is called a sample 7 4 point. Therefore, spinning the spinner gives ε = {1, 2, 3, 4, 5, 6, 7, 8}. A subset of the sample space is known as an event. For the example 5 6 above, if event A is defined as ‘an odd number when the spinner is spun’, then A = {1, 3, 5, 7}. If event B is defined as ‘a number less than 5 when the spinner is spun’, then B = {1, 2, 3, 4}. The union (symbol ∪) of the two events A and B above implies a combined event, that is, either event A or event B or both occurring. Therefore the set A ∪ B = {1, 2, 3, 4, 5, 7}. (Note: Common elements are written only once.)

484


The intersection (symbol ∩) of the two events A and B above is represented by the common sample points of the two events. Therefore the set A ∩ B = {1, 3}.

Venn diagrams Venn diagrams involve drawing a rectangle that represents the sample space and a series of circles that represent subsets of the sample space. They provide a visual representation of the information at hand and clearly display the relationships between sets. The Venn diagrams below illustrate an alternative way of presenting information regarding the circular spinner shown on the previous page. ε

ε

A 5 7

1 3

2 4

ε

A

B

5 7

6 8

A

B 1 3

2 4

7

6 8

A∩B

A∪B

B 5

1 3

2 4

6 8

A

Note: Sample points not belonging to either set are placed outside the circles but remain inside the rectangle. Venn diagrams can also assist in determining whether or not two sets are equal, that is, whether they contain the same elements. As the examples below show, the equality of two sets may not be obvious from the set notation but is often easier to see in a diagram. A

A

B

B

A′ ∪ B′ = (A ∩ B)′

A′ ∩ B′ = (A ∪ B)′

Probability Probability deals with the likelihood or chance of some event occurring. The probability of a specific event, say A, occurring is defined by the rule: Pr(A) =

number of favourable outcomes total number of possible outcomes

Its probability lies within the restricted interval 0 ≤ Pr(A) ≤ 1. A probability of zero implies that the event cannot occur, while a probability of 1 implies that the event will most certainly occur. The individual probabilities of a particular experiment will sum to a value of 1 and can be denoted as follows.

∑ p( x) = 1 If Pr(A) is defined as the probability of an event occurring, then its complement, Pr(A′), is defined as the probability of an event not occurring. Therefore, it can be stated that Pr(A) + Pr(A′) = 1 which can be transposed to Pr(A′) = 1 − Pr(A).

Chapter 10 Discrete random variables

485

Worked Example 1

Two fair dice are rolled simultaneously and the sum of the two numbers appearing uppermost is recorded as shown below. (1, 1)

(1, 2)

(1, 3)

(1, 4)

(1, 5)

(1, 6)

(2, 1)

(2, 2)

(2, 3)

(2, 4)

(2, 5)

(2, 6)

(3, 1)

(3, 2)

(3, 3)

(3, 4)

(3, 5)

(3, 6)

(4, 1)

(4, 2)

(4, 3)

(4, 4)

(4, 5)

(4, 6)

(5, 1)

(5, 2)

(5, 3)

(5, 4)

(5, 5)

(5, 6)

(6, 1)

(6, 2)

(6, 3)

(6, 4)

(6, 5)

(6, 6)

Find the probability that the sum will be: a 6 b 10 c a number less than 5 d at least 9 e an odd number. Think a

b

c

d

Write

1

List all the possible outcomes.

2

Define the event.

3

Substitute the values into the probability rule.

1


2

Define the event.

3


4

Simplify.

1


2

Define the event.

3


4

Simplify.

1


a (5, 1) (4, 2) (3, 3) (2, 4) (1, 5)

Let A = the sum of 6. number of favourable outcomes Pr(A) = total number of possible outcomes = 5 36 b (6, 4) (5, 5) (4, 6)

Let A = the sum of 10. number of favourable outcomes Pr(A) = total number of possible outcomes = 3 36 1 = 12 c

(1, 1) (2, 1) (3, 1) (1, 2) (2, 2) (1, 3) Let A = a number less than 5. number of favourable outcomes Pr(A) = total number of possible outcomes 6 = 36 1 = 6

d (6, 3) (5, 4) (6, 4) (4, 5) (5, 5)

(6, 5) (3, 6) (4, 6) (5, 6) (6, 6)

486

2

Define the event.

3



Let A = at least 9. number of favourable outcomes Pr(A) = total number of possible outcomes 10 = 36

e

4

Simplify.

1


=

5 18

e (2, 1) (4, 1) (6, 1) (1, 2) (3, 2) (5, 2)

(2, 3) (4, 3) (6, 3) (1, 4) (3, 4) (5, 4) (2, 5) (4, 5) (6, 5) (1, 6) (3, 6) (5, 6) 2

Define the event.

3


4

Simplify.

Let A = an odd number. number of favourable outcomes Pr(A) = total number of possible outcomes 18 = 36 1 = 2

Worked Example 2

A bag contains 15 marbles comprising 5 black, 3 red, 4 blue, 2 white and 1 green. One marble is drawn randomly from the bag. a Determine the probability of each of the coloured marbles being drawn: i black ii red iii blue iv white v green. b Show that the probabilities sum to 1. c What is the probability that the marble drawn is: i not black? ii either black or white? iii neither blue nor green? Think a

i

ii

iii

iv

Write 1

Define the event.

2


3

Simplify.

1

Define the event.

2


3

Simplify.

1

Define the event.

2


1

Define the event.

a

i Let B = a black marble.

number of favourable outcomes

Pr(B) =

total number of possible outcomes 5 = 15 1 = 3

i i Let R = a red marble.


Pr(R) =

total number of possible outcomes 3 = 15 1 = 5 i i i Let Bl = a blue marble. Pr(Bl) =


total number of possible outcomes 4 = 15 i v Let W = a white marble.


487

v

b

2


1

Define the event.

2


Add each of the probabilities.


Pr(W) =

total number of possible outcomes 2 = 15

v Let G = a green marble.


Pr(G) =

total number of possible outcomes 1 = 15

b Sum of probabilities = 1 + 3

1 5

4

2

1

+ 15 + 15 + 15

=1 c

i

ii

iii

i Pr(B′) = 1 − Pr(B)

1

Write the appropriate rule: Pr(A′) = 1 − Pr(A).

2

Substitute the known values into the rule.

=1−3

3

Evaluate.

=

1

Add each of the probabilities together.

2


= 3 + 15

3

Evaluate.

= 15

1

Write the appropriate rule.

2


= 1−

3

Evaluate.

= 1 − 15

1

2 3

i i Pr(B ∪ W) = Pr(B) + Pr(W) 1

2

7

i i i Pr(Bl′ ∪ G′) = 1 − [Pr(Bl) + Pr(G)]

(

4 15

1

+ 15

)

5

10

= 15 4

Simplify.

2

=3

The addition rule of probability The addition rule of probability states that Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B).

Worked Example 3

a If Pr(A) = 0.4, Pr(B) = 0.7 and Pr(A ∩ B) = 0.2, find Pr(A ∪ B). b If Pr(A) = 0.6, Pr(B) = 0.8 and Pr(A ∪ B) = 0.9, find Pr(A ∩ B). Think a

488

Write a Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)

1

Write the addition rule.

2


= 0.4 + 0.7 − 0.2

3

Evaluate.

= 1.1 − 0.2 = 0.9


b

b Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)

1


2


3

Transpose the equation to make Pr(A ∩ B) the subject.

4

Evaluate.

0.9 = 0.6 + 0.8 − Pr(A ∩ B) 0.9 = 1.4 − Pr(A ∩ B) Pr(A ∩ B) = 1.4 − 0.9 = 0.5

Venn diagrams may also be used to display the probabilities rather than just the outcomes, as shown in the diagram below. ε

A

B

Pr Pr (A∩B′) (A∩B) Pr (A′∩B)

Note: Pr(A ∪ B) is represented by the shaded section. The probabilities given and calculated in worked example 3(a) and 3(b) can be displayed as follows. ε

ε

A

B 0.2

0.2

A

B 0.1

0.5

0.5

0.3 0.1

0.1

Worked example 3(a)

Worked example 3(b)

Mutually exclusive events If two or more events cannot occur simultaneously, they are said to be mutually exclusive or dist; that is, they have nothing in common. In set notation this may be expressed as Pr(A ∩ B) = { } or Pr(A ∩ B) = 0. Therefore for mutually exclusive events the addition rule becomes: Pr(A ∪ B) = Pr(A) + Pr(B)

Independent events Two events A and B are independent if one event does not influence the other event from occurring. The mathematical definition of independence is given by: Pr(A ∩ B) = Pr(A) × Pr(B)

Worked Example 4

Two fair dice are rolled with S representing the event of obtaining a number less than 4 on the first die and T the event of obtaining a number greater than 4 on the second die. Find: a Pr(S) b Pr(T) c if events S and T are mutually exclusive d if events S and T are independent.


489

Think a

b

Write

1

Refer to the dice results recorded in the question in worked example 1.

2

Define the event.

3

Determine the probability.

4

Simplify.

1

Define the event.

a

Let S = obtaining a number less than 4 on the first die. 18 Pr(S) = 36 1 2

=

b Let T = obtaining a number greater than 4 on

the first die.

2


3

Simplify.

Pr(T) = =

12 36 1 3

c Answer the question with reasoning.

c

d

d From the dice results recorded in the question

1

Answer the question using the dice results in worked example 1.

Events S and T are not mutually exclusive since they have common points; that is, (1, 5) (1, 6) (2, 5) (2, 6) (3, 5) (3, 6). in worked example 1, Pr(S ∩ T) = 6 =

2

36 1 6

Using the rule Pr(S ∩ T) = Pr(S) × Pr(T) 1 =1×3

Check with answer obtained using the rule.

2

=1 6 Since both methods give the same answer, S and T are independent events.

Worked Example 5

Two fair dice are rolled with U representing the event of obtaining a 5 on the first die and V the event of the sum of numbers on the two dice exceeding 10. Find: a Pr(U) b Pr(V) c if events U and V are independent. Think a

b

Write a

1

Refer to the dice results recorded in the question in worked example 1.

2

Define the event.

Let U = obtaining a 5 on the first die.

3


Pr(U) = 36

4

Simplify.

1

Define the event.

6

= 16 b Let V = the sum of the numbers on the two

dice exceeds 10.

490

2


3

Simplify.


3

Pr(V) = 36 1

= 12

c

1

Answer the question using the dice results.

2

Check with answer obtained using the rule.

c From the dice results, Pr(U ∩ V) =

1 36

Using rule Pr(U ∩ V) = Pr(U) × Pr(V) 1

1

= 6 × 12 1

= 72 Since the two methods do not give the same answer, U and V are not independent events.

Karnaugh maps and probability tables Karnaugh maps and probability tables summarise all combinations of two events (for example A and B) and their complements (for example A′ and B′ ). B

B′

A

A∩B

A ∩ B′

A′

A′ ∩ B

A′ ∩ B′

B

B′

A

Pr(A ∩ B)

Pr(A ∩ B′)

Pr(A)

A′

Pr(A′ ∩ B)

Pr(A′ ∩ B′)

Pr(A′)

Pr(B)

Pr(B′)

1

Karnaugh map

Probability table

Worked Example 6

For the probability table shown, A is the event Column 1 Column 2 Column 3 ‘not more than 17 years of age’ and B is the B B′ event ‘has a learner permit’. a Complete the probability table at right. Row 1 A 0.12 b What do the following probabilities Row 2 A′ 0.5 represent? Row 3 0.63 1 i Pr(A ∩ B) ii Pr(A′ ∩ B′) c What is the probability that: i a person over the age of 17 does not have a learner permit? ii a person has a learner permit and is older than 17? iii a person over the age of 17 has a learner permit or a person at or under the age of 17 does not have a learner permit? Think a

1

2

Write

Calculate the values of the cells in row 1, column 3 and row 3, column 1.

a Pr(A) = 1 − Pr(A′)

= 1 − 0.5 = 0.5

Enter the values into the appropriate cells. Row 1

A

Row 2

A′

Row 3 3

Calculate the value of the cell in row 1, column 2.

Pr(B) = 1 − Pr(B′) = 1 − 0.63 = 0.37 Column 1

Column 2

B

B′

0.12

Column 3 0.5 0.5

0.37

0.63

1

Pr(A ∩ B′) = Pr(A) − Pr(A ∩ B) = 0.5 − 0.12 = 0.38


491

4

Enter the value into the appropriate cell. Row 1

A

Row 2

A′

Row 3 5


6

Enter the value into the appropriate cell.


8

Enter the value into the appropriate cell.

i

Explain what Pr(A ∩ B) represents in this example.

b

State the appropriate probability from the table.

ii State the appropriate probability from

the table. iii State the appropriate probabilities

from the table and evaluate.

0.12

0.38

0.5 0.5

0.63

1

Column 1

Column 2

Column 3

B

B′ 0.38

Row 1

A

0.12

Row 2

A′

0.25

0.5 0.5

0.37

0.63

1

Column 1

Column 2

Column 3

B

B′

Row 1

A

0.12

0.38

0.5

Row 2

A′

0.25

0.25

0.5

0.37

0.63

1

i Pr(A ∩ B) represents the probability of a person

at or under the age of 17 having a learner permit. In this case the probability of the given event occurring is 0.12. i i Pr(A′ ∩ B′) represents the probability of a person

in this example.

i

B′

Pr(A′ ∩ B′) = Pr(B′) − Pr(A ∩ B′) = 0.63 − 0.38 = 0.25

ii Explain what Pr(A′ ∩ B′) represents

c

B

Column 3

Pr(A′ ∩ B) = Pr(B) − Pr(A ∩ B) = 0.37 − 0.12 = 0.25

Row 3 b

Column 2

0.37

Row 3 7

Column 1

over the age of 17 not having a learner permit. In this case the probability of the given event occurring is 0.25. c

i Pr(A′ ∩ B′) = 0.25 i i Pr(A′ ∩ B) = 0.25 i i i Pr(A′ ∩ B) + Pr(A ∩ B′) = 0.25 + 0.38

= 0.63

Conditional probability Conditional probability deals with an event which has previously occurred and has an effect on an event we are interested in. Due to the initial condition (or restriction) imposed, the number of Pr ( A ∩ B) possible events is reduced. Conditional probability is defined by the rule Pr ( A | B) = ,, Pr(B) where Pr(B) ≠ 0, and can be transposed to Pr(A ∩ B) = Pr(A | B) × Pr(B). The latter is called the multiplication rule. Pr(A | B) is read as ‘the probability of A given B’. 492


WorkeD examPle 7

eBook plus

If Pr(A) = 1 , Pr(B) = 1 and Pr(A ∩ B) = 1 , find: 5 10 20 a Pr(A ∪ B) b Pr(A | B) c Pr(B | A) d if events A and B are mutually exclusive e if events A and B are independent. Think a

b

c


WriTe a Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B)

1


2


= 15 + 101 −

3

Evaluate.

=

5 20

4

Simplify.

=

1 4

1


2


3

Evaluate.

4

Simplify.

1


2


3

Evaluate.

4

d

Tutorial

Simplify.

State the answer, showing reasoning.

b Pr(A | B) =

=

Pr( A ∩ B) Pr( B) 1 20 1 10

=

1 20 1 20

=

10 20

=

1 2

=

c Pr(B | A) =

1 20

÷ 101 × 101

Pr( B ∩ A) Pr( A)

=

1 20 1 5

=

1 20

÷

1 5

=

1 20

×

5 1

=

5 20

=

1 4

d Events A and B are not mutually exclusive

since they have common events, that is, Pr(A ∩ B) =

e

Compare the given value with the answer obtained using the rule.

e Pr(A ∩ B) =

1 . 20 1 . 20

Using rule Pr(A ∩ B) = Pr(A) × Pr(B) = 15 × 101 =

1 50

Since the two methods do not give the same answer, A and B are not independent events. From worked example 7(b) and 7(c) it can be seen that Pr(A | B) ≠ Pr(B | A).

Chapter 10

Discrete random variables

493

Tree diagrams Tree diagrams are a useful tool in solving probability tasks as they display each of the possible outcomes along with their respective probabilities.

Worked Example 8

Nadia knows that if her car starts, she has an 80% chance of getting to work on time. However, if her car doesn’t start, her chance of arriving on time is 50%. If Nadia’s car starts only 70% of the time, what is the probability that: a her car starts and she gets to work on time? b she arrives at work late? c she arrives at work on time? d her car starts, given that she arrives at work on time?

Think a

1

2

Write

Define the events.

a Let C = car starts

Assign probabilities to each event.

Let C′ = car doesn’t start Let O = Nadia arrives at work on time Let L = Nadia arrives at work late

70 Pr(C ) = 100

= 107 Pr(C') = 103

80

If car starts, Pr(O) = 100

=

If car starts, Pr(L) =

4 5 1 5 50

If car doesn’t start, Pr(O) = 100 1

=2 1

If car doesn’t start, Pr(L) = 2 3

4– 5

Draw a tree diagram with each branch assigned the appropriate probability.

7– 10

C

3– 10

1– 5

O 1– 2

C′ 1– 2

4

5

494

Calculate the required probability.

Simplify.


L O

Pr(CO) =

7 10

=

=

28 50 14 25

×4

5

L

b


c


d

b Pr(L) = Pr(CL) + Pr(C′L) 7 10 7 50

=

=

=

14 100

=

29 100

×1 + +

5 3 20

+

3 10

×1

2

15 100

c Pr(O) = Pr(CO) + Pr(C′O)

=

=

=

=

7 ×4+ 3 10 5 10 28 3 + 50 20 56 + 15 100 100 71 100

d Pr(C | O) =

1

Write the appropriate rule. Note: ‘Given’ implies conditional probability.

2


=

3

Evaluate and simplify.

=

×1

2

Pr(CO) Pr(O) 14 25 71 100 14 71 ÷ 25 100

= 14 × 100 25 71 = 56

71

Worked Example 9

A fair coin is tossed three times. Find the probability of obtaining two heads given the first toss resulted in a tail. Think 1

Write

Draw a tree diagram and list all of the possible outcomes.

H H

H T

T

H T

H

H T

T

H T

T

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT 2


3

Calculate the probability of each event.

Pr(2H | tail first toss) =

Pr (2H ∩ tail first toss) Pr ( tail first toss)

Pr(2H ∩ tail first toss) = 1

Pr(tail first toss) =

=

8 4 8 1 2


495

4


5

Evaluate.

Pr(2H | tail first toss) =

1 8 1 2 1

1

=8÷2 =1×2 =

6

Simplify.

8 2 8

1

=1 4

Combinations In mathematics, a combination deals with the number of ways items may be selected from a set of elements where the order is not important. For example, in how many ways can 3 numbers be selected from the set {1, 2, 3, 4}, taking into that order is not important? The following selections can be made: 1, 2, 3 1, 2, 4 2, 3, 4 3, 4, 1 Hence 4 selections could be made. If order was important, there would be a greater number of possibilities since each of the above selections could be arranged in 6 ways. For example, the selection (1, 2, 3) could be arranged as: 1, 2, 3 2, 1, 3 2, 3, 1 3, 1, 2 3, 2, 1 1, 3, 2. A combination is also referred to as a selection or choice, and is defined by the rule nCr. nC = the number of selections of n different objects taken r at a time r n! = − n ( r )!r ! nC r

()

may also be expressed as n and is read as ‘n over (above) r’. r

WorkeD examPle 10

eBook plus

A drawer contains 7 T-shirts of which 3 are white and the rest are black. If 2 T-shirts are randomly selected from the drawer simultaneously, find the probability that they are: a both black b both white c different colours d the same colour. Think a

b

496

Tutorial


WriTe

1

Calculate the number of selections of taking 2 black T-shirts from a total of 4.

2

Calculate the number of selections of taking 2 T-shirts from a total of 7.

3

Calculate the probability using the rule.

4


5

Simplify.

1

Calculate the number of selections of taking 2 white T-shirts from a total of 3.

a

2 = 6; that is, there are 6 ways of selecting 2 black T-shirts from a total of 4. 4C

2 = 21; that is, there are 21 ways of selecting 2 T-shirts from a total of 7. 7C

Pr(both black) =

2 black T-shirts from 4 2 T-shirts from 7

6 = 21 2 = 7 b


2 = 3; that is, there are 3 ways of selecting 2 white T-shirts from a total of 3. 3C

c

d

2


7C = 21; that is, there are 21 ways of selecting 2 2 T-shirts from a total of 7.

3


Pr(both white) =

4


5

Simplify.

1

Calculate the number of selections of taking 1 black T-shirt from a total of 4.

2

Calculate the number of selections of taking 1 white T-shirt from a total of 3.

= 3; that is, there are 3 ways of selecting 1 white T-shirt from a total of 3.

3


4


5


6

Evaluate.

7

Simplify.

= 21; that is, there are 21 ways of selecting 2 T-shirts from a total of 7. 1 black × 1 white Pr(different colours) = 2 T-shirts from 7 4×3 = 21 12 = 21 4 = 7

1


c

2 white T-shirts from 3 2 T-shirts from 7

3 21 1 = 7 =

= 4; that is, there are 4 ways of selecting 1 black T-shirt from a total of 4. 4C 1 3C 1 7C 2

d Pr(same colours) = Pr(both black)

+ Pr(both white)

2


3

Evaluate.

=2+1 7

=

7

3 7

1. Outcomes are results of experiments. 2. The set of all possible outcomes of an experiment is called the sample space and is denoted by ε, and each possible outcome is called a sample point. 3. A subset of the sample space is known as an event. 4. The union (symbol ∪) of two events A and B implies a combined event, that is, either event A or event B or both occurring. Common elements are written only once. 5. The intersection (symbol ∩) of two events is represented by the common sample points of the two events. 6. Venn diagrams involve drawing a rectangle that represents the sample space and a series of circles that represent subsets of the sample space. They provide a visual representation of the information at hand and clearly display the relationship between sets. 7. The probability of an event occurring is defined by the rule Pr(A) =



497

8. The probability of an event occurring lies within the restricted interval 0 ≤ Pr(A) ≤ 1 9. The individual probabilities of a particular experiment will sum to 1; that is,

∑ p( x) = 1 10. The addition rule of probability is defined by the rule Pr(A ∪ B) = Pr(A) + Pr(B) − Pr(A ∩ B) 11. If two events, A and B, are mutually exclusive then Pr(A ∩ B) = 0, and therefore the addition rule becomes Pr(A ∪ B) = Pr(A) + Pr(B). 12. If two events, A and B, are independent then Pr(A ∩ B) = Pr(A) × Pr(B). 13. Karnaugh maps and probability tables summarise all combinations of two events (for example A and B) and their complements (for example A′ and B′). 14. Conditional probability is defined by the rule Pr(A | B) =

Pr(A ∩ B) , where Pr(B) ≠ 0. Pr ( B)

This can be transposed to Pr(A ∩ B) = Pr(A | B) × Pr(B). 15. Tree diagrams are useful tools in solving probability tasks because they display each of the possible outcomes along with their respective probabilities. 16. A combination is defined by nCr; that is, the number of selections of n different objects taken r at a time.

Exercise

10A

Probability revision 1 WE1 Two fair dice are rolled simultaneously and the sum of the two numbers appearing uppermost is recorded. Find the probability that the sum will be: a 3 b 12 c 7 d greater than 4 e at least 7 f an even number g a prime number. 2 WE2 A bag contains 12 marbles comprising 3 black, 5 red and 4 green. One marble is drawn randomly from the bag. a Determine the probability of each of the coloured marbles being drawn: i black ii red iii green. b Show that the probabilities sum to 1. c What is the probability that the marble drawn is: i not green? ii either black or red? iii neither red nor green? iv either black, red or green? 3 A fair coin is tossed three times. Find the probability of obtaining: a three heads b two heads c one head d no heads e at least two heads. 4 A circular spinner is divided into 8 equal sectors and numbered as shown in the diagram at right. If the spinner is spun once, find the probability of obtaining: a a one b a two c a three or a four or a five d a one or a two.

498


1

1

1

2

5

2 4

3

5 WE3a If Pr(A) = 0.3, Pr(B) = 0.6 and Pr(A ∩ B) = 0.2, find Pr(A ∪ B).

6 WE3b If Pr(A) = 0.5, Pr(B) = 0.4 and Pr(A ∪ B) = 0.8, find Pr(A ∩ B). 7 If Pr(A) = 4 Pr(B), Pr(A ∪ B) = 0.8 and Pr(A ∩ B) = 0.2, find: a Pr(B) b Pr(A). 8 Of the 200 students studying VCE at Merlynston Secondary College, 80 study Maths Methods, while there are 65 Physics students. If there are 85 students who don’t take either Maths Methods or Physics, find the probability that a randomly selected student: a studies Maths Methods b studies Physics c studies neither Maths Methods nor Physics d studies Maths Methods and Physics e studies Physics, given that the student studies Maths Methods. 9 WE4,5 Two fair dice are rolled, with F representing the event of obtaining a number greater than 4 on the first die and G the event of obtaining an even number on the second. Find: a Pr(F) b Pr(G) c if events F and G are mutually exclusive d if events F and G are independent. 10 For two events P and Q, Pr(P) = 0.72, Pr(Q) = 0.25 and Pr(P ∪ Q) = 0.91. Are P and Q mutually exclusive events? 11 For two events X and Y, Pr(X) = 0.4, Pr(Y) = 0.5 and Pr(X ∩ Y) = 0.2. Are X and Y independent events?

12 WE6 For the probability table shown, A is the event ‘is unfit’ and B is the event ‘is a smoker’. a Complete the probability table at right.

Row 1

A

Row 2

A′

Column 1

Column 2

B

B′

0.22 0.60

Row 3 b c

0.68 1

What do the following probabilities represent? i Pr(A ∩ B) ii Pr(A′ ∩ B′) What is the probability that: i a person is unfit and a non-smoker? ii a person is a smoker and fit? iii a person is unfit and a smoker or is unfit and a non-smoker? iv a person is a non-smoker?

If Pr(A) = 12 , Pr(B) = 13 and Pr(A ∩ B) = a Pr(A ∪ B) b Pr(A | B) d if events A and B are mutually exclusive e if events A and B are independent.

13 WE7

14

Column 3

1 6

find:

If Pr(A) = 0.4, Pr(B) = 0.5 and Pr(A ∩ B) = 0.2 find: a Pr(A ∪ B) b Pr(A | B)

c Pr(B | A)

c Pr(B | A).

15 WE8 A recent study has shown that 60% of people who don’t wear glasses get regular headaches, while only 30% of people who wear glasses are headache sufferers. If 35% of people wear glasses, find the probability that a randomly selected person: a wears glasses and gets headaches b does not wear glasses and suffers from headaches c suffers from headaches d wears glasses, given that the person suffers from headaches.


499

16

Jemma knows that if her alarm goes off, she has a 90% chance of getting to school on time. However, if the alarm does not ring, her chance of arriving on time is only 40%. If Jemma’s alarm clock works only 60% of the time, what is the probability that: a she gets to school on time b she arrives late to school c her alarm rang, given that she arrived on time?

17

A bag contains 5 red marbles and 3 green marbles. A marble is selected at random, its colour is observed and it is then replaced. A second selection is then made. Find the probability that the two marbles chosen were: a both red b both green c different colours d the same colour.

18 MC Two fair dice are rolled. The probability of the numbers showing uppermost on both dice being the same is: A

1 36

B

1 18

C 1 6

1

D 3

E

1 2

19 MC If Pr(S) = 0.2, Pr(T) = 0.5 and Pr(S ∪ T) = 0.6, which one of the following is not true? A Pr(S ∩ T) = 0.1 B Pr(S | T) = 0.2 C Pr(T | S) = 0.5 D S and T are mutually exclusive. E S and T are independent. 20 MC The probability of picking a red picture card from a standard pack of playing cards is: A 1 B 3 C 2 D 3 E 1 2

13

13

26

26

21 MC If Pr(M) = 0.3, Pr(N) = 0.4 and Pr(M | N) = 0.5 then Pr(M ∩ N) is equal to: A 0.15 B 0.2 C 0.6 D 0.75 E 0.8 22 WE9 A fair coin is tossed three times. Find the probability of obtaining three tails, given that the first toss resulted in a tail. 23

If Pr(A) = 0.6, Pr(B) = 0.5 and Pr(A ∩ B) = 0.36, find: a Pr(A′) b Pr(B′) d Pr(A′ ∩ B′) e Pr(A′ ∪ B) g Pr(B | A) h Pr(A | B′)

c Pr(A ∪ B) f Pr(A | B) i Pr(B | A′).

24 WE10 A drawer contains 6 T-shirts, of which 2 are white and the rest are black. If 2 T-shirts are randomly selected from the drawer simultaneously, find the probability that they are: a both black b both white c different colours d the same colour. 25 A box contains one dozen chocolates, of which 4 are strawberry creams, 3 are orange creams and 5 are peppermint creams. Two chocolates are selected at random. Find the probability that they are both the same type if: a the first chocolate is replaced before the second is drawn b the first chocolate is not replaced before the second is drawn. 26 MC A fair die has its 4-spot changed to a 5-spot and its 2-spot changed to a 3-spot. The probability of getting an even number when the altered die is rolled is: A 1 B 1 C 1 D 2 E 5 6

3

2

3

6

25

5

25

5

20

27 MC A box contains 3 red balls and 2 green balls. Two balls are chosen simultaneously. The probability that they are the same colour is: A 8 B 2 C 13 D 3 E 13 28 A bag contains 5 red cubes and 3 black cubes. Three cubes are chosen at random. Find the probability of at least 2 reds being chosen, given that the first cube was red: a if the cubes are replaced after each draw b if the cubes are not replaced after each draw. 500


29 Jo-anne knows that her chance of winning each tennis match she plays is 0.8. A knockout tournament requires players to win five matches to win the championship. What is the probability that Jo-anne: a wins the tournament? Give your answer to 4 decimal places. b wins the tournament given that she wins her first three matches? 30

In a particular suburb the chances of a woman owning her own home is 0.4, while the probability of a woman owning her own home and being employed is 0.2. Find the probability that a woman who owns her own home is also employed.

31

The probability of Vanessa’s car starting on a cold morning is 0.6, while on a normal morning the chance of it starting is 0.9. The probability of any morning being a cold one is 0.3. If Vanessa’s car starts tomorrow morning, find the probability that the morning is cold.

32 The Roosters know that they will win 80% of their home matches and 40% of their away matches. This season’s fixture has the Roosters playing 55% of their games at home. Given that the Roosters won their last game, what was the probability that it was played at home? 33 Tatiana is trying out for a place on the high jump team. In order to qualify she must clear three of the four heights. She knows that she has a 70% chance of clearing the first height and a 65% chance of clearing any subsequent height. What is the probability, to 4 decimal places, that Tatiana: a clears the first, third and fourth heights only? b clears three heights? c clears three heights, given she did not clear the first height?

10B

Discrete random variables A random variable is one whose value cannot be predicted but is determined by the outcome of an experiment. For example, two dice are rolled simultaneously a number of times. The sum of the numbers appearing uppermost is recorded. The possible outcomes we could expect are {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}. Since the possible outcomes may vary each time the dice are rolled, the sum of the numbers appearing uppermost is a random variable. Random variables are expressed as capital letters, usually from the end of the alphabet (for example, X, Y, Z) and the value they can take on is represented by lowercase letters (for example, x, y, z respectively). The above situation illustrates an example of a discrete random variable since the possible outcomes were able to be counted. Discrete random variables generally deal with number or size. A random variable that can take on any value is defined as a continuous random variable. Continuous random variables generally deal with quantities that can be measured, such as mass, height or time.

Worked Example 11

Which of the following represent discrete random variables? a The number of goals scored at a football match b The height of students in a Maths Methods class c Shoe sizes d The number of girls in a five-child family e The time taken, in minutes, to run a distance of 10 kilometres Think

Write

Determine whether the variable can be counted or needs to be measured. a Goals can be counted. b Height must be measured.

a Discrete b Continuous


501

c The number of shoe sizes can be counted.

c Discrete

d The number of girls can be counted.

d Discrete

e Time must be measured.

e Continuous

Discrete probability distributions When dealing with random variables, the probabilities associated with them are often required. Worked Example 12

Let X represent the number of tails obtained in three tosses. Draw up a table that displays the values the discrete random variable can assume and the corresponding probabilities. Think 1

Write

Draw a tree diagram and list all of the possible outcomes.

H H

H T

T

H T

H

H T

T

H T

T

HHH, HHT, HTH, HTT, THH, THT, TTH, TTT 2 3

Draw a table with two columns: one labelled number of tails, the other probability.

Number of tails (x)

Probability Pr(x)

0

1 8 3 8 3 8 1 8

Enter the information into the table.

1 2 3

The table above displays the probability distribution of the total number of tails obtained in three tosses of a fair coin. Since the variable in this case is discrete, the table displays a discrete probability distribution. In worked example 12, X denoted the random variable and x the value that the random variable could take. Thus the probability can be denoted by p(x) or Pr (X = x). Hence the table in worked example 12 could be presented as shown below. x

0

1

2

3

Pr(X = x)

1 8

3 8

3 8

1 8

Close inspection of this table shows important characteristics that satisfy all discrete probability distributions. 1. Each probability lies in a restricted interval 0 ≤ Pr(X = x) ≤ 1. 2. The probabilities of a particular experiment sum to 1, that is,

∑ Pr (x = x) = 1 If these two characteristics are not satisfied, then there is no discrete probability distribution.

502


Worked Example 13

Draw a probability distribution graph of the outcomes in worked example 12. Think 1

2

3

Write

Draw a set of axes in the first quadrant only. Label the horizontal axis x and the vertical axis Pr(X = x).

Pr(X = x) 3– 8

Mark graduations evenly along the horizontal and vertical axes, and label with appropriate values.

2– 8 1– 8

Draw a straight line from each x-value to its corresponding probability.

0

1

2

3

x

Note: The probability distribution graph may also be drawn as follows. Pr(X = x)

Pr(X = x) 3– 8

3– 8

2– 8

2– 8

1– 8

1– 8

0

1

2

x

3

0

A column graph

1

2

3

x

A dot graph

Worked Example 14

Which of the following tables represent a discrete probability distribution? a b x 0 1 2 3 x 0 2 4 c

Pr(X = x)

0.2

0.5

0.2

0.1

x

−1

0

1

2

Pr(X = x)

0.2

0.1

0.3

d

0.3

Think a

b

1

6

Pr(X = x)

0.5

0.3

0.1

0.1

x

−2

0

5

7

−0.2

0.3

0.5

0.4

Pr(X = x)

Write

Check whether each of the probabilities lie within the restricted interval 0 ≤ Pr(X = x) ≤ 1.

a All probabilities lie between 0 and 1

inclusive.

2

Check that the probabilities sum to 1.

0.2 + 0.5 + 0.2 + 0.1 = 1

3

Answer the question.

Yes, this is a discrete probability distribution since both requirements have been met.

1


b All probabilities lie between 0 and 1

inclusive.

2


0.5 + 0.3 + 0.1 + 0.1 = 1

3


Yes, this is a discrete probability distribution since both requirements have been met.


503

c

d

c

1


2


0.2 + 0.1 + 0.3 + 0.3 ≠ 1 (totals to 0.9)

3


No, this is not a discrete probability distribution since both requirements have not been met.

1


2


−0.2

3


No, this is not a discrete probability distribution since both requirements have not been met.

All probabilities lie between 0 and 1 inclusive.

d

The first probability is a negative value, so not all probabilities lie between 0 and 1 inclusive. + 0.3 + 0.5 + 0.4 = 1

WorkeD examPle 15

Find the value of k for each of the following discrete probability distributions. a x 1 3 5 7 Pr(X = x) b

0.2

k

0.2

0.3

0.1

x

0

1

2

3

4

Pr(X = x)

5k

6k

4k

3k

2k

Think a

b

9

WriTe

1

Add up each of the given probabilities. They should sum to 1.

2

Simplify.

3

Solve to find k.

1

Add up each of the given probabilities. They should sum to 1.

2

Simplify.

3

Solve to find k.

∑ Pr( X = x) = 1 0.2 + k + 0.2 + 0.3 + 0.1 = 1 0.8 + k = 1 k = 1 − 0.8 = 0.2

a

5k + 6k + 4k + 3k + 2k = 1

b

20k = 1 k=

WorkeD examPle 16

eBook plus 1 42 (5x

a Show that the function p(x) = + 3), where x = 0, 1, 2, 3 is a probability function. 1 b Show that the function p(x) = 100 x2 (6 − x), where x = 2, 3, 4, 5 is a probability function. Think a

504

1

1 20

Tutorial


WriTe

Substitute each of the x-values into the equation and obtain the corresponding probability.

a


When x = 0, p(x) = =

3 42 1 14

2

When x = 1, p(x) =


=

When x = 2, p(x) = When x = 3, p(x) = =

b

8 42 4 21 13 42 18 42 3 7

3



4

Check whether the probabilities sum to 1.

1 4 + 14 21

5

State whether the function is a probability function.

Yes, this is a probability function since both requirements have been met.

1


2

13

3

+ 42 + 7 = 1

b When x = 2, p(x) = 16 100

4

= 25 27

When x = 3, p(x) = 100


32

When x = 4, p(x) = 100

8

= 25 25

When x = 5, p(x) = 100

1

=4

3



4

Check whether the probabilities sum to 1.

4 25

5

State whether the function is a probability function.

Yes, this is a probability function since both requirements have been met.

27

8

1

+ 100 + 25 + 4 = 1

Worked Example 17

Three balls are selected from a box containing 6 blue balls and 4 yellow balls. If the ball chosen after each selection is replaced before the next selection, find: a the probability distribution for the number of blue balls drawn: i 0 blue balls ii 1 blue ball iii 2 blue balls iv 3 blue balls b the probability that 3 blue balls are chosen, given that at least one ball was blue. Think a

i

Write/draw 1

Draw a tree diagram and list all the possible outcomes with their respective probabilities.

a

i

6 — 10

4 — 10

6 — 10

B

4 — 10

Y

6 — 10

B

B

Y 4 — 10

6 — 10

4 — 10

6 — 10

4 — 6 10 — 10 4 — 10

Y

6 — 10

4 — 10

Outcomes B Y

BBB BBY

B Y

BYB BYY

B Y

YBB YBY

B Y

YYB YYY

Probability 6 — 10 6 — 10 6 — 10 6 — 10 4 — 10 4 — 10 4 — 10 4 — 10

6 6 = ×— ×— 10 10 6 4 ×— ×— 10 10 4 6 ×— ×— 10 10 4 4 ×— ×— 10 10 6 6 ×— ×— 10 10 6 4 ×— ×— 10 10 4 6 ×— ×— 10 10 4 4 ×— ×— 10 10

216 ––– 1000 = 144 ––– 1000 = 144 ––– 1000 96 = –– 1000 = 144 ––– 1000 96 = ––– 1000 96 = ––– 1000 64 = ––– 1000


505

ii

iii

iv

2

Obtain the probability required.

1

List the required probabilities from the tree diagram obtained in part i . Note: Three outcomes correspond to 1 blue ball.

2


1

List the required probabilities from the tree diagram obtained in part i . Note: Three outcomes correspond to 2 blue balls.

2


1

Obtain the probability required.

2

Place all of the information in a table.

3

b

Pr(0 blue balls) =

i i Pr(1 blue ball) = Pr(BYY) + Pr(YBY)

+ Pr(YYB)

96 1000 288 = (or 0.288) 1000 = 3 ×

i i i Pr(2 blue balls) = Pr(BBY) + Pr(BYB)

+ Pr(YBB)

144 1000 432 = (or 0.432) 1000 = 3 ×

i v Pr(3 blue balls) =

x Pr(X = x)


1

Define the rule.

2

Determine each of the probabilities.

3

Substitute values into the rule.

4


64 (or 0.064) 1000

b

216 (or 0.216) 1000

0

1

2

3

0.064

0.288

0.432

0.216

ΣPr(X = x) = 0.064 + 0.288 + 0.432 + 0.216 =1 Pr ( X = 3 ∩ X > 1) Pr ( X = 3 | X > 1) = Pr ( X > 1) Pr(X = 3 ∩ X > 1) = Pr(X = 3) = 0.216 Pr(X > 1) = 0.432 + 0.216 = 0.648 0.216 Pr(X = 3 | X > 1) = 0.648 1 = 3

1. Discrete random variables generally deal with number or size and are able to be counted. 2. A discrete probability distribution exists only if the following two characteristics are satisfied. (a) Each probability lies in a restricted interval 0 ≤ Pr(X = x) ≤ 1. (b) The probabilities of a particular experiment sum to 1, that is, Σ Pr(X = x) = 1.

506


exerCiSe

10B

Discrete random variables 1 We11 Which of the following represent discrete random variables? a The number of people at a tennis match b The time taken to read this question c The length of the left arms of students in your class d The shoe sizes of twenty people e The weights of babies at a maternity ward f The number of grains in ten 250-gram packets of rice g The height of jockeys competing in a certain race h The number of books in Melbourne libraries


Spreadsheet 100 Probability distribution

2 We12,13 a If X represents the number of heads obtained in two tosses of a coin, draw up a table that displays the values that the discrete random variable can assume and the corresponding probabilities. b Draw a probability distribution graph of the outcomes in part a. 3 A fair coin is tossed three times and a note is taken of the number of tails. a List the possible outcomes. b List the possible values of the random variable X, representing the number of tails obtained in the three tosses. c Find the probability distribution of X. d Find Pr(X ≤ 2). 4 Draw graphs for each of the following probability distributions. a

x Pr(X = x)

b

x Pr(X = x)

c

x Pr(X = x)

d

x Pr(X = x)

1

2

3

4

5

0.05

0.2

0.5

0.2

0.05

5

10

15

20

0.5

0.3

0.15

0.05

2

4

6

8

10

0.1

0.2

0.4

0.2

0.1

1

2

3

4

0.1

0.2

0.3

0.4

5 We14 Which of the following tables represent a discrete probability distribution? a

x Pr(X = x)

b

x Pr(X = x)

c

d

x

1

3

5

7

9

0.2

0.3

0.2

0.2

0.1

1

2

3

4

5

0.1

0.1

0.1

0.1

0.1

3

6

9

12

15

Pr(X = x)

0.3

0.2

0.4

0.2

−0.1

x

−4

−3

−1

1

2

Pr(X = x)

0.1

0.1

0.4

0.2

0.2

Chapter 10


507

6 WE15 Find the value of k for each of the following discrete probability distributions. a x Pr(X = x) b x Pr(X = x) c x Pr(X = x) d x Pr(X = x) 7

1

2

3

4

5

0.3

0.2

0.2

k

0.1

2

4

6

8

10

0.1

0.1

0.1

0.1

k

0

1

2

3

4

k

2k

3k

4k

k

−2

−1

0

1

2

k

0.2

3k

0.3

0.1

Find the value of k for the following discrete probability distribution. a x Pr(X = x)

1 3k 13

2

3 5k − 4 13

k2 13

4 k2 13

5 7 13

b Explain why one of the values of k had to be discarded. 8

Two fair dice are rolled simultaneously, and X, the sum of the two numbers appearing uppermost, is recorded. a Draw up a table that displays the probability distribution of X, and find: b Pr(X > 9) c Pr(X < 6) d Pr(4 ≤ X < 6) e Pr(3 ≤ X ≤ 9) f Pr(X < 12) g Pr(6 ≤ X < 10).

9 A spinner is numbered from 1 to 5, with each number being equally likely to come up. If X is the random variable representing the number showing on the spinner, find: b the probability of getting an even number a the probability distribution of X c Pr(X > 2). 10 A fair die is rolled and X is the square of the number appearing uppermost. a Draw up a table that displays the probability distribution of X, and find: b Pr(X < 30) c Pr(X > 10). 11

A fair die is altered so that the 1 is changed to a 5. If X is the random variable representing the number uppermost on the die, find: a the probability distribution of X b the probability of a number bigger than 2 appearing uppermost c Pr(X = 5 | X > 2).

12 WE16a function.

Show that the function p(x) =

13 WE16b Show that the function p(x) =

1 160

1 90

(8x + 2), where x = 0, 1, 2, 3, 4 is a probability

x2 (x + 2), where x = 1, 2, 3, 4 is a probability function.

14 WE17 Three balls are selected from a box containing 4 red balls and 5 blue balls. If the ball chosen after each selection is replaced before the next selection, find, correct to 4 decimal places: a the probability distribution for the number of red balls drawn: i 0 red balls ii 1 red ball iii 2 red balls iv 3 red balls b the probability that three reds are chosen, given that at least one ball is red. 508


15

A circular spinner, divided into five equal sectors numbered 1, 2, 3, 4, 5, is spun twice, and the sum of the numbers the pointer lands on is recorded. The following events are then defined. A = ‘an odd number on the first spin’ B = ‘an even number on the second spin’ C = ‘the sum of the two numbers is odd’ D = ‘the sum of the two numbers is at most 7’ a List each of the possible outcomes. b Find: i Pr(A) ii Pr(B) iii Pr(C) iv Pr(D) c Find: i Pr(A | B) ii Pr(B | C) iii Pr(C | D)

16

A biased coin is tossed twice. If the probability of obtaining a head is 3: 5 a find the probability distribution of the number of heads in 2 tosses b show that the sum of the probabilities is 1.

17

A discrete random variable has the following probability distribution: x Pr(X = x)

1

2

3

4

5

6

7

0.2

0.11

0.15

0.09

0.17

0.13

0.15

Find: a Pr(X > 3) d Pr(2 < X < 5) g {x: Pr(X ≥ x) = 0.54}

b Pr(X ≤ 4) e Pr(X < 3 | X < 5)

c Pr(3 ≤ X ≤ 6) f {x: Pr(X < x) = 0.46}

18 MC Which one of the following random variables is not discrete? A The price, in cents, of a loaf of bread at the local supermarket B The number of runs scored by a batsman in each innings over a season C The weight of a baby as he grows over a one-year period D The number of houses sold by a real estate agent each month for a year E The number of newspapers recycled by a family each month. 19 MC What is the value of k which will make this table a probability distribution table? x

1

2

3

4

Pr(X = x)

2k

3k

4k

k

A 0

B 1

C 0.1

D 1

20 MC Examine the following probability distribution table. x Pr(X = x)

E −0.1

9

4

9

16

25

36

0.16

0.21

0.35

0.08

0.2

Pr(X ≥ 10) is equal to: A 0.38 B 0.84

C 0.35

D 0.28

E 0.63

21 MC The following table represents a discrete probability distribution for a random variable, Y. x

4

7

10

13

Pr(X = x)

d

4d

5d

2k

The value of d is: B A 1 9

1 10

C

1 11

D

1 12

E

1 13


509

22 mC A coin is biased so that the probability of obtaining a head is 3 . If the coin is tossed 7 3 times the probability of obtaining exactly 2 heads is: 27 108 144 135 A B C D E 64 343

343

343

343

343

23 mC Which of the following is a probability function? 1 A p(x) = 0.1, 0.3, 0.4, 0.2, 0.1, x = 0, 1, 2, 3, 4 B p(x) = 66 (3x + 7), x = 0, 1, 2, 3, 4 1

1

C p(x) = 40 (5x − 1), x = 1, 2, 3, 4 E p(x) = 24

x2 20

D p(x) = 20 x2 (4 − x), x = 1, 2, 3

(3x − 1), x = 1, 2, 3

eBook plus

If the random variable X represents the number of boys in a four-child family: a write the values that X may take b assuming that the Pr(boy) = 1 , find the probability distribution of X.

Digital doc

WorkSHEET 10.1

2

10C

measures of centre of discrete random distributions The expected value of a discrete random variable, X, is the average value of X. It is also referred to as the mean of X or the expectation. The expected value of a discrete random variable, X, is denoted by E(X) or the symbol µ (mu). It is defined as the sum of each value of X multiplied by its respective probability; that is, E(X) = x1Pr(X = x1) + x2Pr(X = x2) + x3Pr(X = x3) + … + xnPr(X = xn) = ∑ x Pr(X = x) all x

Note: The expected value will not always assume a discrete value. WorkeD examPle 18

Find the expected value of a random variable that has the following probability distribution. x

1

2

3

4

5

Pr(X = x)

2 5

1 10

3 10

1 10

1 10

Think 1

WriTe

E(X) = ∑ x Pr ( X = x )

Write the rule for the expected value.

all x

2

Substitute the values into the rule.

3

Evaluate.

2

1

3

1

1

E(X) = 1 × 5 + 2 × 10 + 3 × 10 + 4 × 10 + 5 × 10 2

2

9

4

5

= 5 + 10 + 10 + 10 + 10 2

= 25 WorkeD examPle 19

Find the unknown probability, a, and hence determine the expected value of a random variable that has the following probability distribution. x Pr(X = x)

510

2

4

6

8

10

0.2

0.4

a

0.1

0.1


Think

Write

1

Determine the unknown value of a using the knowledge that the sum of the probabilities must total 1.

0.2 + 0.4 + a + 0.1 + 0.1 = 1 0.8 + a = 1 a = 1 − 0.8 = 0.2

2


E(X) = ∑ x Pr ( X = x ) all x

3


E(X) = 2 × 0.2 + 4 × 0.4 + 6 × 0.2 + 8 × 0.1 + 10 × 0.1

4

Evaluate.

= 0.4 + 1.6 + 1.2 + 0.8 + 1 = 5

Worked Example 20

Find the values of a and b of the following probability distribution if E(X) = 4.29. x Pr(X = x)

1

2

3

4

5

6

7

0.1

0.1

a

0.3

0.2

b

0.2

Think

Write

1

Write an equation for the values of a and b using the knowledge that the sum of the probabilities must total 1. Call this equation [1].

0.1 + 0.1 + a + 0.3 + 0.2 + b + 0.2 = 1 0.9 + a + b = 1 a + b = 1 − 0.9 a + b = 0.1 [1]

2


E(X) = ∑ x Pr(X = x ) all x

3


4.29 = 1 × 0.1 + 2 × 0.1 + 3 × a + 4 × 0.3 + 5 × 0.2 + 6 × b + 7 × 0.2

4

Evaluate and call this equation [2].

= 0.1 + 0.2 + 3a + 1.2 + 1 + 6b + 1.4 4.29 − 3.9 = 3a + 6b 3a + 6b = 0.39 [2]

5

Solve the equations simultaneously. Multiply equation [1] by 3 and call it equation [3]. Subtract equation [3] from equation [2]. Solve for b. Substitute b = 0.03 into equation [1]. Solve for a.

a + b = 0.1 [1] 3a + 6b = 0.39 [2] 3 × (a + b = 0.1) 3a + 3b = 0.3 [3] [2] − [3]: 3b = 0.09 b = 0.03 a + 0.03 = 0.1 a = 0.1 − 0.03 = 0.07

6


a = 0.07 and b = 0.03

Worked Example 21

Niki and Melanie devise a gambling game based on tossing three coins simultaneously. If three heads or three tails are obtained, the player wins $20. Otherwise the player loses $5. In order to make a profit they charge each person $2 to play. a What is the expected gain to the player? b Do Niki and Melanie make a profit? c Is this a fair game?


511

Think a

1

2

WriTe a

List the possible outcomes and place all of the information in a table.


S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} x

0

1

2

3

Pr(X = x)

1 8

3 8

3 8

1 8

Gain ($)

20

−5

−5

20

E(X) =

∑ x Pr( X = x)

all x

3


= 20 × 1 + −5 × 3 + −5 × 3 + 20 × 1

4

Evaluate.

= 20 − 15 − 15 + 20

8

8 10 8

8

8

8

8

8

8

= = $1.25 5


The player’s expected gain per game is $1.25; however, as each game incurs a cost of $2, the player in fact loses 75c per game.

b Answer the question using the results from a .

b

The girls make a profit of 75c per game.

c Answer the question using the results from a .

c

No, this is not a fair game, since the cost to play each game does not equal the expected gain of each game.

Note: In a fair game E(X) = 0.

It is important to understand that the expected value signifies the average outcome of an experiment and can be used to determine the feasibility of a situation. The previous worked example illustrates that, in the long run, the player will lose on average 75 cents per game; it does not mean the player will lose 75 cents each time the game is played.

expectation theorems WorkeD examPle 22

eBook plus

A random variable has the following probability distribution. x Pr(X = x) Find:

1

2

3

4

0.25

0.26

0.14

0.35

a E(X) c E(2X − 4)

1


b E(3X) d E(X2).

Think a

Tutorial

WriTe


a


512

2


3

Evaluate.

E(X) = 1 × 0.25 + 2 × 0.26 + 3 × 0.14 + 4 × 0.35 = 0.25 + 0.52 + 0.42 + 1.4 = 2.59


b

c

1


b E(3X) =

∑ 3x Pr(X = x)

all x

2


E(3X) = (3 × 1) × 0.25 + (3 × 2) × 0.26 + (3 × 3) × 0.14 + (3 × 4) × 0.35

3

Evaluate. Note: 1. The probability remains the same. 2. Each x-value is multiplied by 3 because of the new function, 3x.

= 3 × 0.25 + 6 × 0.26 + 9 × 0.14 + 12 × 0.35 = 0.75 + 1.56 + 1.26 + 4.2 = 7.77

1


c E(2X − 4) =

∑ (2 x − 4) Pr(X = x)

all x

d

2


= (2 × 1 − 4) × 0.25 + (2 × 2 − 4) × 0.26 + (2 × 3 − 4) × 0.14 + (2 × 4 − 4) × 0.35

3

Evaluate. Note: 1. The probability remains the same. 2. Each x-value is multiplied by 2 and then 4 is subtracted from the result, because of the new function, 2x − 4.

= − 2 × 0.25 + 0 × 0.26 + 2 × 0.14 + 4 × 0.35 = −0.5 + 0 + 0.28 + 1.4 = 1.18

1


2


= (12) × 0.25 + (22) × 0.26 + (32) × 0.14 + (42) × 0.35

3

Evaluate. Note: 1. The probability remains the same. 2. Each x-value is squared because of the new function, x2.

= 1 × 0.25 + 4 × 0.26 + 9 × 0.14 + 16 × 0.35 = 0.25 + 1.04 + 1.26 + 5.6 = 8.15

d E(X 2) =

∑ x 2 Pr(X = x)

all x

The above worked example displays some important points that shall be investigated. For this example, E(X) = 2.59 from part (b) E(3X) = 7.77 note that 3E(X) = 3 × 2.59 = 7.77 from part (c) E(2X − 4) = 1.18 note that 2E(X) − 4 = 2 × 2.59 − 4 = 1.18 Hence if X is a random variable and a is a constant, its expected value is defined by E(aX) = aE(X). Furthermore, if X is a random variable where a and b are constants, then the expected value of a linear function in the form f (X) = aX + b is defined by E(aX + b) = aE(X) + b If a = 0 then E(aX + b) = aE(X) + b becomes E(0X + b) = 0E(X) + b =b These rules are called expectation theorems and are summarised below. E(aX) = aE(X) where X is a random variable and a is a constant. E(aX + b) = aE(X) + b where X is a random variable and a and b are constants. E(b) = b where b is a constant. E(X + Y) = E(X) + E(Y) where X and Y are both random variables.


513

These theorems make it easier to calculate the expected values. Finally, from part (d) of the above example it can be seen that E(X 2) ≠ [E(X)]2

Worked Example 23

Casey decides to apply for a job selling mobile phones. She receives a base salary of $200 per month and $15 for every mobile phone sold. The following table shows the probability of a particular number of mobile phones, x, being sold per month. What would be the expected salary Casey would receive each month? x Pr(X = x)

50

100

150

200

250

0.48

0.32

0.1

0.06

0.04

Think

Write

Method 1 1

Define a random variable.

Let X = the number of mobile phones sold by Casey in a month.

2

Write the rule for the expected salary.

E(15X + 200) =

∑ (15x + 200)Pr(X = x)

all x

3


= (15 × 50 + 200) × 0.48 + (15 × 100 + 200) × 0.32 + (15 × 150 + 200) × 0.1 + (15 × 200 + 200) × 0.06 + (15 × 250 + 200) × 0.04

4

Evaluate.

= 950 × 0.48 + 1700 × 0.32 + 2450 × 0.1 + 3200 × 0.06 + 3950 × 0.04 = 456 + 544 + 245 + 192 + 158 = 1595

5


The expected salary Casey would receive each month would be $1595.

Method 2 Using the expectation theorem: 1

Write the rule for the expected salary.

E(X) = ∑ x Pr (X = x ) all x

2


= 50 × 0.48 + 100 × 0.32 + 150 × 0.1 + 200 × 0.06 + 250 × 0.04

3

Evaluate.

= 24 + 32 + 15 + 12 + 10 = 93

4

Using the fact that E(aX + b) = aE(X) + b find E(15X + 200).

E(15X + 200) = 15E(X) + 200 = 15 × 93 + 200 = 1595

Note: Using the expectation theorem is quicker because it is easier to evaluate aE(X) + b than E(aX + b).

514


Median and mode The median is the middle value of the distribution. It is the value such that 50% of the distribution lies to the left of this value, and 50% of the distribution lies to the right. For a random variable, X, the mode is the most commonly occurring value, that is, it is the variable with the highest probability. Worked Example 24

For the following probability distributions, calculate: i the median ii the mode.

a

x Pr(X = x)

3

5

6

8

0.25

0.4

0.15

0.2

5

8

11

14

0.18

0.32

0.2

0.3

b x

Pr(X = x) Think a

i

ii

1

i

Write the answer.

1

To calculate the mode, identify the highest probability in the table.

1

i Pr(X ≤ 3) = 0.25, which is ≤ 0.5

Pr(X ≤ 5) = 0.25 + 0.4 = 0.65, which is ≥ 0.5 The median value is 5.

i i The highest probability is 0.4.

That is, Pr(X = 5) = 0.4.

Write the answer.

The mode is 5. b

To calculate the median, work from the left of the data. Add up the probabilities until the total is 0.5 or greater.

i Pr(X ≤ 5) = 0.18, which is ≤ 0.5.

Pr(X ≤ 8) = 0.18 + 0.32 = 0.5 Similarly Pr(X ≥ 11) = 0.2 + 0.3 = 0.5 8 + 11 2 = 9.5

Calculate the median by calculating the mean of 8 and 11.

Median =

3

Write the answer.

The median is 9.5.

1

To calculate the mode, identify the highest probability in the table.

2

ii

a

To calculate the median, work from the left of the data. Add up the probabilities until the total is 0.5 or greater.

2

2

b

WRITE

2

i i The highest probability is 0.32.

That is, Pr(X = 8) = 0.32.

Write the answer.

The mode is 8.

1. The expected value of a discrete random variable, X, is defined by the rule E(X) =

∑ x Pr(X = x). Also E(X2) = ∑ x2 Pr(X = x).

all x

all x

2. A game is considered fair if the cost to play the game is equal to the expected gain. 3. A fair game is one in which E(X) = 0.


515

4. The expected value of a linear function can be calculated using the expectation theorems: E(aX) = aE(X) E(aX + b) = aE(X) + b E(b) = b E(X + Y) = E(X) + E(Y) 5. E(X2) ≠ [E(X)]2 6. The median is the middle value of a distribution. 7. The mode is the variable with the highest probability.

Exercise

10c

Measures of centre of discrete random distributions 1 WE18 Find the expected value of a random variable that has the following probability distribution. x Pr(X = x) 2

0

3

6

9

12

0.21

0.08

0.19

0.17

0.35

Find the expected value of a random variable that has the following probability distribution. x

-2

-1

0

1

2

3

4

Pr(X = x)

1 18

1 3

1 18

2 9

1 6

1 18

1 9

3 WE19 Find the unknown probability, a, and hence determine the expected value of a random variable that has the following probability distribution. x Pr(X = x)

1

3

5

7

9

11

0.11

0.3

0.15

0.25

a

0.1

4 Find the unknown probability, a, and hence determine the expected value of a random variable that has the following probability distribution. x

-2

1

4

7

10

13

Pr(X = x)

5 18

a

1 9

5 18

1 18

2 9

5 Find the unknown probability, b, and hence determine the expected value of a random variable that has the following probability distribution. x

0

1

2

3

4

5

Pr(X = x)

b

0.2

0.02

3b

0.1

0.08

6 Find the value of k, and hence determine the expected value of a random variable that has the following probability distribution.

7

516

x

4

8

12

16

20

Pr(X = x)

6k

2k

k

3k

8k

If X represents the outcome of a fair die being rolled, find: a the probability distribution of each outcome


b E(X).

8 Two fair dice are rolled simultaneously. If X represents the sum of the two numbers appearing uppermost, find: a the probability distribution of each outcome b E(X). 9 A fair coin is tossed 4 times. If X represents the number of tails obtained, find: a the probability distribution of each outcome b E(X). 10 We20 x Pr(X = x) 11

Find the values of a and b of the following distribution if E(X) = 1.91. 0

1

2

3

4

5

6

0.2

0.32

a

0.18

b

0.05

0.05

Find the values of a and b of the following distribution if E(X) = 2.41. x Pr(X = x)

0

1

2

3

4

5

0.2

a

0.23

0.15

b

0.12

12 We21 Lucas contemplates playing a new game which involves tossing three coins simultaneously. He will receive $15 if he obtains 3 heads, $10 if he obtains 2 heads and $5 if he obtains 1 head. However, if he obtains no heads he must pay $30. He must also pay $5 for each game he plays. a What is Lucas’s expected gain? b Should he play the game? Why? c Is this a fair game? Why? 13

Angie plays a game based on tossing three coins simultaneously. She will receive $10 if she obtains 3 tails, $5 if she obtains 2 tails and $5 if she obtains 1 tail. However, if she obtains no tails she must pay $40. a What is Angie’s expected gain? b Should she play the game? Why? c Is this a fair game? Why?

14

X is a discrete random variable with the following probability distribution. x Pr(X = x)

2

4

7

k

0.3

0.2

0.4

0.1

Find the value of k if the mean is 5.3. 15 X is a discrete random variable with the following probability distribution. x

−2

3

8

10

14

k

Pr(X = x)

0.1

0.08

0.07

0.27

0.16

0.32

Find the value of k if the mean is 10.98.


SkillSHEET 10.1 Expected value of a function of a random variable

16 A coin is biased such that the probability of obtaining a tail is 0.6. If X represents the number of tails in three tosses of the coin, find: a the probability distribution of X b E(X) c the mode. 17 We22 x Pr(X = x) Find: a E(X)

A random variable has the following probability distribution. 1

2

3

4

2 15

7 15

1 3

1 15

b E(4X)

c E(2X + 1)

Chapter 10

d E(X2).


517

18 A random variable has the following probability distribution. x Pr(X = x) Find: a E(X)

1

2

3

4

0.33

0.25

0.27

0.15

c E(X 2 + 1)

b E(4X − 6)

d E(3X 2).

19 We23 Christian decides to apply for a job selling mobile phones. He receives a base salary of $180 per month and $12 for every mobile phone sold. The following table shows the probability of a particular number of mobile phones, x, being sold per month. What would be the expected salary Christian would receive each month? x Pr(X = x)

50

100

150

200

250

0.32

0.38

0.2

0.06

0.04

20 We24 For the following probability distributions, calculate: i the median ii the mode. eBook plus

a

Pr(X = x)

Digital doc

WorkSHEET 10.2

x

b

x Pr(X = x)

c

x Pr(X = x)

10D

1

2

3

4

5

0.25

0.15

0.1

0.1

0.4

4

8

9

13

17

0.06

0.36

0.17

0.29

0.12

−2

−1

0

1

2

3

4

1 4

1 16

3 16

1 8

1 8

1 16

3 16

measures of variability of discrete random distributions


int-0255

Measures of variability of discrete random distributions

Variance

Variance is an important feature of probability distributions as it provides information about the spread of the distribution with respect to the mean. If the variance is large, it implies that the possible values are spread (or deviate) quite a distance from the mean. A small variance implies that the possible values are close to the mean. Variance is also called a measure of spread or dispersion. The variance is written as Var(X) and denoted by the symbol σ2 (sigma squared). It is defined as the expected value (or average) of the squares of the spreads (deviations) from the mean. The rule for variance is given by: Var(X) = E(X − µ)2 = ∑(X − µ)2 Pr(X = x) Although this rule clearly demonstrates how to obtain the variance, performing the calculation is quite a lengthy process. Hence an alternative rule is used for calculating the variance: Var(X) = E(X − µ)2 = E(X2 − 2µX + µ2) = E(X2) − E(2µX) + E(µ2) = E(X2) − 2µE(X) + E(µ2) = E(X2) − 2µ2 + µ2 since E(X) = µ (the mean) = E(X2) − µ2 = E(X2) − [E(X)]2 518


Worked Example 25

Find the expected value and variance of the following probability distribution table. x Pr(X = x)

1

2

3

4

5

0.15

0.12

0.24

0.37

0.12

Think

Write

Method 1: Using the rule 1



2


= 1 × 0.15 + 2 × 0.12 + 3 × 0.24 + 4 × 0.37 + 5 × 0.12

3

Evaluate.

= 0.15 + 0.24 + 0.72 + 1.48 + 0.6 = 3.19

4

Calculate E(X 2).

E(X 2) =

∑ x 2 Pr(X = x)

all x

= (12) × 0.15 + (22) × 0.12 + (32) × 0.24 + (42) × 0.37 + (52) × 0.12 = 1 × 0.15 + 4 × 0.12 + 9 × 0.24 + 16 × 0.37 + 25 × 0.12 = 0.15 + 0.48 + 2.16 + 5.92 + 3 = 11.71

5

Calculate [E(X)]2.

[E(X)]2 = 3.192 = 10.1761

6

Calculate Var(X) using the rule for variance.

Var(X) = E(X2) − [E(X)]2 = 11.71 − 10.1761 = 1.5339

Method 2: Using a CAS calculator 1

Open a Lists & Spreadsheet page. Label column A as x and column B as probx. Enter the data into the relevant columns. Press: • MENU b • 4:Statistics 4 • 1:Stat Calculations 1 • 1:One-Variable Statistics 1 The number of lists is 1. The X1 List is x and the frequency is probx.

2

The expected value and the variance (SSD) can be read from the screen. You will need to scroll down to get the value of the variance.

E(X) = 3.19 Var(X) = 1.5339


519

WorkeD examPle 26

eBook plus

Find the variance of 2Y + 1 for the following probability distribution table. y Pr(Y = y)

0

1

2

3

0.25

0.35

0.2

0.2

Think

Tutorial


WriTe

E(2Y + 1) = ∑ (2 y + 1) Pr(Y = y )

1


2


= (2 × 0 + 1) × 0.25 + (2 × 1 + 1) × 0.35 + (2 × 2 + 1) × 0.2 + (2 × 3 + 1) × 0.2

3

Evaluate.

= 1 × 0.25 + 3 × 0.35 + 5 × 0.2 + 7 × 0.2 = 0.25 + 1.05 + 1 + 1.4 = 3.7

4

Calculate [E(2Y + 1)]2.

5

Calculate E(2Y + 1)2.

6

Calculate Var(2Y + 1) using the rule.

all x

[E(2Y + 1)]2 = 3.72 = 13.69 E(2Y + 1)2 = 12 × 0.25 + 32 × 0.35 + 52 × 0.2 + 72 × 0.2 = 0.25 + 3.15 + 5 + 9.8 = 18.2 Var(2Y + 1) = E(2Y + 1)2 − [E(2Y + 1)]2 = 18.2 − 13.69 = 4.51

The variance of a linear function can also be calculated by the following rule: Var (aX + b) = a2Var (X) For worked example 26, given that Var(Y) = 1.1275, Var(2Y + 1) can be determined using the above rule: Var(2Y + 1) = 22Var(Y) = 4 × 1.1275 = 4.51 as before. WorkeD examPle 27

X is a discrete random variable with the following probability distribution. x Pr(X = x)

3

4

6

k

0.15

0.3

0.45

0.1

Find the value of k, a positive integer, if the variance is 1.7475. Think 1


WriTe


520

2


= 3 × 0.15 + 4 × 0.3 + 6 × 0.45 + k × 0.1

3

Evaluate.

= 0.45 + 1.2 + 2.7 + 0.1k = 4.35 + 0.1k


4

E(X2) =

Calculate E(X2).

∑ x 2 Pr(X = x)

all x

= (32) × 0.15 + (42) × 0.3 + (62) × 0.45 + (k2) × 0.1 = 9 × 0.15 + 16 × 0.3 + 36 × 0.45 + k2 × 0.1 = 1.35 + 4.8 + 16.2 + 0.1k2 = 22.35 + 0.1k2

5

Calculate [E(X)]2.

[E(X)]2 = (4.35 + 0.1k)2 = 0.01k2 + 0.87k + 18.9225

6

Calculate Var(X) using the rule and equate it to the given value of the variance; that is, σ 2 = 1.7475.

Var(X) = E(X2) − [E(X)]2 1.7475 = 22.35 + 0.1k2 − (0.01k2 + 0.87k + 18.9225) 1.7475 = 0.09k2 − 0.87k + 3.4275

7

Solve for k.

0.09k2 − 0.87k + 3.4275 − 1.7475 = 0 0.09k2 − 0.87k + 1.68 = 0 (9k − 24)(k − 7) = 0 k=

8

24 9

2

(or 2 3 ) or k = 7

k = 7. Reject the other value of k since the variable is discrete.


Standard deviation Another important measure of spread is the standard deviation. It is written as SD(X) or denoted by the symbol σ (sigma). The standard deviation is the positive square root of the variance. It is defined by the rule: SD(X) =

Var ( X )

= σ2 =σ Variation and standard deviation are used extensively in many real-life applications involving statistics. Analysis of data would be useless without any information about the spread of the data. Worked Example 28

A random variable has the following probability distribution. x

0

1

2

3

Pr(X = x)

1 4

3 8

1 8

1 4

Calculate the expected value, the variance and the standard deviation. Think

Write

Method 1: Using the rule 1

2

Calculate the expected value.

Calculate [E(X)]2.

1

3

1

1

E(X) = 0 × 4 + 1 × 8 + 2 × 8 + 3 × 4 3

2

3

=0+8+8 +4

= 18

3

( )

[E(X)]2 = 1 83

2

57

= 164 (≈1.890 625)


521

3

4

1

3

1

1

E(X2) = 02 × 4 + 12 × 8 + 22 × 8 + 32 × 4

Calculate E(X2).

3

4

9

=0+8+8 +4

= 38

1

Var(X) = E(X2) − [E(X)]2 1 57 = 38 − 164

Calculate Var(X).

15

= 164 (≈ 1.234 375)

5

Calculate the standard deviation.

SD(X) = 1.234 375

6

Round the answer to 4 decimal places.

= 1.1110

Method 2: Using a CAS calculator 1

Open a Lists & Spreadsheet page. Label column A as x and column B as probx. Enter the data into the relevant columns. Press: • MENU b • 4:Statistics 4 • 1:Stat Calculations 1 • 1:One-Variable Statistics 1 The number of lists is 1. The X1 List is x and the frequency is probx.

2

The expected value and the variance (SSD) and the standard deviation can be read from the screen. You will need to scroll down to get the value of the variance.

E(X) = 1.375 Var(X) ≈ 1.2344 SD(x) ≈ 1.1110

Worked Example 29

In order to encourage car pooling, a new toll is to be introduced on the Eastgate Bridge. If the car has no engers, a toll of $2 applies. Cars with one enger pay a $1.50 toll, cars with two engers pay a $1 toll and cars with 3 or more engers pay no toll. Long-term statistics show that the number of engers (X) follows the probability distribution given below. x (no. of engers) Pr(X = x)

0

1

2

≥3

0.4

0.35

0.2

0.05

a Construct a probability distribution of the toll paid. b Find the mean toll paid per car. c Find the standard deviation of tolls paid. Think a Construct a table of values of toll information.

Probabilities remain the same.

Write a Let Y = toll to be paid.

y Pr(Y = y)

522


2

1.5

1

0

0.4

0.35

0.2

0.05

b

c

1

b E(Y) =


∑ y Pr(Y = y)

all y

2


= 2 × 0.4 + 1.5 × 0.35 + 1 × 0.2 + 0 × 0.05

3

Evaluate.

= 0.8 + 0.525 + 0.2 + 0 = 1.525

4


= 1.53

5


The mean toll is $1.53.

1

Calculate

c

E(Y2).

E(Y2) =

∑ y 2 Pr (Y = y)

all y

= (22) × 0.4 + (1.52) × 0.35 + (12) × 0.2 + (02) × 0.05 = 4 × 0.4 + 2.25 × 0.35 + 1 × 0.2 + 0 = 1.6 + 0.7875 + 0.2 = 2.5875

2

Calculate [E(Y)]2.

[E(Y)]2 = 1.5252 = 2.325 625

3

Calculate Var(Y).

Var(Y) = E(Y2) − [E(Y)]2 = 2.5875 − 2.325 625 = 0.261 875

4


SD(Y) =

5


6


The standard deviation of tolls paid is $0.51.

0.261 875

= 0.51

Interpreting the standard deviation A characteristic of many distributions is that approximately 95% of the spread or distribution lies between 2 standard deviations of the mean; that is, Pr(µ − 2σ ≤ X ≤ µ + 2σ) ≈ 0.95 It is important to note that when calculating the Pr(µ − 2σ ≤ X ≤ µ + 2σ) for a specific distribution an exact value of 0.95 will not always be achieved, but should be close to it. For many random variables, approximately 95% of the spread of the population lies between 2 standard deviations of the mean, that is, Pr(µ − 2σ ≤ X ≤ µ + 2σ) ≈ 0.95. Worked Example 30

A probability distribution is shown below. Given that µ = 3.38 and σ = 0.946, calculate Pr(µ - 2σ ≤ X ≤ µ + 2σ). x Pr(X = x)

2

3

4

5

6

0.14

0.5

0.23

0.1

0.03

Think 1

Calculate µ - 2σ

Write

µ - 2σ = 3.38 - 2 × 0.946 = 3.38 - 1.892 = 1.488


523

2

Calculate µ + 2σ.

µ + 2σ = 3.38 + 2 × 0.946 = 3.38 + 1.892 = 5.272

3

Substitute the values obtained in steps 1 and 2 into the given interval. Since X represents discrete values, Pr(1.488 ≤ X ≤ 5.272) becomes Pr(2 ≤ X ≤ 5).

Pr(µ − 2σ ≤ X ≤ µ + 2σ) = Pr(1.488 ≤ X ≤ 5.272) = Pr(2 ≤ X ≤ 5) = 0.14 + 0.5 + 0.23 + 0.1 = 0.97 Note: In this example, 97% of the distribution lies within 2 standard deviations of the mean, which is close to the estimated 95%.

WorkeD examPle 31

The table below represents the probability distribution of the number of accidents per week in a factory. x Pr(X = x)

1

2

3

4

5

6

7

8

9

0.02

0.22

0.18

0.16

0.14

0.07

0.13

0.03

0.05

Given that µ = 4.36 and σ = 2.105 find Pr(µ − 2σ ≤ X ≤ µ + 2σ). Think

WriTe

1

Calculate µ − 2σ.

µ − 2σ = 4.36 − 2 × 2.105 = 4.36 − 4.21 = 0.15

2

Calculate µ + 2σ.

µ + 2σ = 4.36 + 2 × 2.105 = 4.36 + 4.21 = 8.57

3

Substitute the values obtained in steps 1 and 2 into the given interval. Since X represents discrete values, Pr(0.15 ≤ X ≤ 8.57) becomes Pr(1 ≤ X ≤ 8). Note: In this example, Pr(1 ≤ X ≤ 8) = 1 − Pr(X = 9).

Pr(µ − 2σ ≤ X ≤ µ + 2σ) = Pr(0.15 ≤ X ≤ 8.57) = Pr(1 ≤ X ≤ 8) = Pr(X = 1) + Pr(X = 2) + Pr(X = 3) + Pr(X = 4) + Pr(X = 5) + Pr(X = 6) + Pr(X = 7) + Pr(X = 8) = 1 − Pr(X = 9) = 1 − 0.05 = 0.95 Note: The answer is the estimated one of 95%. In this case, 95% of the distribution lies within 2 standard deviations of the mean.

WorkeD examPle 32

The probability distribution of X is given by the formula: x2 Pr(X = x) = where x = 2, 3, 4, 5. 54 Find: a the probability distribution of X as a table b the expected value of X, correct to 4 decimal places c the standard deviation of X, correct to 4 decimal places d Pr(µ − 2σ ≤ X ≤ µ + 2σ), correct to 3 decimal places. 524


eBook plus Tutorial


Think a

1

WriTe


a When x = 2, p(x) =

= When x = 3, p(x) = = When x = 4, p(x) = = When x = 5, p(x) =

2

b

1

Enter the information into a table.

Calculate the expected value.

x

2

3

4

5

Pr(X = x)

2 27

1 6

8 27

25 54

= c


1

Calculate [E(X)]2.

2 +3×1+ 27 6 4 + 3 + 32 + 125 27 6 27 54 4 4 27

b E(X) = 2 ×

=

2

4 54 2 27 9 54 1 6 16 54 8 27 25 54

4×

8 27

+5×

25 54

= 4.1481 c [E(X)]2 = (4 4 )2 27 151

= 17 729 (≈ 17.2071) 2

Calculate E(X2).

2

1

8

25

E(X2) = 22 × 27 + 32 × 6 + 42 × 27 + 52 × 54 8

9

= 27 + 6 +

128 27

+

625 54

1

= 18 9 (≈ 18.1111) 3

Calculate Var(X).

Var(X) = E(X2) − [E(X)]2 1

151

= 18 9 − 17 729 =

d

659 729

(≈ 0.9040)

SD(X) = 0.9040 = 0.9507776039

4


5


1


d µ − 2σ = 4.1481 − 2 × 0.9508

2

Calculate µ + 2σ.

3

Substitute the values obtained in steps 1 and 2 into the given interval. Since X represents discrete values, Pr(2.2465 ≤ X ≤ 6.0497) becomes Pr(3 ≤ X ≤ 6).

µ + 2σ = 4.1481 + 2 × 0.9508 = 4.1481 + 1.9016 = 6.0497 Pr(µ − 2σ ≤ X ≤ µ + 2σ) = Pr(2.2465 ≤ X ≤ 6.0497) = Pr(3 ≤ X ≤ 6) = 1 − Pr(X = 2) =1− 2

= 0.9508 = 4.1481 − 1.9016 = 2.2465

27

= 25 27

Chapter 10


525

4

= 0.926


Note: In this example, 92.6% of the distribution lies within 2 standard deviations of the mean, which is close to the estimated value of 95%.

1. The variance, Var(X) or σ 2 is defined by the rule: Var(X) = E(X2) − [E(X)]2 2. The variance of a linear function can also be calculated by the following rule: Var (aX + b) = a2Var(X). 3. The standard deviation, SD(X) or σ, is defined by the rule: SD(X) = Var ( X ) = σ2 4. Approximately 95% of the spread of the population in many distributions lies between 2 standard deviations of the mean; that is, Pr (µ − 2σ ≤ x ≤ µ + 2σ) ≈ 0.95. exerCiSe

10D eBook plus Digital doc

Spreadsheet 100 Probability distribution

measures of variability of discrete random distributions 1 We25 Find the expected value and variance of the following probability distribution table. x Pr(X = x)

1

2

3

4

0.2

0.4

0.3

0.1

2 A random variable has the following probability distribution. x

2

4

6

8

Pr(X = x)

1 8

3 16

9 16

1 8

Find: a the expected value, E(X)

b the variance of X, Var(X).

3 The cost of a loaf of bread is known to vary on any day according to the following probability distribution. x

$1.20

$1.25

$1.30

$1.35

$1.60

Pr(Y = y)

0.05

0.2

0.1

0.25

0.4

Find: a the expected cost of a loaf of bread

b the variance of the cost.

4 We26 Find the variance of 2Y − 1 for the following probability distribution table. x Pr(Y = y)

0

1

2

3

0.3

0.2

0.3

0.2

5 A random variable has the following probability distribution. x Pr(X = x) Find: a Var(X) 526

2

4

6

8

0.15

0.3

0.42

0.13

b Var(2X)


c Var(3X + 1)

d Var(−5X + 7).

6 A random variable has the following probability distribution. x Pr(X = x)

0

1

3

5

7

0.27

0.15

0.13

0.1

0.35

Find: a Var(X) 7 WE27

d Var(5X − 2).

Let X be a discrete random variable with the following probability distribution.

x Pr(X = x)

c Var(10X − 5)

b Var(3X) 2

4

6

k

0.3

0.1

0.5

0.1

Find the value of k, a positive integer, if the variance is 5.8. 8 Let X be a discrete random variable with the following probability distribution. x Pr(X = x)

1

k

7

10

0.1

0.2

0.3

0.4

Find the value of k, a positive integer, if the variance is 7.96. 9 WE28 A random variable has the following probability distribution. x

1

2

3

4

Pr(X = x)

1 4

1 3

1 4

1 6

Calculate the expected value, the variance and the standard deviation. 10 A random variable has the following probability distribution. x Pr(X = x)

6

7

10

12

0.3

0.3

0.2

0.2

Find: a the expected value, E(X) c the standard deviation of X, SD(X), to 2 decimal places.

b the variance of X, Var(X)

11

For a random variable, X, E(X) = 12 and E(X2) = 340. Find the standard deviation of X. 12 For a random variable, X, E(X) = 20 and E(X2) = 529. Find the standard deviation of X, to 2 decimal places. 13 WE29 In order to encourage car pooling, a new toll is to be introduced on the International Gateway. If the car has no engers, a toll of $2 applies. Cars with one enger pay a $1.50 toll, cars with two engers pay a $1 toll and cars with 3 or more engers pay no toll. Long-term statistics show that the number of engers follows the probability distribution given below. x Pr(X = x)

0

1

2

≥3

0.5

0.3

0.15

0.05

a Construct a probability distribution of the toll paid. b Find the mean toll paid per car. c Find the standard deviation of tolls paid. 14 WE31 The table below represents the probability distribution of the number of accidents per week in a factory. x Pr(X = x)

1

2

3

4

5

6

7

8

9

0.03

0.21

0.16

0.18

0.14

0.07

0.15

0.01

0.05

Given that µ = 4.35 and σ = 2.08, find Pr(µ − 2σ ≤ X ≤ µ + 2σ).


527

x2 15 WE32 The probability distribution of X is given by the formula, Pr(X = x) = where 35 x = 1, 3, 5. Find: a the probability distribution of X as a table b the expected value of X c the standard deviation of X, to 4 decimal places d Pr(µ − 2σ ≤ X ≤ µ + 2σ).

16 The probability distribution of X is given by the formula, Pr(X = x) = Find: a the probability distribution of X as a table c the standard deviation of X

x2 − 1 where x = 2, 3, 4, 5. 50

b the expected value of X d Pr(µ − 2σ ≤ X ≤ µ + 2σ).

17 A random variable has the following probability distribution. x Pr(X = x)

1

2

3

4

0.4

0.2

0.2

k

Find: a the value of the constant k c E(X), the mean of X e SD(X), the standard deviation of X, to 4 decimal places

b the most likely value of X d Var(X), the variance of X f Pr(µ − 2σ ≤ X ≤ µ + 2σ).

18 Calculate the values between which 95% of the distribution would be expected to lie where: a µ = 4, σ = 2 b µ = 10, σ = 3 c µ = 35, σ = 7 1 1 d µ = 21.6, σ = 5.2 e µ = 9.7, σ = 0.7 f µ = 17 2 , σ = 23.

19 Two fair dice are rolled and the outcomes are noted. If X represents the sum of the two

numbers showing, find: a the expected value of X b the variance of X, to 2 decimal places c Pr(µ − 2σ ≤ X ≤ µ + 2σ), to 2 decimal places.

20 For the spinner shown, X represents the number obtained. Find: a the probability distribution of X b the expected value of X c the standard deviation of X, to 2 decimal places d the probability that the number is 4, given that it is not 1.

21 MC The following table represents a discrete probability

2 1

0

1

2

3

Pr(X = x)

k

2k

3k

4k

4

1

distribution for a random variable, X. x

3

2 2

3

The standard deviation of x is: A 1.0 B 1.2 c 1.8 d 2.0 e 2.2 Questions 22 and 23 refer to the following information. The probability distribution of X is given in the table below. x Pr(X = x)

0

3

6

9

0.4

0.3

0.1

0.2

22 MC The variance and standard deviation of X, respectively, are: A 3.9, 15.21

B 26.1, 3.3

23 MC Var (6X − 3) is equal to: A 91.26

528

B 541.56

c 26.1, 3.9

d 26.1, 5.11

e 11.61, 3.41

c 417.96

d 939.6

e 140.4


Summary Probability revision

• Outcomes are results of experiments. • The set of all possible outcomes of an experiment is called the sample space and is denoted by ε, and each possible outcome is called a sample point. • A subset of the sample space is known as an event. • The union (symbol ∪) of two events A and B implies a combined event, that is, either event A or event B or both occurring. Common elements are written only once. • The intersection (symbol ∩) of two events A and B is represented by the common sample points of the two events. • Venn diagrams involve drawing a rectangle that represents the sample space and a series of circles that represent subsets of the sample space. They provide a visual representation of the information at hand and clearly display the relationships between sets. • The probability of an event occurring is defined by the rule Pr(A) =


.

• • • •

The probability of an event occurring lies within the restricted interval 0 ≤ Pr(A) ≤ 1. The individual probabilities of a particular experiment will sum to 1; that is, Σ p(x) = 1. The addition rule of probability is defined by the rule Pr(A ∪ B) = Pr(A) + Pr(B) − Pr (A ∩ B). If two events A and B are mutually exclusive, then Pr(A ∩ B) = 0 and therefore the addition rule becomes Pr(A ∪ B) = Pr(A) + Pr(B). • If two events A and B are independent, then Pr (A ∩ B) = Pr(A) × Pr(B). • Karnaugh maps and probability tables summarise all combinations of two events (for example A and B) and their complements (for example A′ and B′). Pr ( A ∩ B) , where Pr(B) ≠ 0. This can be transposed • Conditional probability is defined by the rule Pr(A | B) = Pr ( B) to Pr(A ∩ B) = Pr (A | B) × Pr(B).

• Tree diagrams are useful tools in solving probability tasks as they display each of the possible outcomes along with their respective probabilities. • A combination is defined by nCr, that is, the number of selections of n different objects taken r at a time. Discrete random variables

• A random variable is one whose value is determined by the outcome of an experiment. • Discrete random variables generally deal with number or size and are able to be counted. • Two important characteristics satisfy all discrete probability distributions: 1. Each probability lies in a restricted interval 0 ≤ Pr(X = x) ≤ 1. 2. The probabilities of a particular experiment sum to 1; that is,

∑ Pr( X = x) = 1 If these two characteristics are not satisfied, then there is no discrete probability distribution. Measures of centre of discrete random distributions

• The expected value of a discrete random variable, X, is denoted by E(X) or the symbol µ (mu). It is defined by the rule: E(X) =

∑ x Pr (X = x)

all x

• A fair game is one in which E(X) = 0.


529

• The expected value of a linear function can be calculated using the expectation theorems: E(aX) = aE(X) E(aX + b) = aE(X) + b E(b) = b E(X + Y) = E(X) + E(Y) Note: E(X2) ≠ [E(X)]2 • The median is the middle value of a distribution. • The mode is the variable with the highest probability. Measures of variability of discrete random distributions

• The variance is denoted by Var(X) or the symbol σ2 (sigma squared). • It is defined by the rule: Var(X) = E(X2) − [E(X)]2. • The variance of a linear function can also be calculated by the following rule: Var(aX + b) = a2Var(X) • The standard deviation is written as SD(X) or denoted by the symbol σ. • It is defined by the rule: SD(X) = Var( X ) = σ2 • Approximately 95% of the spread of the population in many distributions lies between 2 standard deviations of the mean, that is, Pr(µ − 2σ ≤ X ≤ µ + 2σ) ≈ 0.95

530



1 The Santaroos have 3 soccer teams, A, B of and C, entered in the interschool championships. Each team is entered in a separate division. The probability of each team winning their particular final is given as follows: 1 1 Pr(Team A wins) = 2 Pr(Team B wins) = 3 3

Pr(Team C wins) = 4 . Find the probability that: a none of the Santaroos teams win b one of the Santaroos teams wins c two of the Santaroos teams win d each of the Santaroos teams win their particular final. 2 The probability of event A occurring is 0.25 and the probability of event B occurring is 0.5. a Calculate Pr(A ∩ B′) when Pr(A ∩ B) = 0.1. b Calculate Pr(A ∩ B′) when events A and B are mutually exclusive. Exam tip Many students confuse independent

events with mutually exclusive events.


[© VCAA 2007] 3 Thirty students were required to complete a logic puzzle. The time taken to complete the puzzle was recorded in the table below. Time taken (whole number of minutes)

3

4

5

6

7

Number of students

4

8

10

2

6

For the information given: a what proportion of students completed the puzzle in less than 5 minutes? b what proportion of students took more than 5 minutes to complete the puzzle? 4 The probability distribution for the service time at a bakery is given below. Service time (whole number of minutes) Probability

1

2

3

4

5

0.2

0.3

0.2

0.2

0.1

a What is the probability that the service time for a customer is: i 2 minutes? ii 2 minutes or less? iii more than 2 minutes? iv not more than 4 minutes, given it is more than 1 minute? b What is the expected value for the service time? c If 50 customers were served at the bakery in a morning, how many of these would you expect to take 4 minutes to be served? 5 A LUCKYDIPZ is a toy that is packaged inside an egg-shaped chocolate. A certain manufacturer produces 4 different types of LUCKYDIPZ toy — car, ship, plane and ring, in the proportions given in the table below. Car

4k2 + 4k

Ship

5k2 + 2k

Plane

k2 + k

Ring

2k

a Show that k must be a solution of the quadratic equation 10k2 + 9k – 1 = 0. b Find the exact value of k. [© VCAA 2004] 6 The probability distribution of X is given by the x2 formula, Pr(X = x) = , where x = 1, 2, 3, 4. Find: 30 a the probability distribution of X as a table b the expected value of X. 7 A player rolls a fair die. If the player gets a 1 on the first roll, she rolls again and her score is the sum of the two results, otherwise her score is the result of the first roll. The die cannot be thrown more than twice. Find: a the probability distribution b the expected score c Pr(X < µ). 8 Tayah and Sandy are playing a game where a biased die is rolled. The probabilities of rolling each number are Pr(1) = 0.2, Pr(2) = Pr(3) = Pr(5) = 0.1, Pr(4) = 0.3 and Pr(6) = 0.2. They have to pay $1.00 to play. If a 2, 3 or 5 is rolled, they win $3. If a 1 or 6 is rolled, they get


531

their money back, and if a 4 is rolled, they do not receive any money. Is this a fair game to play? 9 A game of ‘three-up’ is played where three coins are tossed simultaneously. A player must pay $2 to play the game. If three heads come up, the player collects $6. If two heads come up, the player collects $3. Is it a fair game? 10 Kylie is about to 2 points 30 cm compete in her club’s archery 5 points finals. If she is 10 cm equally likely to 11 points hit any point on the board and 20 cm never misses the target, find: a her expected score from i 1 shot at the target shown ii 5 shots at the target shown b her probability of getting five points on every one of her 5 shots.

x

3

4

5

6

7

0.1

0.45

0.09

0.26

0.1

x

1

4

6

8

11

Pr(X = x)

1 5

1 5

1 10

2 5

1 10

Pr(X = x) b

12 At Fast Eddy’s Drive-In Theatre the cost is $5 per car, plus $1 per occupant. The variable X represents the number of people in any car and is known to follow the probability distribution below. x Pr(X = x)

2

3

4

5

0.4

0.2

0.3

0.1

Find: a the expected cost per car b Fast Eddy’s expected profit if 100 cars enter and costs for wages, electricity, etc. are $300 c the mode. 13 Let X be a discrete random variable with the following probability distribution. x Pr(X = x)

1 0.1

3 0.25

5 0.35

Find the value of n if the mean is 4.7.

532

15 A raffle is to be drawn from 500 tickets. Each ticket was purchased for $1, with first prize being $200, second prize $150 and third prize $100. Find: a the expected loss per ticket b the profit made by organisers, that is, the house c the house percentage. x ( x + 1) for 0 ≤ x ≤ n 40 a Find the value of n. b Find the probability distribution for X as a table. c Find the expected value of X.

16 Pr(X = x) =

Multiple choice

11 For the following probability distributions, calculate: i the median ii the mode. a

14 Five thousand ‘scratch-and-match’ tickets are to be sold for $2 each. The tickets offer the following prizes: 1 prize of $5000 2 prizes of $500 20 prizes of $50 100 prizes of $10. Find: a the expected loss per ticket b the profit made by organisers, that is, the house c the house percentage.

1 If Pr(A) = 0.65, Pr(B) = 0.37 and Pr (A ∩ B) = 0.28, then Pr(A ∪ B) is equal to: A 0.93 B 0.56 C 0.09 D 0.74 E 1.02 2 If Pr(A) = 0.47, Pr(B) = 0.27 and Pr (A ∩ B) = 0.19, then Pr(A | B) is equal to: A 0.57 B 0.40 C 0.70 D 0.43 E 0.30 3 The probability that Fiona attends an aerobics class is 0.60, and the probability that Kath attends an aerobics class is 0.85. If these two events are independent, the probability of one of these two people attending an aerobics class is: A 0.60 B 0.85 C 0.51 D 0.34 E 0.43 4 A bag contains 3 white balls and 7 yellow balls. Three balls are drawn one at a time from the bag without replacement. The probability that they are all yellow is: A 3 B 27 c 21 D

500 7 24

e

1000 243 1000

100

Exam tip Students need to be reminded that material from Units 1 and 2 of the study can be drawn on for the end of year exams.

n 0.3


[© VCAA 2006]


5_61_60255_MQ12MM_CAS_10.indd 532

7/15/10 9:24:36 AM

5 Which of the following random variables is discrete? A The number of runs scored by Sir Donald Bran in his cricketing career B The weight of people in an elevator C The life span of a fly D The volume, in litres, of water in the Yarra River E The time, in hours, for a student to complete a Mathematical Methods test 6 Which of the following does not represent a probability distribution? A x

2

4

6

8

10

0.2

0.3

0.2

0.2

0.1

1

2

3

4

5

0.1

0.24

0.03

0.56

0.07

3

6

9

12

15

0.36

0.12

0.4

0.02

0.1

−4

−3

−1

1

1

−0.1

0.2

0.4

0.2

0.3

−7

−5

−3

−1

1

0.09

0.12

0.41

0.18

0.2

Pr(X = x) B

x Pr(X = x)

C x Pr(X = x) D x Pr(X = x) E

x Pr(X = x)

7 X is a discrete random variable with the following probability distribution. x Pr(X = x)

0

1

2

3

4

5

0.1

0.2

0.3

0.1

0.2

0.1

The probability that the variable is an odd number, given that it is less than 4 is: A 1 B 7 c 4 2

10

d 3

e

7

5

1 3

8 The value of k for the following probability distribution is: x Pr(X = x) A 0.15 D 0.4

1

2

3

4

5

0.3

5k

0.2

3k

0.1

B 0.05 e 1

c 0.25

9 A die is biased so that Pr(X = 1) = Pr(X = 2) = 0.1 and Pr(X = 3) = Pr(X = 4) = Pr(X = 5) = Pr(X = 6) = 0.2. A game is played where a player rolls the die. The player receives $5 if a number greater than 4 is obtained but must pay $2 if a number less than

or equal to 4 comes up. The expected result for the player for each roll is: A a loss of 80c B a loss of $1.40 C a win of 80c d a loss of $1.40 E a win of $1.20 10 Gertrude’s Gambling House offers patrons a card game which uses a deck comprising four aces, three kings, two queens and one jack. A player draws a card at random. If a jack is drawn the player wins $5, while a queen results in a win of $2. However, if the player draws an ace, a loss of $1 is incurred. On average, Gertrude’s Gambling House wins 40c each time the game is played. If a king is drawn, a player must pay: A $1.00 B $3.00 c $1.50 D $2.00 e $2.50 11 Sam’s chance of getting a bullseye while playing darts is 0.1 and his chance of missing the board altogether is 0.2. Sam collects $2 for a bullseye or 20c for hitting any other part of the board. If the game is to be fair, missing the board altogether means Sam must pay: A $1.70 B $2.70 c $2.20 D $1.50 e $1.85 12 The toll for a new freeway is $2 per car and 50c per occupant. Long-term surveys show that X, the number of occupants per car, is distributed as follows. x Pr(X = x)

1

2

3

4

5

0.3

0.3

0.2

0.1

0.1

The expected toll for each car is: A $3.20 B $3 c $3.50 D $2.95 e $3.65 Questions 13 and 14 refer to the following probability distribution. x Pr(X = x)

1

2

3

4

5

0.3

0.3

0.2

0.1

0.1

13 E(X) is equal to: A 2.5 B 3.1 c 5.2 D 2.4 e 2.6 14 E(5X − 8) is equal to: A 18 B 4.5 c 4 D 7.5 e 5 Questions 15, 16 and 17 refer to the following probability distribution. x

−2

−1

0

1

2

Pr(X = x)

0.1

0.25

0.2

0.15

0.3

15 E(X) is equal to: A 1.2 B 0.5 D 1.5 e 0.3

c 1.4


533

16 Var(X) is equal to: A 2 B 1.7 D 0.61 e 2.11

c 1.91

17 Var(3X + 2) is equal to: A 8 B 17.19 D 15.3 e 2.7

c 7.73

C µ = 7.6 and σ = 3.3 D µ = 7.5 and σ = 3.3 E µ = 6 and σ = 10.89 19 Let X be a random variable with the following probability distribution. x

18 Let X be a discrete random variable with the following probability distribution. x Pr(X = x)

Pr(X = x)

3

6

9

12

0.21

0.35

0.17

0.27

The mean and standard deviation are: A µ = 6 and σ = 3.3 B µ = 7.5 and σ = 10.89

1

2

3

4

0.2

0.2

0.4

0.2

The values between which 95% of the distribution lie for the discrete random variable X are: A (1, 3) B (1, 4) c (2, 4) D (1, 2) e (2, 3)

Extended Response

Probability

Jam

$14.40

1 4

Iced

$15.60

1 4

Cinnamon

$12.00

3 10

Iced jam

$18.00

1 5

12

35

3 26 0 32 15 1

9

4 21

5 17 34 6 27 1 22 33

Find: a the mean price per box b the standard deviation per box c the average price per doughnut. The canteen wants to make a 14% profit on costs. Find: d the cost of a doughnut at the canteen, if all doughnuts are to be sold for the same amount e the average profit per box.

7

Cost per box (2 dozen)

1 9 22 18 29

Doughnut type

28

1 Bob’s Bakery makes four different types of doughnut, each at a different price, depending on the ingredients used. A school canteen buys all its doughnuts from Bob and is currently estimating budgets for the financial year. Types of doughnut and their prices are listed below, along with their popularity (expressed as a probability).

14 3

20

1

3 11

23 10 5 24 1 63 0 8 3

534


6

2 Amina plays roulette, a game where a wheel containing 37 slots numbered 0–36 is spun and the winning number is the one in which a ball lodges when the wheel stops spinning. Amina plays three different games: a First she bets $20 on her favourite number coming up at Casinonominated odds of 35:1 against. i How much would Amina collect if her number came up? ii Find her expected win or loss for the game. iii Is this game fair? b In the second game Amina bets $20 on an even number coming up at Casino-nominated odds of 1:1 (even money chance). i How much would Amina collect if an even number came up? ii Find her expected win or loss for the game. iii Is this game fair?

c In game number three, Amina bets $20 on a line of 12; that is, if numbers 1–12 come up, she wins. The Casino-nominated odds for this game are 2:1 against. i How much would Amina collect if one of her numbers came up? ii Find her expected win or loss for the game. iii Is this game fair? d What is the house percentage for these games? 3 A door-to-door telecommunications representative has recorded her day-by-day sales figures over a period of time. She knows that her probability of selling X packages on any one day follows the probability distribution shown in the table. x Pr(X = y)

0

1

2

3

4

5

>5

2t2

3t

2t2

2t

4t2 + t

t

0

a b c d e f g

Find the value of t. Find the probability that she sells at least 2 packages on any one day. Find the probability that she sells at most 4 packages on any one day. Find the number of packages she can expect to sell each day. Calculate the Var(X) and standard deviation of X, correct to 4 decimal places. Find Pr(µ − 2σ ≤ X ≤ µ + 2σ). If the representative receives a commission of $25 per package sold and a bonus of $200 if she sells 4 or more packages in one day, find her expected daily earnings from commissions and bonuses. h Given that the representative will sell at least two packages tomorrow, find the probability that she will get her $200 bonus. eBook plus Digital doc


Chapter 10


535

eBook plus

aCTiViTieS


• 10 Quick Questions: Warm up with ten quick questions on applications of discrete random variables. (page 484) 10A

Probability revision

Tutorials

• We 7 int-0570: Watch a worked example on calculating conditional probability. (page 493) • We 10 int-0571: Watch a worked example on considering combinations when conducting probability experiments. (page 496) 10 B


Tutorial

• We 16 int-0572: Watch a worked example on ing probability functions. (page 504) Digital docs

• Spreadsheet 100: Investigate probability distributions. (page 507) • WorkSHEET 10.1: Determine probabilities given discrete data and distributions including conditional probability and graphs. (page 510) 10C

Measures of centre of discrete random distributions

Measures of variability of discrete random distributions


• Measures of variability of discrete random distributions: Consolidate your understanding of measures of centre and variability. (page 518) Tutorials

• We 26 int-0574: Watch a worked example on calculating variance. (page 520) • We 32 int-0575: Watching a worked example on constructing a probability distribution table. (page 524) Digital doc

Tutorial

• Spreadsheet 100: Investigate probability distributions. (page 526)

Digital docs

Digital doc

• We 22 int-0573: Watch a worked example on calculating measures of centre. (page 512) • SkillSHEET 10.1: Practise finding the expected value of a function of a random variable. (page 517) • WorkSHEET 10.2: Calculate probabilities, expected values and variance for discrete probability distributions. (page 518)

536

10D

Chapter review



11

11A The binomial distribution 11B Problems involving the binomial distribution for multiple probabilities 11C Markov chains and transition matrices 11D Expected value, variance and standard deviation of the binomial distribution

The binomial distribution areaS oF STudy

• Random variables, including: – calculation and interpretation of the expected value, variance and standard deviation of a random variable – Bernoulli trials and two-state Markov chains, including the length of run in a sequence, steady values for a Markov chain (familiarity with the use of transition matrices to compute values of a Markov chain will be assumed) • Discrete random variables, including: – specification of probability distributions for discrete random variables using graphs, tables and probability functions

– interpretation and use of mean (µ), median, mode, variance (σ 2) and standard deviation of a discrete random variable – the binomial distribution, Bi(n, p), as an example of a probability distribution for a discrete random variable (students are expected to be familiar with the binomial theorem and related binomial expansions) – the effect of variation in the value(s) of defining parameters on the graph of a given probability function for a discrete random variable – probabilities for specific values of a random variable and intervals defined in of a random variable, including conditional probability eBook plus

11a

The binomial distribution

Digital doc

10 Quick Questions

The binomial distribution is an example of a particular type of discrete probability distribution. It has relevance and importance in many real-life everyday applications. This particular branch of mathematics moves away from the textbook and the classroom and into the areas of medical research, simulation activities and business applications such as quality control. The binomial distribution may be referred to as a Bernoulli distribution, and the trials conducted are known as Bernoulli trials. They were named in honour of the Swiss mathematician Jakob Bernoulli (1654–1705).

Bernoulli trials and sequences A Bernoulli trial is an experiment in which the outcome is either a success or a failure. A Bernoulli sequence is a sequence of Bernoulli trials in which: 1. the probability of each possible outcome is independent of the results of the previous trial 2. the probability of each possible outcome is the same for each trial. Note: In a Bernoulli sequence, the number of successes follows the binomial distribution.

Chapter 11


537

Worked Example 1

Determine which of the following sequences can be defined as Bernoulli sequences. a Rolling an 8-sided die numbered 1 to 8 forty times and recording the number of 6s obtained b Drawing a card from a fair deck with replacement and recording the number of aces c Rolling a die 60 times and recording the number that is obtained Think

Write

a Check that all the characteristics have been

a

Yes, this is an example of a Bernoulli sequence, as there are two possible outcomes for each trial (success is ‘obtaining a 6’ and failure is ‘not obtaining a 6’), the outcome of each trial is independent of the outcome of previous trials, and the probability of success is the same for each trial.

b

Yes, this is an example of a Bernoulli sequence, as there are two possible outcomes for each trial (success is ‘obtaining an ace’ and failure is ‘not obtaining an ace’), the outcome of each trial is independent of the outcome of previous trials, and the probability of success is the same for each trial (as each time a card is drawn it is replaced).

c

This is not an example of a Bernoulli sequence, since ‘success’ has not been defined.

satisfied for a Bernoulli sequence.

b Check that all the characteristics have been


c Check that all the characteristics have been


If X represents a random variable that has a binomial distribution, then it can be expressed as: X ∼ Bi(n, p) or X ∼ B(n, p) Translated into words, X ∼ Bi(n, p) means that X follows a binomial distribution with parameters n (the number of trials) and p (the probability of success). Consider the experiment where a fair die is rolled four times. If X represents the number of times a 3 appears uppermost, then X is a binomial variable. Obtaining a 3 will represent a success and all other values will represent a failure. The die is rolled four times so the number of trials, n, equals 4 and the probability, p, of obtaining a 3 is equal to 16 . Using the shorthand notation, X ∼ Bi(n, p) becomes X ~ Bi(4, 16 ). We will now determine the probability of a 3 appearing uppermost 0, 1, 2, 3 and 4 times. Obtaining 3 is defined as a success and is denoted by S. All other numbers are defined as a failure and are denoted by F. The possible outcomes are listed in the table below.

538

Occurrence of 3

Possible outcomes

0

FFFF

1

SFFF FSFF FFSF FFFS


Probability 1× 4×

( 65 )

4

625 = 1296

500 ( 65 ) ( 16 ) = 1296 3

q4 4q3p

Occurrence of 3

Possible outcomes

Probability

2

SSFF SFSF SFFS FSSF FSFS FFSS

6×

( 65 ) ( 16 )

3

SSSF SSFS SFSS FSSS

6×

( 65 )( 16 )

4

SSSS

2

1×

3

( 16 )

4

2

150 = 1296

20 = 1296

1 = 1296

6q2p2 4qp3 p4

It is interesting to note that the binomial probability distribution is closely related to the binomial theorem (see far right-hand column). Furthermore, if we examine the coefficients of the — that is, 1, 4, 6, 4, 1 — it is evident that they are the entries of Pascal’s triangle. This procedure for determining the individual probabilities can become tedious, particularly once the number of trials increases. Hence if X is a binomial random variable, its probability is defined as follows. Pr(X = x) = nCx pxqn − x where x = 0, 1, 2, . . . n. That is: x = the occurrence of the successful outcome. The formula may also be written as Pr(X = x) = nCx px(1 − p)n − x where x = 0, 1, 2, . . . n Here, the probability of failure, q, is replaced by 1 − p. nC represents the number of ways that x different outcomes can be obtained from n trials. It can x  n n! also be written as   and has the formula nCx = . x !( n − x )!  x Since this is a probability distribution, we would expect that the sum of the probabilities is 1. Therefore, for the previous example: Pr(X = x) = Pr(X = 0) + Pr(X = 1) + Pr(X = 2) + Pr(X = 3) + Pr(X = 4) 625 500 150 20 1 = 1296 + 1296 + 1296 + 1296 + 1296

=1 Worked example 2

eBook plus

A binomial variable, X, has the probability function Pr(X = x) = 6Cx(0.4)x(0.6)6 − x where x = 0, 1, . . . 6. Find: a n, the number of trials b p, the probability of success c the probability distribution for x as a table. Think

Tutorial


WriTe

Method 1: Using the rule a Obtain the relevant information from

the given function. The number of trials, n, is the value of the number located at the top left-hand corner of C. b Obtain the relevant information

from the given function. The probability of success, p, is the value in the first bracket.

a n=6

b p = 0.4

Chapter 11


539

c

c

Pr(X = x) = 6Cx(0.4)x(0.6)6 − x

1

Write the rule of the given probability function.

2

Substitute x = 0 into the rule.

Pr(X = 0) = 6C0(0.4)0(0.6)6

3

Evaluate.

4


Pr(X = 1) = 6C1(0.4)(0.6)5

5

Evaluate.

6


Pr(X = 2) = 6C2(0.4)2(0.6)4

7

Evaluate.

8


Pr(X = 3) = 6C3(0.4)3(0.6)3

9

Evaluate.

10


Pr(X = 4) = 6C4(0.4)4(0.6)2

11

Evaluate.

12


Pr(X = 5) = 6C5(0.4)5(0.6)

13

Evaluate.

14


Pr(X = 6) = 6C6(0.4)6(0.6)0

15

Evaluate.

16

Place values in a table. Note: The table at right clearly displays the probability distribution of the given function. It also shows that the individual probabilities sum to 1.

= 1 × 1 × 0.046 656 = 0.046 656 = 6 × 0.4 × 0.077 76 = 0.186 624 = 15 × 0.16 × 0.1296 = 0.311 04 = 20 × 0.064 × 0.216 = 0.276 48 = 15 × 0.0256 × 0.36 = 0.138 24 = 6 × 0.010 24 × 0.6 = 0.036 864 = 1 × 0.004 096 × 1 = 0.004 096 x

540

1

2

3

4

5

6

Pr(X = x) 0.0467 0.1866 0.3110 0.2765 0.1382 0.0369 0.0041

Method 2: Using a CAS Calculator 1

0

A CAS calculator can also be used to find the probability distribution. n=6 p = 0.4 ∴X ~ Bi(6, 0.4) On a Calculator page, press: • MENU b • 5:Probability 5 • 5:Distributions 5 • D:BinomialPdf D Enter values of n and p, using Tab e to move between fields.


Press ENTER ·. Scroll across to see all the probabilities.

2

Write the distribution in a table.

x

Pr(X = x)

Pr(X = 0)

0.046 656

Pr(X = 1)

0.186 624

Pr(X = 2)

0.311 04

Pr(X = 3)

0.276 48

Pr(X = 4)

0.138 24

Pr(X = 5)

0.036 864

Pr(X = 6)

0.004 096

Worked Example 3

A fair die is rolled five times. Find the probability of obtaining: a exactly four 5s b exactly two even numbers c all results greater than 3 d a 5 on the first roll only e a 5 on the second and third roll only. Think a

1

Write

Check that all the characteristics have been satisfied for a binomial distribution.

a Binomial distribution — n independent trials

and two outcomes, fixed p and q.

2

Write the rule for the binomial probability distribution.

Pr(X = x) = nCx pxqn − x

3

Define and assign values to variables. The number of 5s obtained is exactly four.

n=5 Let X = the number of 5s obtained. That is, x = 4 p = probability of a 5 = 16 q = 65

4


Pr( X = 4) = 5C4

5

Evaluate.

( 16 ) ( 65 ) 4

1 = 5 × 1296 ×

=

5 6

25 7776

Chapter 11 The binomial distribution

541

b

1

Define and assign values to variables. Two even numbers means we have x = 2.

b n = 5

Let X = the number of even numbers. That is, x = 2 p = probability of an even number = 12 q=

c

2


3

Evaluate.

1 2

Pr( X = 2) = 5C2

= 10 ×

4

Simplify.

1

Define and assign values to variables. We require 5 occasions when results are greater than 3.

3

Evaluate.

1 8

= 165

1 2

Pr( X > 3) = Pr( X = 5) = 5C5 =1× =

1 32

( 12 ) ( 12 ) 5

0

×1

1 32

d Pr(5 on the first roll only)

binomial rule is not required.

= Pr(SFFFF)

= 16 × 65 × 65 × 65 × =

e Since a specific order is required here, the

×

Let X = values greater than 3. That is, x = 5 Three outcomes — 4, 5, 6 out of six are greater than 3. p = probability of number greater than 3 = 12

d Since a specific order is required here, the

1 4

3

c n = 5

q= Substitute the values into the rule.

2

= 10 32

2

( 12 ) ( 12 )

5 6

625 7776

e Pr(5 on the second and third roll only)

binomial rule is not required.

= Pr(FSSFF)

= 65 × 16 × 16 × 65 × =

5 6

125 7776

Note: If the rule for the binomial probability distribution were to be used in part d, it would 625 provide an answer of 5 × 7776 = 3125 . This answer gives the probability of obtaining a 5 once on 7776 any of the five trials, not necessarily on the first roll only. Hence, if a specific order is required the rule for the binomial probability distribution should not be used. Similarly in part e the rule would produce an answer of 10 × of obtaining a 5 twice on any of the five trials.

542


125 7776

= 1250 , giving the probability 7776

Worked Example 4

A new drug for hay fever is known to be successful in 40% of cases. Ten hay fever sufferers take part in the testing of the drug. Find the probability, correct to 4 decimal places, that: a four people are cured b no people are cured c all 10 are cured. a

Think

Write

1

a This is a binomial distribution with n independent Check that all the characteristics have been satisfied for a binomial distribution. trials and two outcomes, p and q.

2 3

Write the rule for the binomial probability distribution. Define and assign values to variables.

4


5

Evaluate.

6


7

Note: A CAS calculator can also be used to calculate the probability for a particular x-value. Find the number of trials, n, the probability of success, p, and the x-value you want to find the probability of. On a Calculator page, press: • MENU b • 5:Probability 5 • 5:Distributions 5 • D:Binomial Pdf D Enter values of n, p and x, using Tab e to move between fields.

8

Pr(X = x) = nCx pxqn − x n = 10 Let X = the number of people cured, therefore x = 4 p = 0.4 q = 0.6 Pr(X = 4) = 10C4(0.4)4 (0.6)6

= 210 × 0.0256 × 0.046 656 = 0.250 822 656 ≈ 0.2508

n = 10 p = 0.4 ∴X ~ Bi(10, 0.4) X = the number of people cured; therefore x = 4.

Press ENTER ·.


543

9 10

b

c

1

Write the solution.

Pr(X = 4) = binomPdf(10, 0.4, 4) = 0.250 823

Answer the question and round to 4 decimal places.

The probability that 4 people are cured is 0.2508. b ∴ X ~ Bi(10, 0.4)

Define and assign values to variables.

As X = the number of people cured, therefore x = 0.

2

Repeat as above using the CAS calculator to find Pr(X = 0).

Pr(X = 0) = binomPdf(10, 0.4, 0) = 0.006 046 6

3


The probability that no people are cured is 0.0060.

1


c ∴ X ~ Bi(10, 0.4)

As X = the number of people cured, therefore x = 10.

2

Repeat as above using the CAS calculator to find Pr(X = 10).

Pr(X = 10) = binomPdf(10, 0.4, 10) = 0.000 104 86

3


The probability that no people are cured is 0.0001.

Worked example 5

eBook plus

Grant is a keen darts player and knows that his chance of scoring a bullseye on any Tutorial one throw is 0.3. int-0577 a If Grant takes 6 shots at the target, find the probability, correct to 4 decimal Worked example 5 places, that he: i misses the bullseye each time ii hits the bullseye at least once. b Find the number of throws Grant would need to ensure a probability of more than 0.8 of scoring at least one bullseye. Think a

i

1

2 3

544

WriTe

Check that all the characteristics have been satisfied for a binomial distribution. Write the rule for the binomial probability distribution. Define and assign values to variables.

a

i This is a binomial distribution with n independent

trials and two outcomes, p and q. Pr(X = x) = nCx pxqn − x

4


n=6 Let X = the number of bullseyes, therefore x = 0, 1, 2, 3, 4, 5, 6 p = 0.3 q = 0.7 Pr(X = x) = nCx pxqn − x

5

Evaluate and round answer to 4 decimal places. Note: Check the answer using binomPdf(6,0.3,0).

Pr(X = 0) = 6C0(0.3)0(0.7)6 = 1 × 1 × 0.117 649 = 0.117 649 ≈ 0.1176

6


The probability that Grant misses the bullseye each time is 0.1176.


ii

1

2

b

Define and assign values to variables. Note: Pr(X ≥ 1) would involve adding probabilities from Pr(X = 1) to Pr(X = 6). Using the fact that Pr(X ≥ 1) = 1 − Pr(X = 0) allows us to solve the problem using fewer . Substitute the values into the rule.

ii n = 6 Let X = the number of bullseyes, therefore x = 0, 1, 2, 3, 4, 5, 6 p = 0.3 q = 0.7 Pr(X ≥ 1) = Pr(X = 1) + Pr(X = 2) + . . . + Pr(X = 6) = 1 − Pr(X = 0) = 1 − 6C0(0.3)0(0.7)6 = 1 − 0.117 649 = 0.882 351 ≈ 0.8824

3

Evaluate and round answer to 4 decimal places. Note: Check the answer using 1 – binomPdf(6,0.3,0).

4


The probability that Grant hits the bullseye at least once is 0.8824.

1


2

Write the rule as required.

3


1 – nC0(0.3)0(0.7)n > 0.8

4

Evaluate by solving for n.

1 – 0.7n > 0.8 1 – 0.8 > 0.7n 0.2 > 0.7n loge (0.2) > loge (0.7n) loge (0.2) > n × loge (0.7) log e (0.2) 4.512 338 026

5

Interpret the results and answer the question.

Grant would need to take 5 shots to ensure a probability of 0.8 of scoring at least one bullseye.

6

Alternatively, a CAS calculator can be used to solve the inequation. On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as: solve(1 – 0.7n > 0.8,n) Then press ENTER ·.

b

n=? Let X = the number of bullseyes, therefore x = 0, 1, 2, 3, 4, 5, 6 p = 0.3 q = 0.7 Pr(X ≥ 1) > 0.8 Pr(X ≥ 1) = Pr(X = 1) + Pr(X = 2) + . . . + Pr(X = 6) = 1 – Pr(X = 0) ∴ 1 – Pr(X = 0) > 0.8


545

7

Write the solution.

Solving 1 – 0.7n > 0.8 for n implies n > 4.51234

8

Interpret the results and answer the question.

Grant would need to take 5 shots to ensure a probability of 0.8 of scoring at least one bullseye.

Graphs of the binomial distribution We will now consider the graph of a binomial distribution. If we refer to the example of obtaining a 3 when rolling a die four times (see the table on pages 538–9) we note that X ~ Bi(4, 16 ). The probability distribution of the random variable, X, is given in the table and graph over the page. Pr(X = x) 0.5

x

Pr(X = x)

0

0.4823

1

0.3858

0.3

2

0.1157

0.2

3

0.0154

0.1

4

0.0008

0.4

0.0

01234

x

The effect of changing n and p on binomial distribution graphs The effect of increasing p can be seen in the series of graphs below. Pr(X = x) 0.4

Pr(X = x) 0.4 0.3

X ~ Bi(8, 0.2)

Pr(X = x) 0.4 X ~ Bi(8, 0.5)

0.3

0.3 X ~ Bi(8, 0.8)

0.2

0.2

0.2

0.1

0.1

0.1

0.0

0.0

01234 56 78 x

The graph is positively skewed.

01234 56 78 x

The graph is symmetrical. It is also called a normal distribution curve.

0.0

01234 56 78 x

The graph is negatively skewed.

We will now keep p fixed and vary n. Pr(X = x) 0.34 0.32 0.30 0.28 0.26 0.24 0.22 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00

546

X ~ Bi(5, 0.5)

012 34 5

x

Pr(X = x) 0.34 0.32 0.30 0.28 0.26 0.24 0.22 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00


X ~ Bi(15, 0.5)

0 1 2 3 4 5 6 7 8 9 101112131415

x

Pr(X = x) 0.34 0.32 0.30 0.28 0.26 0.24 0.22 0.20 0.18 0.16 0.14 0.12 0.10 0.08 0.06 0.04 0.02 0.00

X ~ Bi(25, 0.5)

0 1 2 3 4 5 6 7 8 9 10111213141516171819202122232425

x

From the preceding graphs it can be seen that: 1. when p = 0.5, the graph is symmetrical 2. as n increases, and the interval between the vertical columns decreases, the graph approximates a smooth hump or bell shape. The effects that the parameters n and p have on the binomial probability distribution curve can be summarised in the following way. If p < 0.5, the graph is positively skewed. Pr(X = x)

If p = 0.5, the graph is symmetrical.

If p > 0.5, the graph is negatively skewed.

Pr(X = x)

x

Pr(X = x)

x

x

When n is large and p = 0.5, the vertical columns are closer together and the line graph becomes a bell-shaped curve or a normal distribution curve.

1. The binomial distribution is an example of a discrete probability distribution. 2. The binomial distribution may be referred to as a Bernoulli distribution, and the trials conducted are known as Bernoulli trials. 3. For a sequence to be defined as a Bernoulli sequence, each of the following characteristics must be satisfied. (a) Each trial in the sequence must be a Bernoulli trial, that is, there must be only two possible outcomes for each trial (success and failure). (b) The probability of success, p, is the same for each trial, and is independent of the outcome of previous trials. (The probability of failure, q, is equal to 1 – p.) 4. X ~ Bi(n, p) denotes that X follows a binomial distribution with parameters n (the number of independent trials) and p (the probability of success). 5. If X is a binomial random variable, its probability is defined as Pr(X = x) = nCx pxqn − x where x = 0, 1, 2, . . . n. Note: x represents the occurrence of the successful outcomes.


547

6. The parameters n and p affect the binomial probability distribution curve as follows: (a) If p < 0.5, the graph is positively skewed. (b) If p = 0.5, the graph is symmetrical or is a normal distribution curve. (c) If p > 0.5, the graph is negatively skewed. (d) When n is large and p = 0.5, the interval between the vertical columns decreases and the graph approximates a smooth hump or bell shape. exerCiSe

11a


Spreadsheet 003 Binomial distribution

The binomial distribution 1 We1 Determine which of the following sequences can be defined as a Bernoulli sequence: a Rolling a die 10 times and recording the number that comes up b Rolling a die 10 times and recording the number of 3s that come up c Spinning a spinner numbered 1 to 10 and recording the number that is obtained d Tossing a coin 15 times and recording the number of tails obtained e Drawing a card from a fair deck, without replacement, and recording the number of picture cards f Drawing a card from a fair deck, with replacement, and recording the number of black cards g Selecting three marbles from a jar containing three yellow marbles and two black marbles, without replacement. 2 Evalute the following, correct to 4 decimal places: b 9C3 (0.1)3 (0.9)6 a 7C2 (0.4)2 (0.6)5 d 8C5 (0.2)5 (0.8)3

e

9C

( ) ( 23 )

1 7 3

7

c

2

f

10C (0.5)5 (0.5)5 5 10C (0.15)0 (0.85)10. 0 x) = 5Cx (0.3)x (0.7)5 − x

3 We2 A binomial variable, X, has the probability function Pr(X = where x = 0, 1, . . ., 5. Find: a n, the number of trials b p, the probability of success c the probability distribution for x as a table. 4

Twenty per cent of items made by a certain machine are defective. The items are packed and sold in boxes of 5. What is the probability of 4 items being defective in a box?

exam Tip nC n−1

Students should know that nCn = 1 and = n. Similarly nC0 = 1 and nC1 = n.

[Assessment report 1 2007 © VCAA]

5

Alex lives so close to where she works that she only has a 0.1 chance of being late. What is the probability that she is late on 3 out of 4 days?

6

Ange has four chances to knock an empty can off a stand by throwing a ball. On each 1 throw, the probability of success is . Find the probability that she will knock the empty can 3 off the stand: a once

b twice

c at least once.

7 We3 A fair coin is tossed four times. Find the probability of obtaining: a heads on the first two tosses and tails on the second two b heads on every roll c two heads and two tails. 8

548

Peter is quite poor at doing crossword puzzles and the probability of him completing one is 0.2. Find the probability that: a of the next three crossword puzzles that he attempts, he is successful in completing two b he successfully completes the first three crossword puzzles that he tries, but has no luck on the next one


c he successfully completes the first three crossword puzzles that he tries d he successfully completes the first three puzzles that he tries given that he was successful in completing his first two. 9 A weighted coin is biased such that a tail comes up 60% of the time. The coin is tossed five times. Find the probability of obtaining: a tails on the first four tosses only b four tails. 10 WE4 Fifty-five per cent of the local municipality the local council. If eight people are selected at random, find the probability, correct to 4 decimal places, that: a half the council b all eight the council c five the council d three oppose the council. 11 The probability of Colin beating Maria at golf is 0.4. If they play once a week throughout the entire year and the outcome of each game is independent of any other, find the probability that they will have won the same number of matches, correct to 4 decimal places. 12 It is known that 5 out of every 8 people eat Superflakes for breakfast. Find the probability that half of a random sample of 20 people surveyed eat Superflakes, correct to 4 decimal places. 13 On a certain evening, during a ratings period, two television stations put their best shows on against each other. The ratings showed that 39% of people watched Channel 6, while only 30% of people watched Channel 8. The rest watched other channels. A random sample of 10 people were surveyed the next day. Find the probability, correct to 4 decimal places, that exactly: a six watched Channel 6 b four watched Channel 8. 14 Three per cent of items produced by a certain machine are defective. A random sample of 10 items is taken. Find the probability that exactly 10% are defective, correct to 4 decimal places. 15 A new drug being trialled gives 8% of the patients a violent reaction. If 200 patients trial the drug, find the probability that 12 patients have a violent reaction to the drug, correct to 4 decimal places. 16 A pen company produces 30 000 pens per week. However, of these 30 000 pens, 600 are defective. Pens are sold in boxes of 20. Find the probability, correct to 4 decimal places, that: a two defective pens are found in one box b a box is free from defective pens. 17 A box contains 5 red marbles, 3 blue marbles and 2 yellow marbles. A marble is chosen at random and replaced. This selection process is completed eight times. Find the probability, correct to 4 decimal places, that: a exactly 4 reds are selected b exactly 2 blues are selected c no yellows are selected. 18 MC Claire’s position in the netball team is goal shooter. The probability of her shooting a goal is 78%. If Claire has 10 attempts at scoring, the probability she will score fewer than 3 goals is: A 10C3 (0.78)3(0.22)7 B 10C3 (0.78)3(0.22)7 + 10C4 (0.78)4(0.22)6 + . . . + (0.78)10 C 10C2 (0.78)2(0.22)8 + 10C1 (0.78)1(0.22)9 + (0.22)10 D 10C2 (0.78)2(0.22)8 + 10C1 (0.78)1(0.22)9 + (0.78)10 E 10C2 (0.22)2(0.78)8 + 10C1 (0.22)1(0.78)9 + (0.78)10


549

19 mC The probability that the temperature in Melbourne will rise above 25 °C on any given summer day, independent of any other summer day, is 0.6. The probability that four days in a week reach in excess of 25 °C is: A 0.64 × 0.43 B 7 × 0.64 × 0.43 C 47 × 0.43 D 0.64 E 35 × 0.64 × 0.43 20 mC Rachel sits a multiple-choice test containing 20 questions, each with four possible answers. If she guesses every answer, the probability of Rachel getting 11 questions correct is: A D

( ) ( 43 ) 9 11 1 3 20 C 11 ( 4 ) ( 4 ) 1 20 C 11 4

20

9

B E

( ) ( 43 ) 9 11 1 3 20 C 10 ( 4 ) ( 4 ) 1 20 C 11 4

11

9

C

( ) ( 43 )

1 20 C 10 4

11

9

21 mC A smoke detector has a probability of failing 2% of the time. If a shopping complex has installed 40 of these smoke detectors, the probability that at least one fails is given by: A 1 − 40C1 (0.98)1(0.02)39 B 1 − 40C1 (0.02)1(0.98)39 C 1 − (0.02)40 D 1 − (0.98)40 E (0.98)40 22 mC It is found that 3 out of every 10 cars are unroadworthy. Ten cars are selected at random. The probability that 3 are unroadworthy is: A 0.009 B 0.2601 C 0.2668 D 0.5 E 1 23 We5 Grant is a keen darts player and knows that his chance of scoring a bullseye on any one throw is 0.6. a If Grant takes 5 shots at the target find the probability that he: i misses the bullseye each time ii hits the bullseye at least once. b Find the number of throws Grant would need to ensure a probability of more than 0.7 of scoring at least one bullseye. eBook plus Digital doc

SkillSHEET 11.1 Solving indicial equations

24 The chance of winning a major prize in a raffle is 0.05. Find the number of tickets required to ensure a probability of more than 0.6 of winning a major prize at least once. 25 A darts player knows that her chance of scoring a bullseye on any one throw is 0.1. Find the number of turns she would need to ensure a probability of 0.9 of scoring at least one bullseye. 26 In Tattslotto, your chance of winning first division is 1 . Find: 8 145 060

a the number of games you would need to play if you wanted to ensure a more than 50% chance of winning first division at least once b the number of tickets you would need to buy for part a if there are 16 games on each ticket c the cost of buying these tickets, if they cost $4.10 each. 550


27 The following probability distribution is for p = 0.2 and n = 10. Pr(X = x) 0.3 0.2 0.1 0.0

0 1 2 3 4 5 6 7 8 9 10

x

a Find the most likely outcome for x. b Describe the plot. eBook plus Digital doc

Spreadsheet 003

28 From the following binomial distribution tables: i draw a graph of the probability distribution ii describe the skewness of each graph. a X ∼ Bi(10, 0.3) b X ∼ Bi(10, 0.5)

Binomial distribution

c X ∼ Bi(10, 0.8)

x

Pr(X = x)

x

Pr(X = x)

x

Pr(X = x)

0

0.028 25

0

0.000 98

0

1 × 10−7

1

0.121 06

1

0.009 77

1

4.1 × 10−6

2

0.233 47

2

0.043 95

2

0.000 07

3

0.266 83

3

0.117 19

3

0.000 79

4

0.200 12

4

0.205 08

4

0.005 51

5

0.102 92

5

0.246 09

5

0.026 42

6

0.036 76

6

0.205 08

6

0.088 08

7

0.009 00

7

0.117 19

7

0.201 33

8

0.001 45

8

0.043 95

8

0.301 99

9

0.000 01

9

0.009 77

9

0.268 44

10

5.9 × 10−6

10

0.000 98

10

0.107 37

29

a Describe the plots of the following binomial probability distributions, without drawing the graphs. i n = 25, p = 0.1 ii n = 50, p = 0.5 iii n = 30, p = 0.9 b What effect does p have on the graph of a binomial probability distribution?

30

a Describe the plot of the binomial probability distribution, X ∼ Bi(60, 0.5), without drawing the graph. b Suggest how the graph might look for a binomial probability distribution with the same p, but double the value of n.

31

a Describe the plot of the binomial probability distribution, X ∼ Bi(100, 0.4), without drawing the graph. b Suggest how the graph might look for a binomial probability distribution with the same n, but double the value of p.

Chapter 11


551

32 Describe the skewness of the graphs of the following binomial probability distributions. a Pr(X = x)


WorkSHEET 11.1

Pr(X = x)

0.4

0.4

0.2

0.2

0.0

c

b

0.0

x

1 2 3 4 5

d Pr(X = x)

Pr(X = x) 0.4

0.4

0.2

0.2

0.0

5 10 15 20 x

0.0

x

2 4 6 8 10

1 2 3 4

x

problems involving the binomial distribution for multiple probabilities

11B

We shall now work with problems involving the binomial distribution for multiple probabilities.

Worked example 6

The binomial variable, X, has the following probability table. x Pr(X = x)

0

1

2

3

4

5

0.2311

0.3147

0.3321

0.1061

0.0112

0.0048

a Pr(X > 3)

b Pr(X ≤ 4).

Think a

b

552

WriTe

1

Pr (X > 3) means Pr(X = 4) or Pr(X = 5). Add these probabilities.

2

Evaluate.

1

Pr(X ≤ 4) would involve adding the probabilities from Pr(X = 0) to Pr(X = 4). Using the fact that Pr(X ≤ 4) = 1 − Pr(X > 4) allows us to solve the problem using fewer .

2

Substitute the value into the rule.

= 1 − 0.0048

3

Evaluate.

= 0.9952


a

Pr(X > 3) = Pr(X = 4) + Pr(X = 5) = 0.0112 + 0.0048 = 0.0160

b

Pr(X ≤ 4) = Pr(X = 0) + . . . + Pr(X = 4) = 1 − Pr(X = 5)

Worked example 7

Find Pr(X ≥ 3) if X has a binomial distribution with the probability of success, p, and the number of trials, n, given by p = 0.3, n = 5. Think

WriTe

1

State the probability distribution.

X ∼ Βi(n, p) X ∼ Βi(5, 0.3)

2

Write what is required.

Pr(X ≥ 3) = Pr(X = 3) + Pr(X = 4) + Pr(X = 5)

3

Substitute the values for n, p and q into the rule Pr(X = x) = nCx pxqn − x.

= 5C3(0.3)3(0.7)2 + 5C4(0.3)4(0.7) + 5C5(0.3)5(0.7)0

4

Evaluate.

= 10 × 0.027 × 0.49 + 5 × 0.0081 × 0.7 + 1 × 0.002 43 × 1 = 0.1323 + 0.028 35 + 0.002 43 = 0.163 08

Worked example 8

eBook plus

So Jung has a bag containing 4 red and 3 blue marbles. She selects a marble at random and then replaces it. She does this 7 times. Find the probability, correct to 4 decimal places, that: a at least 5 marbles are red b greater than 3 are red c no more than 2 are red. Think a

1

Tutorial


WriTe

State the probability distribution and define and assign values to variables.

a Let X = number of red marbles selected

n=7 p = 47

∴X ~ Bi(7, 47 ) As X = number of red marbles selected, therefore x = 5. We want at least 5 red marbles, ∴ Pr(X ≥ 5) 2

On a Calculator page, press: • MENU b • 5:Probability 5 • 5:Distributions 5 • E:Binomial Cdf E Enter values of n and p, using Tab e to move between fields. Enter lower bound as 5 and upper bound as 7, as the number of trials is 7.

Chapter 11


553

Press ENTER ·.

3

Pr(X ≥ 5) = binomCdf(7, 47 , 5, 7)

Write the solution.

b

c

4


1

Repeat as above, using the CAS calculator to find Pr(X > 3). Note: Only inclusive values can be entered into the CAS calculator, ∴ Pr(X > 3) = Pr(X ≥ 4) that only discrete values are possible for a binomial distribution.

2


1

Repeat as above, using the CAS calculator to find Pr(X ≤ 2). Note: this time the upper bound will be 2 and the lower bound will be 0.

2

Write the solution.

The probability that at least 5 red marbles are chosen is 0.3593. b Pr(X > 3) = binomCdf(7, 4 , 4, 7) 7

554


= 0.653 100 08

The probability that greater than 3 marbles are chosen is 0.6531. c

Pr(X ≤ 2) = binomCdf(7, 47 , 0, 2)

3

= 0.359 345

= 0.126 584

The probability that no more than 2 marbles are chosen is 0.1266.


Worked Example 9

X follows a binomial distribution with n = 9, p = 0.4. Find, correct to 4 decimal places: a Pr(X ≥ 7) b the probability that X is greater than 7 given it is greater than 5; that is, Pr(X > 7X > 5). Think a

1

write


X ~ Bi(9, 0.4)

2


Pr(X ≥ 7) = Pr(X = 7) + Pr(X = 8) + Pr(X = 9)

3

Substitute the values for n, p and q into the rule Pr(X = x) = nCx pxqn − x.

Pr(X ≥ 7) = 9C7(0.4)7(0.6)2 + 9C8(0.4)8(0.6) + 9C9(0.4)9(0.6)0

4

Evaluate and round the answer to 4 decimal places.

= 36 × 1.6384 × 10−3 × 0.36 + 9 × 6.5536 × 10−4 × 0.6 + 1 × 2.621 44 × 10−4 × 1 = 0.021 23 + 0.003 54 + 2.621 44 × 10−4 = 0.025 034 752 ≈ 0.0250 Pr(X ≥ 7) = binomCdf(9, 0.4, 7, 9) = 0.0250

Part a can also be solved with a CAS calculator. (Refer to worked example 8). b

a X ~ Bi(n, p)

1


2

Evaluate each probability individually. Pr(X > 7) can be obtained from results in a.

3

Evaluate

4

Pr(X > 7) = Pr(X = 8) + Pr(X = 9) = 3.801 088 × 10−3 Pr(X > 5) = Pr(X = 6) + Pr(X = 7) + Pr(X = 8) + Pr(X = 9) = 9C6(0.4)6(0.6)3 + 0.025 034 752 = 84 × 4.096 × 10−3 × 0.216 + 0.025 034 752 = 0.993 525 76 Pr( X > 7) 3.801 088 × 10 −3 = Pr( X > 5) 0.993 525 76

Pr( X > 7) . Pr( X > 5)

Round the answer to 4 decimal places. Part b can also be solved with a CAS calculator.

Pr(( X > 7) ∩ ( X > 5)) Pr( X > 5) Pr( X > 7) = Pr( X > 5)

b Pr(X > 7 | X > 5) =

= 0.038 258 75

≈ 0.0383

Pr(X > 7 | X > 5) =

Pr( X > 7) Pr( X > 5) binomCdf(9, 0.4, 7, 9) binomCdf(9, 0.4, 7, 9 )

=

= 0.0383


555

Worked example 10

eBook plus

Seventy per cent of all scheduled trains through Westbourne station arrive on time. Tutorial If 10 trains go through the station every day, find, correct to 4 decimal places: int-0579 a the probability that at least 8 trains are on time Worked example 10 b the probability that at least 8 trains are on time for 9 out of the next 10 days. Think a

1 2 3

4

WriTe

State the probability distribution. Write what is required. Substitute the values of n, p and q into the rule Pr(X = x) = nCx pxqn − x. Evaluate and round the answer to 4 decimal places.

Part a can also be solved with a CAS calculator. (Refer to worked example 8). 5 Answer the question. b 1 State the probability distribution. 2 Write what is required. 3

Substitute the values of n, p and q into the rule Pr(X = x) = nCx pxqn − x.

4

Evaluate and round the answer to 4 decimal places. Part b can also be solved with a CAS calculator.

5


a X ∼ Βi(n, p)

X ∼ Βi(10, 0.7)

Pr(X ≥ 8) = Pr(X = 8) + Pr(X = 9) + Pr(X = 10) = 10C8(0.7)8(0.3)2 + 10C9(0.7)9(0.3) + 10C10(0.7)10(0.3)0 = 45 × 0.057 648 01 × 0.09 + 10 × 0.040 353 607 × 0.3 + 1 × 0.028 247 524 9 × 1 = 0.233 474 440 5 + 0.121 060 821 + 0.028 247 524 9 = 0.382 782 786 4 ≈ 0.3828 Pr(X ≥ 8) = binomCdf(10, 0.7, 8, 10) = 0.3828 The probability that at least 8 trains are on time is 0.3828. b X ∼ Βi(n, p)

X ∼ Βi(10, 0.3828) Pr(X = 9) = 10C9(0.3828)9(0.6172)

= 10 × 0.000 176 501 08 × 0.6172 = 0.001 089 364 7 ≈ 0.0011 Pr(X = 9) = binomPdf(10, 0.3828, 9) = 0.0011 The probability that at least 8 trains are on time for 9 out of the next 10 days is 0.0011.

When solving problems dealing with the binomial distribution for multiple probabilities always: 1. define the distribution 2. write what is required 3. write the rule for the binomial probability distribution 4. substitute the values into the given rule and evaluate.

556


exerCiSe

11B

problems involving the binomial distribution for multiple probabilities 1 Find: a 4C3(0.4)3 (0.6) + 4C4(0.4)4 (0.6)0 b 5C3(0.6)3 (0.4)2 + 5C4(0.6)4 (0.4) + 5C5(0.6)5 (0.4)0. 2 We6

The binomial variable, X, has the following probability table.

x Pr(X = x) Find: a Pr(X ≥ 4)


SkillSHEET 11.2 Multiple probabilities

0

1

2

3

4

5

0.15

0.3

0.1

0.22

0.15

0.08

b Pr(X > 0)

c Pr(X ≤ 4)

3 A binomial variable, X, has the probability function Pr(X = x) = the probability of success and x = 0, 1, . . ., 5. Find: a Pr(X ≥ 2) b Pr(X < 4).

d Pr(X < 2). 5C (0.3)x x

(0.7)5 − x, where x is

4 We7 Find Pr(X ≥ 4), correct to 4 decimal places, if X has a binomial distribution with the probability of success, p, and the number of trials, n, given by: a p = 0.6, n = 5 b p = 0.5, n = 6 c p = 0.2, n = 7. 5

6

7

A fair coin is tossed 5 times. Calculate the probability of obtaining: a at least one tail b greater than three tails c greater than three tails, given that at least one tail was obtained.

1 Marco has a faulty alarm clock and the probability that it sounds in the morning is . 3 Calculate the probability, for the next 4 mornings, that his alarm clock: a works at least 3 times b works fewer than 2 times c works at least 3 times, given that it works at least once. Generally, 10% of people who enter a modelling contest are male. For a particular competition, three winners were chosen. What is the probability that less than two females were chosen?

8 We8 A bag contains 4 red and 2 blue marbles. A marble is selected at random and replaced. The experiment is repeated 6 times. Find the probability that: a all 6 selections are red b at least 2 are red c not more than 1 is red. 9 It is known that 40% of Victorians play sport regularly. Ten people are selected at random. Calculate the probability, correct to 4 decimal places, that: a at least half play sport regularly b at least nine don’t play sport regularly. 10 Surveys have shown that 8 out of 10 VCE students study every night. Six VCE students are selected at random. Find the probability, correct to 4 decimal places, that, on any one day: a at least 50% of these students study every night b less than 3 students study every night. 11 A die is weighted such that Pr ( X = 6) = 12 , Pr ( X = 2) = Pr ( X = 4) = 16 and Pr (X = 1) = 1 Pr ( X = 3) = Pr ( X = 5) = 18 . The die is rolled five times. Calculate the probability, correct to 4 decimal places, of obtaining: a at least three 6s b at least two even numbers c a maximum of two odd numbers.

Chapter 11


557

12 WE9 If X is binomially distributed with n = 8 and p = 0.7, find, correct to 4 decimal places: a Pr(X ≥ 7) b Pr(X > 7 | X > 5). 13 A survey shows that 49% of the public the current government. Twelve people are selected at random. Calculate, correct to 4 decimal places: a the probability that at least 8 the government b the probability that at least 10 the government, given that at least 8 do. 14 MC When Graeme kicks for goal, the probability of his kicking a goal is 0.7. If he has five kicks at goal, the probability that he will score fewer than two goals is: A 5C1(0.7)1 (0.3)4 + (0.3)5 B 5C2(0.7)2 (0.3)3 C 5C2(0.7)2 (0.3)3 + 5C1(0.7)1 (0.3)4 D 5C2(0.7)2 (0.3)3 + 5C1(0.7)1 (0.3)4 + (0.3)5 E 1 − 5C2(0.7)2 (0.3)3 15 MC The proportion of patients who suffer a violent reaction from a new drug being trialled is p. If 80 patients trial the drug, the probability that one-quarter of the patients have a violent reaction is: A 80C25 (p)25(1 − p)55 B 80C20 (1 − p)20(p)60 C 80C25 (1 − p)25(p)55 D 80C20 (p)20(1 − p)60 E 80C20 (p)20 16 MC If X is a random variable, binomially distributed with n = 10 and p = k, Pr(X ≥ 1) is: A 1 − (1 − k)10 B (1 − k)10 C 10(k)(1 − k)9 D (k)10 E 1 − (k)10 17 MC Three per cent of items made by a certain machine are defective. The items are packed and sold in boxes of 10. If 3 or more are defective, the box can be returned and money refunded. The chance of being eligible for a refund is: A 0 B 0.0002 C 0.0036 D 0.0028 E 0.9972 18 MC Long-term statistics show that Silvana wins 60% of her tennis matches. The probability that she will win at least 80% of her next 10 matches is: A 0.0061 B 0.0464 C 0.1673 D 0.8327 E 0.9536 19 MC Nineteen out of every 20 cricketers prefer ‘Boundary’ cricket gear. A squad of 12 cricketers train together. The probability that at least 11 use Boundary gear, given that at least 10 use it, is: A 0.5404 B 0.6129 C 0.8816 D 0.8992 E 0.9804 20 A school council, comprising 15 of the school community, requires a minimum twothirds majority to a motion. It is known that 50% of the school community favour a new uniform. Calculate the probability that the school council will a motion in favour of a new uniform, correct to 4 decimal places. 21 A car insurance salesman knows that he has a good chance of finding customers in the age group from 18 to 20, as people often buy their first car at this age. Five per cent of all people in this age group are looking to purchase a car. The salesman questions 30 people in this age group. Calculate the probability, correct to 4 decimal places, that he will get: a no more than 3 sales b at least 3 sales. 22 Police radar camera tests have shown that 1% of all cars drive at over 30 km/h above the speed limit, 2% between 10 km/h and 30 km/h above the limit and 4% below 10 km/h over the limit. In one particular hour, a radar camera tests 50 cars. Caculate the probability, correct to 4 decimal places, that: a at most, one car is over 30 km/h above the limit b at most, two cars are between 10 km/h and 30 km/h above the limit c at most, two cars are below 10 km/h above the limit d at most, three cars are above the limit. 558


23 An Australian cricketer scores 50 or more in one-third of all his test match innings. The Australian selectors are aiming to predict his next 10 innings. Calculate, correct to 4 decimal places, the probability that he will score 50 or more on: a no occasions b exactly four occasions c at least two occasions. 24 Two dice are rolled simultaneously and their difference is recorded. Find the probability, correct to 4 decimal places, that in 5 rolls: a a difference of zero occurs at least once b a difference of 1 occurs at least twice c a difference of 5 occurs at least once. 25 We10 Eighty per cent of all scheduled trains through Westbourne station arrive on time. If 10 trains go through the station every day, find: a the probability that at least 8 trains are on time b the probability that at least 8 trains are on time for 9 out of the next 10 days. 26 Seventy-five per cent of all scheduled trains through Westbourne station arrive on time. If 15 trains go through the station every day, find, correct to 4 decimal places: a the probability that at least 10 trains are on time b the probability that at least 10 trains are on time for 8 out of the next 10 days. 27 An experiment involves rolling a die 6 times. Calculate, correct to 4 decimal places: a the probability of obtaining at least four prime numbers b the probability of obtaining at least four prime numbers on 5 occasions if the experiment is repeated 8 times. 28 Tennis balls are packed in cans of 6. Five per cent of all balls are made too flat (that is, they don’t bounce high enough). The cans are then packed in boxes of two dozen. Calculate the probability, correct to 4 decimal places, that: a a can contains, at most, one flat ball b a box contains at least 22 cans with a maximum of one flat ball. eBook plus

11C

markov chains and transition matrices

Interactivity

int-0256 Markov chains and transition matrices

Andrei Markov was a Russian mathematician whose name is given to a technique that calculates probability associated with the state of various transitions. It answers questions such as, ‘What is the probability that James will be late to work today given that he was late yesterday?’ or ‘What can be said about the long-term behaviour of James’ punctuality?’. A Markov chain is a sequence of repetitions of an experiment in which: 1. The probability of a particular outcome in an experiment is conditional only on the outcome of the experiment immediately before it. 2. The conditional probabilities of each outcome in a particular experiment are the same every single time. A two-state Markov chain is one in which there are only two possible outcomes for each experiment.

Chapter 11


559

Consider a leisure centre that offers aerobics classes and has a gym. Records show that 20% of the who use the gym on a particular day will participate in an aerobics class the next day and 70% of the who participate in an aerobics class on a particular day will use the gym the next day. It is also known that 200 use the leisure centre each day and they all participate in aerobics classes or use the gym, but not both. On a particular day 150 use the gym and 50 attend an aerobics class. The possible outcomes may be illustrated on a tree diagram as shown below. Day 1

Day 2 0.8

Gym

0.2

Aerobics

0.7

Gym

0.3

Aerobics

Gym

Aerobics

Outcome Interpretation GG The member uses the gym on day 2 given the member used the gym on day 1. GA The member attends an aerobics class on day 2 given the member used the gym on day 1. AG The member uses the gym on day 2 given the member attended an aerobics class on day 1. AA The member attends an aerobics class on day 2 given the member attended an aerobics class on day 1.

The tree diagram can also be used to calculate how many use the gym or attend an aerobics class. From the tree diagram below, it can be seen that on the second day 155 use the gym and 45 attend an aerobics class. Day 1 Gym Aerobics

Day 2 0.8

Gym

0.2

Aerobics

0.7

Gym

0.3

Aerobics

Outcome Number of GG 150 × 0.8 = 120 GA 150 × 0.2 = 30 AG 50 × 0.7 = 35 AA 50 × 0.3 = 15

The tree diagram may be extended to display the possible outcomes and their respective probabilities for the third day. Day 1

Day 2 0.8

Gym

0.2

Aerobics

0.7

Gym

0.3

Aerobics

Gym

Aerobics

Day 3 0.8

Gym

0.2

Aerobics

0.7

Gym

0.3

Aerobics

0.8

Gym

0.2

Aerobics

0.7

Gym

0.3

Aerobics

Outcome GGG GGA

Probability 0.8 × 0.8 = 0.64 0.8 × 0.2 = 0.16

GAG GAA

0.2 × 0.7 = 0.14 0.2 × 0.3 = 0.06

AGG AGA

0.7 × 0.8 = 0.56 0.7 × 0.2 = 0.14

AAG AAA

0.3 × 0.7 = 0.21 0.3 × 0.3 = 0.09

GGG represents the member using the gym on all three days; Pr(GGG) = 0.64. If you want to find out the probability of the member using the gym on two out of three days, three outcomes need to be considered: GGA, GAG and AGG. ∴ Pr(gym 2 out of 3 days) = Pr(GGA) + Pr(GAG) + Pr(AGG) = 0.16 + 0.14 + 0.56 = 0.86 The tree diagram may be further extended to display the possible outcomes and their respective probabilities on the fourth day.

560


1 Day

Gym Aerobics

Day 2

0.8

0.2

Gym

Aerobics

Day 3 0.8 0.2

Aerobics

0.7

Gym

0.3

0.7

0.2 0.7 0.3 0.8 0.2

Gym Aerobics

0.7 Aerobics 0.3

0.8

Gym

0.2

Aerobics

Gym

0.8

Gym

0.8

Gym

0.2

Aerobics Gym

0.7 0.3 0.8

0.3

Aerobics

0.7 0.3

Gym Aerobics

0.2 0.7 0.3

Day 4 Gym Aerobics Gym Aerobics Gym Aerobics

Aerobics Gym Aerobics Gym Aerobics

Outcome GGGG GGGA GGAG GGAA GAGG GAGA GAAG GAAA AGGG AGGA AGAG AGAA AAGG AAGA AAAG AAAA

Probability 0.8 × 0.8 × 0.8 = 0.512 0.8 × 0.8 × 0.2 = 0.128 0.8 × 0.2 × 0.7 = 0.112 0.8 × 0.2 × 0.3 = 0.048 0.2 × 0.7 × 0.8 = 0.112 0.2 × 0.7 × 0.2 = 0.028 0.2 × 0.3 × 0.7 = 0.042 0.2 × 0.3 × 0.3 = 0.018 0.7 × 0.8 × 0.8 = 0.448 0.7 × 0.8 × 0.2 = 0.112 0.7 × 0.2 × 0.7 = 0.098 0.7 × 0.2 × 0.3 = 0.042 0.3 × 0.7 × 0.8 = 0.168 0.3 × 0.7 × 0.2 = 0.042 0.3 × 0.3 × 0.7 = 0.063 0.3 × 0.3 × 0.3 = 0.027

As we extend the analysis to cover a greater number of days, setting up the tree diagram becomes very messy and time consuming. The original information can be set up as a pair of recurrence relationships. That is, if 20% of the using the gym attend an aerobics class the next day, 80% will use the gym the next day. Furthermore, if 70% of the attending an aerobics class use the gym the next day, 30% will attend an aerobics class the next day. Let gi = the number of who use the gym on day i. Let ai = the number of who attend an aerobics class on day i. gi + 1 = 0.8gi + 0.7ai and ai + 1 = 0.2gi + 0.3ai Number of gym s

Number of aerobics participants

Day i

Day i

gi = 150

ai = 50

Day i + 1

Day i + 1

gi + 1 = 0.8gi + 0.7ai

ai + 1 = 0.2gi + 0.3ai

= 0.8 × 150 + 0.7 × 50

= 0.2 × 150 + 0.3 × 50

= 155

= 45

Day i + 2

Day i + 2

gi + 2 = 0.8gi + 1 + 0.7ai + 1

ai + 2 = 0.2gi + 1 + 0.3ai + 1

= 0.8 × 155 + 0.7 × 45

= 0.2 × 155 + 0.3 × 45

= 155.5

= 44.5

Day i + 3

Day i + 3

gi + 3 = 0.8gi + 2 + 0.7ai + 2

ai + 3 = 0.2gi + 2 + 0.3ai + 2

= 0.8 × 155.5 + 0.7 × 44.5

= 0.2 × 155.5 + 0.3 × 44.5

= 155.55

= 44.45


561

Day i + 4

Day i + 4

gi + 4 = 0.8gi + 3 + 0.7ai + 3

ai + 4 = 0.2gi + 3 + 0.3ai + 3

= 0.8 × 155.55 + 0.7 × 44.45

= 0.2 × 155.55 + 0.3 × 44.45

= 155.555

= 44.445

This method allows us to clearly see how many (when rounded to integer values) are using the gym or attending an aerobics class each day. Often we are interested in the long-term behaviour (or the steady state, as it is often called) of a particular situation, in this case how many will use the gym or attend an aerobics class. We can determine this by using the following information. Let g = the number of who use the gym. Let a = the number of who attend an aerobics class. Total number of = 200 This gives the equation g + a = 200 which when rearranged is equal to a = 200 − g [1] Also g = 0.8g + 0.7a [2] and a = 0.2g + 0.3a [3] Rearranging equation [2] g − 0.8g = 0.7a 0.2g = 0.7a Substituting [4] into [2] and solving gives

0.2 g 0.7a = 0.2 0.2 g = 3.5a a = 200 − 3.5a a + 3.5a = 200 4.5a = 200 400 a= 9

[4]

a = 44.4444 (correct to 4 decimal places) Substituting a = 44.4444 gives g = 155.5556. In the long term, 156 will use the gym and 44 will attend an aerobics class.

Worked Example 11

The Nee Islands are very wet. If it is raining on a particular day, the chance that it will rain the next day is 60%. If it is not raining on a particular day, the chance that it will rain on the following day is 45%. a If it is raining on Tuesday, draw a tree diagram to represent the next two days. b Extending the tree diagram, calculate the probability that, if it is raining on Tuesday, it will also be raining on Friday of the same week. Think a Draw a tree diagram labelling each

Write Wednesday

a

branch and place the appropriate probability along the relevant branch.

0.60

Rain

0.40

Dry

Tuesday

Thursday 0.60

Rain

0.40

Dry

0.45

Rain

0.55

562


Dry

b

1

Draw a tree diagram labelling each branch and place the appropriate probability along the relevant branch. Note: The tree diagram will start at Tuesday and extend to Friday.

b

Wednesday 0.60

Rain

Tuesday 0.40

Dry

Thursday Rain 0.60

0.40

Dry

0.45

Rain

0.55

Dry

0.45 0.55 0.45

0.60 0.40 0.60 0.40

0.55

Friday Rain Dry Rain Dry Rain Dry Rain Dry

2

List each of the outcomes and calculate the probability for each individual outcome.

Outcome RRR RRD RDR RDD DRR DRD DDR DDD

Probability 0.6 · 0.6 · 0.6 = 0.216 0.6 × 0.6 × 0.4 = 0.144 0.6 · 0.4 · 0.45 = 0.108 0.6 × 0.4 × 0.55 = 0.132 0.4 · 0.45 · 0.6 = 0.108 0.4 × 0.45 × 0.4 = 0.072 0.4 · 0.55 · 0.45 = 0.099 0.4 × 0.55 × 0.45 = 0.121

3

Add the probabilities of the required outcomes.

0.216 + 0.108 + 0.108 + 0.099 = 0.531

4

Interpret the answer.

The probability that it will be raining on Friday if it is raining on Tuesday is 0.531. That is, there is a 53.1% chance of raining on Friday given it will rain on Tuesday.

Worked example 12

eBook plus

Commuters travelling into the centre of Trenchtown use either the bus or the train. Tutorial Research has shown that each month 20% of those using the bus switch to train int-0580 travel and 30% of those using the train switch to bus travel. Worked example 12 If, at the beginning of January, 4800 people were using the bus and 3600 were using the train to get into the city, calculate: a the number of people using the train at the beginning of May b the number of people using the bus and train in the long term. Think a

1

WriTe

Interpret the given information.

a Each month 20% of bus commuters switch to train

travel and 80% will not switch. Each month 30% of train commuters switch to bus travel and 70% will not switch. 2

Define the variables.

Let ti = the number of people using the train at the beginning of month i. Let bi = the number of people using the bus at the beginning of month i.

3

Set up a pair of recurrence relationships.

ti + 1 = 0.7ti + 0.2bi and bi + 1 = 0.3ti + 0.8bi

4

Substitute the given values of bus and train commuters into the recurrence relationships.

Chapter 11


563

b

Train travellers January (month i) ti = 3600

Bus travellers January (month i) bi = 4800

February (month i + 1) ti + 1 = 0.7ti + 0.2bi = 0.7 × 3600 + 0.2 × 4800 = 3480

February (month i + 1) bi + 1 = 0.3ti + 0.8bi = 0.3 × 3600 + 0.8 × 4800 = 4920

March (month i + 2) ti + 2 = 0.7ti + 1 + 0.2bi + 1 = 0.7 × 3480 + 0.2 × 4920 = 3420

March (month i + 2) bi + 2 = 0.3ti + 1 + 0.8bi + 1 = 0.3 × 3480 + 0.8 × 4920 = 4980

April (month i + 3) ti + 3 = 0.7ti + 2 + 0.2bi + 2 = 0.7 × 3420 + 0.2 × 4980 = 3390

April (month i + 3) bi + 3 = 0.3ti + 2 + 0.8bi + 2 = 0.3 × 3420 + 0.8 × 4980 = 5010

May (month i + 4) ti + 4 = 0.7ti + 3 + 0.2bi + 3 = 0.7 × 3390 + 0.2 × 5010 = 3375

May (month i + 4) bi + 4 = 0.3ti + 3 + 0.8bi + 3 = 0.3 × 3390 + 0.8 × 5010 = 5025

5

Evaluate the number of bus and train commuters at the beginning of February.

6

Use the values obtained in step 5 to calculate the number of bus and train commuters at the beginning of March.

7

Repeat the processes involved in step 5 and 6 until the number of bus and train commuters at the beginning of May is obtained.

8


1

Set up and number equations that relate to: (i) the number of train commuters (ii) the number of bus commuters (iii) the total number of commuters. Note: Maintain the variables defined in part a .

2

3

564

Rearrange equation [1] so that t is the subject. Note: Either variable may be the subject. Rearrange equation [3] so that b can be expressed in of t. Note: Again either variable may be transposed.

At the beginning of May, the number of people using the bus and train respectively is 5025 and 3375. b Let t = number of people using the train.

Let b = number of people using the bus. Total = 3600 + 4800 t + b = 8400 [1] t = 0.7t + 0.2b [2] b = 0.3t + 0.8b [3] t + b = 8400 t = 8400 − b

[4]

b = 0.3t + 0.8b b − 0.8b = 0.3t 0.2b = 0.3t 0.2b 0.3t = 0.2 0.2 b = 1.5t

[5]


4

t = 8400 − b t = 8400 − 1.5t t + 1.5t = 8400 2.5t = 8400 2.5t 8400 = 2.5 2.5

Substitute equation [5] into [4] and solve for t.

t = 3360

5

Substitute t into equation [5] and solve for b.

b = 1.5t b = 1.5 × 3360 = 5040

6


In the long term, 3360 commuters will travel by train and 5040 will travel by bus.

Transition matrices Transition matrices are another technique for solving some Markov chain problems. Transition matrices are specifically used for problems where you are only interested in the probability of the final outcome (for example, the probability that a person goes to the gym on the seventh day). Tree diagrams must still be used to solve problems that take into the number of different ways of reaching the final outcome (e.g. the probability that a person goes to the gym 4 out of the next 7 days). Consider again the previous scenario, from page 560, of the probability of using the gym or attending an aerobics class. Day 1 Gym

Aerobics

Day 2 0.8

Gym

0.2

Aerobics

0.7

Gym

0.3

Aerobics

This can also be described in a transition probability table: Gym today

Aerobics today

Gym tomorrow

0.8

0.7

Aerobics tomorrow

0.2

0.3

or

Aerobics today

Gym today

Aerobics tomorrow

0.3

0.2

Gym tomorrow

0.7

0.8

Note: Each column should add up to one, as do the pairs of branches in the given tree diagram. A transition matrix can be used to simplify calculations involving Markov chains. As its name suggests, the matrix assists in the calculation of the transition from one state to the next. We can convert the above transition probability tables into a transition matrix, T, as shown below. A1 G1 G1 A1 G  0.8 0.7  A  0.3 0.2  T = 2 or T = 2   A 2  0.2 0.3  G 2  0.7 0.8 


565

The proportion of the population that use the gym or attend the aerobics class is called the state of the system, S. It can either relate to specific populations, or the probabilities associated with the different outcomes. For our example, the initial system, S0 is given by:  n( using the gym on day 1)  150   n(doing aerobics on day 1)   50   =  or  =   n(doing aerobics onn day 1)   50   n( using the gym onn day 1)  150   n(G 0 )   n( A 0 )  At any time, the state is represented by the column matrix   or  , where n(G) is  n(A 0 )   n(G 0 )  the number of gym s and n(A) is the number of aerobics participants. There are two different transition and initial state matrices for this problem, so which one should we choose to use? It does not matter which one we choose, as long as the correct pairing of T and S0 is selected. 150   0.8 0.7  For example, if we choose T =  .  , S0 must be   50   0.2 0.3  The element in the 1st row 1st column of S0 must represent the initial value for the element in the 1st row, 1st column of the transition matrix.

These elements must match (i.e. the probability about gym s matches with the initial number of gym s).

Similarly, these elements must match (i.e. the probability about aerobics participants matches with the initial number of aerobics participants).

 0.8 0.7  150  i.e. when multiplying T and S:  ×   0.2 0.3   50 

 0.8 × 150 + 0.7 × 50  =   0.2 × 150 + 0.3 × 50 

155  =   45 

They are then multiplied together as part of the 1st element for S0.

 n(G1 )   0.8 0.7  150  Hence, the state after 1 day, S1 is  =    n( A1 )   0.2 0.3   50 

 0.8 × 150 + 0.7 × 50  =   0.2 × 150 + 0.3 × 50 

155  =   45 

Note: these are the same values as we calculated earlier. From the above calculation, we can see that S1 = T × S0. If the conditional probabilities remain the same, then a similar equation will express the transition from any particular state to the next.

566


Therefore, in general, Sn + 1 = T × Sn Also, as, S1 = T × S0 and, S2 = T × S1, then, S2 = T × T × S0 = T 2 × S0 and, S3 = T × T × T × S0 = T 3 × S0 ∴ In general, Sn = T n × S0 To find the long-term behaviour, or steady state, choose n to be a large number, for example n = 50, and find S50. T can be written from the generalised transition probability table below: Event A1

Event A1′

Event A2

Pr(A2 | A1)

Pr(A2 | A1′)

Event A2′

Pr(A2′ | A1)

Pr(A2′ | A1′)

 n( A1 )0   Pr( A1 )0   Pr( A 2 | A1 ) Pr( A 2 | A1 ′)  T=  and S0 =   or  .  Pr( A 2 ′ | A1 ) Pr( A 2 ′ | A1 ′)   n( A 2 )0   Pr( A 2 )0 

Worked Example 13

Using the above data for attending the gym or aerobics class, find: a the proportion of people attending the gym and aerobics class on the 5th day b the number of people attending the gym or aerobics class in the long term. Think a

Write

1

Write down the transition matrix.

2

Write down a suitable initial state matrix. In this case, it is the initial numbers of people attending the gym and aerobics class.

3

Identify which state matrix is required.

4

On a Calculator page, complete the entry lines as:

 0.8 0.7    0.2 0.3 

a T=

150  S0 =    50 

As S0 corresponds to day 1, therefore day 5 corresponds to the state matrix S4.

 0.8 0.7   →t  0.2 0.3  150    → s0  50  Press ENTER · after each entry.


567

5

Calculate the proportion of people attending the gym or aerobics on the 5th day. That is, S4 = T 4 × S0.

6

Write the solution.

S4 = T 4 × S0 4

 0.8 0.7  150  =   ×  0 . 2 0 . 3    50  155.555  =   44.445 

b

7

Answer the question (rounding to the nearest whole number).

1

Repeat as above using the CAS calculator to find the long-term proportion. Choose n = 50. S50 = T 50 × S0

2

Write the solution.

3

568

Answer the question (rounding to the nearest whole number).

On the 5th day, there will be 156 people at the gym and 44 people attending the aerobics class. b

S50 = T 50 × S0

 0.8 0.7  =   0.2 0.3 

155.555  =    44.445 

50

150  ×   50 

In the long-term, 156 people attend the gym and 44 people go to aerobics class.


Worked Example 14

The chance of Jo’s netball team winning a given game depends on how the team performs in the previous game. If her team wins, then the chance that it will win the next game is 0.75. If her team loses, there is only a 0.4 chance that they will win the next match. Given the team wins their first match, find the probability, that: a they win two out of their first three games b they win the 7th game they play, correct to 4 decimal places. Think a

1

Write

Draw a tree diagram labelling each branch and place the appropriate probability along the relevant branch. List the outcomes.

a

Game 2

Won game 1

Won

0.75

Outcomes WWW

0.4

Lost Won

WWL WLW

0.6

Lost

WLL

0.25 0.25

b

Game 3 Won

0.75

Lost

2

Select the appropriate outcomes for the required probability.

Pr(win 2 out of 3 games) = Pr(WWL) + Pr(WLW)

3

Calculate the probabilities by multiplying along the branches.

Pr(win 2 out of 3 games) = 0.75 × 0.25 + 0.25 × 0.4 = 0.1875 + 0.1 = 0.2875

1

Write down the transition matrix, T.

b

W1 L1 W2  0.75 0.4  T=   L 2  0.25 0.6 

 0.75 0.4  T=   0.25 0.6  2

3

Write down a suitable initial state matrix, S0. As the value of the 1st row, 1st column of T is the probability associated with winning consecutive games, the first element of S0 will be the probability that the team has won. We know for certain that Jo’s team won the first match, so the probability must be 1. Therefore, the probability of the second element must be 0. Identify which state matrix is required.

 Pr( Jo’s team wins 1st game)  S0 =    Pr( Jo’s team loses 1st game)  1  S0 =   0 

As S0 corresponds to game 1, therefore game 7 corresponds to the state matrix S6.


569

4

On a Calculator page, complete the entry lines as:  0.75 0.4   →t  0.25 0.6  1    → s0 0  t 6 × s0 Press ENTER · after each entry.

5

Pr(winning the 7th game) = T 6 × S0

Write the solution.

6

 0.75 0.4  1  =  ×   0.25 0.6  0   0.616 092  =   0.383 908  6

As the answer required is the probability of winning the 7th game, we need the first element of the state matrix. Round the answer to 4 decimal places.

Pr(winning the 7th game) = 0.6161

1. A Markov chain is a sequence of repetitions of an experiment in which: (a) the probability of a particular outcome in an experiment is conditional only on the result of the outcome immediately before it (b) the conditional probabilities of each outcome in a particular experiment are the same every single time. 2. A two-state Markov chain is where there are only two possible outcomes for each experiment. 3. Problems involving two-state Markov chains can be solved using: (a) tree diagrams where the probabilities are multiplied along the branches (b) recurrence relationships  Pr( A 2 | A1 ) Pr( A 2 | A1 ′)  (c) transition matrices where T =  ,  Pr( A 2 ′ | A1 ) Pr( A 2 ′ | A1 ′)   S0 = the initial state,  n( A1 )0  n( A 2 ) 0 

  Pr ( A )  1 0  or    Pr ( A 2 )0    

Sn = T n × S0 To find the long-term proportion or steady state, let n = a large number, relative to the problem, and solve for Sn.

570


Exercise

11C

Markov chains and transition matrices 1 WE11 Kelly has developed a method for predicting whether or not the surf will be good on a particular day. If it is good today, there is a 65% chance it will be good tomorrow. If it is poor today, there is a 45% chance it will be poor tomorrow. a If the surf is good on Thursday, draw a tree diagram to represent the next two days. b Extending the tree diagram, calculate the probability that, if the surf is good on Thursday, it will also be good on Sunday. 2 Mio Custor’s films have either been hugely successful or have failed miserably at the box office. It is known that there is a 25% chance of Mio’s next film being successful if his previous film was a success. It is also known there is a 62% chance of his next film failing if the previous film was a failure. He is currently filming a trilogy that will be released over a period of 3 years. a What is the probability that the second instalment of the trilogy will be a failure if his latest film before the trilogy was a success? b What is the probability that the final instalment of the trilogy will be a failure if his latest film before the trilogy was a success? Answer to 4 decimal places. 3 In the local cricket competition, teams can use either of two types of ball — Kingfisher or Best Match. At the end of each season, clubs sometimes decide to change the ball they use. Research suggests that 80% of those using Kingfisher stay with that ball for the next season. An incomplete tree diagram representing this situation is 0.80 shown at right. Complete the diagram and then answer the 0.20 following questions. a If a club chooses Best Match one season, what is the 0.70 probability it will choose Best Match the following 0.30 season? b If a club chooses Best Match one season, what is the probability it will be using Kingfisher in 3 years time? 4

The probability that Alicia is on time to school in the morning is dependent upon if she is asleep by 10 pm. If she is asleep before 10 pm, the probability of her being on time to school is 0.8. If she stays up until after 10 pm the night before, the probability of her being on time to school is only 0.4. The probability that Exam tip Multiplication of decimals has not she is asleep before 10 pm is 0.6. been well answered by students on exam questions. a Calculate the probability that she is on Conditional probability is also an area that needs time to school on any given day. more attention. b Given that she was on time to school, [Assessment report 1 2007 © VCAA] find the probability that Alicia went to bed later than 10 pm.

5

The Lo Schiavo family take a holiday every summer. They choose between a resort in Cairns and visiting relatives in Tasmania. If they fly to Cairns one year, the probability of returning to Cairns the next year is 0.4. If they decide to visit their relatives one summer, there is only a 0.3 chance of a repeat visit the following year. In a particular year, the Lo Schiavos visited their relatives in Tasmania. What is the probability that they were holidaying in Cairns two years later?

6

Every Sunday night, Priya gets takeaway. She only selects from Chinese takeaway or fish and chips. If she eats Chinese takeaway one week, the probability of her eating fish and chips the week after is 0.7. If she ate fish and chips one Sunday, the probability that she eats Chinese the next Sunday is 0.5. Given that she eats Chinese on a particular Sunday, calculate the probability that she eats Chinese takeaway on only one of the next three Sundays.


571

7 WE12 Commuters travelling into the centre of Trenchtown use either the bus or train. Research has shown that each month 30% of those using the bus switch to train travel and 60% of those using the train revert to bus travel. If, at the beginning of January, 5600 people were using the bus and 4900 were using the train to get into the city, calculate: a the number of people using the train at the beginning of May b the number of people using the bus and train in the long term. 8 WE13 Residents of Trenchtown purchase their petrol from either Pete’s Petrol or Slick Sam’s Servo. Research has shown that each month 10% of Pete’s customers switch to Slick Sam’s Servo and 20% of Sam’s customers switch to Pete’s Petrol. If, at the beginning of June, 2800 customers purchased their petrol from Pete and 3100 customers purchased their petrol from Sam, calculate: a the number of people purchasing petrol from Pete at the beginning of October b the number of people purchasing petrol from Pete and Sam in the long term. 9 A leisure centre offers aerobics classes and has a gym. Records show that 10% of the who use the gym on a particular day will participate in an aerobics class the next day, and 70% of the who participate in an aerobics class on a particular day will use the gym the next day. It is also known that 300 use the leisure centre each day and they all participate in aerobics classes or use the gym, but not both. On a particular day 200 use the gym and 100 attend an aerobics class. a In the long term, how many people will use the gym? b In the long term, how many people will attend an aerobics class? 10 MC Records show that if a local football team make the top eight in a particular year, the chance that they make the top eight in the following year is 70%. If they don’t make the top eight in a particular year, the chance that they make the top eight in the following year is 40%. The tree diagram that best represents this situation is: A B 0.3 Make 0.7 Make Make top 8

0.7 , Don t make top 8

C Make top 8

0.4 0.6 0.7 0.3

, Don t make top 8

E Make top 8

0.4 0.6 0.6 0.4

, Don t make top 8

572

0.3 0.7

top 8 , Don t make top 8

Make top 8

Make top 8 , Don t make top 8

, Don t make top 8

Make top 8 , Don t make top 8 Make top 8 , Don t make top 8 Make top 8 , Don t make top 8 Make top 8 , Don t make top 8


0.3

D Make top 8

0.6 0.4 0.4 0.6

, Don t make top 8

0.7 0.3

top 8 , Don t make top 8

Make top 8 , Don t make top 8 Make top 8 , Don t make top 8 Make top 8 , Don t make top 8

11 Miya prefers to shop at either Southland or Chadstone each weekend. The place she shops at depends only on where she shopped the previous time. If she visited Southland one weekend, the next weekend she goes shopping, the probability of her returning to Southland is 1 . The 4

transition matrix for the probabilities of Miya visiting either Chadstone or Southland given 1 2 the shopping centre she visited the weekend before is  4 5  . 3 3  4 5  a If Miya shops at Chadstone one weekend, what is the probability she shops at Southland the following weekend? b Miya does visit Chadstone on a particular weekend. What is the probability, correct to 3 decimal places, that she will be at Chadstone again in four weekend’s time? c In the long term, what proportion of weekends, correct to 3 decimal places, does Miya spend at Chadstone? eBook plus Digital doc

WorkSHEET 11.2

11d

12 We 14 The chance of Paul hitting a bullseye in darts is dependent on the success of his previous throw. If he hits a bullseye, then the chance that his next throw will also be a bullseye is 0.65. If he misses, though, there is only a 0.15 chance that he will get a bullseye on his next throw. Given that Paul’s first throw is a bullseye, find the probability, correct to 4 decimal places, that: a he hits the bullseye on his next two throws, but misses on the third b on his tenth throw, he gets a bullseye.

expected value, variance and standard deviation of the binomial distribution When working with the binomial probability distribution, (like other distributions) it is very useful to know the expected value (mean), variance and the standard deviation. The random variable, X, is such that X ∼ Bi(8, 0.3) and has the following probability distribution.

x Pr(X = x)

0

1

2

3

4

5

6

7

8

0.057 65

0.197 65

0.296 48

0.254 12

0.136 14

0.046 68

0.010 00

0.001 22

0.000 07

In chapter 10 we saw that the expected value, E(X) was defined as E( X ) = ∑ x Pr( X = x ). Hence, the expected value for the above table is: E( X ) = ∑ x Pr( X = x ). = 0 × 0.057 65 + 1 × 0.197 65 + 2 × 0.296 48 + 3 × 0.254 12 + 4 × 0.136 14 + 5 × 0.046 68 + 6 × 0.010 00 + 7 × 0.001 22 + 8 × 0.000 07 = 0 + 0.197 65 + 0.592 96 + 0.762 36 + 0.544 56 + 0.233 400 + 0.060 00 + 0.008 54 + 0.000 56 = 2.4 The variance was defined by the rule Var(X) = E(X2) − [E(X)]2. Hence, the variance for the above table is: Var(X) = E(X2) – [E(X)]2 = 02 × 0.057 65 + 12 × 0.197 65 + 22 × 0.296 48 + 32 × 0.254 12 + 42 × 0.136 14 + 52 × 0.046 68 + 62 × 0.010 00 + 72 × 0.001 22 + 82 × 0.000 07 – (2.4)2 = 0 + 0.197 65 + 1.185 92 + 2.287 08 + 2.178 24 + 1.167 00 + 0.360 00 + 0.059 78 + 0.004 48 – (2.4)2 = 7.440 15 – (2.4)2 = 1.68

Chapter 11


573

The standard deviation was defined by the rule SD( X ) = Var ( X ). Hence the standard deviation for the above table is: SD( X ) = Var ( X ) = 1.68 = 1.30 Since the above method for obtaining the expected value, variance and the standard deviation is tedious and time consuming, a quicker method has been developed to calculate these . It can be shown that if X ∼ Bi(n, p) then: E ( X ) = np

Var ( X ) = npq

SD( X ) = npq

To check that these agree with the previous example, we will substitute the values into the given rules. When X ∼ Bi(8, 0.3), we obtain the following. The expected value: E(X) = np = 8 × 0.3 = 2.4 The variance: Var(X) = npq = 8 × 0.3 × 0.7 = 1.68 The standard deviation:

SD( X ) = npq

= 1.68 = 1.30 As can be seen, these values correspond to those obtained earlier. A great deal of time is saved using these rules and the margin for making mistakes is reduced.

Hence, if X is a random variable and X ∼ Bi(n, p) then: E(X) = np Var(X) = npq Note: The distribution must be binomial for these rules to apply.

SD( X ) = npq

Worked Example 15

The random variable X follows a binomial distribution such that X ∼ Bi(40, 0.25). Determine the: a expected value b variance and standard deviation. Think a

b

574

Write a E(X) = np

1


2

List the values for n and p.

n = 40, p = 0.25

3


E(X) = 40 × 0.25

4

Evaluate.

1

Write the rule for the variance.

2

Write the values for n, p and q.

n = 10, p = 0.25, q = 0.75

3


Var(X) = 40 × 0.25 × 0.75

4

Evaluate.

5

Write the rule for the standard deviation.

SD(X) = npq

6

Substitute the value obtained for the variance and take the square root.

= 7.5

7

Evaluate.

= 2.74

= 10

b Var(X) = npq


= 7.5

Worked Example 16

A fair die is rolled 15 times. Find: a the expected number of 3s rolled b the probability of obtaining more than the expected number of 3s. Think a

b

Write a E(X) = np

1


2

List the values for n and p.

n = 15, p =

3


E(X) = 15 × 1 6

4

Evaluate.

1


1 6

= 2.5

b X ~ Bi(n, p)

1  X ~ Bi 15,   6

2

Write what is required. Note: Since X represents the number of 3s, X can have only whole number values.

Pr(X > 2.5) = Pr(X ≥ 3) = 1 − Pr(X < 3) = 1 − [Pr(X = 0) + Pr(X = 1) + Pr(X = 2)]

3

Write the rule for the binomial probability distribution. Then substitute the values of n, p and q into the rule.

14 2 13 0 15 1   1   5  1   5   1   5 = 1 −  15 C0     + 15C1     + 15C2       6  6  6  6   6  6  

4

Evaluate and round the answer to 4 decimal places.

= 1 − (0.064 905 471 5 + 0.194 716 414 6 + 0.272 602 980 4) = 1 − 0.532 224 866 5 = 0.467 775 133 5 ≈ 0.4678

Part b can also be solved with a CAS calculator.

Pr(X > 2.5) = Pr(X ≥ 3) = binomCdf(15, 1 , 3, 15) = 0.4678

6

Worked Example 17

A binomial random variable has an expected value of 14.4 and a variance of 8.64. Find: a the probability of success, p b the number of trials, n. Think a

1

2

Write

Write what is known and what is required.

a E(X) = 14.4

Substitute the value of np into the variance equation.

npq = 8.64 14.4q = 8.64

Var(X) = 8.64 p=?

so np = 14.4 so npq = 8.64


575

b

8.64 14.4

3

Transpose the equation to make q the subject.

q=

4

Evaluate.

q = 0.6

5

Solve for p using the relationship q = 1 – p.

1

Write what is known and what is required.

2

Substitute the values into the equation.

3

Transpose the equation to make n the subject.

q=1−p 0.6 = 1 − p p = 0.4 b np = 14.4

n=?

where p = 0.4

n × 0.4 = 14.4 14.4 0.4 = 36

n=

Worked example 18

eBook plus

A new test designed to assess the reading ability of students entering high school Tutorial showed that 10% of the students displayed a reading level that was inadequate to int-0581 cope with high school. Worked example 18 a If 400 students are selected at random, find the expected number of students whose reading level is inadequate to cope with high school. b Determine the standard deviation of students whose reading level is inadequate to cope with high school and hence calculate µ ± 2σ. c Discuss the results obtained in part b. Think a

b

576

WriTe a E(X) = np

1


2

Write the values for n and p.

n = 400, p = 0.1

3


E(X) = 400 × 0.1

4

Evaluate.

5

Interpret the answer.

1

Write the rule for the variance.

2

Write the values for n, p and q.

n = 400, p = 0.1, q = 0.9

3


Var(X) = 400 × 0.1 × 0.9

4

Evaluate.

5

Write the rule for the standard deviation.

6

Substitute the value obtained for the variance and take the square root.

= 36

7

Evaluate.

=6

8


= 40 It is expected that 40 students will have a reading level that is inadequate to cope with high school. b Var(X) = npq

= 36 SD(X) = npq

µ − 2σ = 40 − 2 × 6 = 28


9

c

Calculate µ + 2σ.

Interpret the results obtained in part b. Note: Recall that for many random variables, approximately 0.95 of the probability distribution lies within two standard deviations of the mean.

µ + 2σ = 40 + 2 × 6 = 52 There is a probability of 0.95 that between 28 to 52 students (inclusive) will have a reading level that is inadequate to cope with high school.

If X is a random variable and X ∼ Bi(n, p) then: E(X) = np Var(X) = npq where q = 1 − p SD(X) = npq exerCiSe

11d

expected value, variance and standard deviation of the binomial distribution In questions 1, 2 and 3, assume we have a binomial distribution with number of trials, n and probability of success, p, as given. 1 We15a Determine the mean if: a n = 10 and p = 0.6 c n = 100 and p = 0.5 2 We15b Determine the variance if: a n = 20 and p = 0.6 c n = 25 and p = 0.4


Spreadsheet 003

b n = 8 and p = 0.2 d n = 50 and p =

Binomial distribution

3 4

b n = 15 and p = 0.9 d n = 20 and p =

1 4

3 We15b Determine the standard deviation if: a n = 10 and p = 0.2 b n = 30 and p = 0.5 c n = 50 and p = 0.7

d n = 72 and p =

2 5

4 A fair coin is tossed 10 times. Find: a the expected number of heads b the variance for the number of heads c the standard deviation for the number of heads. 5 A card is selected at random from a standard playing pack of 52 and then replaced. This procedure is completed 20 times. Find: a the expected number of picture cards b the variance for the number of picture cards c the standard deviation for the number of picture cards. 6 Six out of every 10 cars manufactured are white. Twenty cars are randomly selected. Find: a the expected number of white cars b the variance for the number of white cars c the standard deviation for the number of white cars.

Chapter 11


577

7 WE16 A fair die is rolled 10 times. Find: a the expected number of 2s rolled b the probability of obtaining more than the expected number of 2s. 8 A fair die is rolled 15 times. Find: a the expected number of prime values b the probability of obtaining more than the expected number of prime values. 9 A mathematics exam contains 27 multiple-choice questions, each with 5 possible answers. Find: a the expected number of correct answers if a student guessed each question b the probability of obtaining more than the expected number of correct answers. 10 A new drug that is being trialled has a success rate of 35%. Of the next 80 patients who trial the drug, find: a the expected number of patients who will be cured b the probability that more than the expected number of patients will be cured. 11

Eighty per cent of rabbits that contract a certain disease will die. If a group of 120 rabbits contract the disease, how many would you expect to: a die? b live?

12 WE 17 A binomial random variable has a mean of 10 and a variance of 5. Find: a the probability of success, p b the number of trials, n. 13

A binomial random variable has a mean of 12 and a variance of 3. Find: a the probability of success, p b the number of trials, n.

14 X is a binomial random variable with a mean of 9 and a variance of 6. Find: a the probability of success, p b the number of trials, n c Pr(X = 10). 15 If X is a binomial random variable with E(X) = 3 and Var(X) = 2.4, find: a the probability of success, p b the number of trials, n c Pr(X = 10) d Pr(X ≤ 2). 16 MC The expected number of heads in 20 tosses of a fair coin is: A

1 2

B 5

C 10

D 15

E 20

17 MC Jenny is a billiards player who knows from experience that 7 out of every 10 shots she plays will result in a ball being potted. If she has 40 shots, the number of balls she expects to pot is: A 7 B 14 c 21 d 25 e 28 18 MC The variance of the number of balls that Jenny pots from her 40 shots in question 17 is: A 2.898 B 7.3 c 8.4 d 22.2

e 28

19 MC Eighty per cent of children are immunised against a certain disease. A sample of 200 children is taken. The mean and variance of the number of immunised children is: A 80 and 5.66 respectively B 80 and 32 respectively C 100 and 50 respectively d 160 and 5.66 respectively E 160 and 32 respectively 20 MC A binomial random variable has a mean of 10 and a variance of 6. The values of n and p respectively are: A 5 and 25

578

B 5 and 35

C 20 and 35


D 25 and 25

E 25 and

3 5

21 MC A Bernoulli trial has a probability of success, p. If 5 trials are conducted, the probability of three successes is: A p3q2 B p2q3 C 10p2q3 D 10p3(1 − p)2 E p3 22 A binomial experiment is completed 20 times, with the expected number of successes being 16. Find: a the probability of success, p b the variance c the standard deviation. 23 A multiple-choice test has 20 questions with five different choices of answer for each question. If the answers to each question are guessed, find: a the probability of getting 50% of the questions correct b the probability of getting at least three correct. 24 Four per cent of pens made at a certain factory do not work. If pens are sold in boxes of 25, find the probability that a box contains more than the expected number of faulty pens. 25 A biased coin is tossed 10 times. Let X be the random variable representing the number of tails obtained. If X has an expectation of 3, find: a the probability of obtaining exactly two tails b the probability of obtaining no more than two tails. 26

Eighty per cent of Melbourne households have DVD players. A random sample of 500 households is taken. How many would you expect to have DVD players?

27 The executive committee of a certain company contains 15 . Find the probability that more females than males will hold positions if: a males and females are equally likely to fill any position b females have a 55% chance of holding any position c females have a 45% chance of holding any position. 28

A statistician estimates the probability that a spectator at a Carlton versus Collingwood AFL match barracks for Carlton is 12 . At an AFL grand final between these two teams there are 10 000 spectators. Find: a the expected number of Carlton ers b the variance of the number of Carlton ers c the standard deviation of the number of Carlton ers.

29

Thirty children are given five different yoghurts to try. The yoghurts are marked A to E, and each child has to select his or her preferred yoghurt. Each child is equally likely to select any brand. The company running the tests manufactures yoghurt B. a How many children would the company expect to pick yoghurt B? b The tests showed that half of the children selected yoghurt B as their favourite. What does this tell the company manufacturing this product?

30 The proportion of defective fuses made by a certain company is 0.02. A sample of 30 fuses is taken for quality control inspection. a Find the probability that there are no defective fuses in the sample. b Find the probability that there is only one defective fuse in the sample. c How many defective fuses would you expect in the sample? d The hardware chain that sells the fuses will accept the latest batch for sale only if, upon inspection, there is at most one defective fuse in the sample of 30. What is the probability that they accept the batch? e Ten quality control inspections are conducted monthly for the hardware chain. Find the probability that all of these inspections will result in acceptable batches.


579


Investigation Winning at racquetball!

580

31 We18 A new test designed to assess the reading ability of students entering high school showed that 10% of the students displayed a reading level that was inadequate to cope with high school. a If 1600 students are selected at random, find the expected number of students whose reading level is inadequate to cope with high school. b Determine the standard deviation of students whose reading level is inadequate to cope with high school, and hence calculate µ ± 2σ. c Discuss the results obtained in part b. 32 The success rate of a new drug that is being trialled is 70%. a If 1800 patients are selected at random, find the expected number of patients cured. b Determine the standard deviation of patients cured, and hence calculate µ ± 2σ. c Discuss the results obtained in part b.


Summary The binomial distribution

• The binomial distribution is an example of a discrete probability distribution. • The binomial distribution may be referred to as a Bernoulli distribution, and the trials conducted are known as Bernoulli trials. • For a sequence to be defined as a Bernoulli sequence, each of the following characteristics must be satisfied. 1. n independent trials must be conducted. 2. Only two possible outcomes must exist for each trial, that is, success (p) and failure (q). 3. The probability of success, p, is fixed for each trial. • If X represents a random variable that has a binomial distribution, then it can be expressed as X ~ Bi(n, p) or X ~ B(n, p). This means that X follows a binomial distribution with parameters n (the number of trials) and p (the probability of success). • If X is a binomial random variable its probability is defined as: Pr(X = x) = nCx pxqn − x where x = 0, 1, 2, . . . n The effects of n and p on binomial distribution graphs

• The parameters n and p affect the binomial probability distribution curve as follows. 1. If p < 0.5, the graph is positively skewed. 2. If p = 0.5, the graph is symmetrical or is a normal distribution curve. 3. If p > 0.5, the graph is negatively skewed. 4. When n is large and p = 0.5, the interval between the vertical columns decreases and the graph approximates a smooth hump or bell shape. Problems involving the binomial distribution for multiple probabilities

• When solving problems dealing with the binomial distribution for multiple probabilities, always: 1. define the distribution 2. write what is required 3. write the rule for the binomial probability distribution 4. substitute the values into the given rule and evaluate. Markov chains and transition matrices

• A Markov chain is a sequence of repetitions of an experiment in which: 1. The probability of a particular outcome in an experiment is conditional only on the result of the outcome immediately before it. 2. The conditional probabilities of each outcome in a particular experiment are the same every single time. • A two-state Markov chain is where there are only two possible outcomes for each experiment. • Problems involving two-state Markov chains can be solved using: 1. tree diagrams where the probabilities are multiplied along the branches 2. recurrence relationships  Pr( A 2 | A1 ) Pr( A 2 | A1 ′)  3. transition matrices where T =  ,  Pr( A 2 ′ | A1 ) Pr( A 2 ′ | A1 ′)   n( A ) 1 0 S0 = the initial state,   n( A 2 ) 0 

  Pr ( A )  1 0  or    Pr ( A 2 )0    

Sn = T n × S0 To find the long-term proportion or steady state, let n = a large number, relative to the problem, and solve for Sn.


581

Expected value, variance and standard deviation of the binomial distribution

• If X is a random variable and X ∼ Bi(n, p) then: E(X) = np Var(X) = npq SD( X ) = npq This applies only to a binomial distribution.

582



1 Zoe tosses a fair coin 5 times. Calculate: a the probability of obtaining 3 heads b the probability of obtaining at least 1 head c the probability of obtaining 4 heads then a tail. 2 Matt is a dedicated football player and knows that his chance of scoring a goal on any one kick is 0.9. If he has three shots at goal in a row, calculate the probability that: a he scores two goals b all of his shots miss c after missing the first, he scores a goal on each of his next two shots. 3 Nisha is pulling marbles out of a bag containing 3 green and 9 red ones. After each selection she replaces the marble. In total she selects four marbles. What is the probability she selects: a two red marbles? b two red marbles, given that at least one red marble was selected? 4 One out of three people read the Bugle newspaper, while one out of five read the Headline. If four people are sampled, what is the probability that: a one of them reads the Headline? b at least three of them read the Bugle? 5 A barrel contains 100 balls, some of which are rainbow-coloured. Four balls are randomly selected from the barrel, with replacement (that is, a ball is selected, its colour noted, and the ball replaced before the next ball is selected). Let p be the proportion of rainbow-coloured balls in the barrel. a Write down an expression for the probability that exactly one of the four balls selected is rainbow-coloured. b Use calculus to find the exact value of p for which this probability will be a maximum. Exam tip Only 44% of students scored the mark for part a, and 75% of students received 0/2 for part b. Students who used q in part a failed to define it. In part b, sometimes the instruction to use calculus was not followed. [Assessment report Mathematical Methods 1 2004]

[© VCAA 2004 ]

6 Statistics gathered at a local bookstore have found that 50% of customers purchase crime novels. If

four customers enter the bookstore, what is the probability that less than two of these customers buy a crime novel? Exam tip This was not well answered. Some students put their probability expressions equal to one (for example) and tried to solve for p. A method mark was awarded for indicating that the derivative of the probability expression was to be found, set to zero, and solved for p. Some students simply wrote down an answer (correct or otherwise) and failed to obtain full marks, as the question required working to be shown. [Assessment report 1 2004 © VCAA]

7 A cafe prepares its macchiatos with either skinny or regular milk. It is known that 20% of customers who have a regular macchiato on a particular day will have a skinny macchiato the next day. Also, 70% of customers who have a skinny macchiato on a particular day will have a regular macchiato the next day. Given a particular customer has a skinny macchiato on Monday, calculate the probability: a they had a regular macchiato on Tuesday and a skinny macchiato on Wednesday b they had a skinny macchiato on Wednesday. 8 Hilary loves going to the movies every Friday night, and prefers either comedy or action films. If she watches a comedy one week, the probability that she sees a comedy the following week is 0.1. If she sees an action movie on a particular Friday, there is a probability of 0.4 that she will see an action movie the following Friday. Given that Hilary saw a comedy on the first Friday in March, calculate the probability: a she sees an action film on the third Friday of the month b she saw a comedy on the 2nd Friday, given that she saw an action film on the third Friday of the month. 9 Maria and Patrick love to play a game of tennis each week. The probability of Maria beating Patrick if she wins the previous week is 0.7; however, if she loses a match, the chance of her beating him the following week is only 0.2. Given that Maria beats Patrick in their first match of the year, calculate the probability that she will beat him exactly one more time in the next three weeks.


583

10 Calculate: i the mean ii the variance for the binomial random variables with n and p given by: a n = 100 and p = 0.5 b n = 50 and p = 0.8. 11 A binomial random variable has a mean of 10 and variance of 8. Calculate: a the probability of success, p b the number of trials, n. 12 A mathematics exam contains 40 multiple-choice questions, each with 5 possible answers. Calculate: a the expected number of correct answers if a student guessed each question b the variance of the number of correct answers if a student guessed each question. Multiple choice

1 Three out of every 7 students ride bikes to school. Twenty students are randomly selected. The probability that 8 of these rode to school today is: A 0.0951 B 0.1426 C 0.1738 D 0.3178 E 0.4916 2 An unbiased 8-sided die is rolled 12 times. The probability of obtaining three results greater than 5 is: A 0.1135 B 0.1688 C 0.2188 D 0.2279 E 0.2824 3 Which of the following does not represent a Bernoulli sequence? A Rolling a die 10 times and recording the number of 5s B Rolling a die 50 times and recording the results C Spinning a 7-sided spinner and recording the number of 4s obtained D Surveying 100 people and asking them if they eat ‘Superflakes’ cereal E Drawing a card with replacement and recording the number of red cards obtained 4 The probability that the temperature in Melbourne will rise above 25 ºC on any given summer day, independent of any other summer day, is 0.65. The probability that 3 days in a week reach in excess of 25 ºC is: A 0.653 B 37 × 0.354 C 7 × 0.653 × 0.44 E 0.653 × 0.354

D 35 × 0.653 × 0.354

5 A fair die is rolled 10 times. The probability of obtaining 4 even numbers is: A 0.0004 B 0.2051 C 0.0009 D 0.7949 E 0.2461 584

6 If the same die from question 5 is rolled 10 times, the probability of obtaining at least 8 even numbers is: A 0.0107 B 0.0390 C 0.0547 D 0.9453 E 0.9893 7 The probability of Sam beating Abby in a game of cards is 0.36. Abby and Sam decide to play a game every day for n days. What is the fewest number of games they need to play to ensure the probability of Sam winning at least once is greater than 0.85? A 3 B 4 C 5 D 6 E 7 8 One in every 50 apples sold at Grubby Granny’s Greengrocers has worms in it. If I buy a box of 100 apples, the probability of at least three apples containing worms is: A 0.3233 B 0.3781 C 0.5 D 0.6219 E 0.6867 9 X is a random variable, binomially distributed with n = 20 and p = 37 . Pr(X ≥ 1) is: A 1 − D

( 37 )

( 47 ) 20

20

B 1 − E

( 37 )

( 47 )

20

C 20

( 37 )( 47 )

19

20

10 Claire’s position in the netball team is goal shooter. The probability of her shooting a goal is 67%. If she has 15 attempts at scoring, the probability she will score at least 7 goals is: A 15C7 (0.67)7(0.33)8 B 15C7 (0.33)7(0.67)8 + 15C8 (0.33)8(0.67)7 + … + (0.33)15 C 15C8 (0.67)8(0.33)7 + 15C9 (0.67)9(0.33)6 + … + (0.67)15 D 15C7 (0.67)7(0.33)8 + 15C8 (0.67)8(0.33)7 + … + (0.67)15 E (0.33)15 + 15C1 (0.67)(0.33)14 + … + 15C7 (0.67)7(0.33)8 11 The proportion of patients that suffer a violent reaction from a new drug being trialled is k. If 60 patients trial the drug, the probability that one-fifth of the patients have a violent reaction is: A 60C5 (1 − k)5(k)55 B 60C12 (k)12 C 60C5 (k)5(1 − k)55 D 60C12 (1 − k)12(k)48 E 60C12 (k)12(1 − k)48 12 Theo es three sets of traffic lights on his way to school each morning. The lights at each intersection operate independently of each other. The probability of him having to stop for a red light is 23 and the probability of ing through the intersection without stopping is 13 . If Theo


encountered a red light at least once during a particular trip to school, what is the probability that he had to stop at exactly two intersections? A

2 13

B

6 19

C 3 5

D

6 13

E

2 9

13 Consider a Markov chain where the nth state is given by Sn = T n × S0. Which one of the following could represent the transition matrix, T, and the initial state matrix, S0, respectively?  0.1 0.9  1  A   and    0.6 0.4  0   0.6 0.1  0.3  B   and    0.4 0.9   0.7  15   0.6 0.1 C   and   30   0.4 0.9  1   0.25 0.75  D   and   0   0.4 0.6   0.25 0.45  E   and  0.85 0.65 

10     20 

14 Kelly has developed a method for predicting whether or not the surf will be good on a particular day. If it is good today, there is an 83% chance it will be good tomorrow. If it is poor today, there is a 65% chance it will be poor tomorrow. The probability that the surf will be poor on Thursday given it was poor on Tuesday is:

A 0.4225 D 0.6620

B 0.4820 E 0.6500

C 0.0595

15 Two fair coins are tossed simultaneously 60 times. The number of times that they both show heads is expected to be: A 10 B 15 C 20 d 30 E 45 16 The variance in the number of heads obtained from 50 tosses of a fair coin is: A

1 50

B

2 25

C

1 4

d

25 2

E 3.53

17 The expected number of heads in 50 tosses of a fair coin is: A

1 2

B 6.25 C 10 d 2.5 E 25

18 A variance of 1.35 occurs when: A n = 20 and p = 0.6 B n = 15 and p = 0.9 C n = 25 and p = 0.4 D n = 20 and p = 14 E n = 50 and p = 0.3 19 The number of defective computer parts in a box of computer parts ready for sale is a random variable with a binomial distribution with mean 10 and standard deviation 3. If a computer part is drawn at random from the box, the probability that it is defective is: A 0.1 B 0.3 C 0.5 D 0.7 E 0.9 Exam tip Only 40% of students answered this question correctly. [Assessment report 1 2002]

[© VCAA 2002]

Extended response

1 Speedy Saverio’s Pizza House claims to cook and deliver 90% of pizzas within 15 minutes of the order being placed. If your pizza is not delivered within this time, it is free. On one busy Saturday night, Saverio has to make 150 deliveries. a How many deliveries are expected to be made within 15 minutes of placing the order? b What is the probability of receiving a free pizza on this night? c If Saverio loses an average of $4 for every late delivery, how much would he expect to lose on late deliveries this night? 2 Ten per cent of all Olympic athletes are tested for drugs at the conclusion of their event. One per cent of all athletes use performance enhancing drugs. Of the 1000 Olympic wrestlers competing from all over the world, Australia sends 10. Find: a the expected number of Australian wrestlers who are tested for drugs b the probability that half the Australian wrestlers are tested for drugs c the probability that at least two Australian wrestlers are tested for drugs d the expected number of drug s among all wrestlers.


585

3 Five per cent of watches made at a certain factory are defective. Watches are sold to retailers in boxes of 20. Find: a the expected number of defective watches in each box b the probability that a box contains more than the expected number of defective watches per box c the probability of a ‘bad batch’, if a ‘bad batch’ entails more than a quarter of the box being defective. 4 Aiko is a keen basketball player and knows that her chance of scoring a goal on any one throw is 0.65. a If Aiko takes 6 shots for a goal, find the probability, correct to 4 decimal places, that she: i misses each time ii scores a goal at least three times iii scores a goal five times, given that she scored a goal at least three times. b Find the number of throws Aiko would need to ensure a probability of more than 0.9 of scoring at least one goal. 5 Sixty-eight percent of all scheduled trains through Westbourne station arrive on time. If 20 trains go through the station every day, calculate the probability, correct to 4 decimal places, that: a no more than 10 trains are on time b at least 12 trains are on time c at least 12 trains are on time for 9 out of the next 10 days. 6 The success rate of a new drug being trialled is 60%. a If 2400 patients are selected at random, find the expected number of patients cured. b Determine the standard deviation of patients cured and hence calculate µ ± 2σ. c Interpret the results found in part b. 7 Phil is running a stall at the local Primary School Fair involving lucky dips. It costs $2 to have a go, and contained in a large box are 80 lucky dips from which to choose. Phil claims that one in 5 lucky dips contains a prize. By the end of the day, all 80 have been sold. Calculate the probability, correct to 4 decimal places, that: a the first four people to select a lucky dip don’t win a prize, but the next two do b there are at least 10 winners c there are no more than 18 prize winners, given that at least 10 people won a prize. 8 Sharelle is the goal shooter for her netball team. During her matches, she has many attempts at scoring a goal. Assume each attempt at scoring a goal is independent of any other attempt. In the long term, her scoring rate has been shown to be 80% (that is, 8 out of 10 attempts to score a goal are successful). a i What is the probability, correct to 4 decimal places, that her first 8 attempts at scoring a goal in a match are successful? ii What is the probability, correct to 4 decimal places, that exactly 6 of her first 8 attempts at scoring a goal in a match are successful? iii What is the probability, correct to 3 decimal places, that her first 4 attempts at scoring a goal are successful, given that exactly 6 of her first 8 attempts at scoring a goal in a match are successful? Assume instead that the success of an attempt to score a goal depends only on the success or otherwise of her previous attempt at scoring a goal. If an attempt at scoring a goal in a match is successful, then the probability that her next attempt at scoring a goal in the match is successful is 0.84. However, if an attempt at scoring a goal in a match is unsuccessful, then the probability that her next attempt at scoring a goal in the match is successful is 0.64. Her first attempt at scoring a goal in a match is successful. b i What is the probability, correct to 4 decimal places, that her next 7 attempts at scoring a goal in the match will be successful? ii What is the probability, correct to 4 decimal places, that exactly 2 of her next 3 attempts at scoring a goal in the match will be successful? [© VCAA 2008] 9 Every afternoon Anna either goes for a run or a walk. If she goes for a walk one afternoon, the probability that she goes for a run the next is 0.45, and if she decides to run one afternoon, then the probability of her going for a walk the next afternoon is 0.8. On Wednesday, Anna decides to go for a walk around the park. a What is the probability that she goes for a run on each of the next three afternoons? b What is the probability, correct to 4 decimal places, that over the next three afternoons, she goes for a run at least once?

586


c What is the probability, correct to 4 decimal places, that on the following Wednesday Anna will decide to go for a run? d In the long term, on what proportion of afternoons will she choose to go for a run? 10 A small town has two restaurants — Kaz’s Kitchen and Al’s Fine Dining. Records show that 30% of customers who eat at Kaz’s Kitchen one week will visit Al’s Fine Dining the next. Also, 40% of customers who eat at Al’s Fine Dining during the week will dine at Kaz’s Kitchen the following week. There are 400 of the town who regularly dine out once a week. During the first week of July, 300 people visit Kaz’s Kitchen and Al has 100 customers at his restaurant. a How many customers will visit each restaurant in the first week of August? b In the long run, how many customers will Kaz and Al each get per week? 11 Two companies are competing for the mobile phone market. At the end of January, market research revealed the following patterns in the subscriptions of mobile phone s. Of the 400 Tellya customers who were interviewed, 340 were staying with Tellya and 60 were changing to Yodacall. Of the 100 Yodacall customers who were interviewed, 90 were staying with Yodacall and 10 were changing to Tellya. a Set up a pair of recurrence relationships that describes the given patterns. b What is the original state of the companies in of market share? c What is the state of each company at the end of the next month (February)? d What is the state, in of market share, of each company 4 months later (May)? e What is the state, in of market share, of each company 7 months later (August)? f What is the state, in of market share, of each company at the end of the following January? g A company will fail to be viable if its market share falls below 25%. Which, if either, of these companies will not achieve this market share in the long run? eBook plus Digital doc


Chapter 11


587

eBook plus

aCTiviTieS


• 10 Quick Questions: Warm up with ten quick questions on applications of the binomial distribution. (page 537) 11A


Tutorials

• We 2 int-0576: Watch a worked example on indentifying the number of trials in an experiment. (page 539) • We 5 int-0577: Watch a worked example on constructing a probability distribution table. (page 544) Digital docs

• Spreadsheet 003: Investigate the binomial distribution. (pages 548 and 551) • SkillSHEET 11.1: Practise solving indicial equations. (page 550) • WorkSHEET 11.1: Recognise Bernoulli sequences and calculate cumulative and non-cumulative probabilities. (page 552) 11B

Problems involving the binomial distribution for multiple probabilities

Tutorials

• We 8 int-0578: Watch a worked example on calculating probabilities using a CAS calculator. (page 553) • We 10 int-0579: Watch a worked example on calculating probabilities using the cumulative binomial distribution. (page 556) Digital doc

• SkillSHEET 11.2: Practise multiple probabilities. (page 557) 11C

Markov chains and transition matrices


Tutorial

• We 12 int-0580: Watch a worked example on calculating long-term probabilities. (page 563) Digital doc

• WorkSHEET 11.2: Identify and perform appropriate techniques to calculate probabilities. (page 573) 11D Tutorial

• We 18 int-0581: Watch a worked example on calculating the mean, median and mode of a probability density function. (page 576) Digital docs

• Spreadsheet 003: Investigate the binomial distribution. (page 577) • Investigation: Winning at racquetball (page 580) Chapter review Digital doc


• The binomial distribution: Consolidate your understanding of the binomial distribution. (page 559)

588

Expected value, variance and standard deviation of the binomial distribution


12

12A Continuous random variables 12B Using a probability density function to find probabilities of continuous random variables 12C Measures of central tendency and spread 12D Applications to problem solving 12E The normal distribution 12F The standard normal distribution 12G The inverse cumulative normal distribution

Continuous distributions AREAS OF STUDY

• Random variables, including: – the concept of discrete and continuous random variables – calculation and interpretation of the expected value, variance and standard deviation of a random variable (for discrete and continuous random variables, including consideration of the connection between these) – calculation and interpretation of central measures (mean, median, mode) – the property that, for many random variables, approximately 95% of their probability distribution is within 2 standard deviations of the mean • Continuous random variables, including: – the construction of probability density functions from non-negative continuous functions of a real variable – the specification of probability distributions for continuous random variables using probability density functions

– calculation using technology, interpretation and use of mean (µ), median, mode, variance (σ 2) and standard deviation of a continuous random variable – standard normal distribution, N(0, 1), and transformed normal distributions, N(µ, σ 2), as examples of a probability distribution for a continuous random variable (use as an approximation to the binomial distribution is not required) – the effect of variation in the value(s) of defining parameters on the graph of a given probability density function for a continuous random variable – probabilities for intervals defined in of a random variable, including conditional probability (students should be familiar with the use of definite integrals, evaluated by hand or using technology, for a probability density function to calculate probabilities; teachers may also choose to relate this to the notion of a cumulative distribution function, although this is not required) eBook plus

12A

Digital doc

Continuous random variables

10 Quick Questions

In previous chapters we worked with discrete random variables and the probability distributions associated with them. We will now focus on continuous random variables and an important distribution associated with them — the normal distribution. Continuous random variables represent quantities that can be measured and thus may assume any value in a given range. They include such variables as time, height and weight.

Chapter 12

Continuous distributions

589

Weight (kg) x

Frequencies f

40− < 50

2

50− < 60

10

60− < 70

19

70− < 80

15

80− < 90

3

90− < 100

1

Number of students ((ff) f)

The weights of 50 Year 12 students are displayed in the table below, alongside a histogram.

Total = 50

f 20 18 16 14 12 10 8 6 4 2 0

40 50 60 70 80 90 100 x Weight (kg)

The frequency of individual weights cannot be determined because the weights have been grouped into class intervals. This limits the information we are able to extract from the histogram. For example, to determine the number of students weighing less than 60 kg we simply add the frequency of the 40− < 50 and 50− < 60 class intervals, that is, 10 + 2 = 12 students. However, we would not be able to determine the weight of students below 75 kg as this value lies within the class interval rather than being an end point. Since the value that a continuous random variable can assume is measured in some way, the exact value cannot be obtained. Hence a weight of 60 kg, if we measure in whole numbers, is actually between 59.5 and 60.5 kg. Therefore, the probability of a continuous random variable assuming an exact value is zero. In order to determine the probability of continuous random variables, a new method must be employed. If we were to increase the size of the sample of students’ weights and make the class intervals very small, then the histogram would become a smooth frequency curve as shown below. A special scaled version of a smooth frequency curve is the f x) f( probability density function or pdf. The scale is such that the probability of the random variable lying between certain values is given by the area between the pdf and the horizontal axis. Hence, sections such as the shaded region between x = a and x = b, as shown below right may be regarded as x 0 probabilities. The curve is positioned above the x-axis since it represents a probability distribution in which individual probabilities f x) f( assume a value between 0 and 1. Hence, the probability of X, a continuous variable, falling between x = a and x = b is represented by the shaded area between a probability density function y = f (x), the x-axis and the lines x = a and x = b. The area is determined by integrating x 0 a b f (x) from x = a to x = b. Using mathematical notation, this may be summarised as b

Pr(a < x < b) = ∫ f ( x ) dx. a

Properties of a probability density function 1. f (x) ≥ 0 for all x, (that is, we never have negative probabilities) and ∞ 2. ∫ f ( x ) dx = 1 (that is, the sum of all probabilities is equal to one). −∞

590


Continuous distributions Probability density functions The domain of a probability density function (pdf) is usually R, that is, the variable is continuous and can assume any real value (at least in theory). If we need our function to be non-zero over a particular interval only, say, over (a, b), then we need to specify that the function is equal to zero everywhere else. In such case, the area under b

the curve between x = a and x = b must be equal to one. That is, ∫ f ( x ) dx = 1. a

WORKED EXAMPLE 1

eBoo k plus eBook

Sketch the graph of each of the following and state whether each function may be a probability density function (pdf).

Tutorial


 4 x, 1 ≤ x ≤ 2  0, elsewhere

a f (x (x) = 

10 x − 4, 0 ≤ x ≤ 1 0, eelsewhere

b f (x (x) = 

 0 . 5, − 1 ≤ x ≤ 1  0, elsewhere

c f (x (x) =  THINK a

1

WRITE

The graph of f (x) is a straight line with gradient 4 over the domain [1, 2]. When x = 1, y = 4 × 1 = 4, and when x = 2, y = 4 × 2 = 8. So the end points of the line are (1, 4) and (2, 8). Everywhere else, the function is zero.

a

y (2, 8)

8 4

(1, 4)

0

1

2

2

For a function to be a pdf, it has to be positive or zero over the entire domain. State whether this condition is satisfied by inspecting the graph of f (x).

f (x) ≥ 0 for all x.

3

The total area under the graph of a pdf must be equal to one. Check whether this condition is observed by finding the area under f (x), and comment on the result. (This can be done by integration.)

A = ∫ 4 xdx xd

x

2

1

2

 4x2  =   2 1 =  2 x 2 

2

1

= 2 × (2)2 − 2 × (1)2 =6 Area under the curve ≠ 1 4

Although the first condition for the function to be a pdf is observed, the second one is not. State your conclusion.

f (x) could not be a probability density function.

Chapter 12


591

b

1

The graph of f (x) is a straight line over the domain [0, 1]. Its gradient is 10 and its y-intercept is at y = −4. When x = 0, y = −4, and when x = 1, y = 10 × 1 − 4 = 6. So the end points of the line are (0, −4) and (1, 6). Everywhere else, the function is zero.

b

y (1, 6)

6

0

1

x

(0, −4)

c

2


f (x) is not positive or zero for all x, as part of the graph is below the x-axis.

3

As the first condition is not observed, there is no need to check the area under the curve. State your conclusion.

f (x) does not represent a probability density function.

1

The graph of f (x) is a horizontal line through y = 0.5 over the domain [−1, 1]. Everywhere else, the function is zero.

c

y (0, 0.5) 0

−1

2


f (x) ≥ 0 for all x.

3

The total area under the graph of the pdf must equal 1. Check whether this condition is observed by finding the area under f (x), and comment on the result.

A = ∫ − 0.5dx 5

4

Both conditions for the function to be a pdf are satisfied. State your conclusion.

1

1

1

1

=  0.5x 5  −1 = 0.5 × (1) − 0.5 × (−1) =1 Area under the curve = 1 f (x) could be a probability density function.

WORKED EXAMPLE 2

Find the value of a so that the given function may be a probability density function (pdf). f [2, 4] → R, f (x f: (x) = a(12 − 3x 3 ) THINK 1

592

For the function to be a pdf, the area under its graph must be 1. Area is found by evaluating the definite integral with terminals being the end points of the domain. Write the appropriate statement, equating area to 1.

WRITE 4

∫2 a(12 − 3x)dx = 1


x

4

2

Take the coefficient, a, out of the integral.

a ∫ (12 − 3 x )dx = 1

3

Antidifferentiate.

 3x 2  a 12 x −  =1 2 2 

4

Substitute the terminals in and simplify the left-hand side of the equation.

2

4

 3 × 42   3 × 22   a  12 × 4 − −  12 × 2 −  =1  2   2    a[(48 − 24) − (24 − 6)] = 1 a(24 − 18) = 1 6a = 1 a=

5

Solve for a.

6

Sketch the graph of the resultant function and check whether f (x) ≥ 0 for all x.

1 6

1 (12 − 3x) 6 x =2− ,2≤x≤4 2

f (x) =

y (2, 1)

1 0

2

4

x

f (x) ≥ 0 for all x

1. If X is a continuous random variable, then the probability of it falling between certain values is given by the area under a frequency curve, known as a probability density function or pdf. 2. A pdf must be greater than or equal to zero for all values of x, and the total area under the curve must be equal to one. That is, f (x) ≥ 0 for all x, and

∞

∫ ∞ f ( x)dx = 1. −

3. The domain of a pdf is usually R, that is, the variable is continuous and can assume any real value (at least in theory). 4. If we need our function to be non-zero over a particular interval only, say, over (a, b), then we need to specify that the function is equal to zero everywhere else. For example, 2  x, 1 ≤ x ≤ 2 f ( x) =  3 0, elsewhere er ere 5. If a pdf is only non-zero over a particular interval (a, b), the area under the curve b

between x = a and x = b must be equal to one. That is, ∫ f ( x )dx = 1. a

Chapter 12


593

EXERCISE

12A

Continuous random variables 1 WE1 Sketch the graph of each of the following functions and state whether each function may be a probability density function. 1 1   4 x − 2, 0 ≤ x ≤ 2, 0 ≤ x ≤ a f (x) =  b f (x) =  2 2 0, elsewhere 0, elsewhere er ere 1 − x, − 1 ≤ x ≤ 1 c f (x) =  0, elsewhere

 3x 2 − , 1≤ x ≤1  d f (x) =  2 0, elsewhere 

− π π  ≤x≤ cos (22 x ), e f (x) =  4 4 0, elsewhere er ere

− 3π  2 sin ((xx ), − 2π ≤ x ≤ f f (x) =  2 0, elsewhere er ere 

1  , 1≤ x ≤ e g f (x) =  x 0, elsewhere er ere

2 x − 1, 1 ≤ x ≤ 2 h f (x) =  er ere 0, elsewhere

2 WE2 For each of the following functions, find the value of a so that the given function may be a pdf. a ff: [1, 2] → R, f (x) = a(2x (2 − 1) (2x b ff: [−1, 1] → R, f (x) = a(10 − 3x2) c ff: [0, π] → R, f (x) =

a sin (x) 2

ax 3 4 − − 4, 1] → R, f (x) = ax 2 + a

e ff: [0, 2] → R, f (x) =

3

d ff: [2, 3] → R, f (x) = aex − 2 f ff: [−7, −2] → R, f (x) = a 2 − x h ff: [3, 4] → R, f (x) = ax(x − 1)

Find the value of a, such that the function may be a probability density function. Sketch the graph of the resultant function. a, 0 ≤ x ≤ 1  f (x) = ax, 1 < x ≤ 2 0, elsewhere er ere 

4 Find the value of a, such that the function may be a probability density function. Sketch the graph of the resultant function. a( x + 2), ), − 1 ≤ x ≤ 0    1 f (x) = a  x + 2 , 0 < x ≤ 2    2 0, elsewher erre  5 Find the value of m, such that the function may be a probability density function. Sketch the graph of the resultant function. mx 2 , 1 ≤ x ≤ 3  f (x) = m(12 − x ), 3 ≤ x ≤ 4 0, elsewhere er ere 

594


6

 x , − 1 ≤ x ≤ k 2 f (x) =  0, elsewhere

7 8 9 10

y

The diagram at right shows the graph of the function (−−_1 , _12) 2

1_ 2

(where k > 0). −_1 2

0

x

k

Find the value of k such that the function may be a probability density function. Give your answer in exact form. x2 The function ff: [−1, n] → R, f (x (x) = is a probability density function. Find the value of n. 4 (Give the answer in surd form.) ), 0 ≤ x ≤ 0 . 5 k sin (2π xx), The function f ( x ) =  is a probability density function. Find 0 , elsewhere er ere  the value of k. ke −x , 0 ≤ x ≤ llog og e (3) The function f ( x ) =  is a probability density function. Find the 0, elsewhere value of k. The diagram at right is a graph of rectangular or uniform distribution. y a Specify the probability density function of this distribution in of a and b. b Hence, or otherwise, find the value of p if the probability density function is defined as 0

1  , p ≤ x ≤ 30 f (x) =  25 . 0, elsewhere

12B

(k, k)

a

b

x

Using a probability density function to find probabilities of continuous random variables If X is a continuous random variable, then the probability of it falling between x = a and x = b is given by the area bounded by the probability density function, the x-axis and the lines x = a and x = b. This area can be y easily calculated by evaluating the definite integral between a and b. That is, Pr(a ≤ x ≤ b) b

Pr(a ≤ X ≤ b) = ∫ f ( x ) dx . a

As mentioned in chapter 9, ∴ Pr(X > a) = Pr(X ≥ a)

a

∫a

f ( x ) dx = 0, so Pr(X = a) = 0.

0

a

b

x

WORKED EXAMPLE 3

If X is a continuous random variable with the probability density function defined as 1  x − 4, 8 ≤ x ≤ 10 f (x (x) =  2  0, elsewhere find: a Pr(X > 9)

b Pr(8.5 < X < 9)

c Pr(X > 9 X > 8.5).

Chapter 12


595

THINK a

1

WRITE

Draw the graph of f (x) and shade the required region. (It is the area under the curve to the right of x = 9.)

a

y

0

2

To find the required probability (that is, to find the shaded area under the curve), integrate f (x) between x = 9 and x = 10.

3

Antidifferentiate and simplify.

(10, 1)

1

(8, 0)

9

10

x

10  1  Pr(X > 9) = ∫  x − 4 ddxx 9 2 

10

 1 x2  − 4x = × 2 2 9 10

 x2  =  − 4x 4 9 4

b

1

 10 2   92  = − 4 × 10 −  − 4 × 9  4   4 

Substitute the terminals in and evaluate to obtain the required probability.

Show the required region on the diagram. (It is the area under the curve between x = 8.5 and x = 9.)

= (25 − 40) − (20.25 − 36) = −15 − (−15.75) = 0.75 b

y (10, 1)

1

0

(8, 0) 8.5 9

To find the required probability, integrate f (x) between x = 8.5 and x = 9.

Pr(8.5 < X < 9) = ∫

3

Antidifferentiate. (Use the result from part a.)

 x2  =  − 4x 4 8.5

4

Substitute the terminals in and evaluate to obtain the required probability.

2

9

(

1 8.5 2

10

)

x − 4 dx

9

c

596

1

The rule for conditional probability Pr ( A ∩ B) is Pr( Pr(A | B) = . Write the Pr ( B) appropriate statement.

 92   8.52  =  − 4 × 9 −  − 4 × 8.5  4   4  = (20.25 − 36) − (18.0625 − 34) = −15.75 − (−15.9375) = 0.1875 c Pr(X > 9 | X > 8.5)

Pr[( X > 9) ∩ ( X > 8.5))] Pr(( X > 8.5) Pr ( X > 9) = Pr ( X > 8.5) =


x

2

The expression in the denominator can be found by adding the probability of X being between 8.5 and 9 and the probability of X being greater than 9.

=

Pr ( X > 9) Pr (8.5 < X < 9) + Pr( Pr ( X > 9)

3

The required probabilities are already found in parts a and b. Substitute the numbers in and evaluate.

=

0.775 0.1875 + 0.775

0.775 0.9375 = 0.8 =

The required probabilities can also be obtained with the aid of a CAS calculator. Where exact answers are not required, it is appropriate to use the CAS. WORKED EXAMPLE 4

 logg e ( 0 . 5 x ) , 2 ≤ x ≤ 2e  A random variable, X, has its frequency curve defined as f ( x ) =  2  0, elsewhere Calculate the probability, correct to 4 decimal places, that X is: a less than 4 b between 2.5 and 3.5. THINK a

1

WRITE

The required probability can be obtained by evaluating the definite integral of f (x) over the interval [2, 4]. Note: Pr(X < 4) = Pr(X ≤ 4) because Pr(X = 4) = 0 On a Calculator page, press: • MENU b • 4:Calculus 4 • 3:Integral 3 Complete the entry line as: 4

∫2 

a

ln((0.5 x )  dx 2 

Press ENTER ·. 2

Write the solution, rounding to 4 decimal places.

Pr(X < 4) = Pr(2 ≤ X < 4) logg e (0.5 x ) dx 2 = 0.3863 =∫

4

2

b

Find the required probability as shown above, this time using the interval [2.5, 3.5]. Write the solution, rounding to 4 decimal places.

b Pr(2.5 ≤ X ≤ 3.5) =

3.5 log g e (0.5 x )

∫2.5

2

dx

= 0.2004

Chapter 12


597

In many instances we need to integrate over intervals involving infinity. In such cases, the limits are evaluated as shown at right.

Interval

Limit lim

k

∫ k f ( x)dx

(−∞, ∞)

k →∞

(−∞, a)

k →− ∞ k

(a, ∞)

k →∞ a

−

lim

∫

lim

∫

a

k

f ( x )dx f ( x )dx

When evaluating limits involving infinity, it is useful to that as a number tends to infinity, its reciprocal tends to zero. For example, lim 1 = 0 and lim 1 = 0. This concept is x →∞ x x →∞ e x illustrated in the worked example that follows. WORKED EXAMPLE 5

eBoo k plus eBook −1

1 x 2 , x>0  A random variable, X, has its frequency curve defined as f (x (x) =  2 e . a Draw the graph of f (x (x).  0, elsewhere  b Show that f (x (x) is a probability density function. c Find the probability, correct to 4 decimal places, that X is: i smaller than 3 ii greater than 2.5 iii greater than 2.5, given that it is smaller than 3. THINK

WRITE/DRAW

a Draw the graph of f (x). ). It is a decreasing

function with a starting point (0, 1 ) and a 2 horizontal asymptote y = 0.

a

y 1

(0, –2)

x

0 b

1

2

A pdf must be greater than or equal to 0 for all values of x. Check whether this condition is observed by inspecting the graph of f (x). Find the total area under the curve by evaluating the definite integral of f (x). Note that the interval over which the integral needs to be evaluated is [0, ∞). So, in this case, evaluate k lim f ( x )dx .

∫

k →∞ 0

b f (x) ≥ 0 for all x

A = lim

∫

k

k →∞ 0

1 e 2

1 = lim  e k →∞  2  = lim [ − e k →∞

−

−1

−1

2

2

1 x 2 d dx x

1 ÷  2  −

x k ]0 k

 −1  = lim  1  k →∞  x  e 2 0 3

598

Substitute the terminals in and evaluate the limit. that  −  lim  1  = 0. k →∞  k   e2 

 − 1   − 1  − = klim →∞  k    e 0   e2  1 =0+ 1 =1


k

0

Tutorial


4

c

Both conditions required for the function to be a pdf are observed. State your conclusion. i The required probability can be

obtained by evaluating the definite integral of f (x) over the interval [0, 3]. Use the expression for the antiderivative found in part b to speed up your calculations. Give your answer correct to 4 decimal places.

ii

1

To find Pr(X > 2.5), evaluate the integral of f (x) over the interval [2.5, ∞]. Alternatively, find Pr(X < 2.5) and subtract it from 1. This method avoids finding limits, as the interval over which the integral needs to be evaluated is [0, 2.5], that is, it does not involve ∞.

Since f (x) ≥ 0 for all x and the total area under the curve is 1, f (x) is a pdf.

c i Pr(X < 3) =

31

∫0 2

−

e

1 x 2

ddxx

3

 −1  = x  e 0 1 2

−  1 − 1  = 3  − 0    e   e2  = 0.7769 ii Pr(X > 2.5) = 1 − Pr(X < 2.5)

=1−

2.5 1

∫0

2

−

e

1 x 2

ddxx

2.5

 −1  = 1 −  1x   2   e 0

 −   −  1 1 = 1 −   2.5  −  0     2   e   e = 1 − 0.7135 = 0.2865

2

The required probability can also be obtained by using the CAS over the interval [2.5, ∞) On a Calculator page, press: • MENU b • 4:Calculus 4 • 3:Integral 3 Complete the entry line as: ∞

1

−1

∫2.5 2 × e 2

x

 ddxx 

Press ENTER · 3

iii

1

Write the solution, rounding to 4 decimal places. Note: The answer is the same as above. Write the appropriate statement for the conditional probability.

Pr(X > 2.5) =

∞

∫2.5 12 e

−1 2

x

ddxx

= 0.2865 iii Pr(( X > 2.5X < 3) =

=

Pr[( X > 2.5) ∩ ( X < 3))] Pr(( X < 3) Pr((2.5 < X < 3) Pr(( X < 3)

Chapter 12


599

2

Using the CAS as shown above, find Pr(2.5 < X < 3), with your interval as [2.5, 3].

3

Fill in the required values into the conditional probability statement. Answer to 4 decimal places.

Pr(2.5 < X < 3) = 0.06337

Pr(( X > 2.5X < 3) =

Pr((2.5 < X < 3) Pr(( X < 3)

0.06337 0.7769 = 0.0816 =

WORKED EXAMPLE 6

If X is a continuous random variable with a probability density function given by  4 − 2 x, 1 ≤ x ≤ 2 f (x (x) =  , find the value of a such that Pr(X ≤ a) = 0.75, where 1 ≤ a ≤ 2.  0, elsewhere THINK 1

WRITE

Pr(X ≤ a) = 0.75 means that the area under the curve of f (x) to the left of a is 0.75, where 1 ≤ a ≤ 2. Draw a diagram to illustrate the given information.

y

2

(1, 2)

0.75

0

1

a

(2, 0) x

2

Copy down the probability statement as given in the question.

Pr(X ≤ a) = 0.75

3

The required probability can be found by integrating f (x) over the interval [1, a]. So replace Pr(X ≤ a) with the appropriate integral.

∫1 ( 4 − 2 x ) ddxx = 0.775 a

a

4


 2x2  4 x −  = 0.775 2 1  a

 4 x − x 2  = 0.75 1 5

Substitute the terminals in.

(4a − a2) − (4 − 1) = 0.75

6

The result is a quadratic equation. Use the quadratic formula to solve for a.

4a − a2 − 3 = 0.75 a2 − 4a + 3.75 = 0 4 ± 4 2 − 4 × 1 × 3.775 2 ×1 4± 1 = 2 5 3 a = or a = 2 2 a=

600


7

Since a must be within the interval [1, 2], discard the first solution.

Since 1 ≤ a ≤ 2, a = 3 . 2

Quantiles and percentiles Quantiles and percentiles both define the value of the random variable X, below which a given proportion of the distribution falls. The only difference is that a quantile is a probability value expressed as a decimal, while a percentile is expressed as a percentage. For example, if Pr(X < k) = 0.6, then k is the 0.6 quantile, or the 60th percentile. Finding a particular quantile or percentile involves solving an equation similar to the one used in worked example 6. For example, if we want to find the 60th percentile of a continuous random variable, X, with a probability density function given by f (x) and a ≤ X ≤ b, solve for k k

the equation of the form ∫ f ( x )dx = 0.6. a

Note that in worked example 6, we found a such that Pr(X < a) = 0.75. In effect, we were finding the 0.75 quantile, or the 75th percentile.

1. If X is a continuous random variable, then the probability of it falling between x = a and x = b is given by the area bounded by the probability density function, the x-axis and the lines x = a and x = b. That is, b Pr(a ≤ X ≤ b) = ∫ f ( x )dx .

y Pr(a ≤ x ≤ b) 0

a

b

x

a

2. Pr(X = a) = 0 ∴ Pr(X > a) = Pr(X ≥ a) 3. The required probabilities can be calculated by using a CAS calculator and selecting the integral function from the maths expression template, where f (x) is the function that needs to be integrated and a and b are the terminals. 4. To integrate over intervals involving infinity, the limits are evaluated as shown. Interval (−∞, ∞) (−∞, a)

Limit k

∫ f ( x)dx k →∞ k lim

−

a

∫ k→ ∞ k lim −

(a, ∞)

k

∫ k →∞ a lim

f ( x )dx f ( x )dx

5. As a number tends to infinity, its reciprocal tends to zero. For example, 1 1 lim = 0 and lim x = 0. x →∞ e x →∞ x 6. Quantiles and percentiles both define the value of the random variable, X, below which a given proportion of the distribution falls. 7. A quantile is a probability value expressed as a decimal, while a percentile is a probability value expressed as a percentage.

Chapter 12


601

EXERCISE

12B

Using a probability density function to find probabilities of continuous random variables 1 WE3 If X is a continuous random variable with the probability density function defined as  1 (2 x + 1), 1 ≤ x ≤ 2 f (x)=  4 , find, correct to 4 decimal places: er ere 0, elsewhere a Pr(X > 1.5)

b Pr(1.2 < X < 1.75)

c Pr(X > 1.5 | X > 1.2).

2 If X is a continuous random variable with the probability density function defined as  1 (4 − x )2 , 1 ≤ x ≤ 4 f (x) =  9 , find, correct to 4 decimal places: er ere 0, elsewhere a Pr(X < 3) d Pr(2 < X < 3 | X < 3)

b Pr(X > 2) e Pr(X > 2 | X < 3).

c Pr(2 < X < 3)

ax ( x − 2), 0 ≤ x ≤ 2 3 A random variable, X, has a pdf given by f (x (x) =  . er ere 0, elsewhere a Find the value of a. b Find the probability, correct to 4 decimal places, that X is: i greater than 1.5 ii less than 0.4 iii greater than 0.4 but less than 1.5 iv greater than 0.4, given that it is less than 1.5. 4 WE4 A continuous random variable, X, has probability density function defined by π 1  x co s ( x − 2 ), 2π ≤ x ≤ 3π f (x) =  5π . Calculate, correct to 4 decimal places: er ere 0, elsewhere a Pr(X < 8) b Pr(X < 8 | X < 8.5)

0 . 25 2 5e 0.25 x , x ≥ 0 5 WE5 A random variable, X, has its frequency curve defined as f (x (x) =  . er ere 0, elsewhere a Draw the graph of f (x). b Show that it is a probability density function. c Find the probability that X is: i smaller than 2 ii greater than 1 iii greater than 1, given that it is smaller than 2. 1  , x ≥1 6 A random variable, X, has its frequency curve defined as f (x (x) =  x 2 .  0, elsewhere er ere a Draw the graph of f (x).  b Show that it is a probability density function. c Find: i Pr(X > 2) ii Pr(X < 3) iii Pr(2 < X < 3) iv Pr(2 < X < 3 | X > 2) v Pr(X < 3 | X > 2). −

7 WE6 If X is a continuous random variable with a probability density function given by 1 1  x+ 2, 1≤ x ≤ 2 f (x) =  3 , find the value of a such that Pr(X ≤ a) = 0.36, where 1 ≤ a ≤ 2. 0, elsewhere er ere 602


If X is a continuous random variable with a probability density function given by 4  , 2≤x≤4 1 f (x) =  x 2 , and Pr(X ≥ b) = 3, find the value of b, where 2 ≤ b ≤ 4. 0, elsewhere er ere  9 If X is a random variable with a probability density function given by 2  (3 − x ), 0 ≤ x ≤ 1 f (x) =  5 , find: 0, elsewhere er ere a the 0.4 quantile b the 0.85 quantile c the 70th percentile. −π 1 π ≤θ ≤  2 cos (θ ), 10 If f (θ) =  2 2 is the probability density function of the 0, elsewhere er ere  random variable θ: a sketch the graph of the probability density function and shade the region corresponding to the equation Pr(−a ≤ θ ≤ a) = 12 b find the exact value of a.

The random variable, X, has a probability density function defined as ax, 1 ≤ x ≤ 1 2 2 f (x) =  er ere 0, elsewhere a Sketch the probability density function. b Hence, or otherwise, find the value of a. c Calculate: i Pr(−0.5 < X < 0) ii Pr(X ≤ 0.5) Pr(−0.5 < X < 0 | X ≤ 0.25) −

The random variable, X, has a probability density function defined as  1 ( x − 2)),, 2 ≤ x ≤ 6  16 f (x) =  1 (10 − x ), ), 6 < x ≤ 110 16  er ere 0, elsewhere a Sketch the probability density function. b Calculate: i Pr(X < 4) ii Pr(3 < X ≤ 8)

EXAM TIP On previous exams, students have chosen a y-scale that was too large, drawn curved graphs where straight lines were required, drawn an unsymmetrical graph, or omitted the horizontal lines. Greater care needs to be taken. [Authors' advice]

13 The random variable, X, has a probability density function defined as   3 x + 1 , − 10 2 ≤ x ≤ 0 8  256 3 f (x) =  1 . 4 − x, 0 ≤ x ≤ 4  16 0, elsewhere ere ere  a Draw the graph of f (x). b Show that the total area under the graph is 1.

Chapter 12


603

c Find: i Pr(X < −1) iv Pr(X < 3)

ii Pr(X > 2) v Pr(−2 < X < 3)

iii Pr(X > −5) vi Pr(−2 < X < 1 | X < 3).

A random variable, X, has a probability density function given by −π  π ≤x≤ coss ((222xx ), f (x) =  4 4. 0, elsewhere er ere 

a Draw the graph of the probability density function. b Show that the area under the graph is 1. c Calculate each of the following (giving answers in exact form). π i Pr(X < ) 12 −π ii Pr(X < ) 6 −π iii Pr(X < | X < 0) 6 A random variable, X, has its frequency function given by   x a sinn   , 0 ≤ x ≤ π  2 f (x) =  . 0, elsewhere er ere  a Find the value of a such that f (x) is a probability density function. b Draw the graph of the probability density function. c Find, in exact form, the probability that X is: π i smaller than 2 π ii greater than 3 π 2π iii between and 3 3 2π π iv less than , given that it is greater than . 3 2 1 a for x ∈R and a > 0. 2 π a + x2 a Sketch the graph of f (x) for a = 1, 2 and 3 on the same set of axes. Comment on the effect of varying a in the shape of the graph. b For a = 1, use a CAS calculator to find the following probabilities to 4 decimal places. i Pr(X < 1) ii Pr(−1 < X < 1) iii Pr(−1 < X < 1 | X > −1)

16 A Cauchy distribution is defined as f (x (x) =

17 a Sketch the graph of g(x ( )= (x

1

e x , x ∈R. −

2

2 π b Find the area bounded by the curve and the x-axis. c Use your answers to a and b to define the probability density function f (x) whose rule is the same as the rule of g(x). d If X is a random variable whose probability density function is given by f (x), find, correct to 4 decimal places: i Pr(X > 0.5) ii Pr(X < 1) iii Pr(X > −0.32) iv Pr(−0.5 < X < 1.09).

604


12C

Measures of central tendency and spread In this section we discuss how to find measures of central tendency (mean, median and mode) and measures of spread (variance and standard deviation) for a continuous random variable.

Measures of central tendency Mean If X is a continuous random variable with a probability distribution function f (x) over the ∞

domain R, the mean of X can be found using the formula µ = E(x) = ∫ − xf ( x ) ddx . ∞

However, if f (x) has a certain rule for a ≤ X ≤ b and is zero elsewhere, then the mean can be b

obtained by using the formula µ = E(x) = ∫ xf ( x )ddx . In the worked examples below and the a

b

exercise that follows, we will mostly deal with the latter case. In general, E[g(x)] = ∫ g( x ) f ( x ) dx. a

Median The median value, m, of the continuous random variable, X, is a value such that Pr(X ≤ m) = 1 . 2 (It follows that Pr(X ≥ m) = 1 as well.) In other words, m is a value of X such that there is a 50% 2 chance that X will be less than or equal to m and a 50% chance that it will be greater than m

or equal to m. Since Pr(X ≤ m) is given by ∫ f ( x )dx, to find the median, solve the equation a m 1 ∫a f ( x)dx = 2 . The solution of equations of this type was discussed in detail in the previous section of this chapter. Mode The mode is a value of X for which f (x) has its maximum. To find the mode of the continuous random variable, it is best to sketch the graph of its probability density function first. If the probability density function has a maximum turning point in the interval [a, b], the mode is given by the x-coordinate of the turning point. It can be easily obtained by finding the derivative of f (x), making it equal to zero, and solving for x. If the probability density function continuously increases or decreases over the interval [a, b], or if it has a minimum turning point, the mode is then given by the end point of the interval. That is, the mode is either a or b (whichever one corresponds to the maximum value of f (x)). Note that it is possible to have more than one mode. f x) f(

f x) f(

0

y

y

y

Mode

x

0

f x) f(

Mode x

Mode 0

WORKED EXAMPLE 7

eBoo k plus eBook

Find the mean, median and mode for the following probability density function.  2 x, 0 ≤ x ≤ 1 f (x (x) =   0, elsewhere THINK 1

Write the formula for finding the mean of the continuous random variable.

x

Tutorial


WRITE b

µ = E(x) = ∫ x f ( x )ddx a

Chapter 12


605

2

Substitute 0 for a and 1 for b and replace f (x) with 22x. Simplify the integrand.

1

∫0 x × (2 x)dx

µ=

1

= ∫ 2 x 2 ddxx 0

1

3

Antidifferentiate.

 2x3  =   3 0

4

Substitute the terminals in and evaluate.

=

2(1)3 2(0)3 − 3 3

= 23 5

State the value of the mean.

The mean is 23 .

6

Write the formula for finding the median, m.

∫a

7

Substitute 0 for a and replace f (x)) with 22x.

m

f ( x )dx = m

∫0

2 x ddx =

1 2 1 2

m

8

 2x2    =  2 0


m

9

Substitute in the terminals and simplify.

10

Evaluate the median and rationalise the denominator. (Note that only the positive square root is required, as 0 ≤ x ≤ 1.)

11

State the value of the median.

12

Sketch the graph of the pdf.

1 2

 x 2  = 0

1 2

(m)2 − (0)2 =

1 2

m2 =

1 2

m=

1

=

2

2 . 2

The median is y

(1, 2)

2

0 13

The mode is the value of X for which f (x) has its maximum. As seen from the graph, the largest value of f (x) is at the right-handside end point of the domain. So this point represents the mode.

2 2

1

x

The mode is 1.

Note: In the worked example above, the value of the mode was obtained directly from the graph. In other cases differentiation may be required.

606


Consider the random variable, X, with the probability density function given by

y (0, 1–)

−

 1 ( x 2 − 2 x − 3), 0 ≤ x ≤ 3 f (x) =  9 . er ere 0, elsewhere

3

0

(3, 0) 1

2

3

x

As can be seen from the graph, the probability density function has a maximum turning point. The x-coordinate of this point represents the mode of X. To find the x-coordinate of the turning point, we need to differentiate, make the derivative equal to zero, and solve for x. f ′(x) =

−1 9

(2 − 2) (2x

f ′(x) = 0 −1 9

(2 − 2) = 0 (2x

2 −2=0 2x x=1 So the mode of X is 1.

Measures of spread Variance If X is a continuous random variable with the probability distribution function f (x) over the domain R, the variance of X can be found using the formula Var(X) X = E(x − µ)2 = X)

∞

∫ ∞( x − µ)2 f ( x) ddxx . −

However, if f (x) has a certain rule for a ≤ X ≤ b and is zero elsewhere, then the variance can be obtained by using the formula Var(X) X = E(x − µ)2 = X)

b

∫a ( x − µ)2 f ( x) ddxx .

This formula may prove to be rather hard to use when f (x) is complicated. The alternative formula for variance is Var(X) X = E(X X) X2) − µ2. This can also be expressed as Var(X) X = X)

b



b

2

 .

∫a ( x)2 f ( x) ddxx −  ∫a x f ( x) ddxx

Note that both formulas require us to find the mean of X first. Standard deviation The standard deviation of X can be easily found by taking a positive square root from the Va X ) . variance of X. That is, SD(X) X = Var( X) Note: Standard deviation is often denoted as σ and variance as σ 2. The calculations of both variance and standard deviation are shown in detail in the worked example below. WORKED EXAMPLE 8

Find the variance and standard deviation for the following probability density function.  x − 1 , 1 ≤ x ≤ 2 2 f (x (x) =  .  0, elsewhere THINK 1

Write down the formula for finding the variance.

WRITE

Var(X) X = E(X X) X2) − µ2

Chapter 12


607

b

2

Find the value of the mean. Write down the appropriate formula.

µ = E(x) = ∫ x f ( x ) ddx

3

Substitute 1 and 2 for a and b respectively, and

µ=∫ x× x−

replace f (x) with x −

1 . 2

Simplify the integrand.

a

( ) dx

2

1

2

= ∫ x2 − 1

1 2

x dx 2 2

4

Antidifferentiate.

5

Substitute the terminals in and evaluate µ.

 x3 x 2  = −  4 1 3

 23 22   13 12  = −  − −  4   3 4  3  8   1 1 5 1 =  − 1 −  −  = −  3   3 4  3 12 = 19 12

6

Find the value of E( E(X X2). Write the appropriate formula.

E(x2) = ∫ x 2 f ( x ) ddx

7

Make all necessary substitutions and simplify the integrand.

E(x2) = ∫ x 2 × x −

b

a

( ) dx

2

1

1 2

2 x2  = ∫  x 3 −  dx 1  2 2

8

Antidifferentiate.

 x 4 x3  = −  6 1 4

9

Substitute the terminals in and evaluate E(X X2).

 24 23   14 13  = −  − −  6   4 6  4 4   1 1  = 4 −  −  −   3  4 6 =

8 1 − 3 12

31 = 12

10

Substitute the values of E(X X2) and µ into the variance formula and evaluate.

31

( )

31

361

Var(X) X = 12 − X)

19 12

2

= 12 − 144 =

608

11 144

11

State the value of the variance.

11 Variance = 144

12

Write the formula for the standard deviation.

SD(X) X = Var( X) Va X )


13

Substitute the value of the variance into the formula and evaluate.

11 144

=

11 12

= 14

Standard deviation =

State the value of the standard deviation.

WORKED EXAMPLE 9

eBoo k plus eBook

 logg ( x ), 1 < x < e . The probability density function for X is given by f ( x ) =  e  0, elsewhere Calculate, correct to 3 decimal places: a the mean c the standard deviation

b the median d Pr(µ – 2σ ≤ X ≤ µ + 2σ).

THINK a

1 2

11 12

Tutorial


WRITE

Write the formula for finding the mean of the continuous random variable. We need to use the CAS calculator to integrate and find the probability. On a Calculator page, complete the entry line as:

e

a µ = E( X ) =

∫0 x loge ( x) ddxx

µ = E( X ) =

∫1 x loge ( x) ddxx

e

n( x )) dx ∫1 ( x × lln( Then press ENTER ·.

3

b

1

Write the solution and round to 3 decimal places. Write the formula for finding the median. On a Calculator page, press: • MENU b • 3:Algebra 3 • 1:Solve 1 Complete the entry line as:  m 1  solvee  ∫ ( lln( x ) ) dx = , m 2   1 Then press ENTER ·.

e

= 2.097 b

m

∫1

logg e ( x ) ddxx =

1 2

Chapter 12


609

2

Write the solution.

m

Solving ∫ logg e ( x ) ddxx = 1

1 2

for m implies

m = 0.1866 or m = 2.1555.

c

3

As 1 ≤ m ≤ e, so m must be the bigger of the two possible solutions. Answer the question and round to 3 decimal places.

1

We first need to find the variance. On a Calculator page, complete the entry line as:

The median is 2.156.

c Var(X) X = X)

e

x dx − µ 2 . ∫1 x 2 loge ( x)

∫1 ( x 2 × ln( x)) dx − (2.09726)2 e


2

d

1

Calculate the standard deviation, correct to 3 decimal places.

Find the intervals µ − 2σ and µ + 2σ.

SD(X) X = Var( X) Va X ) = 0.176047371282 = 0.420 d µ − 2σ = 2.097 − 2 × 0.420

= 1.2581 µ + 2σ = 2.097 + 2 × 0.420 = 2.9364

2

State the interval µ − 2σ ≤ X ≤ µ + 2σ.

µ − 2σ ≤ X ≤ µ + 2σ = 1.2581 ≤ X ≤ 2.9364 = 1.2581 ≤ X ≤ e, since 2.9364 > e (the upper value).

3

Calculate Pr(µ − 2σ ≤ X ≤ µ + 2σ) using the CAS calculator.

Pr(µ − 2σ ≤ X ≤ µ + 2σ) = ∫

e

1.2581

logg e ( x ) ddxx

= 0.969 Note that in this example, 96.9% of the data lies within 2 standard deviations of the mean, which is close to the estimated value of 95%.

Interquartile range The interquartile range (IQR) is the middle 50% of the distribution. In the previous section we discussed percentiles and quantiles. IQR = 75th percentile − 25th percentile. = Q3 − Q1

610


1. If X is a continuous random variable with a probability distribution function f (x) over the domain R, the mean of X can be found using the formula ∞

µ = E(x) = ∫ xf ( x )ddx. −

∞

2. If f (x) has a certain rule for a ≤ X ≤ b and is zero elsewhere, then the mean can be b

obtained by using the formula µ = E(x) = ∫ xf ( x )ddx . a

b

In general, E[g(x)] = ∫ g( x ) f ( x )dx. a

3. The median value, m, of the continuous random variable, X, is a value such that Pr(X ≤ m) = 12 .

m

4. To find the median of X we need to solve for m the equation ∫ f ( x )dx = 12 . a 5. The mode is a value of X for which f (x) has its maximum. 6. To find the mode of the continuous random variable, it is best to sketch the graph of its probability density function first. 7. If the probability density function of X has a maximum turning point in the interval [a, b], the mode is given by the x-coordinate of the turning point. It can be obtained by finding the derivative of f (x), making it equal to zero, and solving for x. 8. If the probability density function continuously increases or decreases over the interval [a, b], or if it has a minimum turning point, the mode is then given by the end point of the interval (a or b) that corresponds to the maximum value of f (x). 9. It is possible to have more than one mode. 10. If X is a continuous random variable with the probability distribution function f (x) over the domain R, the variance of X can be found using the formula ∞

Var(X) X = E(x − µ)2 = ∫ ( x − µ )2 f ( x )ddxx . X) −

∞

11. If f (x) has a certain rule for a ≤ X ≤ b and is zero elsewhere, then the variance can be b

obtained by using the formula Var(X) X = E(x − µ)2 = ∫ ( x − µ )2 f ( x )ddxx . X) a

12. The alternative formula for variance is given by Var(X) X = E(X X2) − µ2. This can also be b

b

expressed as Var(X) X = ∫ ( x )2 f ( x )ddxx − ( ∫ xf ((xx )dx )2 . X) a a

13. The standard deviation of X can be found by taking the positive square root of the variance of X. That is, SD(X) X = Var( X) Va X ) . 14. The interquartile range (IQR) is the middle 50% of the distribution. IQR = 75th percentile − 25th percentile. = Q3 − Q1 EXERCISE

12C

Measures of central tendency and spread 1 WE7 Find the mean, median and mode for each of the following probability density functions.  1 x + 1, − 2 ≤ x ≤ 0 2, − 2 ≤ x ≤ − 1 . 5 a f (x) =  b f (x) =  2 0, elsewhere 0, elsewhere 2 3 x , 0 ≤ x ≤ 1 c f (x) =  er ere 0, elsewhere

Chapter 12


611

2 Find the mean, median and mode for each of the following probability density functions.  − 1 x, − 11 ≤ x ≤ − 1 a f (x) =  5 0, elsewhere

 3 (2 − x )2 , 2 ≤ x ≤ 4 b f (x) =  8 er ere 0, elsewhere

6  x ( x − 1), 1 ≤ x ≤ 2 c f (x) =  5 er ere 0, elsewhere

3 WE8 Find the variance and standard deviation for each of the following probability density functions.  1 , 2 ≤ x ≤ 6 a f (x) =  4 er ere 0, elsewhere

 1 (3 − x ), 0 ≤ x ≤ 2 b f (x) =  4 er ere 0, elsewhere

 4 ( x − x )3 , 0 ≤ x ≤ 1 c f (x) =  er ere 0, elsewhere

4 Find the variance and standard deviation for each of the following probability density functions. 3 2( x − 2), 2 ≤ x ≤ 3  ( x + 1)2 , 0 ≤ x ≤ 1 a f (x) =  b f (x)=  7 er ere er ere 0, elsewhere 0, elsewhere  1 sin ( x ), − 2π ≤ x ≤ −π c f (x) =  2 er ere 0, elsewhere 5 A continuous random variable, X, has a probability density function defined by − 3x 0 . 3e 0.3x , x≥0 f (x) =  . Calculate: er ere 0, elsewhere a the mean of X b the median of X c the mode of X d the variance of X e the standard deviation of X.  a , 2≤x≤3  2 6 WE9 For f (x (x) =  ( x − 1) , find:  er ere 0, elsewhere a the value of a, such that f (x) would represent the probability density function of a random variable, X b the mean of X, correct to 4 decimal places. c the standard deviation of X, correct to 4 decimal places. d Pr(µ − 2σ ≤ X ≤ µ + 2σ), that is, the probability that X is within 2 standard deviations from the mean, correct to 4 decimal places. 7 X is a continuous random variable with a probability density function given by  π  a co s  x −  , 0 ≤ x ≤ π  2 f (x) =  . 0, elsewhere er ere  a Calculate the value of a. b Sketch the graph of the probability density function. c Calculate the exact mean and the variance, correct to 4 decimal places.  π π π  d Find Pr X ≤  . e Find Pr X ≤ X ≤  .  2 2 4   π  −π π  ≤x≤ a sin 2  x +  , For f (x (x) =   4 4 4: 0, elsewhere er ere 

612


a calculate the value of a such that f (x) could be a probability density function b sketch the graph of the probability density function c find the mean, median and mode explain the result obtained in part c. 9

The probability density function of a random variable, X, is given by k ( x + 2)3 , − 2 ≤ x ≤ 0 f (x) =  . er ere 0, elsewhere a Find the value of k. b Calculate the median. c Calculate the 0.25 quantile. d Calculate the 0.75 quantile. e Use the answers to parts c and d to find the IQR (interquartile range).

10 The probability density function of a random variable, X, is given by  π cos (π x ) , wh er eree x ∈[0, 1] 1 f (x) =  2 er ere 0, elsewhere a Sketch the probability distribution function. Calculate: b the mean c the median d the mode. 11 The probability density function of a random variable, X, is given by −5 mx 2 + nx, − 1 ≤ x ≤ 0 f (x) =  . If the mean of X is 7 , find the values of m and n. 0, elsewhere (Hint: Use simultaneous equations.)

12D

Applications to problem solving As was shown in the previous sections of this chapter, if the probability density function of a continuous variable is known, we can use it for finding the probability of that variable falling between particular values. So far, we have been practising finding these probabilities for variables with various probability density functions without concerning ourselves as to what these variables actually represent. In this section, we will look at continuous random variables that represent something of practical interest, for instance, waiting time for a dental appointment, or the weight of fish caught on a trout farm.

WORKED EXAMPLE 10

The local milk bar sells a particular brand of chocolate bar. The number sold per day is given by the probability distribution function  x , 0 ≤ x ≤ 10  f ( x ) =  50  0, elsewhere Calculate: a the maximum number of chocolate bars that are sold each day b the expected number of chocolate bars sold each day c the probability that fewer than 4 chocolate bars are sold each day d the probability that more than 8 chocolate bars are sold each day. THINK a The maximum number of chocolate bars that could

be sold is given by the domain of the function.

WRITE a The maximum number of chocolate bars

that could be sold is 10.

Chapter 12


613

b

1

Write the formula for finding the expected number of chocolate bars sold. Substitute for f (x) and the appropriate interval [0, 10] for a and b.

b µ = E( X ) =

b

∫a x × f ( x)ddx

10

x × x dx 50

10

x 2 dx 50

µ=

∫0

µ=

∫0

10

2

3

 x3  µ=   150  0

Antidifferentiate and substitute in the terminals. Evaluate.

State the expected value.

=

1000 −0 150

=

20 3

The expected number of chocolate bars sold each day is

c

1

Sketch the graph of the pdf curve. Shade the required section on the graph. State the probability that needs to be found.

c

20 . 3

y (10, 1_5 )

4

10

x

Pr(X < 4) = Pr(0 < X < 4) 2

b×h 2 The base of the triangle = 4. The height of the triangle is f (4). 2 4 f (4) = = 50 25 A=

As the shaded region forms a triangle, it is more efficient to find the area by a geometric method, rather than integrating.

2

A= = 3

State the final answer.

4 × 25 2 4 25

The probability that less than 4 chocolate 4

bars are sold is 25. d

1

Shade the required section on the graph. State the probability that needs to be found.

d

y (10, 1_5 )

8

10

Pr(X > 8) = Pr(8 < X < 10)

614


x

2

3

Pr(X > 8) = ∫

This time, we will integrate to find the area under the curve. State the integral that needs to be evaluated to find the probability. Note: Instead of integrating, the area of the trapezium could be found.

10

8

=  x

x dx 50 10

  100  8 2

= 100 − 64

State the final answer.

=

100 100 36 100

=

9 25

The probability that more than 8 9 chocolate bars are sold is 25.

WORKED EXAMPLE 11

The length of time (in minutes) between the successive incoming calls to a certain national helpline is a continuous random variable with a probability density function given by  0.6 e 0.6 x , x ≥ 0 f (x (x) =  .  0, elsewhere Find, correct to 4 decimal places, the probability that the next call will come: a within 30 seconds of the previous call b between 2 and 3 minutes after the previous call c more than 7 minutes after the last call. −

THINK a

1

WRITE

Sketch the graph of the pdf curve. If X represents time (in minutes) between the successive calls, then getting a call within 30 seconds from the previous one would mean that X must be between 0 and 0.5 (as 30 seconds = 0.5 minutes). Shade the required section on the graph. State the probability that needs to be found and the integral that needs to be evaluated in order to find that probability.

a

y (0, 0.6)

0

0.5 1

2

Pr(0 ≤ X ≤ 0.5) = ∫

0.5

3

0.6e

0

−

0.6 x

ddxx

0.5

 0.6e 0.6 x  = −  0.6  0  −

2

x

Evaluate by first antidifferentiating and then substituting terminals in for x. Give the answer correct to 4 decimal places as required.

=  − e

−

0.6 x 

0.5

0

= ( − e 0.6 × 0.5 ) − ( − e = − e 0.3 + e 0 = −0.7408 + 1 = 0.2592 −

−

0.6 × 0 )

−

b

1

X is anywhere between 2 and 3. Show the required section on the graph. State the probability that needs to be found and the integral that needs to be evaluated in order to find that probability.

b

y (0, 0.6)

0

1

2 3

6e Pr(2 ≤ X ≤ 3) = ∫ 0.6e 2

Chapter 12

−

3 0.6 x

x

ddxx


615

2

3

 0.6e 0.6 x  = −  0.6  2  −

Evaluate.

=  − e

−

0.6 x 

3

2

= ( − e 0.6 × 3 ) − ( − e 0.6 × 2 ) − − = − e 1.8 + e 1.2 = −0.1653 + 0.3012 = 0.1359 −

c

1

The probability that X will exceed 7 can be obtained by finding the area under the pdf curve to the right of 7. This is equivalent to finding the area to the left of 7 and subtracting it from the total area under the curve (that is, from 1).

c

−

y (0, 0.6)

x

0 1 2 3 4 5 6 7 8 9 10

Pr(X > 7) = 1 − Pr(X ≤ 7) 7

6e = 1 − ∫ 0.6e

−

0.6 x

0

2

ddxx 7

 0.6e 0.6 x  =1− −  0.6  0  −

Evaluate.

= 1 −  − e

−

0.6 x 

7

0

= 1 − [( − e 0.6 × 7 ) − ( − e = 1 − ( − e 4.2 + e 0 ) = 1 − (−0.0150 + 1) = 0.0150 −

−

−

0.6 .6 × 0 )]

1. Always sketch diagrams to represent the required probability. 2. Find probabilities by integrating the probability density function between the required values. EXERCISE

12D

Applications to problem solving 1

WE10 The time, in hours, that Kathryn spends watching television each day is a continuous random variable with probability density function given by 3

 (t 2 + 1), 0 ≤ t ≤ 2 f ( x ) =  14 0, elsewhere Calculate: a the maximum number of hours that Kathryn could watch television each day b the mean number of hours of television she watches each day, correct to 2 decimal places. c the probability that she watches more than 1.5 hours of television d the probability that she watches up to 1 hour of television each day, correct to 4 decimal places.

616


2 The lifespan (in days) of freshly cut roses is a random variable X with the probability density function given by  1 x − 1 , 2 ≤ x ≤ 4, 2  4 −1 f ( x ) =  x + 3 , 4 ≤ x ≤ 6, 2  4 0 , e lsewhere ree.  a What is the probability that a freshly cut rose will last for: i over 5 days? ii between 3 and 5 days? iii less than 4 days and 6 hours, correct to 3 decimal places? b Leo’s girlfriend Elly likes white roses. For their three-month anniversary Leo bought her a bunch. If the florist said that the freshly cut flowers were delivered to the shop 1.5 days ago, what is the probability that the roses will last for at least 3 more days? 3

WE11 The length (in minutes) of waiting time at the dentist is a continuous random variable

0 . 15 1 5e 0.15 x , x ≥ 0 with probability density function given by f (x) =  . Find, correct to er ere 0, elsewhere −

4 decimal places, the probability that: a a patient will be itted within 10 minutes of arrival b a patient will have to wait anywhere between 10 and 20 minutes from arrival c the waiting time will exceed half an hour. 4 The weight (in grams) of rainbow trout caught on a fish farm is a random variable, X, with ke kx , x ≥ 0 probability density function given by f (x) =  , where k = 10 9000 . 0, elsewhere −

a Find, correct to 4 decimal places, the probability that a randomly caught trout will weigh: i less than 800 g ii between 1 kg and 2 kg iii over 2 kg. b Calculate the mean weight of trout on this farm. Give your answer to the nearest gram. c What proportion of trout on this farm weighs below the mean weight? 5 The life (in hours) of a particular brand of batteries is a random variable with probability x  1  1000 , x ≥ 0 density function given by f (x) =  1000 e . 0, elsewhere −

a Find, correct to 4 decimal places, the probability that a randomly selected battery of this brand will last: i less than 100 hours ii no more than 180 hours iii more than 200 hours. b If after 180 hours of operation a battery is still working, what is the probability, correct to 4 decimal places, that it will last at least another 50 hours? c A manufacturer of the batteries claims that 90% of their batteries will work for at least n hours. Find the largest possible value of n. A group of students is given a maths quiz. The time (in minutes) taken by a student to complete the task is a continuous random variable with probability density function given by 0 . 1, 30 ≤ x ≤ 40 f (x) =  . er ere 0, elsewhere

Chapter 12


617

a State the time limits within which the quiz must be done. b Find the probability that a student will: i complete the task in under 33 minutes ii take between 36 and 38 minutes to finish. c Calculate the mean time taken to complete the quiz. d The fastest 10% of the class will receive a bonus question and thus will have an opportunity to earn extra marks. Find the longest time a student can take to finish the quiz and still qualify for a bonus question. 7 Rachel is buying silk fabric for her formal dress. The dressmaker estimates that she will need 3.9 metres of silk. To be on the safe side, Rachel asks a salesperson to cut off 4 metres from the selected roll. The difference (in centimetres) between the ordered length (4 metres) and the actual length of material cut by the salesperson is a continuous random variable with 0.1 probability density function given by f (x) = . π (0.001 + x 2 ) a Find the probability, correct to 4 decimal places, that the actual length of the material will be: i within 2 centimetres of the ordered length ii within 5 centimetres of the ordered length. b Find the probability that the actual length of the material is at least 4.07 metres, correct to 4 decimal places. c What is the probability that the purchased material will not be long enough to make a dress (that is, will be shorter than 3.9 metres)? Answer to 4 decimal places. 8 An Australian Chamber Orchestra concert is to be broadcasted live on Classic FM. It is scheduled to begin at 7 pm sharp. Although every effort is made to ensure the concert will start on time (due to the live broadcast), it may still start anywhere between 6.55 pm and 7.05 pm. The difference (in minutes) between the d starting time and the actual starting time is a continuous random variable with probability density function given by  π  π x cos   , − 5 ≤ x ≤ 5   20  f (x) =  20 2 . 0, elsewhere er ere  a What is the probability that the concert will start within 30 seconds of the scheduled time? Answer correct to 4 decimal places. b Maya is listening to the concert at home. She turns on her radio at 6.58 pm. What is the probability that Maya will miss the beginning, correct to 4 decimal places? c Patrons who arrive after the concert has started will not be itted until the interval. Lena and Alex are caught in a traffic jam and estimate that they will arrive at the concert hall at 7.03 pm. Assuming that the couple will indeed arrive at their estimated time, what is the probability that they will be itted, correct to 4 decimal places? 9 In a certain bank, the time (in minutes) the customer has to wait to be served is a random variable with a probability density function defined as kx, 0 ≤ x ≤ 10  f (x) =  − k ( x − 20)),, 10 < x ≤ 20, fo r k > 0. 0, else wh ere er  elsewh a Sketch the graph of the probability density function. b Hence, or otherwise, find the value of k. c Find the probability that a customer had to wait more than 10 minutes, if it is known that he was waiting for at least 7 minutes, correct to 4 decimal places. eBoo k plus eBook d Nathan parked his car in a 15-minute parking zone just outside the bank. He needs to deposit a cheque, which usually takes Digital doc 3 minutes. What is the probability that Nathan will return to his WorkSHEET 12.1 Work car on time (that is, within 15 minutes)?

618


12E

The normal distribution

eBoo k plus eBook

Interactivity The normal distribution is an important tool when dealing with the int-0257 probability distribution of a continuous random variable. The frequency The normal distribution curve of the normal distribution is characterised by the symmetrical bell shape called the normal distribution curve or normal curve. The normal curve fairly realistically models many observed frequency distributions such as heights and weights of infants, Mathematical Methods examination results, the intelligence quotient of children in a particular age group, the lengths of battery lives, the diameters of steel cans, etc. If X is a continuous random variable that follows a normal distribution with mean, µ, and variance, σ 2, it is written as X ∼ N( N µ, σ 2).

f x) f(

Properties of the normal distribution

1 σ 2π

1. The normal probability distribution is characterised by a bellshaped curve that is symmetrical about the mean. 2. The equation of the normal curve is given by the probability distribution function (pdf) f ( x) =

1

σ 2π

e

2  σ 

where x ∈ R. 3. From the graph, a maximum value of

1

σ 2π

µ Mean Mode Median

0

− 1 x − µ2

x

is obtained when x = µ.

4. For a normal distribution: (a) the mean, mode and median are the same (b) many of the frequencies are situated near the mean (c) the graph extends indefinitely to the right and left of the mean, but never touches or goes below the x-axis (d) the area between the normal curve and the x-axis is equal to 1 square unit, that is, ∞

(e) Pr(a < x < b) = ∫a f ( x ) dx . b

∫∞σ −

1 2π

−1 x−µ2

e

  2 σ 

=1

Effect of µ and σ on the graph of the normal distribution function The equation of the normal distribution function is given by: f ( x ) =

1

−1  x − µ  2

e

2  σ 

. σ 2π The standard deviation, σ, and the mean, µ, represent the transformations of the graph. The most basic form of the normal distribution function is when the mean is 0 and the standard −1 x 1 deviation is 1. In this case the equation would be: f ( x ) = e 2 . 2π The effect of the mean and the standard deviation on the normal distribution curve is: • dilation factor of σ1 from the x-axis • dilation factor of σ from the y-axis • translation of µ units in the positive x direction (for µ > 0). 2

Common probabilites associated with a normal distribution 1. Approximately 68% of the observations lie within one standard deviation of the mean. This may be written as Pr(µ − σ ≤ X ≤ µ + σ) ≈ 68%.

68%

µ −σ µ µ + σ

Chapter 12

x


619

2. Approximately 95% of the observations lie within two standard deviations of the mean. This may be written as Pr(µ − 2σ ≤ X ≤ µ + 2σ) ≈ 95%. We can also say that a value of x will most probably lie within two standard deviations of the mean. 3. Approximately 99.7% of the observations lie within three standard deviations of the mean. This may be written as Pr(µ − 3σ ≤ X ≤ µ + 3σ) ≈ 99.7%. We can also say that a value of x will almost certainly lie within three standard deviations of the mean.

95%

µ − 2σ

µ + 2σ x

µ

99.7%

µ − 3σ µ µ + 3σ The mean, µ, and the standard deviation, σ, are used when dealing with a population and are thus called population parameters. If these values are unknown then the sample mean, x, and sample standard deviation, s, are used. WORKED EXAMPLE 12

Draw a curve of the normal distribution, showing an appropriate scale, for µ = 25 and σ = 5. THINK

WRITE

1

Calculate µ ± σ.

2

Calculate µ ± 2σ.

3

Calculate µ ± 3σ.

4

Rule up a set of axes.

5

Label the axes, class intervals, frequency values and the µ, µ ± σ, µ ± 2σ, µ ± 3σ values.

6

Draw the normal distribution curve.

7

µ − σ = 25 − 5 = 20 µ + σ = 25 + 5 = 30 µ − 2σ = 25 − 2 × 5 = 15 µ + 2σ = 25 + 2 × 5 = 35 µ − 3σ = 25 − 3 × 5 = 10 µ + 3σ = 25 + 3 × 5 = 40 (25, 5 12π )

10 15 20 25 30 35 40

 1  to add the maximum point:  µ,  σ 2π 

WORKED EXAMPLE 13

Scores from a certain test are normally distributed with mean, µ = 84 and standard deviation σ = 4. Find the percentage of scores which are: a between 80 and 88 b between 72 and 96 c above 88 d above 96. THINK a

1

2

620

x

eBoo k plus eBook Tutorial


WRITE

As the values 80 and 88 are symmetrical about the mean, determine the difference between the mean and the extreme value. Divide the difference by the standard deviation. Note: This gives us the number of standard deviations by which the extreme value differs from the mean.


a

Difference = 88 − 84 =4 Difference 4 = σ 4 =1

x

b

3

Match the number of standard deviations with the appropriate percentage.

µ ± σ ≈ 68%

4


Approximately 68% of the observations lie between the scores of 80 and 88.

1

Determine the difference between the mean and the extreme value.

2

Divide the difference by the standard deviation.

b

Difference = 96 − 84 = 12 Difference 12 = σ 4 =3

c

3

Match the number of standard deviations with the appropriate percentage.

µ ± 3σ ≈ 99.7%

4


Approximately 99.7% of the observations lie between the scores of 72 and 96.

1

Relate the score with the corresponding mean and standard deviation difference. Use part a to assist.

2

Draw a diagram representing the situation.

c

µ + σ ≈ 88

16%

16% 68% 80

d

84

88

x

3

Relate the percentage with the corresponding mean and standard deviation difference.

µ ± σ ≈ 68%. Therefore 32% of the scores lie outside of the µ ± σ values.

4

Answer the question. Note: Since we are looking at values beyond the extreme, we halve the percentage.

16% of the scores will be greater than 88.

1

Relate the score with the corresponding mean and standard deviation difference. Use part b to assist.

2

Draw a diagram representing the situation.

d

µ + 3σ ≈ 96

0.15%

0.15% 99.7% 96 x

72 3

Relate the percentage with the corresponding mean and standard deviation difference.

µ ± 3σ ≈ 99.7%. Therefore 0.3% of the scores lie outside of the µ ± 3σ values.

4

Answer the question. Note: Since we are looking at values beyond the extreme, we halve the percentage.

0.15% of the scores will be greater than 96.

In parts a and b of the above worked example, each end point was symmetric about the mean. Hence, when determining the difference between the mean and extreme value, only one value was required.

Chapter 12


621

WORKED EXAMPLE 14

For a normally distributed variable with mean µ = 70 and standard deviation σ = 18, find: a the range between which 68% of the values lie b the range between which 95% of the values lie c the value which has 2.5% of all values below it. THINK a

b

c

WRITE a

µ ± σ ≈ 68%

1

Relate the given percentage with the appropriate µ and σ.

2

Calculate µ − σ.

µ − σ = 70 − 18 = 52

3

Calculate µ + σ.

µ + σ = 70 + 18 = 88

4


68% of the observations lie between 52 and 88.

1

Relate the given percentage with the appropriate µ and σ.

2


µ − 2σ = 70 − 2 × 18 = 34

3

Calculate µ + 2σ.

µ + 2σ = 70 + 2 × 18 = 106

4


95% of the observations lie between 34 and 106.

1

Draw the normal distribution curve and relate the given percentage with the appropriate mean and standard deviation difference.

b

c

µ ± 2σ ≈ 95%

2.5% 95% 34

2

2.5%

70

106

x

µ ± 2σ ≈ 95%. Therefore, 5% of the observations lie outside of this interval. Furthermore, 2.5% lie above the range and 2.5% lie below the range. 2.5% of the observed values will lie below 34.


1. If X is a continuous random variable that follows a normal distribution with mean, µ, and variance, σ2, it is written as X ∼ N(µ, σ2) 2. The equation of the normal curve is given by the probability distribution function (pdf) f ( x) =

1

−1 x − µ2

2  σ 

where x ∈ R. σ 2π 3. For a normal distribution, the mean, median and mode are the same. 1 4. The maximum value of the graph is when x = µ. σ 2π e

f x) f(

1 σ 2π

0

622

µ Mean Mode Median

x


5. The area between the normal curve and the x-axis is equal to 1 square unit, that is, ∞

∫ ∞σ −

1 2π

−1 x − µ2

e

2  σ 

ddxx = 1

b

6. Pr(a < x < b) = ∫a f ( x ) dx 7. The effect of the mean and the standard deviation on the normal distribution curve is: • dilation factor of σ1 from the x-axis • dilation factor of σ from the y-axis • translation of µ units in the positive x direction (for µ > 0). 8. The common probabilities associated with the normal distribution are: (a) Approximately 68% of the observations lie within one standard deviation of the mean. (b) Approximately 95% of the observations lie within two standard deviations of the mean, or we say that a value of x will most probably lie within two standard deviations of the mean. (c) Approximately 99.7% of the observations lie within three standard deviations of the mean, or we say that a value of x will almost certainly lie within three standard deviations of the mean. EXERCISE

12E

The normal distribution 1 WE12 Draw a curve of the following normal distributions, showing an appropriate scale for each. a µ = 10 and σ = 2 b µ = 20 and σ = 5 c µ = 0 and σ = 1 The three curves on the scale below show three normal distributions.


Spreadsheet 079

ii

Normal curves

i

0 a b c d

3

5

iii

10 15 20 25 30 35 40 45 50 55 60 x

State the mean for each curve. Estimate the standard deviation for each curve. Which two curves have the same mean? Which two curves have the same standard deviation? a Draw normal curves for the following on the one graph.

i µ = 60 and σ = 10 ii µ = 60 and σ = 20 iii µ = 80 and σ = 10 b How does an increase in σ affect the graph? c How does an increase in µ affect the graph? 4

Draw graphs of the following distributions. a X ∼ N(0, 1) b X ∼ N(20, 25) c X ∼ N(50, 100)

EXAM TIP Axes and a scale must be clearly shown as part of the sketch.

Chapter 12

[Authors' advice]


623

5 A large number of azalea bushes are planted, with heights following a normal distribution. Their heights are recorded after 3 months, 6 months and 9 months. These records are shown on the three curves below. a Which curve represents the i 3-month readings? ii b What can be said about iii the mean heights as time progresses? c What can be said about the x standard deviation as time progresses? − 1  x − 65    1 6 The pdf of a normal random variable is given by f ( x ) = e 2 5  . 5 2π a State the mean and the standard deviation of X. b Sketch the graph of the normal distribution. 2

a If a particular normally distributed variable has a mean of 20 and a standard deviation of 2, state what effect (in of transformations) the mean and standard deviation has on the graph of the distribution. b A normal distribution is given by X ~ N(9, 0.52). State what effect (in of transformations), the mean and standard deviation have on the graph of the distribution. 8 WE13 Scores on a certain test are normally distributed with mean µ = 65 and standard deviation 10. Find the percentage of scores which are: a between 55 and 75 b between 45 and 85 c between 35 and 95 d above 75 e below 45 f above 95. 9

The random variable, X, is normally distributed with mean µ = 30 and standard deviation σ = 5. Find the percentage of values which are: a above 35 b above 40 c above 45. If X ∼ N(42, 169), find the percentage of values which are: between 29 and 55 b between 16 and 68 d below 29 e above 55

c between 3 and 81 f below 3.

11 WE14 A normally distributed variable has µ = 40 and σ = 12. Find: a the range between which 68% of the values lie b the range between which 95% of the values lie c the range between which 99.7% of the values lie. A normally distributed variable has µ = 27.2 and σ = 1.4. Find:

12 a b c d e

a b c d e f 624

the range between which 68% of the values lie the range between which 95% of the values lie the range between which 99.7% of the values lie the value which has 16% of all values above it the value which has 2.5% of all values above it the value which has 0.15% of all values above it. A normally distributed variable has µ = 16.6 and σ = 0.6. Find: the range between which 68% of the values lie the range between which 95% of the values lie the range between which 99.7% of the values lie the value which has 16% of all values below it the value which has 2.5% of all values below it the value which has 0.15% of all values below it.


a A normally distributed variable has µ = 102.3 and σ = 21.4. Draw a normal curve marking the positions of µ, µ ± σ, µ ± 2σ, µ ± 3σ. Find the percentage of values between: 102.3 and 123.7 ii 59.5 and 102.3 iii 102.3 and 166.5 iv 80.9 and 145.1. A normally distributed variable has µ = 24 and σ = 3. Find the percentage of values between: a 18 and 27 b 30 and 33. 16

Weights of babies are normally distributed with a mean of 3 kg and a variance of 0.36 kg. Find the weight range over which: a 68% of babies’ weights lie b 95% of babies’ weights lie c 99.7% of babies’ weights lie.

17 MC Scores from a certain test are known to be normally distributed. Approximately 95% of the scores attained were between 36 and 60. The mean and standard deviation respectively are: A 48 and 4 B 48 and 6 C 48 and 12 D 50 and 5 E 60 and 12 18 MC Which of the following is not true for all normal distributions? A The median is equal to the mean. B 95% of the values lie within 2 standard deviations of the mean. C Pr(X > µ + σ) ≈ 0.16. D The mean is greater than the standard deviation. E 68% of the values lie within 1 standard deviation of the mean. 19 MC For the normal distribution below, the shaded area is approximately equal to 0.997.

19

Which of the following is true? A µ = 19, σ = 7 B µ = 19, σ = 40 D µ = 40, σ = 19 E µ = 40, σ = 21

61 x

C µ = 40, σ = 7

20 MC If X ∼ N(9, 9), then approximately 95% of the values would lie in the range: A −9 ≤ X ≤ 27 B 0 ≤ X ≤ 18 C 3 ≤ X ≤ 15 D 5.5 ≤ X ≤ 12.5 E 9 ≤ X ≤ 18 21 MC The heights of a class of 7-year-old girls are normally distributed with a mean of 100 cm and a standard deviation of 9 cm. In this class, 95% of the girls would be between approximately: A 73 cm and 127 cm B 82 cm and 118 cm C 91 cm and 109 cm D 94 cm and 106 cm E 97 cm and 103 cm 22 MC Which one of the following would represent a continuous random variable? A The number of people watching a netball match B The time taken to read this question C The number of 6s obtained in ten rolls of a die D The cost of a new car E The number of heads obtained in twenty tosses of a coin 23 If X is normally distributed with a mean of 12 and a variance of 2, find the range, correct to 2 decimal places, between which: a 68% of the values lie b 95% of the values lie c 99.7% of the values lie.

Chapter 12


625

12F

The standard normal distribution As seen earlier, the position and shape of the normal curve depend on the parameters µ and σ, respectively. Thus, the area below the normal curve and hence the probability for a given interval would vary for each different value of µ and σ. Integrating the equation for the normal distribution each time a value is required can become tedious. This problem can be quickly rectified by introducing a standard normal distribution where µ = 0 and σ = 1. The standard normal variable is denoted by the letter z in order to distinguish it from the normal variable, x. The equation of the normal distribution curve f ( x ) = 1

−1

−1  x − µ  2

1

σ 2π

e

2  σ 

when converted to the

x−µ where z = and g( z ) = σ f ( x ) . 2π σ The standard normal distribution is written as Z ∼ N(0, 12). The graph of the standard normal distribution is shown below. standard normal curve becomes g( z ) =

e

2

z2

y 1 2π

−3 −2 −1

0

1

2

3

z

To convert a normal distribution into a standard normal distribution, the mean, µ, is subtracted from the observed value, x, and the result is divided by the standard deviation, σ.

Calculating probabilities A CAS calculator can be used to calculate the probabilities associated with the normal distribution for any value of µ or σ, using normCdf (x1, x2, µ, σ). x1 and x2 are the lower and upper limits, respectively, of the interval for which you want to find the probability.

WORKED EXAMPLE 15

Calculate the value of the following probabilities, correct to 4 decimal places. a Pr(Z < 2) b Pr(Z ≥ −0.728) c Pr(−2.02 < Z < 1.59) THINK a

1

WRITE

Draw a diagram and shade the region required.

a

0

626


2

z

2

On a Calculator page, press: • MENU b • 5:Probability 5 • 5:Distributions 5 • 2:Normal Cdf 2 Enter the values as shown. Note: As we are dealing with a Z variable, µ = 0 and σ = 1. The lower limit is −∞ as there is no end point for the function.

Press ENTER ·.

b

3

Write the solution and round the answer to 4 decimal places.

1


Pr(Z < 2) = normCdf (−∞, 2, 0, 1) = 0.9772. b

−0.728 0 2

z

For Pr(Z ≥ −0.728), repeat as above on the CAS calculator, this time the lower limit is −0.728 and the upper limit is ∞. : Pr(X = a) = 0 ∴ Pr(X > a) = Pr(X ≥ a)

Chapter 12


627

c

3


1


Pr(Z ≥ −0.728) = normCdf (−0.728, ∞, 0, 1) = 0.7667. c

−2.02 2

For Pr(−2.02 < Z < 1.59), repeat as above on the CAS calculator. This time the lower limit is −2.02 and the upper limit is 1.59.

3


0 1.59

z

Pr(−2.02 < Z < 1.59) = normCdf (−2.02, 1.59, 0, 1) = 0.9224.

WORKED EXAMPLE 16

If X is normally distributed with µ = 50 and σ = 8, calculate, correct to 4 decimal places: a Pr(X > 55) b Pr(28 < X < 65) c Pr(X < 40 | X < 70). THINK a

1

WRITE


a

50 55 2

628

For Pr(X > 55) repeat as shown in worked example 15. Note: As we are dealing with an X variable, the mean and standard deviation have changed and are now 50 and 8, respectively.


x

Press ENTER ·.

b

3


1


Pr(X > 55) = normCdf (55, ∞, 50, 8) = 0.2660 b

28

c

2

For Pr(28 < X < 65), repeat as above on the CAS. This time the lower limit is 28 and the upper limit is 65. Write the solution and round the answer to 4 decimal places.

1

Write the rule for conditional probability. Note: Pr(X < 40 | X < 70) = Pr(X < 40). This is given by the overlapping region in the diagram below.

50

x

65

Pr(28 < X < 65) = normCdf (28, 65, 50, 8) = 0.9666.

c

Pr(( X < 40 | X < 70) =

Pr[( X < 40) ∩ ( X < 70)] Pr(( X < 70)

=

Pr(( X < 40) Pr(( X < 70)

Pr(( X < 40 | X < 70) =

Pr(( X < 40) Pr(( X < 70)

Pr (X < 40) Pr (X < 70) Region required

40 50 70 2

X

Find the individual probabilities of the fraction using the CAS, as seen previously in the example.

=

nor df ( −∞, 40, 50,8 normC , ) nor df ( −∞, 70, 50,8 normC , )

0.105650 = 0.993790

Chapter 12


629

3

Pr(( X < 40 | X < 70) =


=

Pr(( X < 40) Pr(( X < 70) nor df ( −∞, 40, 50,8 normC , ) − nor df ( ∞, 70, 50,8 normC , )

= 0.1063

Symmetry properties The symmetrical nature of the normal distribution curve can sometimes be used to work out probabilities.

0

z

z

−z

z

0

As an example, Pr(Z > z) is shown above on the left. Because of the symmetry of the curve, Pr(Z > z) = Pr(Z < −z) as shown in the graph on the right. Comparing the unshaded regions above reveals another property: Pr(Z > −z) = Pr(Z < z). As the area under the curve adds to 1, Pr(Z > z) + Pr(Z < z) = 1. Therefore, Pr(Z < z) = 1 − Pr(Z > z), as demonstrated by the graph below.

0

z

z

It is also important to be able to find the probability of z falling between two values, say a and b. A diagram is essential as it allows us to see the situation clearly and hence solve the problem. Consider the equation Pr(a < Z < b) = Pr(Z < b) − Pr(Z < a). The figure below clearly demonstrates the above situation. y

y

y

= a

0 b

x

− a

0 b

x

a

0

y Common region is cancelled out = a

0 b

x

y

= a

630

0 b


x

x

WORKED EXAMPLE 17 a If Pr(Z < p) = 0.30, calculate:

i Pr(Z > p) ii Pr(Z > −p) b If Pr(Z < b) = 0.8 and Pr(Z ≤ a) = 0.25, calculate: i Pr(a < Z < b) ii Pr(Z < a | Z < b) THINK a

i

1

WRITE

To calculate Pr(Z Z > p), draw a diagram and shade the region required.

a i

y

z

p0 2

3

ii

1 2

b

i

1

A property of normal distribution curves is: Pr(Z < p) + Pr(Z > p) = 1 : Pr(Z > p) = Pr(Z ≥ p) as Pr(Z = p) = 0 Write the answer. By symmetry: Pr(Z < p) = Pr(Z > −p) Write the answer. To calculate Pr(a < Z < b), draw a diagram and shade the region required.

∴ Pr(Z > p) = 1 − Pr(Z < p) = 1 − 0.30 = 0.70 Pr(Z Z > p) = 0.70 ii

Pr(Z > −p) = 0.30 b i

y

z

a0b 2

3

ii

1

Pr(a < Z < b) = Pr(Z < b) − Pr(Z < a). = 0.8 − 0.25 = 0.55 Pr(a < Z < b) = 0.55

State the probability you want to find and the rule you will use. Pr(a < Z < b) = Pr(Z < b) − Pr(Z < a). State the answer. State the probability you want to find and the rule you will use.

ii

Pr(Z < a | Z < b) = =

2

Substitute in the appropriate probabilities and simplify.

Pr(Z < a | Z < b) =

Pr(( Z < a) ∩ ( Z < b)) Pr(( Z < b) Pr(( Z < a) Pr(( Z < b) Pr(( Z < a) Pr(( Z < b)

0.225 0.8 25 = 80 5 = 16 5 Pr(Z < a | Z < b) = 16 =

3

Write the solution.

Chapter 12


631

WORKED EXAMPLE 18

X is a normal random variable with mean 16 and standard deviation 2, and Z is the standard normal variable. a Find m if Pr(X > 18) = Pr(Z > m). b Find m if Pr(X < 11) = Pr(Z > m). THINK a

b

WRITE a

x−µ σ 18 − 16 = 2 2 = 2 =1 ∴ Pr(X > 18) = Pr(Z > 1) m=1

1

As the probability of the X variable, Pr(X > 18), is in the same form as the probability of the Z variable, Pr(Z > m), m must directly relate to 18. To convert a normal variable to a standard normal variable, we use the x−µ rule z = . σ

2


1

This time the probability of the X variable, Pr(X < 11), is not in the same form as the probability of the Z variable, Pr(Z > m); however, we can x−µ . still link the two via the rule z = σ

2

Use the symmetry properties of the normal distribution to find m.

Pr(Z > −z) = Pr(Z < z). ∴ Pr(Z < −2.5) = Pr(Z > 2.5) ∴ Pr(X < 11) = Pr(Z > 2.5)

3


m = 2.5

b

z=

x−µ . σ 11 − 16 = 2 − 5 = 2 = −2.5 ∴ Pr(X < 11) = Pr(Z < −2.5) z=

WORKED EXAMPLE 19

The lengths of matches made at a certain factory are normally distributed with mean 4.1 cm and standard deviation 0.05 cm. Find the probability, correct to 4 decimal places, that a randomly selected match is: a greater than 4.1 cm b less than 4.13 cm c between 4.08 cm and 4.14 cm. THINK a

1

WRITE


a

4.1 2

632

µ = 4.1, ∴ Pr(X > 4.1) = Pr(X > µ)

Pr(X > 4.1) = 0.5000


x

b

1


b

x

4.14.13

c

2

Pr(X < 4.13) can be found using the CAS as shown in previous examples. Write the solution and round to 4 decimal places.

1


Pr(X < 4.13) = normCdf (−∞, 4.13, 4.1, 0.05) = 0.7257

c

x

4.08 4.14.14

2

Pr(4.08 < X < 4.14) can be found using the CAS as shown in previous examples. Write the solution and round to 4 decimal places.

Pr(4.08 < X < 4.14) = normCdf (4.08, 4.14, 4.1, 0.05) = 0.4436

1. A CAS calculator can be used to calculate the probabilities associated with the normal distribution for any value of µ or σ using normCdf (x1, x2, µ, σ). x1 and x2 are the lower and upper limits, respectively, of the interval for which you want to find the probability. 2. Using the symmetrical nature of the standard normal curve, it can be seen that: Pr(Z > z) = 1 − Pr(Z < z) Pr(Z > −z) = Pr(Z < z). 3. To convert a given normal variable, x, to the standard normal variable, z, use x−µ the rule z = . σ 4. With all normal distribution problems, > is equivalent to ≥ and < is equivalent to ≤, since Pr(Z = z) = 0. EXERCISE

12F

The standard normal distribution 1 WE15 Find, correct to 4 decimal places: a Pr(Z ≤ 1) b Pr(Z ≥ 1.5) d Pr(Z > −2.71) e Pr(0.42 < Z < 1.513) 2

c Pr(Z < −1.75) f Pr(−1.6 ≤ Z ≤ 1.4)

Standardise the following X X-values to Z-values: Z 40 if X ∼ N(25, 25) b X = 12 if X ∼ N(17, 9) 15 if X ∼ N(12, 6.25)

Chapter 12


633

3 The variable X is normally distributed with mean µ = 9 and standard deviation σ = 3. Standardise the following X-values: X a X = 10 b X = 7.5 c X = 12.4 4 WE16 If X is normally distributed with µ = 40 and σ = 7, find, correct to 4 decimal places: a Pr(X > 42) b Pr(X ≥ 30) c Pr(X < 45) d Pr(X ≤ 27) e Pr(25 ≤ X ≤ 45) 5 If X ∼ N(20, 25), find, correct to 4 decimal places: a Pr(X ( > 27) (X b Pr(X ( ≥ 18) (X d Pr(7 ≤ X ≤ 12) e Pr(X ( < 17 | X ≤ 25) (X 6 WE17a a Pr(

If Pr(Z Z > m) = 0.25, calculate: b Pr(Z Z > −m)

c Pr(X ( ≤ 8) (X f Pr(X ( < 17 | X < µ) (X c Pr(Z Z < −m)

7 WE17b If Pr(Z Z ≤ b) = 0.62 and Pr(Z Z < a) = 0.16, calculate: a Pr(a < Z < b) b Pr(Z Z < a | Z < b) If Pr(Z Z ≤ b) = 0.58 and Pr(a < Z < b) = 0.44, find Pr(Z < a). b) = 0.12 and Pr(a < Z < b) = 0.60, find Pr(Z > a). X is a normal random variable with mean 36 and standard deviation 4, and Z is the standard normal variable. a Find m if Pr(X X < 30) = Pr(Z Z < m). b Find m if Pr(X X > 43) = Pr(Z Z < m). 11 WE19 Light bulbs have a mean life of 125 hours and a standard deviation of 11 hours. Find the probability, correct to 4 decimal places, that a randomly selected light bulb lasts: a longer than 140 hours b less than 100 hours c between 100 and 140 hours. 12 The heights jumped by Year 9 high-jump contestants follow a normal distribution with a mean jump height of 152 cm and a variance of 49 cm. Find the probability, correct to 4 decimal places, that a competitor jumps: a at least 159 cm b less than 150 cm c between 145 cm and 159 cm d between 140 cm and 160 cm e between 145 cm and 150 cm, given that she jumped over 140 cm. 13 MC If Z has a standard normal distribution, then Pr(Z > 1.251) is: A 0.1054 B 0.3945 C 0.6055 D 0.6623 14 MC If Z ∼ N(0, 1), then Pr(Z < A 0.0987 B 0.4013

E 0.8945

−0.25)

is: C 0.5987

D 0.7124

E 0.9013

15 MC The variable X is normally distributed with mean µ = 20 and standard deviation σ = 6. The standardised value of Z Z, for X = 29 is: A −1.5 B −1.15 C 0.483 D 1.15 E 1.5 16 MC Tennis balls are dropped from a height of 2 metres. The rebound height of the balls is normally distributed with a mean of 1.4 metres and a standard deviation of 0.1 metres. The probability that a ball rebounds more than 1.25 metres is: A 0.0668 B 0.2826 C 0.4332 D 0.7174 E 0.9332 17 MC The variable X is normally distributed with mean 70 and variance 12. The probability that X is greater than 77 is: A 0.0217 B 0.0429 C 0.0909 D 0.9091 E 0.9783 634


18 MC The life span of dogs is normally distributed with a mean of 12 years and a standard deviation of 2 years. The probability that a dog lives for less than 9 years is: A 0.0668 B 0.2826 C 0.2926 D 0.4332 E 0.9332 19 MC If X ∼ N(16, 4), then Pr( Pr(X > 11.5) equals: A 0.0122 B 0.0836 C 0.7217

D 0.9164

E 0.9878

20 The volume of milk in a 1-litre carton is normally distributed with a mean of 1.000 litres and a standard deviation of 0.006 litres. A randomly selected carton is known to have more than 1.004 litres. Find the probability, correct to 4 decimal places, that it has less than 1.011 litres. 21 Eye fillet steaks are cut with a mean weight of 82 grams and a standard deviation of 5 grams. Steaks are sold at different prices according to their weights, as shown in the table below. Weight (g)

< 70

70–80

80–90

> 90

Cost ($)

1.40

1.60

1.80

2.00

a Find the probability that a randomly selected steak weighs between 80 grams and 90 grams. Give your answer correct to 4 decimal places. b Find the probability that a randomly selected steak costs $2, correct to 4 decimal places. c Copy and complete the following table. Cost ($)

1.40

1.60

1.80

2.00

Probability d Using your answers from part c, find the average price of an eye fillet steak. 22 The length of 6-cm nails is actually normally distributed with a mean length of 6 cm and a standard deviation of 0.03 cm. Only nails which are between 5.93 cm and 6.07 cm are acceptable and packaged accordingly. Find: a the probability of a randomly selected nail being an acceptable length, correct to 4 decimal places b the expected number of acceptable nails in a batch of 1000. 23 The length of fish caught in a certain river follows a normal distribution, with mean 32 cm and standard deviation 4 cm. Fish which are less than 27 cm are considered to be undersized and must be returned to the river. Find: a the probability that a fish is undersize, correct to 4 decimal places b the expected number of fish that a fisherman could take home if he catches 20 fish in one afternoon and follows the rules for undersize fish. 24 The heights of students in Year 12 class were found to be normally distributed with mean of 171.2 cm and standard deviation of 5.2 cm. a What is the probability, correct to 4 decimal places, that a student has a height of: i greater than 168 cm? ii between 164 cm and 174 cm? iii greater than 179 cm, given that they are taller than 168 cm? b If a random sample of 10 students was chosen, what is the probability, correct to 4 decimal places, that 6 of the students would have a height greater than 168 cm? 25 Amy is a keen tennis player. The amount of time, in hours, she spends training each week is normally distributed with a mean of 8.4 hours and standard deviation of 1.8 hours. a Find, correct to 4 decimal places, the probability that: i she spends more than 10 hours training per week ii she spends more than 5 hours training per week, given that she spent less than 10.5 hours training per week. b What is the probability, correct to 4 decimal places, that of the next 5 weeks, Amy will spend at least 1 of them training more than 10 hours a week?

Chapter 12


635

12G

The inverse cumulative normal distribution Up to now, we have focused on finding probabilities associated with either the normal distribution or standard normal distribution. Sometimes, however, we are given the probability and must determine the value of the variable. A CAS calculator, via the function invNorm, can be used to calculate the value associated with a given probability for any value of µ or σ, by invNorm( , µ, σ). A must be the area to the left of the required value, that is, the ‘less-than’ invNorm(A probability (e.g. Pr(X X < c) = 0.47). In this instance A would be 0.47.

WORKED EXAMPLE 20

eBoo k plus eBook

Find the value of c, correct to 3 decimal places, in the following. a Pr(Z < c) = 0.57 b Pr(Z ≥ c) = 0.91 THINK a

1

Tutorial


WRITE

To calculate Pr(Z < c) = 0.57, draw a diagram and shade the region required.

a 57%

0 c 2

As the shaded area is to the left of the unknown value, we can use invNorm straight away. On a Calculator page, press: • MENU b • 5:Probability 5 • 5:Distributions 5 • 3:Inverse Normal 3 Enter the values as shown. Note: As we are dealing with a Z variable, µ = 0 and σ = 1. Press ENTER ·.

3

636

Write the solution and round to 3 decimal places.


c = invNorm(0.57, 0, 1) = 0.176

z

b

1

To calculate Pr(Z ≥ c) = 0.91, draw a diagram and shade the region required.

b 91%

c 2

For this example, the shaded area is to the right of the unknown value, so we subtract the given area from 1.

3

Find the value of c using the CAS as shown above.

4


z

0

Pr(Z < c) = 1 − Pr(Z > c) = 1 − 0.91 = 0.09

c = invNorm(0.09, 0, 1) = −1.341

WORKED EXAMPLE 21

Find the value of c in Pr(−c < Z < c) = 0.9544. Give your answer correct to 3 decimal places. THINK 1

Draw a diagram of the situation. Label the unknown values as c1 and c2. Note: c1 = −c2.

WRITE

0.0228

0.0228 0.9544

c1

0

c2

z

2

Determine the probability of each of the unshaded areas by subtracting the given probability from 1 and dividing the result by 2.

3

Find Pr(Z Z < c2) by adding the probabilities.

Therefore Pr(Z < c2) = 0.9544 + 0.0228 = 0.9772

4

Find the value of c2 using the CAS as shown previously.

c2 = invNorm(0.9772, 0, 1) = 1.999 ∴Pr(−1.999 < Z < 1.999) = 0.9554

5


c = 1.999

1 − 0. 0 9544 2 0.0456 = 2 = 0.0228

Unshaded area =

Chapter 12


637

Percentiles and quantiles that a percentile is a probability value expressed as a percentage while a quantile is a probability value expressed as a decimal; for example, Pr(Z < c) = 0.7 means that c is the 70th percentile or the 0.70 quantile. WORKED EXAMPLE 22

If Z ∼ N(0, 12), find, correct to 3 decimal places: a the 0.30 quantile b the 80th percentile. THINK a

1

WRITE a

Draw a diagram illustrating the information.

0.30

c

b

z

0

2

Find the value of c using the CAS as shown previously. Write the solution.

c = invNorm(0.30, 0, 1) = −0.5244

3

Answer the question, correct to 3 decimal places.

c = −0.524

1

Draw a diagram of the situation.

b The 80th percentile is the 0.80 quantile.

0.80

0

c

2

Find the value of c using the CAS as shown previously. Write the solution.

c = invNorm(0.80, 0, 1) = 0.84162

3

Answer the question, correct to 3 decimal places.

c = 0.842

z

WORKED EXAMPLE 23

X is normally distributed with a mean of 10 and a standard deviation of 2. Calculate x1, correct to 3 decimal places, if: a Pr(X ≤ x1) = 0.65 b Pr(X > x1) = 0.85. THINK a

1

WRITE


a

0.65

10 x1

638


x

2

As the shaded area is to the left of the unknown value, we can use invNorm straight away. Note: This time we are dealing with an X variable, so the mean and standard deviation have changed from 0 and 1, respectively. From the information, µ = 10 and σ = 2. Enter the values as shown.

Press ENTER ·.

b

3


1


x1 = invNorm(0.65, 10, 2) = 10.771 b

0.85

x1

x

10

2

For this example, the shaded area is to the right of the unknown value, so we subtract the given area from 1.

Pr(Z < x1) = 1 − Pr(Z > x1) = 1 – 0.85 = 0.15

3

Find the value of c using the CAS as shown previously. Write the solution and round to 3 decimal places.

x1 = invNorm(0.15, 10, 2) = 7.927

WORKED EXAMPLE 24

If X is normally distributed with a mean of 52.3 and Pr(X < 48) = 0.229, calculate the standard deviation, correct to 2 decimal places.

Chapter 12


639

THINK

WRITE

1

We need to find the standardised Z Z-value for the X-value of 48. As we know µ and σ for Z, we can use X the invNorm function to find the standardised value. Note: Even though z is rounded to 3 decimal places on the page, be sure to use the maximum number of decimal places in your subsequent working.

Pr(X < 48) = Pr(Z < z) = 0.229 z = invNorm(0.229, 0, 1) = −0.742

2

Now we link the z-value to the x-value by x−µ . the rule z = σ

z=

−0.742


The standard deviation is 5.79.

3

x−µ σ

48 − 52.3 σ −0.742 = −4.3 σ − 4.3 σ= − 0.742 = 5.794 =

1. A CAS calculator can be used to calculate the value associated with a given probability for any value of µ or σ, by invNorm( invNorm(A, µ, σ). A must be the area to the left of the required value. 2. A percentile is a probability value expressed as a percentage while a quantile is a probability value expressed as a decimal. 3. When the standard deviation or mean is unknown, the standardised value, z, must be x−µ found. The mean or standard deviation is then calculated using the rule z = . σ EXERCISE

12G

The inverse cumulative normal distribution 1 WE20 Find the value of c in the following, correct to 3 decimal places. a Pr(Z < c) = 0.9 b Pr(Z < c) = 0.2 c Pr(Z > c) = 0.54 d Pr(Z ≥ c) = 0.45 2 WE21 Find the value of c in the following. a Pr(−c < Z < c) = 0.6826 c Pr(−c < Z < c) = 0.2

b Pr(−c ≤ Z ≤ c) = 0.5 d Pr(−c ≤ Z ≤ c) = 0.38

3 WE22 If Z ∼ N(0, 1), find: a the 0.25 quantile c the 0.72 quantile

b the 40th percentile d the 0.995 quantile.

4 WE23 X is normally distributed with a mean of 10 and a standard deviation of 2. Find x1 if: a Pr(X ( ≤ x1) = 0.72 (X b Pr(X ( < x1) = 0.4 (X c Pr(X ( > x1) = 0.63 (X d Pr(X ( ≥ x1) = 0.2. (X 5 X is normally distributed with a mean of 34 and a standard deviation of 16. Find c if: a Pr(X ( > c) = 0.31 (X b Pr(X ( ≤ c) = 0.75 (X c Pr(X ( < c) = 0.21 (X d Pr(X ( ≥ c) = 0.55. (X 640


6 Let X ∼ N(22, 25). Find k if: a Pr(22 − k ≤ X ≤ 22 + k) = 0.7 b Pr(22 − k < X < 22 + k) = 0.24 c Pr(X ( < k | X < 23) = 0.32. (X 7 MC If Pr(Z ≤ c) = 0.8, then c equals: A −0.842 B −0.253 C 0.253

D 0.524

E 0.842

8 MC If Pr(Z > c) = 0.7, then c equals: A −0.524 B −0.496 C 0.496

D 0.524

E 0.553

< Z < k) = 0.4, then k equals: B −0.253 C 0.038

D 0.253

E 0.885

10 MC If Z ∼ N(0, 1), then the 0.35 quantile is: A −0.675 B −0.385 C 0.350

D 0.385

E 0.675

11 MC If Pr(Z < k | Z < 0.5) = 0.6, then k equals: A −0.215 B −0.253 C 0

D 0.253

E 0.215

Pr(−1.2

9 MC If A −0.885

12 MC If X is normally distributed with a mean of 20 and a standard deviation of 4 and Pr( < k) = 0.6, then k equals: Pr(X A 18.99 B 19.49 C 20.51 D 21.01 E 21.79 13 MC If X ∼ N(12, 4) and Pr( Pr(X > k) = 0.82, then k equals: A 8.34 B 10.17 C 13.83 D 14.27

E 15.66

14 The height of Year 9 students is known to be normally distributed with a mean of 160 cm and a standard deviation of 8 cm. Answer the following; correct to 2 decimal places. a How tall is Theo if he is taller than 95% of Year 9 students? b How tall is Luisa if she is shorter than 80% of Year 9 students? 15 The length of nails manufactured as 45 mm are actually normally distributed with a mean of 45 mm and a standard deviation of 0.5 mm. The shortest 20% and longest 20% of nails manufactured are discarded before packaging and sold as seconds. Find correct to 2 decimal places: a the minimum length of packaged nails b the maximum length of packaged nails. 16 At a qualifying meeting, the time taken for runners to complete 400 metres follows a normal distribution, with a mean time of 50 seconds and a standard deviation of 2 seconds. If the fastest 25% of runners qualify for the next meeting, how fast would you need to run to qualify? Give your answer correct to 2 decimal places. 17 WE24 If X ∼ N(20, σ 2) and Pr( Pr(X ≥ 19) = 0.7, find the standard deviation, σ. 18 X is normally distributed with a mean of 50 and a standard deviation of σ. Forty per cent of X-values are less than 48. Find σ. X 19 Weights of packaged rice are normally distributed with a mean of 500 grams. Ten per cent of packages are under 485 grams. Find the standard deviation. 20 If X ∼ N(µ, 16) and Pr( Pr(X X ≥ 17) = 0.99, find the mean, µ. 21 X is normally distributed with a mean of µ and a standard deviation of 3. If 35% of X-values are at least 27, find the mean. 22 The time taken for grade 4 students to complete a small jigsaw puzzle follows a normal distribution with a standard deviation of 30 seconds. If 70% of grade 4 students complete the puzzle in 4 minutes or less, find the mean completion time for grade 4 students correct to 2 decimal places. 23 Bridget owns a fruit and vegetable shop. Being a keen mathematician as well, she knows that the weights of the watermelon she sells are normally distributed with a mean weight of 1.5 kg

Chapter 12


641

and a standard deviation of 0.2 kg. She classifies her watermelons as either large, medium or small. The heaviest 15% are classified as large, the lightest 15% are classified as small and the rest as medium. Find the range of weights, correct to 1 decimal place, for which: a a large watermelon is classified b a medium watermelon is classified. eBoo k plus eBook Digital doc

WorkSHEET 12.2

24 Mr Lim, a physics teacher, sets a particularly hard test for his students. He finds that the average mark is 54 and the standard deviation is 8. He decides to award the top 10% an A, and fail the bottom 10%. Find, correct to the nearest whole number: eBoo k plus eBook a the lowest mark required to achieve an A b the range of marks that a student who fails the test could have achieved. Digital doc Investigation Sunflower stems

642


SUMMARY Continuous random variables

• Continuous random variables represent quantities that can be measured and thus may assume any value in a given range. Probability density functions

• If X is a continuous random variable, then the probability of it falling between certain values is given by the area under a frequency curve known as the probability density function or pdf. • The probability density function must be greater than or equal to zero for all values of x, and the total area ∞

under the curve must be 1. That is, f( f x) ≥ 0 for all x and ∫ − f ( x )dx = 1. ∞

• The domain of a probability density function is usually R, that is, the variable is continuous and can assume any real value (at least in theory). • If we need our function to be non-zero over a particular interval only, say, over (a, b), then we need to specify that the function is equal to zero everywhere else. • If a probability density function is only non-zero over a particular interval (a, b), the area under the curve b

between x = a and x = b must be 1. That is, ∫a f ( x )dx = 1. Finding probabilities using a probability density function y

• If X is a continuous random variable, then the probability of it falling between x = a and x = b is given by the area bounded by the probability density function, the x-axis and the lines x = a and x = b. That is, Pr (a ≤ X ≤ b) =

Pr(a < – b) – x<

b

∫a f ( x)dx .

• Pr(x = a) = 0 ∴ Pr(x > a) = Pr(x ≥ a) • To integrate over intervals involving infinity, the limits are evaluated as follows. Interval

0

a

b

x

Limit lim

(−∞, ∞) (−∞, a)

k

∫ k f ( x) dx

k →∞

−

lim

∫

lim

∫

a

k → −∞ k

(a, ∞)

k

k →∞ a

f ( x ) dx f ( x ) dx

1 1 = 0 and lim x = 0. x→∞ x e • Quantiles and percentiles both define the value of the random variable, X, below which a given proportion of the distribution falls. • A quantile is a probability value expressed as a decimal, while a percentile is a probability value expressed as a percentage. • As a number tends to infinity, its reciprocal tends to zero. For example, lim

x→∞

Measures of central tendency

• If X is a continuous random variable with a probability distribution function f (x) over the domain R, the mean of X can be found using the formula

µ = E (x) =

∞

∫ ∞xf ( x)ddx . −

Chapter 12


643

• In general, E[g(x)] =

b

∫a g( x) f ( x)dx.

• If f (x) has a certain rule for a ≤ X ≤ b and is zero elsewhere, then the mean can be obtained by b

using the formula µ = E(x) = ∫ xf ( x )ddx . a 1 • The median value, m, of the continuous random variable, X, is a value such that Pr(X ≤ m) = 2 . • To find the median of X, solve for m the equation

m

∫a

1

f ( x )dx = 2 .

• The mode is a value of X for which f (x) has its maximum. It is possible to have more than one mode. • To find the mode of a continuous random variable: – sketch the graph of the probability density function first – if the probability density function of X has a maximum turning point in the interval [a, b], the mode is given by the x-coordinate of the turning point. It can be obtained by finding the derivative of f (x), making it equal to zero, and solving for x. – if the probability density function continuously increases or decreases over the interval [a, b], or if it has a minimum turning point, the mode is then given by the end point of the interval (a or b) that corresponds to the maximum value of f (x). Measures of spread

• If X is a continuous random variable with a probability distribution function f( f x) over the domain R, the variance of X can be found using the formula ∞

Var(X) X = E(x − µ)2 = ∫ ( x − µ )2 f ( x )ddxx . X) −

∞

• If f (x) has a certain rule for a ≤ X ≤ b and is zero elsewhere, then the variance can be obtained by using the formula Var(X) X = E(x − µ)2 = X)

b

∫a ( x − µ)2 f ( x)ddxx .

• The alternative formula for variance is Var(X) X = E(X X2) − (µ)2. This can also be expressed as 2

 b  ∫a ( x)2 f ( x)ddxx −  ∫a xf ( x)ddx . • The standard deviation of X can be found by taking the positive square root of the variance of Var(X) X = X)

b

Va X ) . X, that is, SD(X) X = Var( X) • The interquartile range (IQR) is the middle 50% of the distribution. IQR = 75th percentile – 25th percentile. Applications to problem solving

• Always sketch diagrams to represent the required probability. • Find probabilities by integrating the probability density function between the required values. This can be done manually or by using a CAS calculator. The normal distribution

• The normal distribution is the most important distribution associated with continuous random variables. • The normal probability distribution is characterised by a bell-shaped curve which is symmetrical about the mean. • The equation of the normal curve is given by the probability distribution function f ( x ) = 1

when x = µ. σ 2π • For the normal distribution, the same rules apply as for all other probability density functions: 1. f (x) ≥ 0 for all x, where x ∈ R. It has a maximum value of

2.

σ 2π

e

x − µ 2  σ 

2

f x) f(

∞

∫ ∞ f ( x)dx = 1, and −

3. Pr(a < X < b) =

644

−1 

1

b

∫a

0

f ( x )dx .


µ a b

x

• The normal distribution with mean, µ, and variance, σ 2, is written as X ∼ N(µ N( , σ 2). • The common probabilities associated with the normal distribution are as follows: 1. Approximately 68% of the observations lie within one standard deviation of the mean. 68% x

µ −σ µ µ + σ

2. Approximately 95% of the observations lie within two standard deviations of the mean, or we say that a value of x will most probably lie within two standard deviations of the mean.

95%

µ − 2σ

3. Approximately 99.7% of the observations lie within three standard deviations of the mean, or we say that a value of x will almost certainly lie within three standard deviations of the mean.

µ + 2σ x

µ

99.7%

• The parameters µ and σ have an effect on the normal distribution µ − 3σ µ µ + 3σ x curve as follows: The mean, µ, controls the position of the curve. The standard deviation, σ, controls the width of the curve. Specifically, the effect of the mean and the standard deviation on the normal distribution curve is: 1. dilation factor of σ1 from the x-axis 2. dilation factor of σ from the y-axis 3. translation of µ units in the positive x direction (for µ > 0).

The standard normal distribution

• The standard normal distribution is written as Z ∼ N(0, 12). x−µ • To convert a given normal variable, x, to the standard normal variable, z, use the rule z = . σ • A CAS calculator can be used to calculate the probabilities associated with the normal distribution for any value of µ or σ using normCdf(x1, x2, µ, σ ). x1 and x2 are the lower and upper limits, respectively, of the interval for which you want to find the probability. • Using the symmetrical nature of the standard normal curve, it can be seen that: Pr(Z > z) = 1 − Pr(Z < z) 0

z

z

0

z

z

• Pr(Z > −z) = Pr(Z < z)

−z

0

z

• With all normal distribution problems, > is equivalent to ≥ and < is equivalent to ≤, since Pr(Z = z) = 0. The inverse cumulative normal distribution

• A CAS calculator can be used to calculate the value associated with a given probability for any value of µ or σ, using invNorm( invNorm(A, µ, σ). A must be the area to the left of the required value. • A percentile is a probability value expressed as a percentage while a quantile is a probability value expressed as a decimal. • When the standard deviation or mean is unknown, the standardised value, z, must be found. The mean or x−µ . standard deviation is then calculated using the rule z = σ Chapter 12


645

CHAPTER REVIEW SHORT ANSWER

2 ( 2 − 4 x ) , 0 ≤ x ≤ 1 2. 1 a Sketch the graph of f (x ( )= er ere 0, elsewhere b State whether or not f (x) could be a probability density function.  x , 0 ≤ x ≤ 4 2 a Sketch the graph of f (x ( )=  er ere 0, elsewhere and explain why it is not a probability density function. b Adjust the function so that it could be a probability density function by adding a dilation coefficient, that is, by changing the rule to f (x) = a x and retaining the given domain. c Now adjust the left-hand-side end point of the domain so that f (x) could be a probability density function. That is, keep the rule unchanged, but replace 0 with an appropriate number in 0 ≤ x ≤ 4. 3 If X is a random variable with a probability density sin (2 x ), 0 < x < π2 function given by f (x ( )= . er ere 0, elsewhere Calculate: π

a Pr(X < 6 )

π

b Pr(X < π X < 3 ). 6

4 Consider the random variable, X, with probability density function given by  x − 1, 1≤ x ≤ 2 − f (x) =  x + 3, 2 < x ≤ 3.  0, elsewhere  Calculate: a Pr(1.5 < X < 2) b Pr(X X < 2.5) c the mode d the median. 5 If X is a random variable with a probability density function given by kx, 0 < x < 4 f (x ( ) = . er ere 0, elsewhere

646

Calculate: a the value of k c the median

b the mean d the mode.

6 The cooking time (in minutes) of a single portion of rice in a particular microwave is a random variable, X, with a probability density function given by 1  , 4 ≤ x ≤ 6 f x) =  2 f( . er ere 0, elsewhere

a Lena’s cookbook recommends to cook one portion of rice for 5.5 minutes. If Lena follows the instructions in her cookbook, what is the probability that the rice will not be ready? b What is the probability that it will take anywhere between 4 minutes and 36 seconds and 5 minutes and 15 seconds for the rice to cook? c If Lena wants to be 90% sure that her rice will be ready, what is the minimum cooking time she has to set on the microwave? 7 Draw a curve of the following normal distributions, showing an appropriate scale for each. a µ = 8 and σ = 1 b µ = 25 and σ = 8 c µ = 0 and σ = 3 8 Draw graphs of the following distributions. a X ∼ N(30, 64) b X ∼ N(8, 4) c X ∼ N(100, 100) 9 A normal distribution is given by X ~ N(22, 9). State what effect (in of transformations) the mean and standard deviation have on the graph of the distribution. 10 Weights of wooden sleepers are normally distributed with mean µ = 60 kg and standard deviation σ = 9 kg. Find the percentage of sleepers that are: a between 51 kg and 69 kg b between 42 kg and 78 kg c between 33 kg and 87 kg. 11 A normally distributed variable has µ = 19.6 and σ = 3.1. Find: a the range between which 68% of the values lie b the range between which 95% of the values lie


c d e f

the range between which 99.7% of the values lie the value which has 16% of all values above it the value which has 2.5% of all values above it the value which has 0.15% of all values above it.

12 Standardise the following X X-values to Z-values. Z a X = 26 if X ∼ N(22, 36) b X = 18 if X ∼ N(16, 16) c X = 56 if X ∼ N(50.9, 100)

14 a If Pr(Z ≤ b) = 0.81 and Pr(a < Z < b) = 0.44, find Pr(Z < a). b If Pr(Z Z > b) = 0.36 and Pr(a < Z < b) = 0.24, find Pr(Z < a). 15 X is a normal random variable with mean 27 and standard deviation 3, and Z is the standard normal variable. a Find m if Pr(X X < 33) = Pr(Z Z < m). b Find m if Pr(X X > 22.5) = Pr(Z Z < m). 16 Let X be a normally distributed random variable with a mean of 72 and a standard deviation of 8. Let Z be the standard normal random variable. Use the result that Pr(Z Z < 1) = 0.84, correct to 2 decimal places, to find: a the probability that X is greater than 80 b the probability that 64 < X < 72 c the probability that X < 64 given that X < 72. EXAM TIP Some students did not clearly understand the relationship between the random variables X and Z. In part c, 47% of students scored 0/2. Most realised that conditional probability was required but were not able to apply it correctly. [Assessment report 1 2006]

[© VCAA 2006]

MULTIPLE CHOICE

1 Which of the following could be a probability density function?

 − 2, − 1 ≤ x ≤ 0 2 ii f (x) =  0, elsewhere 2 x, −1 ≤ x ≤ 2 iii f (x) =  0, elsewhere

B II only E I, II and III

C III only

2 X is a random variable with probability density 2 sinn ((4 x ), a ≤ x ≤ b function defined as f (x) =  . er ere 0, elsewhere The values of a and b respectively are: π 4 π 3π , 2 4

A 0, D

13 If Pr(Z Z > p) = 0.72, calculate: a Pr(Z Z < p) b Pr(Z Z > −p).

1 3, 0 ≤ x ≤ 3 i f (x) =  er ere 0, elsewhere

A I only D I and III

−

B

π π , 4 2

π 2

C 0,

E 0, π

a  , 1≤ x ≤ 2 3 The value of a such that f (x ( ) =  x2 0, elsewhere er ere  could be a probability density function is: 1 2

B 1

D 3

E 4

A

C 2

4 If X is a random variable with a probability density 2  (4 − x )),, 1 ≤ x ≤ 2 function defined by f (x) =  5 , er ere 0, elsewhere then Pr(X X > 1.2) can be obtained by evaluating: A

1.2

∫1

C 1− E

2 (4 − 5

1.2 2 (4 − 5 1

∫

2 2 (4 − 5 1

∫

B

x )ddxx

1.2 2 (4 − 5 2

∫

D 1−∫

x )ddxx

2

2 (4 − 1.2 5

x )ddxx

x )ddxx

x )ddxx

5 If X is a random variable with a probability density  1 , 0 ≤ x ≤ 4 function defined by f( f x) =  4 and er ere 0, elsewhere Pr (X X < a) = 0.625, the value of a is: A 1.5 B 2 C 2.5 D 3 E 3.5 Questions 6 to 9 refer to the random variable, X, with the probability density function given by  4 x 2 , 1 ≤ x ≤ 4 f (x) =  3 . er ere 0, elsewhere −

6 The mean of X correct to 4 decimal places is: A 1.8484 B 2.0000 C 2.5000 D 2.8484 E 3.8484 7 The median of X is: A D

4 3 15 8

E

8 5 8 15

B

1 12

B

C

5 8

8 The mode of X is: A

1 12

D 1

and

4 3

C

4 3

E 4

Chapter 12


647

9 The variance of X can be found by evaluating: A

∫1 3x dx − ( 5 )

B

∫1 3x dx − ( 8 )

C

4

4

4

∫1

4

8

4

5

4 dx − 3x 2 4

4 D  3 x  1

4

4 E  3 x  1

−

4 3

(

− 4 3

X2 ~ N( µ2, σ22)

()

x

O 2

logg e (4) logg e (4)

)

2

Questions 10 to 11 refer to the following information. The length of time (in minutes) of a gentleman’s haircut at the hairdressers is a continuous random variable with a probability density function given 3x , x ≥ 0 0 . 3e 0.3x by f (x) =  . er ere 0, elsewhere −

10 The probability that Alex’s haircut will take longer than a quarter of an hour is: A 0.0521 B 0.0498 C 0.9502 D 0.0111 E 0.9889 11 The probability that Alex’s haircut will take longer than a quarter of an hour, given that it has already taken 10 minutes is: A 0.2231 B 0.0117 C 0.0524 D 0.0112 E 0.0504 12 Scores on a certain test are known to be normally distributed. Approximately 95% of the scores attained are between 26 and 74. The mean and standard deviation respectively are: A 48 and 6 B 48 and 8 C 48 and 12 D 50 and 6 E 50 and 12 13 If X ∼ N(27, 16), then approximately 95% of the values would lie in the range: A −5 ≤ X ≤ 59 B 9 ≤ X ≤ 39 C 0 ≤ X ≤ 54 D 19 ≤ X ≤ 35 E 23 ≤ X ≤ 31 14 The diagram below shows the graphs of two normal distribution curves with means µ1 and µ2 and standard deviations σ1 and σ2 respectively.

648

X1 ~ N( µ1, σ12)

2

2

8 5

y

A B C D E

µ1 > µ2 and σ1 = σ2. µ1 > µ2 and σ1 > σ2 µ1 = µ2 and σ1 > σ2 µ1 = µ2 and σ1 < σ2 µ1 < µ2 and σ1 = σ2

[© VCAA 2003]

15 The random variable X has a normal distribution with mean 20 and standard deviation 4. If the random variable Z has the standard normal distribution, then the probability that X is more than 24 is equal to: A Pr(Z Z < 1) B 1 – Pr(Z Z > 1) C Pr(Z Z < −1) D Pr(Z Z > –1) E 1 – Pr(Z Z < –1) 16 If Z has a standard normal distribution, then Pr(Z > 1.111) is: A 0.1333 B 0.1335 C 0.1357 D 0.8665 E 0.8667 17 X is normally distributed with mean µ = 16 and standard deviation σ = 5. The standardised value for X = 22 is: A −1.2 B −1.0625 C 1.0625 D 1.2 E 1.5 18 The standing long jump distances of NBL basketballers is known to follow a normal distribution with a mean of 1.4 metres and a standard deviation of 0.25 metres. The probability that an NBL player has a standing long jump of at least 1.6 metres is: A 0.0013 B 0.0228 C 0.2119 D 0.3446 E 0.4207 19 If X ∼ N(8, 4), then Pr( Pr(X < 9.25) equals: A 0.5311 B 0.6228 C 0.7340 D 0.9223 E 0.9991 20 If X is normally distributed with a mean of 13 and a standard deviation of 4 and Pr( Pr(x < k) = 0.71, then k equals: A 9.18 B 10.79 C 15.21 D 16.81 E 18.11


EXTENDED RESPONSE

1 Maya lives in Melbourne, and her sister Rosa lives in Perth. Every Saturday evening they talk on the phone. The length of their conversations (in minutes) is a continuous random variable, X, with a probability density function given by − 0 . 09 0 9e 0 . 0 9 x , x ≥ 0 f x) =  f( . er ere 0, elsewhere a What is the probability, correct to 4 decimal places, that next Saturday the two sisters will talk for: i more than 10 minutes? EXAM TIP Students should take care in writing an ii more than 20 minutes? integral that will yield the correct answer as part of iii more than half an hour? their working. Students often do not gain full marks on questions if they do not show any working. [Assessment report 2 2005 © VCAA]

b What is the probability that the sisters will talk for less than three-quarters of an hour, given that they have been talking for at least 20 minutes? Give your answer correct to 4 decimal places. c Maya’s son Michael lives in London. He always calls Maya every Saturday at 7 pm. What is the latest time that Maya should call her sister Rosa to be at least 95% certain that they will have enough time to chat (i.e. that they will finish by 7 pm)? Give your answer to the nearest minute. d Maya’s phone company charges 60 cents per call for all calls to Perth under 10 minutes duration, $1.40 per call for calls over 10 minutes and under 20 minutes duration, and $2 per call for calls between 20 minutes and 30 minutes duration. All calls with duration of over 30 minutes are capped at $2.50. (That is, a call of any length above 30 minutes will cost $2.50.) Find the expected cost of Maya’s call to the nearest cent. 2 Shoelaces sold as 50 cm are manufactured at a certain factory. However, the lengths actually follow a normal distribution with a mean of 50 cm. Quality control managers at the shoelace factory check the lengths of all laces manufactured and follow the rules below: i If a lace is greater than 53 cm, it is classed as oversize and sold as a 55 cm lace. ii If a lace is smaller than 47 cm, it is classed as undersize and sold as a 45 cm lace. iii If a lace is between 47 cm and 53 cm, it is classed as adequate and sold as a 50 cm lace. If 90% of laces are adequate, find: a the standard deviation, correct to 2 decimal places b the percentage of oversize laces c the percentage of undersize laces d the probability of a lace being adequate, given that it is not undersize, correct to 4 decimal places. 3 The time taken for Year 7 students to read a certain page of writing follows a normal distribution with standard deviation of 20 seconds. If 80% of students read the page in less than 3 minutes, find: a the mean time taken to read the page, correct to 1 decimal place b the probability that a student will take longer than 3.5 minutes to read the page, correct to 4 decimal places c the probability that a student will take between 2.5 and 3.5 minutes to read the page, correct to 4 decimal places d the speed with which 90% of students exceed in reading the page, correct to 2 decimal places. 4 A jeweller knows that the diameter of wedding rings follows a normal distribution, with a mean of 18 mm and a standard deviation of 1 mm. Find the probability that a customer requires a ring with a diameter that is: a greater than 20.5 mm b less than 19 mm c greater than 19 mm, given that it is less than 20.5 mm.

Chapter 12


649

5 The lengths of fish caught in a particular river are normally distributed with a mean of 28 cm. Ninety per cent of fish caught are longer than 25 cm. Find: a the standard deviation for the length of fish caught in the river b the probability of catching a fish that is longer than 30 cm from this river c the maximum length of fish that must be thrown back, if the shortest 30% must be returned to the river. 6 The amount of time, in hours, that Emily spends training for the Ironwoman Championships is a continuous random variable, with probability density function given by 4  ( x − 1)3 , 1 < x < 4 f ( x ) =  81 . 0, elsewhere a What is the probability that she spends less than 2.5 hours training per week? b What is the mean number of hours she trains each week? c What is the median time, correct to the nearest minute, she spends training each week? 7 Each night Kim goes to the gym or the pool. a When she goes to the pool, the time she spends swimming is normally distributed with a mean of 60 minutes. Twenty per cent of the time she swims for less than 50 minutes. Find the standard deviation in minutes, correct to 1 decimal place, of the time she spends swimming. When Kim goes to the gym, the time, T hours, that she spends working out is a continuous random variable with probability density function given by 4t 3 − 2 4t 2 + 4 4t − 2 4, if 1 ≤ t ≤ 2 f (t ) =  . er ere 0, elsewhere b Sketch the graph of y = f (t). Label any stationary points with their coordinates, correct to 2 decimal places. c What is the probability, correct to 3 decimal places, that she spends less than 75 minutes working out when she goes to the gym? d What is the probability, correct to 2 decimal places, that she spends more than 75 minutes working out on 4 out of the 5 next times she goes to the gym? e Find the median time, to the nearest minute, that she spends working out in the gym. [[© VCAA 2006]



650


eBook plus

ACTIVITIES


• 10 Quick Questions: Warm up with ten quick questions on continuous distributions. (page 589) 12A

Continuous random variables

Tutorial

• WE1 int-0582: Watch a worked example on ing density functions. (page 591) 12B

Using a probability density function to find probabilities of continuous random variables

Tutorial

• WE 5 int-0583: Watch a worked example on sketching density functions and calculating probabilities. (page 598) 12C

Measures of central tendency and spread

Tutorials

• WE 7 int-0584: Watch a worked example on calculating the mean, median and mode of a probability density function. (page 605) • WE 9 int-0585: Watch a worked example on finding the mean median and standard deviation of a probability distribution using a CAS calculator. (page 609) 12D

Applications to problem solving

Digital doc

• WorkSHEET 12.1: Using the cumulative normal distribution table. (page 618) 12E

The normal distribution


Digital doc

• Spreadsheet 079: Investigate normal curves using a spreadsheet. (page 623) 12G

The inverse cumulative normal distribution

Tutorial

• WE 20 int-0587: Watch a worked example on inverse normal calculations using a CAS calculator. (page 636) Digital docs

• WorkSHEET 12.2: For normal distributions calculate standard deviations, means and probabilities. (page 642) • Investigation: Sunflower stems (page 642) Chapter review Digital doc


• The normal distribution: Consolidate your understanding of the normal distribution. (page 619) Tutorial

• WE 13 int-0586: Watch a worked example on using confidence intervals to determine probabilities. (page 620)

Chapter 12


651

EXAM PRACTICE 4 SHORT ANSWER

25 minutes

1 Student scores in an exam are standardised so that the mean is 30 and the standard deviation is 7. Students are awarded a C if their standardised score lies between 30 and 37. Assuming that the student results are normally distributed, estimate to the nearest 5% the percentage of students who would receive a C. 1 mark

2 Two events A and B from a given event space have 1 1 probabilities Pr(A) = and Pr(B) = . 6 4 1 a Calculate Pr(A ∩ B ′) if Pr(A ∩ B) = . 10 1 b Calculate Pr(A | B ′) if Pr(A ∩ B) = . 10 c Calculate Pr(A ∩ B ′) if A and B are mutually exclusive events. d Calculate Pr(A ∩ B ′) if A and B are independent events. 4 marks

3 The probability of Karen Gouch shooting a basket from the free throw line is 0.8, given her team is winning the game, and 0.6, given her team is losing. The team wins 7 out of 10 games. a What is Karen Gouch’s overall percentage of baskets from the free throw line? b If Karen throws a basket from the free throw line, what is the probability that her team is winning? 2 marks

4 Claude has a coffee shop. He sells coffee and biscotti. He realises that if a person buys a coffee on a particular day, there is a 75% probability that the person will return and buy coffee the next day and a 20% probability that they will buy biscotti the next day. In addition, if a person buys biscotti one day, there is a 60% probability that they will purchase biscotti the next day and a 30% probability that they will buy coffee the next day. On Monday, 90% of Claude’s patrons bought coffee and 40% bought biscotti. a Determine a transition matrix, T, that models this situation. b Determine the initial state matrix, S. 652

CHAPTERS 1 TO 12

c What is the probability that a patron will purchase a coffee on Tuesday? d Write down a matrix expression that could be used to calculate the probability that a patron will purchase a coffee on Friday. 4 marks

MULTIPLE CHOICE

10 minutes

Each question is worth 1 mark. 1 One normally distributed variable has a variance of 4. Another has a variance of 0.25. Which of the graphs below would best represent these variables? A 0

2

4

6

8

10

0

2

4

6

8

10

0

2

4

6

8

10

0

2

0

2

B

C

D 4

6

8

10

E 4

6

8

10

2 Paddy Beam averages one ace in every 5 serves, but each serve is independent of any other. In one particular game Paddy serves 7 times. What is the probability that 3 of the serves are aces?  1 A    5

3

3

 7  1   4  B       3  5   5  3

 7  3   4  C       3  7   7 

E


 5  7  3  1  12  4 

4

D

 7  5  3  1  12  4 

4

3 A discrete random variable X has the following probability distribution. x

0

1

2

3

4

Pr(X = x)

0.1

0.1

0.4

0.3

0.1

The standard deviation of X correct to 2 decimal places is: A 0.80 B 1.04 C 1.08 D 1.16 E 1.35 4 If a random variable X has a probability density ax 2 − 3 if x ∈ [ − 1,3 , ] function f ( x ) =   0, elsewhere then a is:

39 1 C 28 10 1 18 D E 8 26 5 A random variable X has a normal distribution with mean 12 and standard deviation 0.25. If the random variable Z has the standard normal distribution, then the probability that X is less than 12.5 is equal to: A Pr(Z > −2) B 1 − Pr(Z < 2) C Pr(Z > 2) D 1 − Pr(Z > 2) E Pr(Z < −2) A 1.29

B

EXTENDED RESPONSE

25 minutes

The graph of the probability density function for the standard normal distribution (mean = 0, variance = 1) is plotted below. y −x 1 e 2 It has the rule f: R → R, where f ( x ) = . 0.5 2π a Evaluate: −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 i f (−2) −0.5 ii f (1) to an accuracy of 3 decimal places. 2

x

1 + 1 = 2 marks

b Plot and label the corresponding coordinates on the graph above.

1 mark

c Exact integration of f is not possible. However, it is possible to obtain an approximate value by using a rectangular approximation. By using four intervals of equal width, determine a rectangular approximation to the area bound by the graph of f, the y-axis, the x-axis and the line x = 1.

2 marks

d The area under a probability density function represents the probability. Use your answer to part c to determine an approximate value for the probability that a value of x lies within 1 standard deviation from the mean (that is, that x lies in the interval [−1, 1]).

1 mark

e Use your CAS calculator to determine an approximation for the area defined in part c to an accuracy of 3 decimal places.

1 mark

f The infinite series approximation to e−x is e x ≈ 1 + x +

x2

x3

x4

+ + K 2 6 24 Use this approximation to show that f (x) can be approximated by g(x) where g( x ) ≈

1 2π

(1 −

x 2 x 4 x 6 x8 + − + ) 2 8 48 384 .

g Sketch the graph of g for −2 ≤ x ≤ 2 onto the graph of f provided at the start of this question.

3 marks 2 marks

1

h Determine an approximation for 1

∫ g( x) dx.

−1

eBook plus

∫ f ( x) dx by using calculus to evaluate

−1

2 marks

Digital doc

Solutions Exam practice 4

Exam practice 4

653

Answers CHAPTER 1

Graphs and polynomials Exercise 1A — The binomial theorem 1 a x2 + 6x + 9 b x5 + 20x4 + 160x3 + 640x2 + 1280x + 1024 c x8 − 8x7 + 28x6 − 56x5 + 70x4 − 56x3 + 28x2 − 8x + 1 d 16x4 + 96x3 + 216x2 + 216x + 81 e 2401 − 1372x + 294x2 − 28x3 + x4 f 32 − 240x + 720x2 − 1080x3 + 810x4 − 243x5 3 1 2 a x3 + 3x + + 3 x x b 2187x7 − 10 206x5 + 20 412x3 − 22 680x 15 120 6048 1344 128 + x − x 3 + x 5 − x 7 + 729 c x12 + 18x9 + 135x6 + 540x3 + 1215 + 1458 x3 x6 243 810 1080 720 2 5 + 240 x − 32 x d 10 − 7 + 4 − x x x x 3 a i −21 ii 1 iii 0 b i 40 ii 80 iii 80 c i 0 ii 810 iii 0 d i 0 ii −540 iii 0 e i 0 ii 302 526 iii 0 4 A 5 C 6 D 7 D 8 540x3y3 19 683 x 2 9 4 10 4860 11 160 12 54 13 105 14 6 Exercise 1B — Polynomials 1 ii, iii, vi 2 a x5 − 2x4 − 2x2 − 3x + 7 b x5 − 3x4 − 8x3 − 11x2 + 4x − 1 c 3x4 − 16x3 − 8x2 − x + 24 d 17 − 18x + 29x2 + 24x3 + 5x4 − x5 3 a i 6 ii 0 iii 124 b i 7 ii −8 iii 280 c i 6 ii 3 iii 331 d i 4 ii −7 iii −55 4 B 5 a = −4, b = 8 6 a = 3, b = −5 7 a = 0, b = −7 8 a = 2, b = −5 9 a C b D Exercise 1C — Division of polynomials 1 a 50 b −171 c 465 2 a b 3 a

654

i 50 iii 465 The result is the same. No b Yes

Answers

5

4 0

x

6

2

0

1 − 14 7 − 19

1

2 3 4 5 6 7 8

a a D a c a d a d a

y − 3x + 5 = 0 3x + y + 13 = 0

b 2x + y + 5 = 0 b y - x + 2 = 0

i −1 2x + y = 0 iii vi [−2, ∞) [−2, 3]

b y − 3x + 7 = 0

ii 8 b e b e

( 12, 15) y

v i (−5, ∞) R b

5 x

i (−∞, 5) ii (−∞, 0)

3

d Yes

iv ii [−2, 3) (−∞, 6)

2 8 x (12, 3)

i [−12, 12] ii [−3, 15] y c —) (5, 14

0

2

d

y

c f c f

0

(2, 2–3 )

−3 20 27

x

x

1– 2

0

iv iv iv iv

0

y

c

6

ii −171 iv −3 20 27 c No

4 a C b B 5 a (x − 2)(x + 3)2 b (3x − 1)(x + 2)(x − 6) c (x + 2)(x − 2)(x + 3)(x − 1) d 4x(x + 1)(x + 4)(x − 2) 6 a x = −3, 0 or 2 b x = −4, −3, 0 or 2 c x = −2, 32 or 2 d x = ±1 7 a = 2 8 a = 5 9 a = 7 10 a = 10, b = −26 11 a = −9, b = −11 Exercise 1D — Linear graphs y y 1 a b

3– 2

x

i [2, 5) ii [ 23 , 143 )

9 a 2x + y − 9 = 0 b x + 3y − 17 = 0 10 a y − 2x − 7 = 0 b x + 2y + 1 = 0 11 E 12 B Exercise 1E — Quadratic graphs 1 a 0 b 2 c 0 d 2 e 2 f 1

y

2 a

y

b

1

2

8

c

2

0

x

4

1

4

d

y

x

0

1 1–3 5 x

0

b

c (1 12 , 12 14 )

d

4 5 6

a b c d a b c a

i i i i i i i

(0, 2) (6, 0) (−2, 0) (−3, −6) y = x2 − 2x − 1 y = x2 − 4x + 1 y = 8 + 2x − x2 y

ii ii ii ii ii ii ii

1 1–2

2

3

y ≤2 y≥0 y ≤0 y ≥ −6 [−2, ∞) [−3, ∞) [−16, 9] x 2 1–2

x

2 4 6 x

i [−10, 6]

ii [−4 14 , 128] y

d

(1, 8)

x

5 4 3 2 1 0 1 2 3 (3, 4)

50 ( 5, 100)

100 f(x) 5 6x 3x2

12 a 13 a c 14 a d e

15

c

−3 4]

y 100

f(x) x2 3x 2

10 8 6 4 2 0

iii iii iii iii iii iii iii

11–2

ii (−∞,

1 1 50 (1–2 , 4 –4 ) (6, 16)

y 1

y = x2 x 1, x Œ R

x

− ( 25 , 49 ) 1 1 (12 , 12 24 )

R R R R R [−1, ∞) [−4, 4) b

4 x

3

8

12

3 a (3, −1)

2

i R+ c ( 10, 128)

y

10

2

1

1 3

4 (–2 , –4 )

6

4 0

y

b

i [−5, 3) ii [-100, 8] 8 m3 b 80 m3 48 m b 6 s Domain = [0, 6], range = [0, 48] 12 m b 6 s c 48 m Domain = [0, 11], range = [12, 48] h 48 (0, 48)

(11, 37)

36

y

24 12 1 0.09

b C y

b

0

4

x

( 4, 9)

7

11 t

6

Exercise 1F — Cubic graphs 1 a y = x(x + 6)(x − 5) b y = (x + 2)2(x − 1) 2 a v b iv c ii d i e vi f viii g vii h iii y y 3 a b

8

7 a B 8 D 9 A 10 a

(6, 12)

0

2.91 x

1.5

y

12

1 0

2 1 0

x

2

x

1 0

4

7

(2, 4)

c

d

y 3

11 a ( 2, 10)

1 0

18 3

24

x

3

6

x

1 0

(1, 8)

y 10

10

4 a B 5 C 6 A 10 a

2 5 y = x 2x 2 (2, 2) 2 (1, 1)

2 1 0

1

i [−2, 2]

2

4

y

x

7 B

0

b E 8 D

2 3

x

9 E

8

x

ii [1, 10]

y

d

y

x

4

i [2, ∞)

0

1 2

x

Answers 1a ➜ 1f

3 ( 2, 1)

c

y

3

2

ii [0, ∞)

Answers

655

y 4

b

y

c

4

( 1, 0)

1 0

2– 3

0

x

2

3

0

2

( 2, 32)

i [−2, −1] y c 0

y

d

e

ii [−32, 0] 4– 3

(1, 8)

x

2 y 36

3 x

3

36 (0, 36)

3–2 0

2 a

2– 3

2

x

y

ii (−36, −8]

i (0, 1] y d

(1.15, 2.08)

( 1, 4) x

0

(0, 0)

(2, 0)

(3, 0) x

(2, 14)

i [−1, 2) y e

ii (−14, 4]

(2.59, 1.62)

(3, 33)

y

b ( 1, 0)

( 1, 3)

(1, 0) x

0

x

0

( 2, 12)

i [−2, −1) ∪ (0, 3] y f

(0, 1)

ii [−12, −3) ∪ (0, 33]

( 1, 3) 0

x

(1, 3) (2, 18)

c

y ( 0.22, 3.23) (0, 3)

(3, 57)

(−1,

i 1) ∪ [2, 3) ii (−57, −18] ∪ (−3, 3) −24 11 a = 6, b = 12 a = 7, b = −4 13 a a = 3, b = −3, c = 3 and f (x) = 3(x − 3)3 + 3 b g(x) = −3(x + 3)3 + 3, domain = [−4, −2], range = [0, 6] c 7 cm 14 a a = 14 , b = 5 (5 − t ) , domain = [0, 5] b d (t ) = t 2 4 c d (km) d(t) = t2 —— 5 t

( 3, 0) ( 1, 0)

d

0 (1, 0)

x

( 2.28, 9.91) y (0.25, 8.54)

4

5 4 3 (2, 3) 2 1 0 1 2 3 4 5 t (h)

( 2, 0)

d Max. d ≈ 4.6 km when t = 3 h and 20 min Exercise 1G — Quartic graphs y 1 a b

656

Answers

32

2

4

x

(1, 0)

x

y

24

3 1 0

0

4

2 0 1 2

x

3 4

a c e a b

E b B A d D D f C y = (x + 2)(x − 1)(x − 3)(x + 1) − y = 12 (x − 2)2(x + 1)(x − 4)

2

5– 2

3

x

5 a

y

b

0

3 2

x

2

200

i [2, 3] ii

[−30,

100

2 1

(3, 30)

c

y 400 300

( 1, 36)

24

( 2, 400)

0] i

0 1

(−2, −1]

0 −1 −4

2

x

3

x

ii [36, 400) y

d

y

2

2

0

2

x

Multiple choice 1 E 2 5 C 6 9 E 10 13 E 14 17 A 18 21 C

3 7 11 15 19

B C D B E

4 8 12 16 20

A C A C D

Extended response 1 a y = 2.24 (x + 2.5)2 + 4 b Domain [−5, 0], range [4, 18] c (−1.83, 5) d f : [−3.17, −1.83] → R, f (x) = 5 2 a 0 km b d (1000 km)

(1, 4)

( 3, 45)

(−2, −16) 0

i (−∞, −2] ii (−∞, −16] i [−3, −2] ii [−45, 0] 6 a = 4, b = −19 7 a = 1, b = −7 8 a = 3, b = −1 Chapter review Short answer 1 a 32y5 − 240y4x + 720y3x2 − 1080y2x3 + 810yx4 − 243x5 x8 x6 7 x 4 224 448 − + − 14 x 2 + 70 − 2 + 4 − b 256 8 4 x x 512 256 + 8 x6 x 2 a = −15, b = 2 3 a (x − 9)(x + 2)(x − 5) b (2x − 3)(x + 1)(x + 6)(x − 2) 4 a 7x + 6y − 1 = 0 b x + 2y − 9 = 0 y 5 ( 1, 9)

8

2

x

Domain = R, range = (−∞, 9] y 6 0

3

1– 3

x

—) ( 4–3 , 25 3

−

0

c d = t 3 − 6t 2 + 9t d The satellite by 40 000 km e 10 days f Domain = [0, 10], range = [0, 490] a y = 3 – 0.75x2 b y = 1.3125, so cannot fit. c Reduce the height by 0.4 m. x (m)

Finishing line

(2.1, 8.8)

10

(3.8, 5.24)

5 (0, 2.4) (0.3, 2.6) 0

2

(5.1,0) 4

6

2 3

x

Liney: Starts 2.4 m in front of start line moving forwards at a constant 0.75 m/min. ed by Cubric after 0.3 min and 2.6 m from the start line, meets Cubric coming back towards Liney at 3.8 min and 5.24 m from the start line. Quadder: Starts at start line, travelling 1.3 m the wrong way for 2.55 min, stopping momentarily then moving forward with increasing speed. Meets Cubric at the start line after 5.1 min. Cubric: Starts at start line moving very fast towards the finish ing Liney at (0.3, 2.6), then slowing, stopping momentarily at (2.1, 8.8), then moving back towards the start. Meets Liney at (3.8, 5.24) and Quadder at the start line at 5.1 min. Slows down, stopping at (7.37, −5.8) and then speeding for the finish. Q

10.1

y 2

4

x

10 9.9

16

10 t (min)

8

(7.37, 5.8)

18

1 0

C

(2.55, 1.3)

5

x (m) 10.2

8

L Q

(10.1, 10.1) (10.07, 10)

(10.092, 10) (10.0899, 9.97)

(10.04, 9.9) 10

L (10.13, 10)

C

Answers 1g ➜ 1G

Range = [ 25 , 0] 3 7 a a = 9, b = 2 y b

3

4

(3, 0) t (days)

10

0

4

3

D B D B A

10.15 t (min)

Answers

657

The Finish: Quadder overtakes Limey at (10.04, 9.9). Cubric overtakes Limey at (10.0899, 9.97). Quadder finishes the race in 10.07 min, Cubric in 10.09 min, Liney in 10.13 min. Quadder wins by 1.2 s, Cubric second by 2.4 s to Liney. 5 a C is (3, 0) and D is (2.25, −8.54) b y = x4 − 3x3 c 2 km d Yes, because a straight route from O to D to C is approximately 17.4 km and the river course is longer than this. 6 a y = 14 (x + 3)(x + 1)(x − 1)(x − 3) 7 8

b (4, 26 14 ) and (−4, 26 14 ) c Domain [−4, 4] d (2.236, −4) and (−2.236, −4); range [−4, 26 14 ] e y = 14 (x − 2.236)2(x + 2.236)2 f Domain [−4, 4], range = [0, 30 14 ] Teacher to check the model − 2( x x − 75) a y = 6250 b 46.72 m c Seems to be extremely low 2 km from touchdown. Not very accurate. − 3( x − 4)2 ( x + 4)2 a y = 256

b y =

− 3(

x − 4)2 ( x + 2) 32

x2 −3 4 − ( x − 4)2 ii y = ; the gradients are 1 and 2, 2 respectively, so not smooth. iii Teacher to check d Cubic is the closest, 1.6875 m. c

i

CHAPTER 2

Functions and transformations Exercise 2A — Transformations and the parabola 1 a Dilation by a factor of 2 from the x-axis b Dilation by a factor of 13 from the x-axis c Dilation by a factor of 3 from the x-axis, reflection in the x-axis d Translation 6 units down e Dilation by a factor of 12 from the x-axis, reflection in the x-axis, translation of 1 unit up f Translation of 2 units to the right g Reflection in the x-axis, translation of 3 units to the left h Dilation by a factor of 2 from the x-axis, translation of 3 units to the right i Translation of 2 units to the left, translation of 1 unit down j Translation of 0.5 unit to the right, translation of 2 units up k Dilation by a factor of 2, reflection in the x-axis, translation of 3 units to the left, translation of 1 unit up l Dilation by a factor of 12 from the x-axis, translation of 1.5 units to the right, translation of 0.25 units down 2 D 3 a (ii) b (v) c (i) d (iv) e (iii)

658

Answers

4 a c 5 E 6 a c e 7 a b c

y = − 12 (x − 2)2 + 2 y = −3(x − 1)2 + 3

b y = 2(x + 1)2 − 2 d y = (x + 2)2 − 4

y = 12 x2 y = (x − 2)2 − 1 y = −(x + 3)2 y = (x − 3)2 − 4 y = −2(x + 1)2 + 1 y = 13 (x + 3)2 − 4

b y = −x2 d y = 3x2 − 2

−

y = 12 (x − 2)2 + 2 y = 3(x − 1)2 + 6 y = −4(x + 2)2 + 8 y = 2x2 y = –2x2 y = –2(x + 1)2 y = –2(x + 1)2 + 3 (–x, y) (x, – y) (x, 3y) (2x, y) x ( , y) 3 f (x + 2, y) g (x, y − 1) 10 a z = 3 or z = 15 2 b y = 2(x − 3)2 − 8 or y = 25 (x − 15)2 − 8 11 a 3 − b y = 13 (x + 4)2 + 3 c x = −7, x = −1 12 1. f (x + 2) − 3, −4 ≤ x ≤ 0 2. f (x − 2) − 3, 0 ≤ x ≤ 4 3. f (x + 4), −6 ≤ x ≤ −2 4. f (x − 4), 2 ≤ x ≤ 6 5. −f (x + 4) + 6, −6 ≤ x ≤ −2 6. −f (x) + 6, −2 ≤ x ≤ 2 7. −f (x − 4) + 6, 2 ≤ x ≤ 6 8. −f (x + 2) + 9, −4 ≤ x ≤ 0 9. −f (x − 2) + 9, 0 ≤ x ≤ 4 Exercise 2B — The cubic function in power form 1 a Dilation from the x-axis by the factor of 7 b Dilation from the x-axis by the factor of 23 , reflection in the x-axis c Translation by 4 units up d Reflection in the x-axis, translation by 6 units up e Translation by 1 unit to the right f Reflection in the x-axis, translation by 3 units to the left g Dilation from the x-axis by the factor of 4, reflection in the y-axis, translation by 2 units to the right h Dilation from the x-axis by the factor of 6, reflection in the x-axis, reflection in the y-axis, translation by 7 units to the right i Dilation from the x-axis by the factor of 3, translation by 3 units to the left, translation by 2 units down j Dilation from the x-axis by the factor of 12 , reflection in the x-axis, translation by 1 unit to the right, translation by 6 units up k Dilation from the x-axis by the factor of 2, translation by 25 units to the left l Dilation from the x-axis by the factor of 14 , reflection in the x-axis, translation by 8 units to the left, translation by 3 units up 2 a i, iv b iii, v c ii d i, ii, iv e ii, v f iii, iv 8 9

d e f a b c d a b c d e

3 a i (0, 0) ii (0, 0) iii There are no translations, but there is a dilation of 3 4. b i (0, 1) ii (0.8, 0), (0, 1) iii There is a reflection in the x-axis, a translation of 1 unit up, and a dilation of 2. c i (0, −6) ii (2.08, 0), (0, −6) iii There is a translation of 6 down and a dilation of 23 . d i (4, 0) ii (4, 0), (0, −128) iii There is a translation of 4 right and a dilation of 2. e i (2, 0) ii (2, 0), (0, 4) iii There is a reflection in the x-axis, a translation of 2 right, and a dilation of 12 . f i (1, 0) ii (1, 0), (0, 4) iii There is a reflection in the y-axis, a translation of 1 right, and a dilation of 4. g i (1, 2) ii (−0.3, 0), (0, 1) iii There is a translation of 1 right and 2 up. h i (−2, 3) ii (−0.6, 0), (0, −5) iii There is a reflection in the x-axis and a translation of 2 left and 3 up. i i (−1, −6) ii (0.4, 0), (0, −4) iii There is a translation of 1 left and 6 down, and a dilation of 2. y y 3 a iv b iv 1

0.8 x

x

c iv

y

d iv

y 4

2.08

6

x

x

128

e iv

y

f iv

y 4

4

x

2

g iv

y

h iv

y

1

x

0.6

i iv

y

1

0.4

4

6

x

x

2

5

8

c e a c

1 2

x3

5 C

y = (x − 3)3 − 1 y = −(x + 1)3 − 1 y = 2x3 y = −2(x − 2)3

9 a y =

−1 2 −

d y = 2x3 + 3 b y = −2x3 d y = –2(x − 2)3 – 1

x3 + 4

b y = 2(x − 1)3 + 2

c y = 3(x + 1)3 + 1 e y = 4(x + 1)3 − 12 10 E − 12 a y = 12 (2 − x)3 + 1 − 13 a ( 3, 1) or (−1, 27) y b 3 y (x 3) 1

6 B b y = −(x + 5)3

d y =

−1 3

(x − 3)3

11 y = 2(x + 1)3 − 4 b Positive cubic

28 27

y (x 1)3 27

4 3 1

1

x

Exercise 2C — The power function (the hyperbola) 1 a Dilation from the x-axis by the factor of 2 b Dilation from the x-axis by the factor of 3, reflection in the x-axis c Translation by 6 units to the right d Dilation from the x-axis by the factor of 2, translation by 4 units to the left e Translation by 7 units up f Dilation from the x-axis by the factor of 2, translation by 5 units down g Translation by 4 units to the left, translation by 3 units down h Dilation from the x-axis by the factor of 2, translation by 3 units to the right, translation by 6 units up i Dilation from the x-axis by the factor of 4, reflection in the x-axis, translation by 1 unit to the right, translation by 4 units down 2 a v b iii c i d v, iii e v, ii, iii f i, iii g v, i, iv h ii, iv 3 a i x = 0, y = 0 ii Domain: R \ {0} iii Range: R \ {0} b i x = −6, y = 0 ii Domain: R \ {−6} iii Range: R \ {0} c i x = 2, y = 0 ii Domain: R \ {2} iii Range: R \ {0} d i x = 3, y = 0 ii Domain: R \ {3} iii Range: R \ {0} e i x = 0, y = 4 ii Domain: R \ {0} iii Range: R \ {4} f i x = 0, y = −5 ii Domain: R \ {0} iii Range: R \ {−5} g i x = −6, y = −2 ii Domain: R \ {−6} iii Range: R \ {−2} h i x = 2, y = 1 ii Domain: R \ {2} iii Range: R \ {1} i i x = −n, y = −m ii Domain: R \ {−n} iii Range: R \ {−m} 4 a i x = 4, y = 0 ii Domain: R \ {4} iii Range: R \ {0} b i x = 0, y = 2 ii Domain: R \ {0} iii Range: R \ {2} c i x = 3, y = 2 ii Domain: R \ {3} iii Range: R \ {2}

Answers

Answers 2A ➜ 2C

3

2 1

0.3

x

1

4 E 7 a y =

659

d i iii e i iii f i iii 5

x = −1, y = −1 Range: R \ {−1} x = m, y = n Range: R \ {n} x = b, y = a Range: R \ {a} y

ii Domain: R \ {−1} ii Domain: R \ {m}

y4

ii Domain: R \ {b}

–34 x

7 E

3

x– x

x

3

x–

–4 x 3

–1

x

y

6 a

y

b

–1 3

–21

b y =

y

x

1 5

–43

d

c

y

e

2

x

2 –21

y

h

3 2 x 1 1–21

1

x

1

11 1

y

f –2 5

–2 5

2

x

1

1 (0, 1 — ) 2

y1

x 0

y

i

–31

4

k

y 1

1

x

–23

Answers

x

x 2 –1 –3 4 4

x

11

y

Domain: R \ {0}, range: R \ {3}

3 y 10 8 6 4 2

–1 3

y

x

x3 x 2

10 8 6 4 2 0 2 4 6 8 10 x

4 (0, 3 ) 2

6 x 2

8

10

660

(3, 0)

y

j

4 –31

1–85

x

y

x

y 4

–1 2

y

6

1

1

3

1

d

–21

x

–52

y

1 7 –2

3

g

f

x

1

x

y

5

−

–1 2

2

y

3 +1 x

1

1

3 –43

e

−

4 −1 x 6 f y = −1 x +1 y b d y =

2

c

x

x

2 1

x

3

6

8 C

2 9 a y = x−2 − c y = 3 x+4 2 e y = + 2 x−4 y 10 a

–2 x 3 –2 3

(2, 11)

0 2 4

6 4 2 (0, 3) x1

1

x–

y 10 8 6 4 2

l

Exercise 2D — The power function (the truncus) 1 a Dilation from the x-axis by a factor of 2 b Dilation from the x-axis by a factor of 3, reflection in the x-axis c Translation by 2 units to the left d Dilation from the x-axis by a factor of 2, translation by 3 units to the right

e Dilation from the x-axis by a factor of 5, reflection in the x-axis, translation by 4 units to the left f Dilation from the x-axis by a factor of 2, translation by 6 units up g Reflection in the x-axis, translation by 3 units up h Dilation from the x-axis by a factor of 4, translation by 3 units to the right, translation by 1 unit up i Reflection in the x-axis, translation by 2 units to the left, translation by 5 units up 2 D 3 C 4 a i x = 0, y = 0 ii Domain R \ {0} iii Range: y > 0 b i x = 0, y = 0 ii Domain R \ {0} iii Range: y < 0 c i x = 2, y = 0 ii Domain R \ {2} iii Range: y > 0 d i x = −1, y = 0 ii Domain R \ {−1} iii Range: y > 0 e i x = −4, y = 0 ii Domain R \ {−4} iii Range: y < 0 f i x = 0, y = −3 ii Domain R \ {0} iii Range: y > −3 ii Domain R \ {0} g i x = 0, y = 45

iii Range: y >

4 5

h i x = 0, y = 12

iii Range: y <

ii Domain R \ {0}

i i x = 1, y = 4 ii Domain R \ {1} iii Range: y > 4 5 C 6 B 7 B y y 8 a b –1 9

x

c

y

d

–1 8

4

e

y

x

1

x

y

1

x

3

f

y

1

2

0.4

x

2

0.4

x

3

g

h

y

y

2 1

x

x

3.7 3 2.3

2

i

y

j

y

11 — 12 –2 3

2

x

2.7

1

–41

–1 2

0.7

x

y

l

3 2

x

1

1 2 3

1 + n ( x − m) 2 1 c y = − ( x − r )2

1 − b y = ( x + q)2 + p d y =

1 +t x2

1 − b ( x + a)2 1 g y = 2 − k x 2 10 a y = 2 − 2 x

f y =

1 −e ( x − g)2

9 a y =

e y =

c y = e y =

−

9 + 1 ( x + 2)2

8 − 3 ( x − 4)2

3 − 3 ( x + 2)2

x

1 −l x2 − 3 b y = ( x − 2)2 3 d y = +4 ( x + 1)2

h y = −

−

5 −2 ( x − 1)2 3 12 y = +2 ( x − 1)2

f y =

Exercise 2E — The square root function in power form 1 a Dilated from the x-axis by a factor of 2 b Dilated from the x-axis by a factor of 13 , reflected in the x-axis c Dilated from the x-axis by a factor of 3, translated 1 unit to the right d Dilated from the x-axis by a factor of 2, reflected in the x-axis, translated 4 units to the left e Translated 1 unit down f Dilated from the x-axis by a factor of 3, reflected in the x-axis, translated 2 units up g Translated 4 units to the right, translated 3 units up h Dilated from the x-axis by a factor of 2, reflected in the x-axis, translated 3 units to the left, translated 6 units up i Dilated from the x-axis by a factor of 12 , reflected in the x-axis, reflected in the y-axis, translated 2 units to the right and 23 units up 2 a (0, 0) b (0, 0) c (1, 0) d (−4, 0) e (0, −1) f (0, 2) g (4, 3) h (−3, 6) i (2, 23 ) 3 E 4 D 5 a Domain: x ≥ −1, range: y ≥ 0 b Domain: x ≥ 3, range: y ≥ 0 c Domain: x ≥ 0, range: y ≥ −3 d Domain: x ≥ 0, range: y ≥ 4 e Domain: x ≥ 0, range: y ≤ 5 f Domain: x ≥ 1, range: y ≥ 3 g Domain: x ≥ −2, range: y ≥ −1 − h Domain: x ≥ 12 , range: y ≤ 4

i j k l

Domain: x ≥ 43 , range: y ≤ 2 Domain: x ≤ 3, range: y ≥ −7 Domain: x ≤ 2, range: y ≥ 6 Domain: x ≤ 2, range: y ≤ 1

Answers

Answers 2D ➜ 2E

4 3 –97

1 3 –4

11 y =

1 2

y

k

661

6 D 8 a

y

7 D b

1.4

d

2

(6, 1)

x

x

y

f

y

x

3 –43

4

x

3

1 –21

y

h

y 4.4 2

x

–3 2

2 3 4

C a b c d e f a

Domain: R, range: y ≥ 0 Domain: R, range: y ≥ 1 Domain: R, range: y ≤ 4 Domain: R, range: y ≥ −2 Domain: R \ {−1}, range: y > 1 Domain: R \ {0}, range: y ≥ 0 y b x

x

2

i

x

1

–1 2

3.7 2

11

c

d

y = 3 4−x + 3

( 1, 1)

x

3

e

9

y

x

2

b y = −4 x + 1 + 8 d x ≥ −1 y f

12 a p = 8 c x = 3 e y ≤ 8

2

g

2

2.7

x

y

Exercise 2F — The absolute value function 1 a b y

i

x

–41

–1 4

y

d

y

k

y

3

5

6

6

y

x

y

2

1

x 1 3 5

x

1

2

l

y

2 x

35 63

( 1, 6)

 3 x − 1, x ≥ 13 5 a f ( x ) =  − 1  3 x + 1, x < 3

5 4

4

Answers

f

6

y

2

5

3.6 3

x

x

4

x

1

j

x

1 x

y

1

y

x

1

2

6

y

0.7

3 –31 3

–4 3

–43

x

3

2

h

( 1, 8)

4

e

y

2 1

2

–34

(1, 1) x

f

2 (4, 3)

–1 2

y 2

1

b y = 2 x − 1 − 4

y

y 7

9 E 10 a m = 1

x

5 1 1

11

6

x

1 2

1

662

y

y 0.4

c

x

2 1

y

i

y

2 4

x

x

y

g

y

h

7 1

x

e

y

3

2

c

g

y

x

b f (0) =−3 × 0 + 1 = 1, f (2) = 3(2) − 1 = 5

99

x

y 7 6 5 4 3 2 1

c

f(x) |3x 1|

0

3 2 1

1

2

3

x

 x 2 − 3 x + 2, x ≤ 0 ∪ x ≥ 3 6 a f ( x ) =  – 2  x + 3 x + 2, 0 < x < 3 b f (−1) = (-1)2 – 3 × –1 + 2 = 6, f (2) = –22 + 3 × 2 + 2 = 4 c y 2 f(x) |x 3x| 2

7 6 5 4 3 2 1 0

1

7 a y =

3 2

1

2

3

4

5

x

x , −2 ≤ x ≤ 2

b Yellow: y = 6 −

3 2

−

x , 2 ≤ x ≤ 2;

green: y =

3 2

x − 6, −2 ≤ x ≤ 2;

blue: y = −

3 2

x , −2 ≤ x ≤ 2

c Teacher to check Exercise 2G — Transformations with matrices 1 i Reflection in the y-axis, dilation by a factor of 2 from the x-axis ii Dilation by a factor of 12 from the y-axis, dilation by a factor of 4 from the x-axis, reflection in the x-axis iii Reflection in the y-axis, dilation by a factor of 3 from the x-axis, dilation by a factor of 2 from the y-axis iv Dilation by a factor of 12 from the x-axis, reflection in the x-axis − 2 i (3, 10) ii ( 32 , − 20 ) iii (6, 15)

2 x2 –1 ii y = x2 – b i y = 2 x 3 − 10

iv (−3,

−5 ) 2

ii y = x − 2 − 2 6 a i y = x − 3 + 2 b i y = x 2 − 9 x + 20 ii y = x 2 − 7 x + 8 7 C 8 a Reflection in the y-axis, dilation by a factor of 2 from the x-axis, translation horizontally by +6, translation vertically by +1 – 2 +2 f ( x ) = ( x + 6) b Dilation by a factor of 2 from the y-axis, dilation by a factor of 12 from the x-axis, reflection in the x-axis, horizontal translation by +3, vertical translation by −1 – 1 y = −1 x−3 c Horizontal translation by +1, vertical translation by −2, 13 dilation from the y-axis, reflection in the x-axis – 1 +2 y = 3x + 1 0   x   –1   +     y   3 10 a y = – 2 x 3 + 8 x − 1 b (−2, −1); point lies on the curve 11 Dilation by a factor of 3 from the x-axis, reflection in the x-axis, translation by 2 units in the positive direction of the x-axis, translation by 1 unit in the positive direction of the y-axis 1 x  9 T   =  4  y   0

  x   1 T    =    y   0

–1 

0  x

2 +     1 

–3  y 

12 a, b − g(2( x + 1)) + 1 = − 2( x + 1) + 1 13 a, b f (x) = 2x2 – 10x + 6 14 h(x) = x3 – 5 Exercise 2H — Sum, difference and product functions 1 a f (x) = x2 + x y

3 a i y =

ii y =

– 32 x 3

c i y = 2

+ 20

–x

2 1

3

2

1

0

1

1

2

3

1

2

3

x

dom f : R x3 + x b f ( x ) = 2

Answers 2F ➜ 2H

ii y = – 4 2 x 4 i Translation of 3 units in the positive direction of the x-axis, translation of 2 units in the positive direction of the y-axis. ii Translation of 2 units in the positive direction of the x-axis, translation of 2 units in the negative direction of the y-axis. iii Translation of 15 units in the negative direction of the x-axis. 5 i (4, 0) ii (3, −4) iii ( 45 , −2)

3

y 3 2 1

3

2

1

0

1

x

dom f : [0, ∞)

Answers

663

c f ( x ) = 3 x 2 + x

f

(2, 20.5)

y 1

y

1 1 0

2

3 ( 2, 4) 4

2 1

1.5 1 0.5 0

0.5

1

1.5

x

1

6 a

( 2, 8)

6 4 2 2

2 a g( x ) − h( x ) =

–

6 4 2

2

4

x

6

x +1 − x

b c d e f g

0

2

2

4

( 2, 2)

x

6

3

2

6 4 2 0 ( 1, 2) 2

4

2

4

(4, 4.5)

664

4

1.5 1 0.5 0

2

0.5

(2, 2 ) 1

2

4

x

2

3

x

(2, 6) (2, 8)

g(x) 0.5

1

1.5

x

f (x)

g(x) 0.5 f (x)

1

1.5

f (x)g(x)

8 h( x ) = 8 − yx + x + 5 , x ∈[ – 5, 8] 5 4 3 2 1

(2, 4)

a=0 f (0) = 0, g(0) = 0, h(0) = 0 f (1) = –1, g(1) = 1, h(1) = 0 f (2) = –4, g(2) = 2 , h(2) = –4 + 2

Answers

1

y 4 2

6 1

( 2, 4)

b c d e

f ran h: [ −2, ∞) g

x

4

2

(2, 2)

c f (0) = –3, g(0) = 0, h(0) = 0 d f (1) = –2, g(1) = 1, h(1) = –2 − e f (2) = –1, g(2) = 2, h(2) = 2

y

2

2

b R d R

0

1

1

2

3

4

x

y 4

1.5 1 0.5 0

2

2

5 a

1 20

x

6

4

y 4

( 4, 4.5)

10 8 6 4 2

7 a dom h: [0, ∞) b

2

0

3

(2, 8)

4

6

8

y ( 1, 3) 4

2

2

y

( 2, 10) ( 2, 8)

b g( x ) − h( x ) = x − x + 1 + 2

3 a (−∞, 3] c [0, 1] e [−2, ∞)

1

f (–2) = 8, g(–2) = 2, h(–2) = 10 f (0) = 0, g(0) = 0, h(0) = 0 f (1) = –1, g(1) = 1, h(1) = 0 f (2) = –8, g(2) = 2, h(2) = –6 ran h: [−6, 10]

2

4

0

2

4

8

0

y 4 ( 1, 1)

(2, 2)

2

6

dom f : R\{0}

2

(2, 4)

4

3 2 1

6 4 2

4

(2, 20.5 4)

y 8

( 2, 2)

1 x y

4

x

2

6

dom f : R

d f ( x ) = x + 2 +

1

6 5 4 3 2 10 1 2 3 4 5 6 7 8 9 x

x

y h(x) ( 2, 2) ( 2.5, 1.38)

(2.5, 5.38)

6 (0, 6) 4

8 dom w: ( –3, ∞), ran w: (0, ∞), dom v: R+, ran v: (–2, ∞) w  v(x) = x + 1 , where x ∈ R+; v  w(x) = x + 3 − 2,

f(x)

2

5 4 3 2 1 0

2

4

1

2

3

4

5

x

g(x)

Exercise 2I — Composite functions and functional equations 1 a ii f (g(x)) = 2 x + 3 − 1, domain [−3, ∞) 1 b ii f (g(x)) = , domain R |x|+3 c ii f (g(x)) = 3(x2 − 2)3, domain R d ii f (g(x)) = x 3 , domain R

50 40 30 20 10

50 40 30 20 10 0

10

1 2 3 4 5

x

0

10

5 10 15 20 25 x

c f =

340 λ

2

c a = 2, b = −3.2, y = 2x2 − 3.2 5 y = 14 x3 − 12 6 a

f 1000 800 600 400 200 0

b

2 4 6 8 10 L

f 1000 800 600 400 200 1 0 0.5 1 1.5 2 2.5 3 3.5 — L

7 a

b I =

I 250 200 150 100 50 0

1

2

3

270 d2

4d

Answers 2i ➜ 2J

e ii f (g(x)) = (x2 + 1)(x2 + 3), domain R 2 Teacher to check 3 Teacher to check 4 a i f (x − y) ≠ f (x) − f (y) ii f (x − y) ≠ f (x) ÷ f (y) iii f (x) + f (y) ≠ (x2 + y2) f (xy) x f ( x) iv f ( )) = y f ( y) v f (xy) = f (x) f (y) b i f (x − y) ≠ f (x) − f (y) ii f (x − y) ≠ f (x) ÷ f (y) iii f (x) + f (y) ≠ (x2 + y2) f (xy) x f ( x) iv f ( )) = y f ( y) v f (xy) = f (x) f (y) c i f (x − y) ≠ f (x ) − f (y) ii f (x − y) ≠ f (x) ÷ f (y) iii f (x) + f (y) ≠ (x2 + y2) f (xy) x f ( x) iv f ( )) = y f ( y) v f (xy) = f (x) f (y) d i f (x − y) ≠ f (x) − f (y) ii f (x − y ) ≠ f (x) ÷ f (y) iii f (x) + f (y) = (x2 + y2)f (xy) x f ( x) iv f ( )) = y f ( y) v f (xy) = f (x) f (y) e i f (x − y) ≠ f (x) − f (y) ii f (x − y) ≠ f (x) ÷ f (y) iii f (x) + f (y) ≠ (x2 + y2) f (xy) x f ( x) iv f ( )) = y f ( y) v f (xy) = f (x)f (y) f i f (x − y) ≠ f (x) − f (y) ii f (x − y) = f (x) ÷ f (y) iii f (x) + f (y) ≠ (x2 + y2)f (xy) f ( x) x iv f ( )) ≠ f ( y) y v f (xy) ≠ f (x) f (y) 5 a > 0 1 6 f  g does not exist; g  f (x) = +2 x − 2 +1 7 Restricted domain of g(x) = R \(−1, 1)

where x ∈ (–3, ∞) 9 Teacher to check 10 Teacher to check 11 x = 4.4305 or 0.902 832, maximum 4.23 when x = 4.43 12 Reflection in the y-axis, translated +1 parallel to the x-axis Exercise 2J — Modelling 1 a iii b ii c iv d i e v 2 a y = ax3, a = 0.3 b y = ax2, a = −6 a d y = a x , a = 1.6 c y = 2 , a = 2 x a f y = ax3, a = −1.5 e y = , a = 5 x 3 D y y 4 a b

8 y = 3 x + 4 9 a $ 11 10 9 8 7 6 5 4 3 2 1 0

Price

9

2

4

6

b p = 2 m + 4

8 10 Month

c 10.63, 10.93

Answers

665

12 a a = −8, b = 1

Chapter review Short answer 1 a (3, −4) b Domain: R, range: y ≥ −4 c y 14

1− x +2 8 (1 − x )3 1 − x ii y = + +2 512 8

4

x′ y′ 4 x2 2x ,y = b y = 2 − a 2 a a 2 c a = 2, y = x - x 14 g:(− 2, 1) → R, g( x ) = x 3 + 2 x 2 + 1

b i y = 2

13 a x =

3 1.6

x

4.4

2 a b = 2, c = −1 b m = 7 – 3 c y = +3 x−4 7 3 f ( x ) = 2 − 3x + 1 Dilation 13 from the y-axis, dilation from the x-axis by a factor of 7, reflection in the x-axis, translation

1 3

2

x

1

3

6 a Reflection in the x-axis, translation by 2 units to the left and 1 unit down, dilation by the factor of 2 along the y-axis −2 b y = −1 ( x + 2)2 1 7 y = −2 ( x − 1)2 8 4 units to the right and 3 units up; y = 3 − y 9 a

4−x

g(x) ( 2, 1)

2

2.7 2

2

3

0

b Range [0, 20] 11 a = 3, b = −3

666

Answers

0

1

x

1

(− x )+1 f

1

2

1

0

1

2 g

1 fg

2

x

16 a Yes b No c Yes 17 c 100 Multiple choice 1 D 2 E 3 C 4 D 5 C 6 E 7 D 8 D 9 E 10 C 11 A 12 C 13 E 14 B 15 C 16 E 17 C 18 D 19 E 20 E Extended response y 1 a i D(4, 6)

d No

f(x)

B C(2, 3) A 2

2 A' B'

2 C'(2, 3)

x

f(x)

b Domain: R \ {−2}, range: (−∞, 2] 10 a y

f(x)

2

–1 2

1.3

(1, 7) ( f g)(x) (1, 4) (1, 3)

15 g( x) = – x + 1, ( fg)( x ) = x y

( 1, 2) ( 3, 2)

y 7 6 5 4 3 2 1

parallel

to the x-axis in the negative direction, translation + 2 parallel to the y-axis 4 y = 1 − 3(x − 1)3 5 a x = −2, y = −1 b Domain: R \ {−2}, range: R \ {−1} − 4 c y = −1 x+2 y d

6

( f + g)( x ) = 2 x 3 + 4 x 2 + 1

x

D'(4, 6)

A(−2, 0), B(0, −2), C (2, −3), D(4, −6) y ii f(–x) f(x) D'( 4, 6) D(4, 6)

x C'( 2, 3) A

2

B

C(2, 3) A' x 2

A(2, 0), B(0, 2), C(−2, 3), D(−4, 6)

4 a 55 m b Maximum height is 30.01 m above ground and maximum depth is 5.38 m. c i Minimum depths and maximum heights would be greater. ii No change iii Would be greater 5 a y = a(x − b)2 + c b b = 9 c Straight line (negative gradient) d a = −0.55, c = 275 e y = −0.55(x − 9)2 + 275 f No, the prices started going down. g $266 000, $261 000 h About 4 months 6 a a = 40 b i The eagle is 48 m above the ground. ii The eagle is 42 m above the ground. c It takes 24.2 s to reach the ground. d The speed is slow to start but increases rapidly as the eagle approaches the ground. e h

y

iii

f(x 2) D (4, 6) C(2, 3) B2

A

2

f(x) D'(6, 6)

C'(4, 3) B'(2, 2) x

A'

A(0, 0), B(2, 2), C(4, 3), D(6, 6) iv y D'(4, 9)

A'( 2, 3)

C' (2, 6) B' 5 B

2

f(x) 3 A

2

D(4, 6)

C(2, 3) x

f(x)

A(−2, 3), B(0, 5), C(2, 6), D(4, 9) v 2f(x) y D'(4, 12)

D(4, 6)

B'4 A 2B A' 2

50 (0, 48.4) (5, 48) 40

f(x)

C'(2, 6)

C(2, 3)

30

x

20

A(−2, 0), B(0, 4), C(2, 6), D(4, 12) f(x) y vi B 2

A'(–3, 1) A

x

D'(3, –5)

1), C(1, −2), D(3, −5) b Add multiples of 2, for example, f (x) + 2, f (x) + 4, f (x) + 6, f (x) − 2 etc. and keep the domain fixed at [−3, 7]. b p > 1, p <

25

t

g

f x

0

6

y 30 20

1 0

2 3 4

10 1

2

x

1.5 1 0.5 0

10

0.5 1 1.5 2 2.5 3 3.5

4 4.5 x

Answers

Answers 2j ➜ 2j

3 2

1 e y = 2 + 1 4x

20

y

−

c ( 12 , 2), (1, 1.25),( 12 , 2), (−1, 1.25)

1

15

x ≥ 3 or x ≤ − 3 a Minimum value of -1 3 b x = 2 a Let P(x) = x3 - 7x - 6. P(-2) = -8 +14 - 6 = 0 so x - (-2) is a factor. Hence x + 2 is a factor. b Q(x) = x2 - 2x - 4 1 5 c < 4

 1 b  2 0 ,  0     0 1   1 

2

10

Exam practice 1 Short answer y 1 f g

25 27

c i Dilation of k from the y-axis, translation of 1 in the negative direction of the y-axis ii x = k, x = 2k d h ∈ [1, 2) 3 a Dilation from the y-axis by a factor of 12 , vertical translation of +1

d

5

f a = −2.5, c = 10 40  , 0 ≤ t ≤ 24 50 + g f (t ) =  t − 25 −  2.5(t − 24)2 + 10, 24 ≤ t ≤ 26 

B(−1, −1),

2 a a = 1, b = 35

(24.2, 0) 0

C(2, 3) C'(1, –2)

B'(–1, –1)

A(−3,

10

D(4, 6)

1 – f(x + 1)

(20, 42)

667

Multiple choice 1 B 2 C 3 E 4 B 5 D 6 C Extended response 1 a v = 100t - 100 b 2:11 pm c 3:42 pm d It predicts that the wind speed will increase indefinitely. It also predicts a negative speed at noon. These problems could be avoided, for example, simply by restricting the domain to be [1, 3.69]. e i If the wind peaked at 5 pm, then (5, 256) would be a turning point. In turning point form the rule for the quadratic model would be v(t) = a(t - 5)2 + 256. − 256 Since v(1) = 0, 0 = a(1 - 5)2 + 256 ⇒ a = = −16 16 so the model is v(t) = -16(t - 5)2 + 256. Using this model, v(3) = -16 × (-2)2 + 256 = 192 km/hour. This is reasonably close to the recorded value of 200 km/hour. ii The predicted value is v(12) = -16 × 72 + 256 = -528 km/hour. Clearly the model is inadequate. f i v2(t) = -8(x - 1)(x - 13) ii Check at t = 3. v2(3) = -8(3 - 1)(3 - 13) = 8 × 2 × -10 = 160 km/hour. Compared with the actual value of 200 km/hour, this model is not adequate. Maximum wind speed predicted at t = 7 is v2(7) = -8(7 - 1)(7 - 13) = -8 × 6 × -6 = 288 km/hour, which is later and faster than the actual data. g v3(t) = x3 - 27x2 + 195x - 169 h i x = 250 1 875 000 ii v (t ) = ( x − 250)2 − 625 000 iii P (t ) = + 1020 ( x − 250)2

CHAPTER 3

Exponential and logarithmic equations Exercise 3A — The index laws 1 a x7

b x5

e x

f 5x4y3

2 a 9

1 125 8 27

d

9 16

b

e

3 a x6 d 37

5 a

y4

b b

1 y4

Answers

e log2

f logx (27) = 3

2 c 243 x y2

c

3 8y 8

b 36n + 1 d 23 b 1 2x c 1 − x2

( 12 ) = −1

4 a 1 d 4 5 a

1 2

b 1 e −1

d

 x2  c log2    y c 2

b log3 (x)

log2 (x)

c log2 (10) c f c f h

− 0.301 3.727 0 5 log10 (x) log4 (x − 2)

Exercise 3C — Exponential equations 1 a 4 b −3 c −5 d −2 e 15 2 a 4 b 5 c −1 d 3 3 a 3 b 1 c 2 d −1 4 a 2, 0 b 2, 1 c 1, 0 d 3 e 1 5 a 0 b 1, 2 6 a 2.322 b −0.912 c −3.325 d 1.365 − e 0.827 f 0.672 7 a x > 1.465 b x ≤ 1.404 c x < −0.683 d x ≥ −0.356 e x < 1.756 8 D 9 E

v 2

b i 81

d

c 5 f −3

6 a 1 b 1 d 1 e 9 7 a 4 b 3 d 2 e 12 8 a 0.477 b 0.861 d −0.322 e 2.161 9 a 0 b log2 (x) d 0 e 3 log10 (x) g log5 (x + 1)3 or 3 log5 (x + 1) 10 C 11 E 12 31.623

5

6 a 26n − 1 c 2n × 33n + 1 7 a 22n × 32n + 1 1 + x2 ( x + 1)2 b 8 a x x4 9 B

668

b log3 (81) = 4 d log5 (125) = x

Exercise 3D — Logarithmic equations using any base 1 a i 25 ii 18 iii 100 iv 26

1 x6

c 216

5 5 6x 4 y 3

20 x

d

d 1 c 5x = 125

3 a log2 (8) = 3 c log4 (x) = 3

b 8 27 e 64

3 4 a 11 2 x y x6

c x10 10 x 4 y 2 g 3 c 3

Exercise 3B — Logarithm laws 1 a 0 b 0 c 1 2 a 24 = 16 b x2 = 25 − d 35 = x e 5 1 = 15

3 a 9b3 15 7 2a 2 b 2

3

1 d x+y

v 0

2 a i b i 3 a i v b i v 4 a i v b i v 5 a i v b i v 6 B

3 2 3 −1 2 −2 5 4 2 128 5 3 2 3

vi

−1 32

vii −624

ii

1 16 −1 9

iii 16

vi

vii

iv 127

−5 2

ii 125 ii 625 ii −1

iii 2 iii 8 iii 12

ii −2

iii

iv 0

ii 6

iii 10

iv 8

ii 9

iii

ii 6

iii 1

iv 2

ii 8

iii 5

iv 4

7 D

1 3

2 5

iv 6 iv 4 iv 0

iv 500

1 8 a 10, 100

b 1000,

e 3, 27 i −5, 3 9 16 Exercise 3E — 1 a 7.389 e 1.284 i 0.405 2 a 1

f 25, 5 j −2, 5

1 10

1 4

c 16,

d 2,

g 1, 16

Exercise 3H — Literal equations  a 1 1 x = 2 log e   , a > 0  5

1 16

h 1, 27

2 D = Z × 3cy log e (3k ) − n 3 x = m 9− p 4 q = 5 5 Proof:

Exponential equations (base e ) b 54.598 c 1.649 d 1.396 f 1.221 g 1.386 h 1.609 j 1.281 − b 2 c 6 d 12

e 2

f 0

g

− 11 6 −

h 1 14

3 a 0.693 b 1.609 c 0.693 e 0.262 f 0.956 g 1.099 4 a 0 b −0.693, 0.549 c d 0 e 1 f g −0.405, 1.099 h 0.251, 0.693 5 a 0.693 b 1.609 c d −0.405, 1.099 e −0.288, 0.693 f 6 a x > 0 b x < 1 c d x ≥ 0.693 e x ≤ 0.792 f g x < 0.288 7 20.00, 0.01 8 0.04 Exercise 3F — Equations with natural logarithms 1 1 a e b e2 c 2 e 2 a 3.695 b 0.906 c e e 1.221 f 1.350 g 1.368 3 a 2 b 5 c 3 e 10 f 2.207 4 a 6 b 23 c 2.5 e 1 5 B 7 ex2 9 2 10 a 2.5

f 1.5

b e

−kt

d −1.386 h 1.386 0.405, −0.693 0.693, 1.609

 xy   xy  = log10  5  − log10  5  z  z  = log10 (1) = 0 4 − 9a 6 a x = 2b 7 b x = 3 log e 20 + 2b log e 2 7 x = 4 log e 2

(base e ) 1 e d e h −0.432 d 4 d

d 6

8 x =

6 A 8 y = x =

1 2

c t =

 x 1 a y = loge    2

ln (2) d 139 days k

b y = loge (x) − 1 1 + log e ( x ) d y = 2 2 − log e ( x ) f y = 3

c y = loge (x) + 1 e y = 2 − loge (x)

d y = loge (x − 2) − 1

d f ( x ) =

( )

f y = loge 2 − x − 1 3

b f (x) = e x − 1

e +1 e f (x) = 2 − ex 2 x

4 a y = ex − 2

b y = e2 − x

d f (x) = e2 − x + 1 e y 5 C

x−3 = 1+ e 2

6 E

c f (x) = e x + 1 f f ( x ) = c y =

2−e 3

x

x−2 e 3

f f ( x ) = e 7 D

1  y − m log e   R  b

10 a =

16b 1 or a = 5 5 × 16b

11 x =

 3 + 9 + 4a  1 log e   k 2  

a2 e2 a x = 1.512 or x = 0.691 b −0.443 c −1.058 a x = 1.557 b x = 2.159 and x = 5.146 c x = 0.191 and x = 1.433 log ( p ) x = 7 log ( 4 )

1− x 3

−2

14 15

5

e

16

e

Exercise 3I — Exponential and logarithmic modelling 1 a 200 b y = 200e 0.6t c 110 g 1 d 2 4 hours 2 a 10 g d 13 days

b y = 10e

−0.18t

Answers 3a ➜ 3i

c y = loge 1 − x 2 e

 2 log e    3

9 x =

13 b =

b y = loge (2 − x)

( ) e y = 2 + log ( 3 − x ) 2

a log e ( 3) + log e ( 2 )

12 x = 9 y 6

2 a y = loge (x − 2)

3 a f ( x ) =

 z2x5    x2y4    y  = log10  3 5  −  log10  3 4  + log10  4   z x   z   y z x   x2y4   z2x5y  = log10  3 5  − log10  7 4   y z x  z x 

1.099 0, 0.693 x < 0.693 x ≥ −1.303

Exercise 3G — Inverses

x e2

 z2x5   x2y4   y  log10  3 5  − log10  3 4  − log10  4  z x   z   y z x

c 8 g

Answers

669

3 a 3, 2 b y = 3 + 2 loge (x) c 5.197 d 10 4 a 1000, 0.05 b A = 1000e0.05t c 1051, 1649 d 14 5 a 500 b 0.03 c P = 500e0.03t d 2240 e 1896 − 6 a 15 °C, 90 °C, 0.05 b T = 15 + 75e 0.05t c 60 °C d 22 e 17 7 a 50, 0.09 b D = 50e0.09t c 78 cm d 8 years − 8 a 100 b 0.002 c M = 100e 0.002t d 90 g e 347 years 9 a 0.23 b N = N0e0.23t c 1000 d N = 1000e0.23t e 6000 − 10 a 0.1386 b I = I0e 0.1386d c 25% d 16.6 metres Chapter review Short answer 1

33 x5

Exponential and logarithmic graphs Exercise 4A — Graphs of exponential functions with any base 1 a Asymptote y = 0, b Asymptote y = 0, domain R, range R+ domain R, range R+ y

(0, 1)

(0, 1) 01 2 x

3 2 1

f (x) =

2

(1, 2)

3 3

4 1 23

5 8 7 4

6 2 − 5 ln (2)

( )

1 2

f (x) = 10x

c Asymptote y = 0, domain R, range R+

d Asymptote y = 0, domain R, range R+

y

y (0, 3)

2 (0, 2)

( 1, 1–21) 2

f (x) = 2 × 10x e Asymptote y = 0, domain R, range R+

0 1

f (x) = 3 ×

2x

y

(2, 1.2) 0 1

–2 –1

)

f (x) = 0.3 ×

2 (0, –21 )

x

2

y 8

f (x) = 0.5 ×

y (1, 8)

8 6 4 2

4

i

(0, 1) 0 1

(1–2 , 10)

(0, 1) 0 0.5

0.5

x

j

Asymptote y = 0, domain R, range R+ y 6

(0, 5)

4

4 (0, 4)

2

2

3 2 1 0 1 2 3

f (x) = 5 ×

x

f (x) = 4 ×

k Asymptote y = 0, domain R, range R+

l

y 4

x 10 2

x

23x

Asymptote y = 0, domain R, range R+ y 6

(3, 4)

4

2 (0, 2)

f (x) = 2 ×

01 2 3

3 2 1

32x

3 2 1 0 1 2 3

x

f (x) = 102x

f (x) = 23x Asymptote y = 0, domain R, range R+ y 6

x

2

10x

h Asymptote y = 0, domain R, range R+

6

1

0 1

2 1

2x

g Asymptote y = 0, domain R, range R+

2

(1, 5)

4

(0, 0.3)

2 x

Asymptote y = 0, domain R, range R+

2

k ε R+ 8 − b 13 ,a ≠ 0 a 14 z × 2y, x, z ε R+ Multiple choice 1 E 2 A 3 D 4 B 5 A 6 D 7 B 8 C 9 B 10 E 11 B 12 D 13 B 14 C 15 E 16 B 17 E Extended response 1 −0.41 x 4 x 5 x6 2 a + + 4! 5! 6! 1 1 1 1 1 b e = 1 + 1 + + + + + 2 6 24 120 720 3 4.75 4 7.8 − 5 a 2.5 b e kt = 0.5 ln (2) d 139 days c t = k 6 a 30, 0.4055 b N = 30e0.4055t c 228 d 9 years − e 1000, 0.1054 f E = 1000e 0.1054t g 590 h 7 years i 485 1 7 x = ln 2 + 4 + k k

Answers

f

2 1

y

k 11  6 − b + ln (2k ) 12 , a ε R \{0} a bεR

(

x

01 2

3 2 1

 1 ln 3  

01 2 x

3 2 1

2x

( 1, 0.2)

loge x + a 3 9 y = 1 − e(x − 3) 2e 2 − 5 10 x = 3

670

y

2

2 −0.849

14 y 15

8 y =

CHAPTER 4

2 (0, 2) x

3 2 1 0 1 2 3

f (x) = 2 ×

x 23

x

2 a y-intercept is 2, no x-intercepts, horizontal asymptote y = 1, domain R, range (1, ∞) f (x) = 2x + 1 y 6

h y-intercept is 18 , no x-intercepts, horizontal asymptote

4

x

b y-intercept is 3, no x-intercepts, horizontal asymptote y = 2, domain R, range (2, ∞)

y 6 4 2

f (x) =

3x

+2

3

4

x

-6,

i y-intercept is x-intercept is 2, horizontal asymptote y = −8, domain R, range is (−8, ∞) f (x) = 2x + 1 − 8 (2, 0)

3 2 1 0 1 2 3 x

5

x

c y-intercept is 2, x-intercept is 1, horizontal asymptote y = −3, domain R, range (−3, ∞) f (x) = 3x − 3 y 4 (1, 0) 2

(0, 6)

4 2

(2, 0)

3 2 1 0 1 2 3 x

2 (0, 3) y 4

4

asymptote y = 1, domain R, range (1, ∞) f (x) = 10x − 2 + 1 y 2 (0, 1.01)

y1 0

y 6

4 (0, 4)

−1,

3 a y-intercept is no x-intercepts, horizontal asymptote y = 0, domain R, range (−∞, 0) f (x) = −2x y x

2

b y-intercept is −1, no x-intercepts, horizontal asymptote y = 0, domain R, range (−∞, 0) f (x) = −10x y 0

1 (0, 1)

2

2

4 x

2

0

2 (0, 1)

2

e y-intercept is 4, no x-intercepts, horizontal asymptote y = 0, domain R, range R+ f (x) = 2x + 2

y 8

1 j y-intercept is 100 , no x-intercepts, horizontal

1

1 2 3 x (0, 2) y 3

d y-intercept is −3, x-intercept is 2, horizontal asymptote y = −4, domain R, range (−4, ∞) f (x) = 2x − 4 y

( 2, 1)

2

y

−

3 2 1 0

f (x) = 2x − 3

(3, 1)

0 1

(1, 5) (0, 3) y 2 01 2 3

3 2 1

y

3 2 1 (0, 1– ) 8

(1, 3) 2 (0, 2) y 1

3 2 1 0 1 2 3

y = 0, domain R, range R +

x

1

3 2 1 0 1 2 3 x

f y-intercept is 10, no x-intercepts, horizontal asymptote y = 0, domain R, range R+ f (x) = 10x + 1 y (0, 10)

y

5 01 2 3 x

1 g y-intercept is 81 , no x-intercepts, horizontal asymptote

y = 0, domain R, range R+ y 1

(4, 1)

f (x) =

3x − 4

2 1 (1, — 10 ) (0, 1)

2 1 0 1 2

x

d y-intercept is 1, no x-intercepts, horizontal asymptote y = 0, domain R, range (0, ∞) − f (x) = 2 x y

4

0.5 (0,

1 –– 81 )

0 1

(–1, 2) 2

3

4

x

2 1

2

(0, 1) 0 1 2

Answers 4A ➜ 4A

4

( 1, 1)

3 2 1

c y-intercept is 1, no x-intercepts, horizontal asymptote y = 0, domain R, range (0, ∞) − f (x) = 10 x

x

Answers

671

e y-intercept is 0, x-intercept is 0, horizontal asymptote y = 1, domain R, range (−∞, 1) f (x) = 1 − 3x y

(1, 2)

10

y 10

(0, 9)

g y-intercept is 3, no x-intercepts, horizontal asymptote y = 2, domain R, range (2, ∞) − f (x) = 2 + 10 x y

4

0 1

y2 x

2

h y-intercept is 2, no x-intercepts, horizontal asymptote y = 1, domain R, range (1, ∞) − f (x) = 1 + 2 x (0, 2) 1

(1, 1–2 )

2

0 1

2 1

y1 x

2

i y-intercept is 0, x-intercept is 0, horizontal asymptote y = 2, domain is R, range (−∞, 2) f (x) = 2 − 21 − x y

y2

2

2 1

(1, 1) 0 1 2

x

j y-intercept is −8, x-intercept is 2, horizontal asymptote y = 1, domain is R, range (−∞, 1) y

(2, 0) 0 1

2

y1 3

f (x) = 1 − 32 − x

x

4

5 (0, 8)

4

f (x) = 2 ×

31 − x

4

1

2

3

4 x

y-intercept is 6, no x-intercept. Horizontal asymptote is the x-axis, domain is R, range R+.

672

Answers

(0.8, 0) 2 3

4 x

(0, 7)

y-intercept is −7, x-intercept is 0.8. Horizontal asymptote is y = 5, domain is R, range is (−∞, 5). 7 a Dilation by factor 13 from the y-axis b Dilation by factor

1 4

from the y-axis

c Dilation by factor 2 from the x-axis d Dilation by factor 3 from the x-axis e Reflection in x-axis f Reflection in y-axis g Translation of 1 unit up h Translation of 3 units down i Translation of 1 unit right j Translation of 5 units left 8 a 4 units up b 2 units down c 2 units to the left d 3 units to the right e 3 units down f 2 units up g 1 unit to the right h 2 units to the left i 5 units up and 1 unit to the left j 2 units up and 4 units to the right k 3 units to the right and 4 units down l 2 units to the left and 3 units down 9 a R, R+, y = 0 b R, (1, ∞), y = 1 c R, (3, ∞), y = 3 d R, (−∞, 1), y = 1 e R, (2, ∞), y = 2 f R, (−∞, 1), y = 1 10 f (x) = 1 − 2x 11 f (x) = 1 − 3 × 2x + 1 12 E 13 B 14 C 15 a 50 b 400 c Asymptote y = 0 p (3, 400)

200

2 1

(1, 2) 0

1

400

y 6 (0, 6)

2

5

5 (1, 0) 0 1 2 x

1 (1, 2— 10 )

(1, 3)

y-intercept is 1 12 , no x-intercepts. Horizontal asymptote is the x-axis, domain is R, range is (−∞, 0). f (x) = 5 − 4 × 31 − x 6 y y5 0

5

(0, 3) 2

0 1 2 x (0, 11–2)

4

f y-intercept is 9, x-intercept is 1, horizontal asymptote y = 10, domain R, range (−∞, 10) f (x) = 10 − 10x y

y 4

2

x

1

2

2 1

2 1

f (x) = −3 × 21 − x

y

y1

1 0

2 1

5

(0, 50) 0 1 2 3 t

Exercise 4B — Logarithmic graphs to any base 1 a y-axis b y-axis c y-axis d y-axis e y-axis f y-axis g y-axis h y-axis

2 a

b

y 2

0

8 10 x

y

d

(6, 1)

(10, 2) 8 10 x

2

(2, 5)

e

f

y 2

y

( 2–3 , 1) 2 x

0 ( 1– , 0) 1 3

0

i

g

1

2 3

4 5 x

Asymptote x = 0 f (x) = log10 (4x) R+, R

y (5, 3)

2 ( 1–2 , 0)

0

1

2 3

h

0

Asymptote x = 0 f (x) = 3 log10 (2x) R+, R 3

a d g j

4 a

b (0, 1)

x3 (4, 0) (5, 1) 2

4

6

f (x) = log2 (x − 3) Domain (3, ∞), range R

y

0

1

1 0

4

6

f (x) = log2 (x − 4) Domain (4, ∞), range R

2

4

8

Asymptote x = 0 R−, R, in y-axis y 2 (0.1, 0)

(1, 1) 6

0

(10, 0) x 8 10

g

1

Asymptote x = 0 R−, R, in y-axis

x

y 2

1

5 4 3 2 1

4

h

y 1 ( —– , 0) 100

3

Asymptote x = 0 R+, R, no reflection

2 x

2

2

Asymptote x = 0 R+, R, in x-axis

(6, 1) (5, 0) 2

f

y

x

2

Asymptote x = 0 R−, R, in y-axis

x4

y 2 ( 1, 0)

5 4 3 2 1

2

1 x

d

1 ( 1, 0) 0 x

5 4 3 2 1

1

f (x) = log2 (x + 1) Domain (−1, ∞), range R d

y

0 x

( 100, 0)

( 10, 1)1

100 80 60 40 20 0

x

Asymptote x = 0 R−, R in x- and y-axis

Answers

Answers 4B ➜ 4B

f (x) = log2 (x + 2) Domain (−2, ∞), range R

y

2

e

x


c

(1, 1) x

1 2 3 4 (3, 0)

2


1 0 1

1

2 1 0 1 2 3 4 x

1 0

x 1 y 1

y x 1

1 0

1

(1, 0) 0 0.5 1 1.5 2 2.5 3 3.5x

x

2

c 3 right f 3 right, 2 up i 3 left, 2 down

( 7–8 , 0)

2 (1, 0) 01 2 3 4

x

2

0 1 2 3 4 x

f (x) = log2 (x + 1) − 2 Domain (−1, ∞), range R

2

Asymptote x = 0 f (x) = 2 log2 (3x) R+, R

x 2 y

y

1

x

2 (0, 2)

b

y

2)

( 1–3 , 0)

2 left b 1 left 4 right e 5 right, 1 up 1 left, 2 up h 1 left, 3 up 1 left, 2 down

( 1, 0)

c

5 a ( 2–3 ,

8

f (x) = 3 + log2 (x + 1) Domain (−1, ∞), range R j

x 3 y

6

(0, 3)

2

x

f (x) = log2 (x + 3) − 2 Domain (−3, ∞), range R

y 2

4 5 x

2

(1, 0) x 0 1 2 3 4

3 2 1

2

–1

Asymptote x = 0 f (x) = log2 (3x) R+, R

1

2 ( 1–4 , 0)

y x 1 4

2

f (x) = 2 + log2 (x + 1) Domain (−1, ∞), range R

(2 1–2 , 1)

1

4

2

2 (0, 2) 1

Asymptote x = 0 f (x) = 5 log2 (x) R+, R

(3.25, 0) 2

f (x) = log2 (x − 3) + 2 Domain (3, ∞), range R

(1, 3)

0

Asymptote x = 0 f (x) = 2 log10 (x) R+, R

(4, 2)

1

h

y x 13

x

2

x3

0

(1, 0)

0

(5.5, 0) 6 8 x

4

y 2

f (x) = 1 + log2 (x − 5) Domain (5, ∞), range R g

y 6 4 2

f

x5

0

1

Asymptote x = 0 f (x) = log10 (x) R+, R

y

(1, 0) 2 4 6

(1, 0) 2 4 6

0

Asymptote x = 0 f (x) = log2 (x) R+, R

2

e

(10, 1)

1

(2, 1) (1, 0) 0 1 2 3 4 5 x

c

y 1

673

6

y 2 1

h 2, dilation by factor 4 from the x-axis, translation 2 units to the right and 1 unit down i 2, dilation by factor 2 from the x-axis, and by factor 13 from the y-axis, translation 1 unit up j 2, dilation by factor 4 from the x-axis and by factor 12 from the y-axis, translation of 3 units up and reflection in the x-axis y 9 x2

1

x –2 (1.5, 1)

1 0 1

2

3

2 3 (1, 0)

4

x

Domain ( 12, ∞), range R, dilation by factor 12 from the y-axis, horizontal translation of 12 unit to the right 7 a

b

y (1, 3)

2

13

– (2 2, 0)

2

x

c

d

y 2 x1 (0, 0)

f

y 2 0 1 2 (2, 1)

4

3

h

x 1 y 2 (3.6, 0) 0 1

2

3

4

8

674

x

4

x2

0 1 2

j

4

(0.24, 0)

6

8

3 4 (3, 1)

x

2

0 2

2 (0, 2)

Asymptote x = 0 11 B

x-axis, translation of 1 unit up i Dilation by factor 3 from the x-axis, translation of 2 units to the left and 1 unit down 2 a b y f(x) ex y f(x) ex 4

4 6 (5, 1)

8

x

Asymptote x = 0 Asymptote x = 0 R+, R R+, R a 2, dilation by factor 2 from the x-axis, translation of 3 units up b 10, dilation by factor 3 from the x-axis, translation of 1 unit up c 10, reflection in the y-axis, translation of 1 unit to the right d 10, reflection in the y-axis, translation of 3 units to the right e 2, reflection in the x-axis, dilation by factor 2 parallel to the y-axis, translation of 1 unit up f 10, reflection in the x-axis, dilation by factor 5 from the x-axis, translation 2 units up g 10, dilation by factor 3 from the x-axis, translation 1 unit to the left, translation 2 units down

Answers

x

h Dilation by factor 13 from the y-axis, reflection in the

(2.8, 0)

x

8 10

g Dilation by factor 12 from the y-axis, reflection in the x-axis, translation of 3 units up

y 4 2

3

5

3

(2, ∞), R

( 2 , 3) 2

2

(3.2, 0)

(0, 2)

6

b Dilation by factor 12 from the y-axis c Dilation by factor 4 from the x-axis and by factor 2 from the y-axis d Dilation by factor 2 from the x-axis and by factor 3 from the y-axis e Translation of 2 units to the right and 1 unit up f Translation of 5 units to the left and 2 units up

Asymptote x = 0 R+, R

x

(−1, ∞), R

0

(1, 2) 2 (105 , 0)

y 5

x

Exercise 4C — Graphs of exponential functions with base e 1 a Dilation by factor 13 from the y-axis

y

0 1


y 5

x3

(−∞, 3), R

2

2

(10, 1.1) (3.6, 0) 0 2 4

4 x

–2

(1.2, 0) 0 1

2

2

( 2, 0)

i

y 2

–1

(−∞, 2), R 10 y

(2, 0) 0 0.5 1 1.5 2 2.5 3 x

(−∞, 1), R

2

4


x

2

g

1 12 3 – (10 3 , 0)

x

(0, log103)

3 2 1 0 1 2

2

(1, 1)

0


e

2

1 0

4 (0, 4)

y 2

4

f(x) ex 1

f(x) ex 3 2 (0, 1) (0, e 3) (3, 1) x

3 2 1 0 1 2 3

2 (0, 1)

(1, 1) (0, 1e ) x

3 2 1 0 1 2 3

Asymptote y = 0 Translated 1 unit to the right, R, R+ c

y f(x) 2ex 4 f(x) ex

(0, 2) (0, 1) x

3 2 1 0 1 2 3

Asymptote y = 0 Dilated by factor 2 from the x-axis, R, R+

d

Asymptote y = 0 Translated 3 units to the right, R, R+ y f(x) 3ex 4 f(x) ex (0, 3) 2 (0, 1) x

3 2 1 0 1 2 3

Asymptote y = 0 Dilated by factor 3 from the x-axis, R, R+

e

f(x) 1 ex f(x) ex

y

4 (0, 2)

f

y f(x) = e2 x 8 (0, e2) f(x) ex

h

(0, 1) 0 1

2 1

i

y 4

f(x)

(0, e) 0 1

2 1

x

2

2

y3

c

y (0.5, e1.5)

4

y 4

Asymptote y = 0 Dilation by factor from the y-axis, R, R+

x

1 0.5

1 3

y 4 (0, 4)

d

Asymptote y = 0 Dilation by factor from the y-axis, R, R+

x

1

y 6 4

2 (0, 2)

h

(0, 3.37)

y3

2 x

1.5 1 0.5 0 0.5 1 1.5

x

1.5 1 0.5 0 0.5 1 1.5

f

y 4 2 1

(0, –2 ) x

3 2 1 0 1 2 3

Asymptote y = 0 Dilation by factor 2 from the x-axis, R, R+

i

y 2

x

01 2 3

Translation of 1 unit to the right and 3 units up, R, (3, ∞)

x

y 4 2

y=2

Translation of 2 units up, R, (2, ∞) y

8 6 4 2

(0, 8.39)

y 8 6 4 2

3 2 1

y5 (0, 5.14) 01 2 3 4 x

Translation of 2 units to the right and 5 units up, R, (5, ∞)

y (0, 8.39) 6

1 1 (0, –4 )

1

4

Answers 4C ➜ 4C

Asymptote y = 0 Dilation by factor 4 from the x-axis, R, R+

3 2 1

2

Translation of 2 units to the left and 1 unit up, R, (1, ∞)

Translation of 1 unit to the left and 2 units up, R, (2, ∞) g

3

1

y=1 x

3 2 1 0 1 2 3

x

3 2 1 0 1 2 3

1 2

0

x

3 2 1 0 1 2 3

y2

2

y 4

2

e

d

f

(0, 4.72)

(0.5, e) (0, 1) 0 0.5

(3, 1)

(0, 3)

y

4

(0, 0.05)

y=1

Translation of 1 unit up, R, (1, ∞) 4

2

1 0.5

c

e

y

b

y 2

Asymptote y = 0 Translation of 3 units to the right, R, R+

x

3 2 1 0 1 2 3

2 (0, 1) 0 0.5 1

0 1 x

2 (0, 2)

Translated 3 units up, reflected in the x-axis, R, (−∞, 3) 3 a

b

1

Asymptote y = 0 Translation of 3 units to the left, R, R+

0 1 2 x f(x) 3 ex

2 1

Asymptote y = 0 Dilation by factor 3 from the x-axis and by factor 14 from the y-axis, R, R+

10

3 2 1

x

0

1.5 1 0.5

y 20 (0, 20.09)

x

ex

(0, 2)

2

4 a

Asymptote y = 0 Reflected in the y-axis, translated 1 unit to the right, R, R+

Asymptote y = 0 Reflected in the y-axis, translated 2 units to the right, R, R+

(0, 1)

f(x) ex

2

x

1

Asymptote y = 0 Dilation by factor 2 from the x-axis and by factor 13 from the y-axis, R, R+

6 4

0

1

y 2

y f(x) e1 x 8

(0, 1)

y 4 2

x

1 2 3

Translated 2 units down, R, (−2, ∞)

4

h

y 2 (0, 2)

(0, 3)

(0, 1) 2

6 2

g

1

3 2 1 0

Translated up 1 unit, R, (1, ∞) g

f(x) ex f(x) ex 2

2 (0, 1)

y1 (0, 1) 01 2 3 x

3 2 1

y 4

4 0

1

x

Asymptote y = 0 Dilation by factor 12 from the x-axis And by factor 14 from

Asymptote y = 0 Dilation by factor 14 from the x-axis and by factor 12 from

the y-axis, R, R+

the y-axis, R, R+

2 ( 2, 0)

3 2 1

0

x

y 1

Translation of 2 units to the left and 1 unit down, R, (−1, ∞)

Answers

675

5 a

y

b

y 4 ( 1, e) 2

(0, 1) 0 1

2 1

c

2

d

0

1

Reflection in the x-axis, translation of 1 unit up, R, (−∞, 1)

f

y

( 1, 1 e) 4 (0, 2)

x

0 1

Reflection in the y-axis, translation of 1 unit up, R, (1, ∞)

y 2

1 0

g

2

x

2

h

y5

i

0

1

i

1

Reflection in the y-axis, reflection in the x-axis, translation of 3 units up, R, (−∞, 3) y

0

1

x

1

2 (0, 2)

Asymptote y = 0 Reflection in the y-axis, reflection in the x-axis, dilation by factor 2 from the x-axis, R, R− y

6

1

(1–2 , 0) 0 1 2

2 1

1

x y 1

(0, 0.95)

R, (−1, ∞) y 7 (2, 0) 0 1 2

1

1

(0, 0.86)

[0, ∞), [−0.86, ∞)

676

Answers

(0, 1)

x y 1

0

1

h

(0.69, 0) x 1

y 2

(0, e)

1 (0, e 2) 0 1

1

j

y 2 (0, 2) 1

y1

( ln5, 0)

1

0

1

x

Reflection in the y-axis, reflection in the x-axis, translation of 5 units up, R, (−∞, 5)

1

0

1

x

x

Asymptote y = 0

1 (0, 1) x

y2

2

x

y 2

(0, 2) ( ln3, 0)

x

y

f

1

1

Asymptote y = 0

y1

Asymptote y = 0

4 (0, 4)

0

1

x

3 2 1 0 1 2 3

y

y3

1

y 4

y1

Translation of 1 unit up, R, (1, ∞)

y 4

y

(0, –2 )

Asymptote y = 0 e

x

1

1

y 2

2

d

x

3 2 1 0 1 2 3

y2

(0, 2) 0

x

1

y 4

1

2

2 1 (0, 1)

2 (0, 2)

Reflection in the y-axis, translation of 2 units up, R, (2, ∞)

y1

(0.69, 0) 0 1 2 x

y1

y 4

c

4 (0, 3) ( 1, 2 e) 2

1 (1, 1 e)

0 1

2 1

y

x

1

(1, e)

y

b

2 (0, 2)

4

y1

0

1

y

8 a

x

Asymptote y = 0 Reflection in the x-axis, R, R−

y 2

2

g

x

Asymptote y = 0 Reflection in the y-axis, R, R+

1

e

1 (0, 1)

2

0

1

0

1

x

Asymptote y = 0 9 D 10 E 11 a 1, 3 b y = e x Exercise 4D — Logarithmic graphs to base e 1 a Dilation by factor 5 from the x-axis and by factor 1 2 from the y-axis b Dilation by factor 2 from the x-axis and by factor 14 from the y-axis c Translation of 1 unit to the left and 3 units down d Translation of 2 units to the right and 1 unit up e Reflection in the x-axis and in the y-axis f Reflection in the x-axis and dilation by factor 1 from 2 the y-axis g Translation of 2 units to the right and 1 unit up, reflection in the x-axis h Translation of 3 units to the left and 2 units up, reflection in the x-axis i Translation of 4 units to the left and 1 unit down, dilation by factor 3 from the x-axis j Translation of 4 units to the right and 1 unit up, reflection in the x-axis k Translation of 1 unit to the right and 3 units up, dilation by factor 2 from the x-axis, reflection in the y-axis l Translation of 2 units to the right and 1 unit down, dilation by factor 3 from the x-axis, reflection in the y-axis

y 1

2 a

d

y 1

Asymptote x = 0 R+, R y

(3, 0)

0

0.5 0.5 1 1.5 2 2.5 3

1

f


b

y

d

x1


e

0

1

x 2

2

4

3

y

y 2

( 1, 0)

2

(0, loge2) 0 2

y 4

2

i

0

2

x 3 y

4

6

0

4

2

0

2

j

x

3 2 1 0 1 2 3 4 5

1 (0, 0.9)

y x 1

1

2

2

4

2

x

8

c

b

( 1, 0) x 0

2 1.5 1 0.5 0.5

2

Asymptote x = 0 (−∞, 0), R

b

0.5 1 1.5 2 2.5 3

x

2

d

y 5 0 0.5

1

1.5

x

e

y 2

2 (1, 0) 0 1 2

3

4


x

(7.39, 0) x

1 2 3 4 5 6 7 (1, 2)


(3, 0) 0 0.5 1 1.5 2 2.5 3 x

2

4

(0, 2)

y

y 0

5

x3

b f (x) = ln (x) − 2  x d f (x) = ln    3 f f (x) = ln (−x)  x h f (x) = ln   + 1  4


Asymptote x = 0 R+, R y 2

y 2

(2, 0) x

3 2 1 0 1 2 3 (0, 1.10)

2

x

x1

(1, 0) (6.4, 0) 01 2 3 4 5 6 7 x

(−∞, 2), R

(−∞, 3), R

(2,0)

2

4 a

1

(1, ∞), R 6

x

(1, 0) 0 0.5 1 1.5 2 2.5 3 x

2


f

y 2 (–1, 0) x 0

2.5 2 1.5 1 0.5 0.5

2

4

Answers 4D ➜ 4d

(4.4, 0)

y 2 0

(1.4, 0)

x

j

x2

x

2 8

y 2

y 4 x1

2 (3.1, 0)

1), R

e f (x) = −ln (x)

6 a

(0, loge3)

3

2

i f (x) = −2 ln (x) − 3

2

2

x3

(−∞,

x2

(0, 0.69) (1, 0)

3 2 1 0 1 2

0 1 2 3 x

x 3 y 2

h

y 2

h

x1

g f (x) = −ln (x + 3)

2

g

y

2

( 2, 0)

x

2

x

Asymptote x = 0 R−, R

c f (x) = 5 ln (x)

0 12 3 4 5 6 7x

f

2

(−∞, 2), R 5 a f (x) = ln (x − 1)

x2

y 5 1 ( –4 , 0) 0

1

5

(1, 0)

(3, 0)

x

x

0


2 1 0 (0, 0.69)

2

x

Asymptote x = 0

(2, 0)

1

1

3 2 1

i

y

2

1

f

5

5

y


y 5

2

x

0

1

2

2

0

0.5 0.5 1 1.5 2 2.5 3

Asymptote x = 0

x

1

g

(e 2, 0)

5

4

( –2, 0)

5

(e 1, 0) x 0

0.5 0.5 1 1.5 2 2.5 3

c

e

(0.5, 0) x 0

0.5 0.5 1 1.5 2 2.5 3

2

3 a

y 1

3


(1, 0) x

1.5 1 0.5 00.5 1 1.5 2

1

x


2 1 ( –2 , 0)

(1, 0) 0 1 2



y

d

y 2

1

(1, 0) x

1.5 1 0.5 00.5 1 1.5 2

1

e

c

( –3 , 0) x

1.5 1 0.5 00.5 1 1.5 2

1

(0.5, 0) x

1.5 1 0.5 00.5 1 1.5 2

1

c

y 1

b


Answers

677

g

y

x 3

h

2 x

(−3, ∞), R y 2

i

b (1.47, 0)

( 2, 0)

2 1 0 (0, 1.10) 2

3

y 2 0

0.5 1 1.5 2 2.5 3

x

2

R+, R

(0.22, 0)

x 0

2 0.5 1 1.5 2 2.5 3

y1 e x 8 6 4 ( 2, 5.4) y x 2 2

6

x 2 y

( 2, 2)

8

0

2

2 4 6 ( 1.45, 0)

y

(1, 0) 0

1

2

1

e

1 2,

( 2, 4.14) ( 2, 4) ( 2, 0.14)

(0, 2.08)

1, f (x) =

1 2

678

Asymptote y = 0

Answers

y2 0

1

1

2

(2, 4) x

Asymptote y = 0 f ( 2, 11.4) y 10

(3, 9.05) f(x) (3, 9)

5

y2 x2

( 2, 7.4) ( 2, 4)

y1 e x (3, e 3)

2 1

0 1 2 3 x

Asymptote y = 0 4 a y 2

(1, 1) 0

0.5

1

2

1.5 (1, 0)

h(x) g(x) f(x) x

Asymptote x = 0 Domain R+, range R h(x) b y 2

(1, 1)

g(x)

x

0.5 0 0.5 1 1.5

2 f(x) 3 loge(x)

Asymptote x = 0 Domain R+, range R c y 5

6

f(x) ( 1, 5.4) (1, 5.4) 4 (0, 4) y1 2ex y2 2e x 2 (1, 0.74) ( 1, 0.74) (0, 2) 0 0.5 1 x

1 0.5

(2, 11.4) f(x) (2, e2) y1

5

2

× 2x + 1

6 E 7 −2, 3 8 4, −1 9 2.9, 2.0, 2.0, y = 2.9 loge (x + 2.0) + 2.0 10 2.0959, 3 11 1.5, 2, 3 Exercise 4F — Addition of ordinates 1 a R \{0} b [0, ∞) c [0, ∞) d [−2, ∞) e R f R \{3} g R \{−1} h (−∞, 1] i R \{0} j [−1, 3] 2 C ( 1, 6.2) y (1, 6.2) 3 a

1

y 10

x

4

5

1 2 0

(2, 2.14) (2, 2) (2, 0.14) x 2

y (2, 9.4) f(x) (2, 7.4) 8 y1 ex 6 4 (0, 1) y x 2 ( 2, 0.14) 2 (2, 2) ( 2, 1.9) x

1 2 0 1 2 ( 2, 2)

x

9 D 10 B 11 C 12 a f: (2, ∞), g(x) = a − b loge (x − 2) b Domain is (2, ∞) and range is R. c h(x) = 4a − 4b loge (−x − 2) or 4a + 4b loge (x + 2) Exercise 4E — Finding equations for graphs of exponential and logarithmic functions 1 0.58, 2.42, f (x) = 0.58ex + 2.42, y = 2.42 2 0.84, −0.33 3 −1.60, −0.92 32 4 A = 15 , B = −19 15

f(x)

Asymptote y = 0 d

2 (0, 2.58)

2

f(x) (1, 9.3) 8 (0, 6) (1, 8.2) 6 y1 3ex y2 3e x 4 (0, 3) 2 (1, 1.1) ( 1, 1.1) x

1 0.5 0 0.5 1

Asymptote y = 0 c ( 2, 7.4) y

4

Asymptote x = 0 R+, R 7 Dilation by factor 2 from the x-axis, translation of 2 units to the left and 1 unit down

y ( 1, 9.3) ( 1, 8.2)

(1, 2)

0 0.5 1 1.5 x

1

5 f(x) log (x) e

Asymptote x = 0 Domain R+, range R

h(x) g(x) 2x

d

y 2

0 0.5 1 1.5 (1, 0)

1

2

8

h(x) g(x) 2–1 x f(x) loge(x) x

1

(1, –2 )

y y = x2 5x 6

Asymptote x = 0 Domain R+, range R e y

0.5 0 0.5 1 1.5

2 (1, 0)

x

3

f(x) loge(x)

2

5

x

0 0.5 1 1.5

y = x3 x2 1

3 2

h(x)

2 g(x) y

10 a

f(x)

y x 2 x

2 y 2 x

y x

1

(2, 3.76)

1 y

b

2 0

1

1

2

x

c

y

y x 3 x

x y 2x x

Domain [−2, 2], range [1, 3.76] 6 E y 7 a h(x)

x

–1 4

y x

f(x)

d

y x2

y x 3 x2

y

g(x) x

b

2

y 2x

(0, 1)

2

x

2

f(x)

Asymptote x = 0 Domain R+, range (0.6, ∞)

y ( 2, 3.76) 4

x

1

g(x)

(1, 1)

0.5

2

y

9

h(x)

Asymptote x = 0 Domain R+, range [1, ∞) f y 2

f(x)

6 5 4 3 2 1 0 1 2 g(x)

(1, 1) x g(x)

2

8 7 6 5 4 3 2 1

y 5 x

y

2 5 y x5 5 x y x5

5

5

x

11 E x f(x) g(x) h(x)

c

y

12 y

d x

1

2

3

4

5

x

2

h(x) y

y loge(x)

1

x

g(x)

e

y x

1

h(x)

Domain x > 0

Answers 4e ➜ 4f

f(x)

y loge(x) x

2

f(x) g(x)

y 4 3

13 f

h(x)

y

f(x) g(x) x

y f(x)

2 1 x f(x) g(x) h(x)

1

1

2

3

4

g(x) 1

2

3

x

h(x)

Answers

679

14 a

y

c

y 0.3 0.2 0.1

4

3

2

1

0

0

x

1

Domain R, range [0, ∞) Asymptote x = 0 b y

0.6

0.2

1 2 3 4 5 6 7 8

6 (0, 3loge 2 3) 5 4 3 2 1

1 0 1 2

2 1 (2 –e , 0) x2

x

Domain R, range [0, ∞) Asymptote x = 0

c

 −1  Domain (0, ∞), range  , ∞  e  x-intercept (1, 0)

Exercise 4G — Exponential and logarithmic functions with absolute values 1 a y

0.4

1 0

y

( 1 e 12 , 0)

1

3 2

0

x

1

2 1

( 1 e 12 , 0)

1 1 0 (0, 2)

3 x = 1

1

2

(3.1, 0)

(2.9, 0) 0

1

b

2

1

x

0

0 x

2

3

4

x

5

1

( 1– , 0) 3 1

1

Domain R\{3}, range R 2 a y

 1  Domain (0, ∞), range  − , ∞  2e  x-intercept (1, 0) y

x

x3

y

0

2

1

Domain R\{−1}, range R c y (0, log (3) 1)

Domain R, range [1, ∞) y-intercept (0, 1)

10

15 a

x 3 4 1 (2 –e , 0)

Domain R\{2}, range R y b

2

1

x

1

1

2 (3, 0) 4 x2

5

x

Domain (2, ∞), Range [0, ∞) b y 4 3

680

1   Domain (0, ∞), range  − ∞,  18e 

x-intercept  1 , 0  3 

Answers

(0, 2) 1 1

0

1 (1 e 12 , 0) x=1

Domain (−∞, 1), range [0, ∞)

x

y

c

c

y 5 4 3 (0, 2) 1

3 2 1

( 2, 1)

3 2 x 3

(0, log10 (3) 1)

0

1

2 1 1 0 1

x

1

y 10 9 8 7 (0, e1 4) 65 4 3 2 1

3

x

(15, 92)

80

(10, 75) 60 (0, 50) (2, 54) 40

(1, 5)

20 0

1

2

Domain R, range [2, ∞) Exercise 4H — Exponential and logarithmic modelling using graphs 1 a 50 koalas b 54 koalas, 75 koalas c d 92 koalas K

Domain (−3, ∞), range [1, ∞)

3 a

y3

1

x

2

0 4

t

12 16

8

2 a 150 wallabies b 162 wallabies, 220 wallabies c W

Domain R, range [5, ∞) y b (0, e4 4)

500 400 300 200 x

(4, 2)

0

(4 ln (3), 0)

(4 ln (3), 0)

Domain R, range [−2, ∞) c y ( 1, 1)

( ln (2) 1, 0)

3

1

2

1

1 (0, 2 e1)

2

3

4

(ln (2) 1, 0) 0

800 (0, 500)

(1, 167)

y2

1 1 0 1 2

2 (ln (6), 0)

3

Domain R, range

(−∞,

4]

x

0 2 4 6 8 10121416 t

5

d e f a c e

7 days No, because the graph approaches the line E = 0. 4 days − T = 80e kt + 18 b 0.277 − d 9 minutes T = 80e 0.277t + 18 f 19 °C T 80 60

(5, 38) (9, 25)

40 20 0

T 18 4

8

12 16

Answers 4g ➜ 4H

y

4 (0, 3) 2 1

(5, 81) (10, 33)

x

Domain R, range [0, ∞) b

8 t

b 167, 81

100 1 2 (ln (3), 0)

6

4

200

1

2

(1, 530.9)

400

e $810 4 a 0.18 c E

(0, 2)

0

(5, 675)

600

0 2

y3

3

1

e 9 years b $530.90

200

y

2

t

d 475 wallabies a $500 c $174.95 d A

x

1

Domain R, range (−∞, 1]

4 a

3

(5, 220) (1, 162) (0, 150) 0 4 9 15

t

Answers

681

5 a f (x) = 2 log10 (x) y

g The egg will never reach 18 °C because the line T = 18 is an asymptote. 6 a A = −1242.67, B = 10 000 b 7139 c 279 weeks d P 10 000 6300 5000

7 8

(1, 0) 0

2

(1, 10 000) (5, 8000) (10, 7139)

0

4

e 6300 C a C = 20n c R → 2000 d y

2 4

6 8 10

2 (1, 0) ( 1, 0)

3 2 1 0 1 2 3 −20

6 120 e 7 a

y

g(x) 2e x 1 4

R 2000 (55, 1992)

R

(0, 2e 4)

(20, 1729)

( 1 loge 2, 0)

P(n) (20, 1329)

(55, 892) y 4

(20, 400)

e f g h i

20

x

55

Teacher to check. − P = 2000 (1 − e 0.1n) − 20n 23 jackets, profit $1339 100 No. Not enough jackets are made before profits begin to decrease.

2

(1, 0) x 2

6 4 2 0

2 (0, 2) y 4

4

y 4

Domain = (−4, ∞), range = R y

x

(5, 3)

2 0

Domain R, range (−4, ∞) 3 f (x) = 3ex − 4 (f g) (x) g(x) f(x) x

b

8

0 (0.33, 0)

2

y

 x + 4 − b g 1(x) = loge  −1  2 

y = h(x)

Chapter review Short answer 1 y

4 a

x

0

(55, 1100)

C

x

2

b Teacher to check

2000

0

x

y

b

12 16 18 20 t

8

(10, 2)

2

y

1 ( –2

1

, 0) 2

3

4

x

Domain R+, range R

Domain = R+, range = R 9 Dilation by a factor of 2 from the x-axis, reflection through the y-axis, a translation of 3 units right and a translation of 5 units up 10 a On 1 January 2006, there are 50 tigers and 400 elephants. b On 1 April 2007, there are 3200 tigers and 50 elephants. c The numbers of tigers and elephants is equal on 1 June 2006 (there are 100 of each). d The elephant population is at risk from 1 September 2007, as the model predicts their population will reach 25 on this date. 11 y ( 2, 0)

(f g) (x)

3

1

1 0

2

g(x) f(x)

3

1

2 3 4 5

x

(0, loge (3))

4 x 3 0

682

Answers

x

Domain (−3, ∞), range (−∞, 0]

12 a R\{−3}

−1

b x-intercepts: x = − 3 + e −1 2

∴ (− 3 + e

y-intercept:

∴ (0, 2 log e (3) + 1)

, 0),

(− 3 − e

2

−1 2

, x = −3− e

−1 2

, 0)

y = 2 log e (3) + 1

c

6 a a = 1 because it is where the horizontal asymptote cuts the y-axis. (0, 0) satisfies the equation, so 0 = 1 + be0 ∴ b = −1. − b Range of h: y ∈ [0, 1 − e 2] −1 −2 − − c h : [0, 1 e ] → R, where h 1(x) = −loge (1 − x) y 7 a

y x 3

4 3 2 1

Multiple choice 1 D 2 4 C 5 7 C 8 10 D 11 13 C 14 16 A 17 19 B 20 22 C 23 Extended response 1 y

1 2,

2 x

0)

B 3 A C 6 D B 9 E D 12 D A 15 B B 18 B A 21 D B

1 2 3 4 5 6 x

2 1 10

2

3

(0, 2loge (3) 1)

6 5 4 3 2 1 1 0 1 1

2 ( 3 e 2, 0)

3

4 ( 3 e

(0, loge (5) 1)

3 2 1

8 9 10 11

− f 1 ( x) =

(5 e , 0)

x 5

5 − ex − 1

b a $5000 b $5256.35 c $6420.15 d $6410.20 e 13.95 years f The first investment, because the interest is compounded continuously. g $9.95 h About $20 a 1 b 0.52, 516 metres c 16 metres d 3.6 kilometres − a P = 101.3e 0.125x b 76.7 kPa a b = 5, c = 6 b x = loge (2) and loge (3) c (0.92, −0.25) d x = (loge (5), 6) e g: R → R, g(x) = −e2x + 5ex − 6 −2x − f h: R → R, h(x) = e − 5e x + 6 g (log (2), 0) y e

(loge(3), 0)

x

0 2

3 2 1

(0, 2)

(0, 1.74) y1 0 1 2 3 x

y 6

Domain R, range (1, ∞) 2 y x2

−

Domain 3 y

(−∞,

(1, 0.3) 0

2), range R

Inverse functions

f(x) y log 2x

Exercise 5A — Relations and their inverses y y b 1 a

y log x x 3 4

3 2 1 10

2

7

5 4 3 2 1 0 123

Domain R+ range R 4 a 10 000 5 y (0, 4) 3 2 1

CHAPTER 5

b 15 000 c

y 3

x

8

{0, 1, 3}, {1, 2, 5} y

d

5

y

y 2x 5 0

1 2

x

(1, 1) 0 1

x

x

{−8, −5, −2, 1}, {−1, 1, 2, 7}

yx 1

2 1

2 0 1

1

5– 2

x

5

Answers 5a ➜ 5A

2

1 2 (1, 0)

−

k(x) = −e 2x + 5e x − 6, domain R, range (−∞, 0.25] − − h y = e 2x + 4 − 5 x + 2 + 7 or y = e2(2 − x) − 5e(2 − x) + 7, domain R, range [0.75, ∞)

4 2 ( 0.7, 0) (0, 0.3) x

3 2 1 0 1 2 3

(loge (5), 0)

Domain R, range (−∞, 4]

R, R

R, R

Answers

683

y

e

y

f 2x 4y 8

2 0

4

y

x

0

5 2

y

y x2 4x x

4

d y = 12 x +

c y=x

5– 2

(1, 1) x

0

x

0

5

( 2, 4)

g

R, R

1

y

h

0

e

2

yx 1

R, [−1, ∞)

j

y

2

1

y (x 1)2

1 0

x

R, [0, ∞)

2 x

y2

m

x

2

4

h y = −1 ± x

g y = ± x +1

y

y 1

y –3 x

1

0

1

x

0

y

[−1, ∞), R i

0

x2 + y2 = 4

j

y=

3 x y

y

x

2 (3, 1)

{−4}, R

2

y

2

0

y 2x3

x

( 3, 1)

(1, 2) x

0 1

[−2,

2], [−2, 2]

R\{0}, R\{0}

k x=2

l

y = −4

y

R, R 2 a {(1, 0), (2, 1), (2, 3), (5, 3)}

0 x

2

4

3 2 1 0 12345

y

0

y

R, {−4}

{2}, R x

{1, 2, 5}, {0, 1, 3} b {(7, −8), (2, −5), (−1, −2), (1, 1)}

m y=

3

1

1 0 1 2 3 4 5 6 7 x

2

{−1, 1, 2, 7}, {−8, −5, −2, 1}

(2, 1)

1 0

5

8

x 2 y

y

Answers

x

0

2

2

684

x

[0, ∞), R

x 4

4

1

1

R\{0}, R\{0}

x

R, {2}

( 4, 2)

[−4, ∞), R

x

1

2 0

x

0

4

(1, 3)

1 0

l

y

y=± x+4−2 y

0

( 1, 3) 3

[−2, 2], [−2, 2]

f

R, R

2

k

+4 y

x2 y2 4

0

−2x

y 3

2

R, R

R, R

y

x

1

1

i

R, [−4, ∞)

R, R 3 A 4 C

2

x

x

5 a

y

x 2

y

b

yx

3

2

2 y

c

y3

x

0

y 2

0

yx

6

y (4, 4)

(3, 1)

4

y

4 ( 2, 0) 0 –2 ( 4, 2)

4

yx

y

f (1, 1)

6

x

2

x

2

0

2

(0, 3)

yx y

h

2

1

(1, 1) x

0

yx

y

i

1

2

yx

yx

y

j

d

( 1, 1)

3

4

l

yx

y

(1, 4)

4

x

0

y

8

yx

x

–4

(–1, –4)

( 4, 16)

n

y

h

yx

(1, –1 )

–1 2

x

3

y

2

0 1

2

x

x

0

o

y

p

yx

2

4

2

4

2 0 2

x

4 2 0

2

4

R, R+

R, R

y

yx 1

j

k

y

y 2

2 3

x 0 –1 2

R+, R

x

2

0

2

x

Answers 5B ➜ 5B

0

i

y

3 x

R, [−16, ∞)

R\{0}, R\{0} yx

2 0 2

x

0

x

1

(3, 3)

y

m

g

y

–1 0 3

x

0

R, [0, ∞)

f

3 0

4

R, [−9, ∞)

x

4

0

( 3, 3)

3

3 x

0

9 x

y

3

y

2

4

k

e

y

yx

4

x

5– 2

0

R, R

(1, 1) 0

x

(4, 0)

R, R

x

0

y 5

0

6

y

g

x

2 3

{−4, −2, 0, 2, 3}, {−2, 0, 1, 4, 6} b c y

2

0

(2, 4)

(0, 1)

x

0 2

yx

yx

(3, 6)

4

2 x

0

e

x

3

d

(1, 3)

Exercise 5B — Functions and their inverses 1 a Function b Function c Function d Function e Not a function f Function g Function h Function 2 a y

x3

[−2, 2], [0, 2]

Answers

685

y

3 a

3 2

2

0 (1, 0) 4 (0, 2)

−

−

2

y

c

y (0, 4)

k

–5 2

y –25 –21 x

0

x

5

2

x

0

4 a i

3

2 1 0

x

2

ii R−, R+ iii R+, R− y d i

yx

4

y

3

x y o x 2 [0, c), R

2

0

ii R, [4, ∞) iii [4, ∞), R e i y

f

3

ii R, R+ iii R+, R i

0

x

3

4

x

3

4 y –x

3

y1 x

0

ii [−3, 3], [0, 3] iii [0, 3], [−3, 3] y g i

(4, 1)

1

y yx

3

y

x

3

yx

3

y=± x −2 [0, ∞), R

0

1

x

4

0

4

x

0 1

ii R, R iii R, R y c i

y=± x+9 [−9, ∞), R

0

yx

yx y o x9 0 x [ 9, c), R

9

0

x1

ii (1, ∞), R iii R, (1, ∞) h i

yx 3

( 4, 1)

x

4

3

R\{0}, R\{0}

y

4

0

x

i

ii R, R iii R, R i

y = o x 16 4 [ 16, c), R

y 5

yx

–8

5

y = ± x + 16 − 4 [−16, ∞), R

Answers

0

5

5 x

ii iii ii iii

yx

1 0 1

y

( 16, 4)

686

y o 4 x2 [0, 2], [ 2, 2]

y = ± 4 − x2 [0, 2], [−2, 2] b i y

y

x

2

0

R, R +

3

g

y 2

y

4 ( 4, 1)

R+, R

y 1–2 ex

(0, 1– )

R, R

d

f

2x

2

R, R

e

x

3

(3, 0) x

0

x

2

y loge ( 2 )

y

j y

4x 3y 12

0

R, R

{−2, 0, 1, 4, 6}, {−4, −2, 0, 2, 3} b

x

0 1 –

( 2, 4)

4

y

2

{( 2, 4), (0, 2), (1, 0), (4, 2), (6, 3)}

x

6

i

( –1 , 1) 1

(4, 2)

2

y

h

(6, 3)

R, [1, ∞) [1, ∞), R [−5, 5], [−5, 0] [−5, 0], [−5, 5]

(1, 4)

x

j

ii [2, 6], [−2, 0] iii [−2, 0], [2, 6]

y 6

i

yx

2

2

6 x

2

0

iii iv v c i d i e i f i ii

(0, 0) R, R R, R − f 1 does not exist − f 1 does not exist − f 1 does not exist − one-to-one, f 1 exists y

y (x 1)3

2

k i

ii ∞), R iii R, (−3, ∞)

yx

y

x 3

(0, 1) ( 1, 0)

2

3 2

0

2

yx

(−3,

y

x 1

3

x

(1, 0) (0, 1)

x

2

3

l

i

y

(3, 4)

4 3 2

3 2

ii [−2, 3), [0, 4) iii [0, 4), [−2, 3)

yx (4, 3)

0

1 2 3 4 5

iii iv v g i ii

(−2.3, −2.3) R, R R, R − one-to-one, f 1 exists y

x

yx

2

5 6 7 8 9

D A E B − − a f 1 : [−3, ∞) → R, f 1 (x) = ±

b f

−1

: (2, ∞) → R, f

−1

x

( x + 3) + 2

( )

= loge x − 2 + 1 3

Exercise 5C — Inverse functions − 1 a i one-to-one, f 1 exists y ii y 4x 1 yx 1

(0, 1) 1 4

( —, 0)

(1, 0)

iii iv v h i i i j i ii

None R \{0}, R \{0} R \{0}, R \{0} − f 1 does not exist − f 1 does not exist − one-to-one, f 1 exists x 2 y

y e4x 2 yx

1 4

y— x — 4 (0, 0.17)

x

1 4

(0, – —)

( 1, 0) x

(0.17, 0) (0, 1) − 1, − 1 3 3)

iii (

(1, 6)

(0.2, 0.2) R, (−2, ∞) (−2, ∞), R − one-to-one, f 1 exists y

yx

x1

Answers 5c ➜ 5c

iii iv v k i ii

iv R, R v R, R − b i one-to-one, f 1 exists y 6x y ii

y 2

y e0.5x 1 yx

(0, 2) (6, 1) (0, 0) y

y1 x

(2, 0)

x

1 — x 6

y 2 loge (x 1)

Answers

687

iii None iv (1, ∞), R v R, (1, ∞) 2 i a, e, g, i, j, k, l, m, n, o − ii a f 1(x) = x 4

−

−

−

−

m f 1(x) =

x e2

b

n f (x) = − o f 1(x) = 12 (e3 − x − 3)

3

Function Domain a b c d e f g h

R [1, ∞) [−3, 3] R R R− [−5, 5] R+

f(x) (0, 3)

Range

Domain

R− R [0, 3] R+ [−10, ∞) (0, ∞) [0, 8] R

−1

R− R [0, 3] R+ [−10, ∞) (0, ∞) [0, 8] R b b b b d f h

b f

−1

c f

−1

d f

−1

e f

−1

: (1, ∞) → R, f

−1

(x) =

f f

−1

: R\{0} → R, f

−1

(x) =

g f

−1

: [−5, 0] → R, f

−1

(x) = 25 − x 2

(x) =

0 (0, 1)

: [−2, ∞) → R, f

−1

(x) = x + 2 − 3

: (−3, ∞) → R, f

−1

(x) =

1 x+3 1 +4 x −1 1 +2 x

x

(3, 0)

yx

c

y

f(x)

x

0 f 1(x)

yx y

d

3 f 1(x) 0

3

x

yx y

e

(0, 1) y x (1, 0) ( 1, 0)

x +1 3

: [−1, ∞) → R, f

(x) =

f 1(x)

x−3

−1

Answers

R [1, ∞) [−3, 3] R R R− [−5, 5] R+

functions b [0, ∞) or (−∞, 0] d (0, ∞), or (−∞, 0) f R\{2} h [−1, 0] or [0, 1] j (−∞, 1] l (5, ∞)

−1

( 1, 0)

Range

E D C [−3, ∞) R+ [−9, 0] or [0, 9] R− or R+

2 a f

: [3, ∞) → R, f

f(x)

y

Inverse of function

4 a B 5 a C 6 a D 7 a (−∞, 0] or [0, ∞) c [−1, ∞) e R g [1, 5] or [5, 9] i (−∞, 5) or (5, ∞) Exercise 5D — Restricting 1 a [0, ∞) or (−∞, 0] c [−3, ∞) e (−∞, 4) or (4, ∞) g [−5, 0] or [0, 5] i [4, ∞) k R

x

6

f 1(x)

yx

+4

5

0

( x −1) e 2

−1

x

−

6 5

−  ( x + 2)  k f 1(x) = −loge    3  x e − l f 1(x) = 3

−

l f 1 : R → R, f 1 (x) = e 2 + 5 3 a B b C c D 4 a y

( )

−

j f 1 : [4, ∞) → R, f 1 (x) = 1 + x − 4 − − k f 1 : R+ → R, f 1 (x) = −2 + loge (x)

e f 1(x) = 3 2 x − g f 1(x) = x2 + 2, where x ∈ [0, ∞) − i f 1(x) = loge x − 1 2 − j f 1(x) = 2 + loge (5 − x)

688

−

h f 1 : [0, 1] → R, f 1 (x) = 1 − x 2 − − i f 1 : [0, ∞) → R, f 1 (x) = x2 + 4

0 f 1(x)

(0, 1)

f

x

y f 1(x) ( 5, 2)

5

2 0

f(x)

yx (2, 5)

x

d A

y

g y4

−

−

−

0

x

1

4 − ( x − 3)2 or

4 x

0

f(x)

yx

−

g 1 : [3, 5] → R, g 1 (x) = 4 − ( x − 3)2 Chapter review Short answer y y b 1 a

1

f 1(x)

−

10 g 1 : [3, 5] → R, g 1 (x) =

3

x4

0

x

2

4

3

(4, 4)

y

h

2 a

(2, 5)

5 3

f 1(x)

(5, 2)

y

b

y

6

6 x

0

6

0 1

4

3

5

x

0

x

2

[−6, 6], [−6, 0] R, R+ 3 a b y

f(x)

yx

2

4

5 a [1, ∞) − − b f 1 : [3, ∞) → R, f 1 (x) = 1 + (x − 3)2 y

x

y 2 y 8

c

f 1(x)

0

1

4 ( 4, 1)

1 x

0

1

2

y

1 0

x

3

( 4, 4)

6 a 4 − − b f 1 : [−5, 31] → R, f 1 (x) = 4 − x + 5 y

( 2, 31) f(x)

yx

10

4

4 x

0

(−4,

4 a ∞) b (−∞, 2] or [2, ∞) 5 a i One-to-one x +1 iii y = 2 iv y 6

4

4

f 1(x) 0

5

4

10

7 a f

−1

: R− → R, f

b x 8 a

−1

c x

y

(x) = 12 x

7 6 5 4 3 2 1 0

2

f

yx

3

4

5

6

7

x

6

iii y =

x

1(x)

1– 2

d Teacher to check

iv

3

x −1 +1 2 y 2

y

((x 2 1))(13) 1 y 2 (x 1)3 1

f(x) 0

2

x

9 a 0 − − b g 1 : [−55, 9] → R, g 1 (x) = 9 − x c [0, 8]

0.4 0.2 0

0.2 0.4 0.6 0.8

1

1.2 1.4 1.6 1.8

x

2

Answers 5d ➜ 5d

f(x)

–1 2

1

b i One-to-one 0

y

y x 1 2

4

d Teacher to check b (−∞, 0] or [0, ∞)

–1 2

c

y 2x 1

2

x (31, 2)

5

c R− or R+

4

Answers

689

c i Many-to-one ii x ≥ 0

y

6 a

−

b f 1(x) = ex − 1 + 2 c m = 1

x2

iii y = x + 1 − 1, x ≥ 0 iv

y y |(x

1.5

x

2 2.4

0 2x

)|

(2, ∞), R y d

1 0.5

R, (2, ∞) yx

y ”x 1 1

0.8 0.6 0.4 0.2 0

0.2 0.4 0.6 0.8

1.2 1.4 1.6 1.8 x

1

(1, 3)

2.4

0.5

4

x

2.4

0

d i One-to-one iii y = -x2 - 6x - 7, [-3, ∞) y iv e (3, 1)

y

2

yx 0.5 1 1.5 2 x

3 2.5 2 1.5 1 0.5 0

2 y x2 6x 7

4

2.4

y (2 x)(.5) 3

(3, 1) 0

6

x

2.4

8

( )

iii y = log e x − 1 3

e i One-to-one iv x

y 3e 1

y 4 3

7

−5 2

8 a b = –1 c

− h 1 :[0,

d

2 1

y ln

y

0

x 1 3

f i One-to-one 2 +2 iii y = 3( x + 1) iv y

9 a f b

1 2

(x 3)

4 1 1 2

3 2 1 0

2

690

Answers

−

R, f 1 ( x ) = 12 log e ( x + 1)

y y x

x

0 1 2 3 4 5 6 7 x 2

y (3(x 2)) 1

ii Restrict domain to (-3, ∞).

g i Many-to-one iii y = 1 − 3 x −1 y iv y

x

1

−1 − :( 1, ∞) →

8 6 2 y (3(x 1)) 2 4 2

0

5 4 3 2 1

2

4

6

8

h

y

1 2 3 4 x

3 2 1 10

2

3

1 − 12 ] → R, e

b Range h( x ) = [0, 1 − 12 ] e log e (1 − x )

− h 1 ( x) = −

y

1 (.5)

(x 1)

1 2 3 4

3

x

( 1, 1) −

c f (− f 1 (2 x )) = Multiple choice 1 B 2 5 D 6 9 C 10 13 B 14 17 E Extended response

−

2x 2x + 1 E A D A

3 7 11 15

C C D D

4 8 12 16

C A C A

± 852 000 x + 3 079 777 − 1127 426 b The inverse is not a function because it is a one-tomany relation (does not the vertical line test).

1 a y =

c a = –2.65 − − d Domain f 1 ( x ), [− 3.61, ∞), range f 1 ( x ), [− 2.65, ∞) e x = 2.87 − 2 2 a f(x) = 3 x + 4 2 4 4 b ( 3 , 3 ) c Domain = [−2, 43 ], range = [−2, 4] y

d

yx

4

A ( –4 , –4 ) 3 3

4 x

0

( 2, 2)

e Heart-shaped 4

f i [−2, 3 ) −1

−

−

ii f : [ 2, g i

[ 43 ,

4 3)

−1

→ R, where f (x) =

− 2( x − 4)

3

c

π 2 a    36  c  13π  d    8  3 4 5 6

g j a d g j a d g j a d g a b

7

8 15

4]

ii Inverse of f (x) is y = ± 3 a (−2, 0.54) and (0, 0) b R, [0, ∞) c y

−

 2π  c    3

e 1.12c

f 1.38c

5

4 f(x) exx2

3 4

A

9 D

10 B

11

13.5 °C, 15.8 °C

Exercise 6B — Symmetry and exact values 3 2

1 a

y0

x (0.54, 2)

−

d

f 1

−

g

x0

[0, ∞), R (0.57, 0.57) (−∞, −2] [0, 0.54]

−

−

e

2 2

h

2 2

2 a

−

b

2 2

3 2

j

y 10 8 y 6 ln (\x 3\) 6 4 2

2 2 1 2

1 2

k −

b −

−

3 3

c −

3 3

f i 1

2 2

l

3 2

c

2 2

f

3 −

3

1 2

e

g

1 2

h

1 2

i

b a = 3 c x = 5.50 d a = 4.97

j

2 2

k

1 2

l

3 3

c

−2.4

CHAPTER 6

d

1

2

3

4

d −

5

6 x

Circular (trigonometric) functions Exercise 6A — Revision of radians and the unit circle 1 a 171.89° b 286.48° c 275.02° d 146.68° e 63° f 54° g 150° h 225°

3 a 0.3 91 10

g −0.3 j

91 10

m 0.3

b −0.5 e

3 2

h 0.5

3 3 −1

f 2.4 i

−2.4

3 2

l 2.4

n −0.5

o −2.4

k

Answers 6a ➜ 6B

2 1 20

4

c

3

8

( 2, 0.54)

d e f g a

c

4.10c h 4.54c i 5.41c 6.11c 0.389 b 0.717 c 0.170 − 0.129 e −0.246 f −0.916 0.966 h 0.927 i −0.940 − 0.996 k 11.430 l 1.732 0 b 0 c 1 − 1 e Undefined f Undefined 1 h 0 i −1 1 k Undefined l 0 1 b 1 c 1 1 e 1 f 1 2 h 5 sin (240°), sin (150°), sin (35°), sin (120°), sin (70°) cos (3.34), cos (1.5), cos (5.3), cos (0.2), cos (6.3)

2( x − 4) 3

yx

4

π b    12 

Answers

691

2

4

2

7  7 49 576 625  24  +  = + = =1  25  24  24  625 625 625 −

−

5 a

3 3 , 2 3

b

,− 3

d

−

2 −

1 2

c 6 a

−

−

d

1 2

or

−

e −

−

2 ,1 2

3 −1 , 2 2

2 2

b

3 2

1

c

−

3

3 2

f 1

3 2

i

−

g

2 2

h

j

3 2

k

2 2

l

3 2

n

3 2

o −1

3 2

b

3 2

c

−

m 7 2

−

8 a −

d

−

2 2

e −

g

1 2

h

j

1 2

k

−

3 2

f

2 2

i

1 2

l

π 5π π 3π , b , 6 6 2 2 4π 5π π 7π c , d , 3 3 4 4 6 a 0°, 180°, 360° b 105°, 165°, 285°, 345° c 45°, 75°, 165°, 195°, 285°, 315° d 20°, 40°, 140°, 160°, 260°, 280° e 62.40°, 117.60°, 182.40°, 237.60°, 302.40°, 357.60° f 39.44°, 140.56°, 219.44°, 320.56° g 26.39°, 333.61° h 101.22°, 258.78° π 7 a b 0, 2π c 2.7184 2 π 5π 13π 17π 25π 29π , , , , , d 0.9772 e 18 18 18 18 18 18 π 7π 9π 15π 17π 23π , , , , , f 12 12 12 12 12 12 5π 7π 17π 19π g , , , h 1.2579, 5.0253 12 12 12 12 3π 7π b 8 a 2 2 16π 11π c d 3 3 e 19.7766 f 6.9266 9 a The particle is 3 metres from O. b It takes the particle 1.99 seconds to reach O for the first time. − 11π − π π 11π 10 a x = , , , 6 6 6 6

5 a

3 3

−1

3 3 −

3

−1

b x =

2

2  − 1   1  1 1 9  +  = + =1 2 2  2  2  10 a 0.7 b 0.3 d −0.3 e −0.7 g 0.7 h −0.3 j 0.3 k −0.7 11 C 12 D 13 a 10 cm/s b 11 cm/s 14 a 0.9 m b 0.3 m

c x = c 2.5 f −2.5 i 2.5 l −2.5 c 12 cm/s c 0.7 m

Exercise 6C — Trigonometric equations 1 a d 2 3 4

692

a d a b c d a b c d

π 3π 5π 7π , b , 4 4 2 2 3π 5π 7π e , 2 6 6 90° b 60°, 300° 180° e 45°, 135° 2.2904, 3.9928, 8.5736, 10.2760 1.1442, 1.9973, 7.4274, 8.2805 1.0701, 5.2130, 7.3533, 11.4962 3.5217, 5.9031, 9.8049, 12.1863 15.58°, 164.42° 137.91°, 222.09° 212.90°, 327.10° 78.83°, 281.17°

Answers

π 7π c , 4 4

c 60°, 120°

11 x = 12 x =

−

−11π − 5π π 7 π , , , 6 6 6 6 − 7 π − 5π

4

,

4

π 3π , , 4 4

11π − 7π π 5π , , , 12 12 12 12

− 7π

9

,

− 5π

9

,

−π

9

,

π 5π 7π , , 9 9 9

13 a x =

π 7π , 6 6

b x =

π 5π 9π 13π , , , 8 8 8 8

5π 11π 17π 25π 29π 35π , , , , , 18 18 18 18 18 18 4 x = –2.0944, –1.5116, 0.1799, 1.0472 1 c x =

Exercise 6D — Trigonometric graphs 1 a 2π, 1 d 2π, 13 g 4π, 3 2π j ,4 3

b 2π, 1 2π e , 2 3 h 6π, 2

c 2π, 4 f π, 3 i π,

1 3

6 a 1, 2π, 0 to 2 y

−2 2 2 , 2π, to 3 3 3

2 a

2

y 0.8 0.6 0.4 0.2

1 0 –1 P

–P 2

3P — 2

b 1, 2π, −2 to 0 y

2P Q

0.5

b 4, 2π, −4 to 4 y

y 3 2 1

2 –P 2

Q

3P — 2

P

2π − , 2 to 2 3

–P 4

3P — 4

–P 2

0

1

2

3

4

PQ

5 4 3 2 1

2 1 0

1

–P 6

–P 3

–P 2

0

2P Q — 3

1 2

2π − 1 , to 3 2

1 2

y 0.6 0.4 0.2 0

0.2

0.4

0.6

–P 6

–P 3

–P 2

π,

−1 3

to

–P 4

–P 2

3P — 4

P Q

0.4

2 P 2P

3P

0.5

1

1.5

2

2.5

4 2π − 3 , 4

y

to 4

4 2 3P P — 2

2P Q

–P 3

2P x — 3

–P 2

2.5 2 1.5 1 0.5

y 3 2 1

0.5 P 2P 3P 4P 5P 6P Q

0 –P

–P 2

4

3P — 4

P

x

2π − , 5 to 1 3

g 3,

y 1

4 2, π, −2 to 2

1

2

3

4

5

y 2 1

–P 2

–P 6

0 –P

P –P 2 1 2

2

π π 5 a to the left, up 3 b to the right, up 1 3 2 π π to the right, down 2 d to the left, down 1 c 4 3

P x

h

1 , 2

0 –P 6

–P 3

–P 2

Answers 6c ➜ 6d

3 4,

0

f 1, π, 0 to 2 y

h 3, 6π, −3 to 3

0

1 –2 –3

4P Q

2P x

3P — 2

y 0.5

1 3

y 0.4

0

0.2

2P Q — 3

4

0

2

4

1 3,

0.2

g 4, 4π, −4 to 4 y

0

2

f

P

–P 2

2π − , 2 to 0 3

e 1,

2

2P x

P

d 2, 2π, 1 to 5 y

y

e

2P x

P

c 2, 2π, −4 to 0 y

0 –1 –2 –3

4

0

0.5

1

1.5

2

c 3, π, −3 to 3

4

d 2,

2P x

3P — 2

–2

0 –0.2 –0.4 –0.6 –0.8

0

2

P

–P 2

2P x — 3

π, 2.5 to 3.5

y 3 2 1 0

–P 4

–P 2

3P — 4

P x

Answers

693

2π g ,1 3 y

i 3, 4π, 1 to 7 y 6 4

2.5 2 1.5 1

2 0

P

2P

4P x

3P

1 0

1.5

2

2.5

j 2, 6π, −3 to 1 y 0 P 2P 3P 4P 5P 6P x

1

2

3

1

2

3

4

1

0 P P 3P P — 5P — 3P — 7P 2P Q

0.5 –4 –2 — 4 4 2 4

1

1.5

1

2

3

b 2π, 1 y

P

–P 2

3P — 2

2.5 2 1.5 1

2P Q

1 0

1.5

2

2.5

c 2π, 3 y 3 2 1

4P — 5P P — 7P — 4P — 9P — 5P11 –P –P –P — —P 2P 6 3 2 6 6 6 3 6 3 6

d 2π, 2 y 3 2 1

–P 2

P

3P — 2

2P Q

–P 2

P

3P 2P Q — 2

2π ,3 3 y

694

Answers

9

g a d g

b 3, 4π, −3 to 3

d 1, 6π, −1 to 1 π f 5, , −7 to 3 2 1, 2π, −1 to 1 h 1, 2π, −1 to 1 1, −1 b 1, −1 c 3, −3 2, −2 e 2, −2 f 3, −3 − − − 5, 3 h 1, 3 i −1, −5

j 5, −3 10 a 2π, 1, −1 to 1 y

0

–P 2

k 2 12 , 1 12

P

3P — 2

2P x

P

3P — 2

2P x

b 2π, 1, −1 to 1 y 1.5 1 0.5 0

0.5 –1 –1.5

–P 2

c 2π, 2, −2 to 2 y

3 2 1

8 a 4, π, −4 to 4 2π − c 2, , 2 to 2 3 e 3, π, −2 to 4

0.5

1

1.5

3 2 1

0

1

2

3

Q

2P Q

P

1.5 1 0.5

e π, 2 y

f

0 –P –P — 3P P — 5P — 3P — 7P 2P Q 4 2 4 4 2 4

j 4π, 1 y

1.5 1 0.5 0

0.5

1

1.5

0

1

2

3

0 P P 3P P 5P 3P 7P 2P Q – – — ——— 4 2 4 4 2 4

i 2π, 2 y

1.5 1 0.5

0

2P Q

1

7 a 2π, 1 y

1

2

3

3P — 2

h π, 2 y

1

0

1

2

3

P

–P 2

–P 2

P

3P — 2

2P Q

2 1

0

1

2

–P 2

P

3P — 2

2P x

l

−3 2 , −4 1 3 3

d 2π, 3, −3 to 3 y

Exercise 6E — Graphs of the tangent function 1 a Period = π y

3 2 1 –1 –2 –3

4 2 0

P

–P 2

3P — 2

2P x

b Period =

e π, 3, −3 to 3 y

π 2

3 2 1

0 P P 3P P — 5P — 3P — 7P 2P x –1 –4 –2 — 4 4 2 4 –2 –3

f

c Period =

2π , 1, −1 to 1 3 y

d Period =

1

0

P

–P 2

3P — 2

2P

π 3

π 4

e Period = 4π

g 2π, 4, −3 to 5 y f Period = 3π

0

2

P

–P 2

3P — 2

2P x

–P 2

P

3P — 2

2P x

g Period = 2π

2π 1 1 i , , to 1 12 3 2 2 y h Period =

1.5 1 0.5 0

0.5

2P P –P — 3 3

4P — 5P 2P x — 3 3

j 2π, 1, 1 to 3 y

P

3P — 2

π π ) + 1 12 y = 2 cos 3 (x − ) + 1 3 4 2π )−2 13 y = 3 sin (x − 3 4 a 1 1 b 12 c 3 metres d 1 metre e 3.00 am and 3.00 pm  πx f 9.00 am and 9.00 pm g y = sin   + 2  6 11 y = sin (x +

0

1

2

j Period = 2π

P–

P– x

P–

P–

6

x

2

3

y 2 1 8

x

4

y 2 1 2P

4P x

3P — 2

3P

y 2 1 x

y 3 2 1 P

0 –1 –2 –3

2P x

y 5

5

x

P– 4

P– 2

P– 8

P– x 4

P

2P x

y 2 1 0

1

2

2P x

P–

Answers 6e ➜ 6e

–P 2

P–

y 2 1

0

π i Period = 4

3 2 1

0

π 2

P x

4

0

1

2

h π, 2, −4 to 0 y 0

1

2

3

4

0

1

2

0

1

2

4 2

–P 2

y 2 1

0 –1 –2

x

1

0

2

4

y 1 1– 3

1–3

0

1

Answers

695

π 3 ii Dilated by factor of 2 from the x-axis and by factor of 13 from the y-axis π b ii Period = 4 ii Dilated by factor of 14 from the y-axis; reflected in the x-axis; translated 1 unit up π c i Period = 2 ii Dilated by factor of 3 from the x-axis and by factor of 12 from the y-axis; translated 4 units down d i Period = 2π ii Dilated by factor of 2 from the y-axis; reflected in the x-axis; translated 2 units down e i Period = π ii Dilated by factor of 5 from the x-axis; reflected in π the x-axis; translated units to the left 2 π f i Period = 2 ii Dilated by factor of 12 from the y-axis; translated π units to the right 4 π g i Period = 3 ii Dilated by factor of 14 from the x-axis and by factor π of 13 from the y-axis; translated units to the left 6 h i Period = 3π 1 ii Dilated by factor of 2 from the x-axis and by factor of 3 from the y-axis; reflected in the x-axis; translated 6 units up π i i Period = 4 π ii Dilated by factor of 14 from the y-axis; translated 12 units to the right and 13 units down j i Period = π π ii Dilated by factor of 2 from the x-axis; translated 3 units to the left and 5 units up y 3 a Period = π 2 a i Period =

2 1 0

1

2

b Period =

π 2

Answers

P– 4

P– 2

3— P 4

P x

P – 4

P– 2

P 3— 4

P x

y 2 1 0

1

2

696

P x

2

y 2 1 0

1

2

3

c Period = π

P–

d Period = π

y 2 1 P –2

0 –1 –2

e Period = 4π

P

x

x =P

y 1 0

f Period = 6π

P– 2

P x

y 1 0

g Period = 2π

P– 2

2.78 P x

y 1 0 1

h Period =

i Period =

π 2

π 4

j Period = 2π

P– 2

y 4 3 2 1

2.21 P x x P

0 P P 5P 3— P 7P P x

1 P–8 P–4 3— 8 –2 — 8 4— 8

2

3

4 y 3 2 1 0 P–

1 24

2

3 y 3 2 1 0

1

2

3

P 7— 24

P – 2

13 —P 24

19 —P P 24

x

P x

Exercise 6F — Finding equations of trigonometric graphs π π 1 a 3, 2, y = 3 sin (2x) b 2, , y = 2 sin ( x) 6 6 π π 2 a 2, 4, y = 2 cos (4x) b 2, , y = 2 cos ( x) 3 3 3 a 12 , 1, 1, y = 12 sin (x) + 1 π π b 3, , −2, y = 3 sin ( x) − 2 4 4 π π 4 a 4, 2, , y = 4 cos 2(x − ) 2 2 π π b 5, 2, , y = 5 cos 2(x − ) 4 4 π π 5 a 2, 1, , −1, y = 2 sin (x + ) − 1 3 3 π π b 3, 1, , 2, y = 3 sin (x + ) + 2 2 2 π π 6 a −1, , 3, y = 3 − cos ( x) 6 6

π π b −1, , 2, y = 2 − cos ( x) 4 4 7 D 8 E Exercise 6G — Trigonometric modelling 1 a 8.5 m, 3.00 pm b 12 hours c 3.5 m d d(t) 8 6 4 2

Exercise 6H — Further graphs 1 a y 1.5 1 0.5 0

0.5

1

1.5

b

0 2 4 6 8 10 12 14 16 18 20 22 t

t

0

2 1.5 1 0.5

d

e

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 t

0.5

5 a 2, 1

f

b h(t) = 18 − 16 cos ( c 5, 34 m e 3 minutes

2π t) 3 d 2 m f 26 m

0

1

2 y 2 1

1

2 y 2 1

g

0

–P 2

P

3P — 2

2P x

–P 2

P

3P — 2

2P x

0 –P 2

P

x

3P — 2

0 –P 2

3P 2P x — 2

P

y 2 1

1

2

h

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 x

π  T = 20 − 10 cos  (t + 21)  12 

2 a

2

0 –P 2

b

h 13°C

2P x

3P 2P x P — 2

y 2P P 0

π  or T = 20 + 10 cos  (t − 15)  12 

3P P — 2

Answers 6f ➜ 6h

c 20 e 3.00 am, 3.00 pm π  f T = 20 − 10 cos  (t − 3) or  12 

0 –P

y 2 1

1

2

b 24

π  or T = 20 + 10 cos  (t + 9)  12  g 10 °C, 20 °C, 30 °C, 20 °C

x

y

1

2

30 25 20 15 10 5

3P — 2

P

–P 2

2 1

 2π  g h(t) = 18 − 16 cos  (t + 1)  3  h y

7 a 10 d 20

0

y 2 1

1

2

b 8 m, 8

2π 6 a 18, 16, 3

2P x

y

0.5

1

1.5

c

2 4 to 6 85.00 10 12 14am 16 18 20 22 d 1.00 am Saturday, 1.00 pm to 5.00 pm Saturday, 1.00 am to 5.00 am Sunday morning e 3.00 am to 5.00 am Sunday 3 a 18 °C, 14.5 °C b 25 °C, January, December c 21.5 °C d 8 months 4 a 0.9, π, 0.9, h = 0.9 sin (π t) + 0.9 b c 1.3 seconds h

3P — 2

1.5 1 0.5

e 1.00 pm and 5.00 pm, 1.00 am and 5.00 am 2 a 10 m, 4 m b d c 3.00 am 10 8 6 4 2

P

–P 2

2P x

P

y 1 0 –1 –2 –3 –4 –5

P– 2

P

P 3— 2

2P

x

Answers

697

c

y 0

2

4

6

8

10

12

d

y 10 8 6 4 2

e

–2

P– 2

2P x

P 3— 2

P

e

0

f

0

P

P– 2

2P

P 3— 2

x

0

5

15

20

25

30

35

P

P– 2

2P x

3— P 2

g

h

y

P

–2P

2P

P 3— 2

x

P 3— 2

2P x

P– 2

P

P 3— 2

2P x

P– 2

P

P 3— 2

2P x

y 3 2 1 0

f

P

P– 2

y 6 4 2 0

y

0

y

y 5 4 3 2 1 0 P– P– 3—P P 5—P 3—P 7—P 2P x 4 2 4

g

y 5

4 a

0

5

10

15

20

h

P

P –2

3— P 2

2P

x

b

–10 –2 –3 –4 –5

3 a

P

P– 2

P 3— 2

2P x

c

y

P 3— 2

2P x

P

P 3— 2

2P x

1

698

Answers

2P

x

P

P– 2

P 3— 2

2P x

(2.03, 0.91) P– 2

P

P 3— 2

2P x

b Domain: R P– 2

P

P 2P x 3— 2

y 3 2 1 0

1

3—P 2

P

(4.91, 2.41)

1

d

0

1

2

y

0

P– 2

y 3 2 1

3 2 1 P– 2

2P x

5 a Domain: R y

y 2 1 0

c

P

P– 2

P 3— 2

y 3 2 1

0

0

1

P

P– 2

0

1

1

b

y 3 2 1 0

y 3 2 1

4 2 4

P– 2

P

P 3— 2

2P x

y 5 4 3 2 (2.03, 0.91) 1 0

1

P– 2

P

P 3— 2

2P x

(3.52, 2.34)

c Domain: R+

y 3 2 1 0 –1 –2 –3 –4 –5

c

y

(2.13, 1.92)

P

P– 2

x

0 (1, 0)

2P x

P 3— 2

x0

y

d

(4.84, –4.69)

d Domain: R+ ∪ {0}

y 3 2 1 0

1

2

3

4

5

( 1, 0) 0 (1, 0) x ( 0.58, 0.38) (0.58, 0.38)

(1.65, 1.28)

P

P– 2

2P x

P 3— 2

e

y ( 1.7, 2.7)

(3.29, 3.59)

( 2, 0)

x

0

e Domain: R y 4 2 0

2

4

P

P– 2

f

2P x

P 3— 2

y (0, 1)

f Domain: R y

(5.57, 7.07) 8 6 4 (.61, 2.61) (2.67, 1.08) 2

0

2

4

6

P

P– 2

y

g

2P x

P 3— 2

1 (0.5, 0.5)

(4.05, 3.98) (1, 0) x

0

g Domain: R y 3 2 (.18, 1.02) 1 0

1

2

(1, 0) x

0

P

P– 2

h

(2P, 3.14)

2P x

P 3— 2

(1, 0) 0 (0.61, –0.18) x

(3.32, 1.8)

h Domain: R (−∞, 2] y

2P 3—P P P–2 10 2

2 ( 3.5, 2.3)

2 x

1

0

1

b

y

P– 2

P

P 3— 2

2P x

b ii f (g(x)) = sin (2 x ); domain f (g(x)) = R+ ∪ {0} iii y

y

2 (0.62, 1) 1

0

( 2, 0) 0 ( 11–3 , 1.09)

x

(0, 2)

(2, 0) x (1, 2.7)

0

1

2

P– 2

P

P 3— 2

2P x

(5.55, 1)

c ii f (g(x)) = 2 sin (

x2 ); domain f (g(x)) = R 4

Answers

Answers 6h ➜ 6h

7 a ii f (g(x)) = cos (loge (x)); domain f (g(x)) = R+ iii y (1, 1)

1 (1.27, .51)

6 a

y

699

y 2 (2.51, 2) 1

iii

0

1

2

P

P– 2

b i ii iii

(5.6, 2) 2P x

P 3— 2

(4.34, –2)

Both f (g(x)) and g( f (x)) are defined. f (g(x)) = x 2 − 1 g( f(x)) = x 2 − 1 Domain is R Domain is R − Range is [0, ∞) Range is [ 1, ∞) y

f(g(x))

d ii f (g(x)) = 2 cos ( x ) + 2 ; domain f (g(x)) = R iii y 2 1 0

1

(0, 1)

P

P– 2

2P x

P 3— 2

( 1, 0) 0

x

(1, 0)

y

g(f(x))

e ii f (g(x)) = (sin (x))2; domain f (g(x)) = R iii y 2 1

0

1

P

P– 2

2P x

P 3— 2

( 1, 0) 0

f ii f (g(x)) = iii y

2cos (2x);

domain f (g(x)) = R

2 1

0

1

P

P– 2

2P x

P 3— 2

x

(1, 0) (0, 1)

c i ii iii

Both f (g(x)) and g( f (x)) are defined. f (g(x)) = 1 − e2x g( f (x)) = e(1− x Domain is R Domain is R Range is (− ∞, 1) Range is (0, e) y

2

)

y1

x2

(0, 0) x

g ii f (g(x)) = cos ( ); domain f (g(x)) = R 8 iii y 2 1

0

1

P

P– 2

P 3— 2

2P x

f(g(x))

(5.01, 1)

y (0, 2.7)

h ii f (g(x)) = 2 sin ( x − 3) + 1; domain f (g(x)) = R+ ∪ {0} iii y

g(f(x))

2 1 0

1

P

P– 2

(2.04, 1)

P 3— 2

2P x

8 a i Both f (g(x)) and g( f (x)) are defined. ii f (g(x)) = ex − 2 g( f (x)) = ex − 2 Domain is R Domain is R Range is (−2, ∞) Range is (0, ∞) iii y f(g(x))

(0.69, 0) 0

(0, 1)

x

y 2

y

d i Only g( f (x)) is defined. ii g( f (x)) = sin ( x ) Domain is [0, ∞), Range is [−1, 1] iii y (2.47, 1) g(f(x)) ( 2, 0)

e i Both f (g(x)) and g( f (x)) are defined. g( f (x)) = cos (ex) ii f (g(x)) = ecos (x) Domain is R Domain is R Range is [ 1, e] Range is [−1, 1] e y iii

g(f(x))

700

Answers

x

x

0

( , 1– ) e

2 –

(0, 0.14) 0

x

0

(0, e) (2 , e) f(g(x)) ( , –1 ) e x 0 2

y 1

g(f(x))

0

1

x

b Amplitude = 3, period = 2π c y

f i Only g( f (x)) is defined. ii g( f (x)) = sin (loge (x)) Domain is (0, ∞), Range is [−1, 1] iii y (4.8, 1) 1

0

1

5

g(f(x))

4

x

3

5 10 15 20

2 1

Exercise 6I — Trigonometric functions with an increasing trend 1 a T(m) P 20 18 16 14 12 10 8 6

7

0

d −

10

20

1 2

30

40

π 2  x 8 a = 3, n = 12 , c = −1; y = 3 sin   − 1  2 9 a y

50

60 t

8

b $20 d April 2008

5 2 0

π 6

c

e 2π

f

b

5π 3 7π k 6 h

b

2 2

2 2

e 0

2 2

h

j 1 3 a 0.85 c 0.85 (8n + 1)π (8n − 1)π 4 , 4 4 5

x

P

3 3

i l

π 4 3π 2 5π 4 7π 4

0

11

P – 2

x

P

3—P 2

x

2P

y 12 y 2x

10

6

f

3 2

i

2 2

6 a y = sin (x) is dilated 3 units from the x-axis, and π translated 4 units to the right and 2 units up.

24

2 1

3 3

(6 n + 1)π π 7π 13π 19π ,x=, , , , 12 12 12 12 12

18

b 10 am, 2 pm, 10 pm, 2 am c Water sports may run between 10 am and 2 pm. 10 y

c

b −0.85 d −0.85

12

6

8

y sin (x) 2x

4 2 0

2

12

y sin (x) P

2P

x

Answers 6i ➜ 6i

g

x

Period =

Chapter review Short answer π 1 a 3 π d 2 5π g 6 2π j 3

−

2P

2

18 964 December 2009 2% $20.12

−

0 P P 3P – – — 4 2 4

c December 2008

25000 20000 15000 10000 5000

2 a

P 3— 2

y 4

2 4 6 8 10 12 14 16 18 20 22 24 m

b 20.9 °C 2 a 100 b M

c d a c

P – 2

2

0

3

0

1

T(m) 12 0.2m 5 cos (m) 6

y 6 4

yx

2 0

2

4

6

y sin x P

2P

x

y x sin x

Answers

701

13 a b c 14 a b

Teacher to check. Show that range g ⊆ domain f . f (g(x)) = sin (x2 + 2) Domain f(g(x)) = R, range f(g(x)) = [−1, 1] amp = 5 period = p y 5 4 3 2 1

5 4 3 2 1 0

1

2

3

4

5

Multiple choice 1 E 3 D 5 B 7 C 9 E 11 A 13 C

1 2 3 4 5 x

2 4 6 8 10 12 14

A C D B D E B

Extended response π π  1 a 5, 5, 6 and , D(t) = 5 − 5 sin  (t + 6) 12  12  b 5°, 1.46° c 8.30 am and 3.30 pm. 8.30 am the next day. d 8.27 am or 8.46 hours after midnight and 3.32 pm or 15.54 hours after midnight. Differences occur due to inaccuracy of graphical methods. π e D(t) = 10 − 5 sin (t + 6) 12 2 a 1.5 m b 19.5 m c 18 m d 60 s 3π e i m/s ii 0.94 m/s 10 f f (m) 19.5 10.5 1.5

0

1

2

30

60

Exam practice 2 Short answer 1 x = 4 12 2 x = -2 3 a f (x) = e-x - 2 + 3 1 b ( 2 + 3, 4) e 1 4 a Dilation of 2 vertically and 2 horizontally, translation π of left 3 π 1 b y = sin 2( x + ) 2 3 π 5π 9π  3π  or 5 x = , 12 12 12  4  6 [3, ∞) 7 x = loge (2) 8 a y f(x) xsin (x)

1

y sin (x)

0.8 0.4 0

1

0.4 0.8 1.2 1.6

Differentiation Exercise 7A — Review — gradient and rates of change y y 1 a b

c

1

f '(x)

0

x

d

y 0

x

– 1–

f '(x)

2

6000

0

e

702

Answers

y

2 4 6 8 10 12 14 16 18 20 22 24 t hours

c 0 at t = 10 and t = 14 d 9:20 am e 2 h 40 min

x

CHAPTER 7

8000

2000

2.4 2.8

b 0 c 1.8 Multiple choice 1 E 2 C 3 C 4 B 5 D 6 B 7 E Extended response a 200 birds b 12 months c 300 birds π d 7 months e a = 200, b = , c = 2, d = 300 6 f 473 g A = 500, k = 0.005 018 P ( x )Q( x ) h Consider the graph of . 500

10 000

4000

2

2

90 120 150 t (sec)

π g d = 10.5 − 9 cos ( (t − 6)) 30 h 17.78 m 3 a m = 8000 b C

y x

2

f '(x) 0

x

0

x

3

f '(x)

y f '(x) 5– 2

0

x

2 a B 3 a

y

g'(x)

b C b

x

0

i i n/a j i n/a 8 a

y

x

0

g'(x)

c

y

g'(x)

d

y

0

x

0

g'(x)

ii (− ∞, 0) iii (0, ∞) ii (− ∞, 0) iii (0, ∞) y b

y 1

f '(x)

0

x

c

x

0

y

e x

0

0

b A b

2

x

0

f '(x)

1

g

f

5

f '(x)

i

f

y

x

y

f '(x)

0

x

y

g'(x)

2

1 0

x

j

y f '(x)

2

x

0

f '(x)

ii iv iii iii iii iii ii iv iii ii

c E R n/a R iv (−1, ∞) iv (−∞, 4) iv R\{0} iv (−2, 3) n/a n/a iv (− ∞, −1) ∪ (2,

n/a n/a n/a n/a n/a ∞)

9

a d g j

x

y 1

R\{2} R\{−1, 2} R\{−2} R\{−5, 4}

x

b R\{0} e R\{3} h R\{2}

0

1

x f '(x)

c R f R\{0} i R\{5}

Exercise 7B — Limits and differentiation from first principles 1 a 6 b −7 c 8 d 4 e −6 f 14 g 24 h 3 i 2 2 a 3 b 3 c −2 d 4 e 1 f 7 g 12 h 27 i −6 3 a 5 b −6 c 5 d 35 e 48 f −10 g 9 h 0 i −1 4 4 + h 5 E 6 a h b 0 7 4

Answers

Answers 7a ➜ 7B

D b C i Not applicable (n/a) iii n/a i n/a ii n/a i x = −1 ii (− ∞, −1) i x = 4 ii (4, ∞) i x = 0 ii n/a i x = −2 and x = 3 iii (− ∞, −2) ∪ (3, ∞) i x = 2 ii R\{2} i x = −1 and x = 2 iii (−1, 2) iv n/a

x

3

f '(x)

h

f '(x)

0

x

0

f '(x)

6 a 7 a b c d e f g h

y

x

4

y

3

g

x

4

x

y 0

y

0

f '(x)

e

g'(x)

g'(x) y

d

y

0

y

x

0

f '(x)

c

f'(x)

2 0

y

3

3

0

g'(x)

4 a B 5 a

x

g'(x)

y

2

0

x

3

1

e

d

y

iv x = 0 iv x = 0

703

8 a 9 a 10 C 11 a d 12 a d

Approx. −4 Approx. −10

b −4 b −10 b e b e

3 2x − 2 −4 3x2 − 4

c 2x + 6 f 3x2 c 6x + 8 f −2x − 2

2x −6x 2x + 3 5 − 6x2

Exercise 7C — The derivative of x n 1 a c e g i 2 a c

dy = 6x5 dx dy = 20 x 3 dx dy − = 12 x 2 dx dy 3 2 = x dx 2 dy =0 dx f ′( x) = 12x2 + 5 h′( x) = 35 x2

b d f h j b d

e g′( x) = 77x10 + 30x4 g f ′( x) = −6 + 6x − 12x2 i h′( x) = 2x + 3 − 6 3 a x4 1

c

2 3x 3

1

e g

2 x x

2

2

−1

2 + x 3 3

i 1 + 2 x −

k

4 5x3

m x 2 − 4 − 3 x 10x4

4 a 5 D

−4

− 10

8 a

−5 8

9 a

1 3 2

−

g h

10 a g′ ( x) =

b Undefined b 1 2

15 8716

+ 4

53

c 467 54 b

1 43

d

1 412

3x 3 c

1 412

704

i iii i iii i iii

x2 + 2x + 1 2(x + 1) x3 + 3x2 + 3x + 1 3(x + 1)2 4x2 + 4x + 1 4(2x + 1)

Answers

i

− 4 u5

3 2u 2

3

1 2 −4 −

1

−6 4 3

3 −

1

2

iii

ii 5

iii

ii 3

iii

−

2 (2 x − 5)2

8 (4 − 2 x )5 5 2 5x + 2 −

9 3

2(3 x − 2) 2

ii 4x + 5 i 15u4 iii 15(4x + 5)(2x2 + 5x)4 − i −2u 3 ii 4 − 6x − − iii 4(2 − 3x)(4x − 3x2) 3 x2 − 1 i i 6u5 ii x2 5 1  6( x 2 − 1)  x +  x  iii 2 x − − j i −16u 5 ii −6 iii 96(5 − 6x) 5 4 C 5 A 6 B 7 a 32(8x + 3)3 b 6(2x − 5)2 3x c −15(4 − 3x)4 d 3x 2 − 4 −2 2 − e ( x − 2)( x 2 − 4 x ) 3 f −2(6x2 + 1)(2x3 + x) 3 3 5

1  1  g 6  1 + 2   x −   x x  8

−2

h −(2x − 3)(x2 − 3x)

−2

ii 3x2 + 6x + 3 ii 8x + 4

c

ii 2x + 2

81x2 + 54x + 9

1 3

18x + 6

iii

2 u −

24x2 + 24x + 6

ii −2

(4 x + 7)3 − 3 9 a 3 (6 x − 5) 2 10 a 8x7(2x + 5)(x + 5)7

11 Teacher to check 12 a b c

e i f

1

7 3 4

d i

1 5x 4

1 + 2x f x2 −3 h x 2 −3 j 4 x 2 −1 6 − 3 l 3 x x2 1 1 − 5 n 2 x 4x 4 b 150 6 C

− 4x

−3

−1

f f ′( x) = 2x4 + x2 h g′( x) = 14x − 4 j f ′( x) = 9x2 − 12 3 b 2 x d

dy = 6x dx dy = 20 x19 dx dy − = 5 dx dy 4 3 = x dx 3 dy = 40 x 4 dx g′( x) = −10x + 6 h′( x) = −3 + 12x + 3x2

d i 8x3 + 12x2 + 6x + 1 ii iii 6(2x + 1)2 e i 9x2 + 6x + 1 ii iii 6(3x + 1) f i 27x3 + 27x2 + 9x + 1 ii iii 9(3x + 1)2 13 na(ax + b)n − 1 Exercise 7D — The chain rule 1 a i 5x − 4 ii b i 3x + 1 ii c i 2x + 3 ii d i 7 − 4x ii e i 5x + 3 ii f i 4 − 3x ii 2 B 3 a i 2u ii iii 6(3x + 2) b i 3u2 ii iii −3(7 − x)2 − 1 c i 2 ii u

3x 2 5( x 3

+

+ 4x

2x2

4 − 7) 5

b 3 x x 2 + 2 b 2x(3x2 − 2)(x2 − 2) d 3x(4x2 − 3) 2 x 4 − 3 x 2 + 1

11 37 500

7 500

12

13 a 2

b

c 1 7 14 4

d 1

15 9

16 4

17 7

2

or 0.014 x −1

x2 − 2x + 1

18 5

Exercise 7E — The derivative of e x

1 2

a d g a d g

3 4 5

j A a d g a

10e10x

b e h b e h

−e−x −2x

12e 6e6x − 2 −8e7 − 2x −90e6 − 9x x

e2

+1

k

x 1 3 3e 6e3x

c f i 22e c 10e5x + 3 f 10e6 − 5x − i −42e 7x

e0.2x −6e8 − 6x −24e8x + 1 −15e3x + 4 −e

2−

x 3

l

x +5 −e 4 −

2

2

2

−2 x

d

e (2 x − 3)e6 −3 x + x g 3(8 x − 7)e 4 x

2

−7 x

− 5e 2 − 5 x 2

+3 x − 2

h 10(1 + 3 x )e1− 2 x −3 x

i 6(2 x + 1)2 e( 2 x +1) 3

j

− 2( x + 2)− 3 e( x + 2)− 2

l

1 3

−

4 ( 4 − x )3 e ( 4 − x )

3e

2

3x + 4

2 3x + 4

2 3

d

8 8x − 1

e

log e (x)

1 x 1 d x 1 h x

d

3x 2 + 4 x − 7 x3 + 2x 2 − 7x

d

4 x3 − 6x 2 + 2x 1 e 2x + 1 x 4 − 2 x3 + x 2

f

2 4x − 3

g

5 3(5 x + 2)

h

i

j

12 3x − 2

k

7 a

4x x2 + 1

b

4x x2 − 3

c

6( x − 1) ( x − 3)( x + 1)

d

6 x+2

9

− 18 7

−

10 5x + 8

c C 3 3x − 4

−

5 5 or 3 − 5x 5 x − 3

1 1 or x−2 2− x

g

− 7 7 or 7x − 4 4 − 7x

16 i 2x − 1

h −

48 j 12 x + 5

1 x+3 −

3 4 + 3x

l

2

d

1 3

 x cos    3

f

2 3

 2x  cos    3

+ 3 x )2

2 a −3 sin (3x) d

e

 πx 4 sin  4  e

a d a a a

1 8

 −x sin    8

30 5x + 2

63 k 8 − 9x

c

−1 3

 x sin    3

f

−2 5

 2x  sin    5

b −4 sec2 (−4x)

3 a 2 sec2 (2x)

4 5 6 7

 −x cos    2

b 2 sin (−2x)

−π

 x sec2    5 A B cos (u) −sin (u) 2 cos (2x + 3)

−1 2

1 5

b e b b

 3x  d − 43 sec2    4 E c D C 4 c 4 cos (4x + 3) 3 c −3 sin (3x + 1) b −7 cos (6 − 7x)  3x + 2  cos   4 

c 5 cos (5x − 4)

d

3 4

e 10π2 cos (2π x)

f

−3 2

8 a −3 sin (3x − 2)

b −4 sin (4x + 7)

c 5 sin (6 − 5x)

d

−

f

−

10 95 11 15 13 1 log ( x ) 12 a e e or 1 x b i 1 ii 1 iii 1 iv 1 The gradient is always 1 since e loge ( x ) = x. 2 log ( x ) 13 a e e or 2x x b i 2 ii 10 iii −4

c

c

3 5(3 x − 2)

e −40π2 sin (10π x)

 3x  cos    8

 2x + 3 sin   3  − f 12 sin (−2x) −2 3

Answers

Answers 7c ➜ 7g

Exercise 7F — The derivative of 1 1 a 4x b 4 c u 1 1 1 2 a b c x x x − 3 1 6 e f g x x x − 4 5 i j x x 3 D 4 a A b D 2 6 5 a b 2x + 5 6x + 1

c

x2

Exercise 7G — The derivatives of sin (x), cos (x) and tan (x) 1 a 8 cos (8x) b −6 cos (−6x) c cos (x)

4

e( x +1) n 2x(2x + 3)(x + 3)e( x 3( x + 1) − 6 B 7 −20e 8 −6e2 9 −25e 3 m

2x +3

b

2

f 3( x 2 + 1)e x

2

4 x

8 E

2ex b 9e3x + 6e2x c −20e 4x + 10 − − 3ex − 2e x e 6e2x − 7e 7x f 36e9x − 2ex ex h 20e5x + 4x (2 x + 3)e x +3 x b (2 x − 3)e x −3 x − 1

c 2( x − 1)e x

k

x 1 4 4e −20e−5x −11x

6 a

705

2 cos (2 x + 1) 5 c cos2 (5 x − 2) 3 e 2 − cos ( x ) 9 a

2

−1

b

cos (8 − x ) 2 d cos2 ( x + 1) 2

2(2 − x) sin (x2 − 4x + 3) (2x − 5) cos (10 − 5x + x2) ex cos (ex) −(2x + 7) sin (x2 + 7x) 4 − 2x e cos2 (4 x − x 2 ) 2x + 3 f cos 2 ( x 2 + 3 x )

10 a b c d

o

n o p q r 5

3 4

e

− 3x

−

− 3 xe 3 x 2 x − − −4 6x sin (2x + 3) + 4x 3 cos (2x + 3) − 6 xe 2 x − −2x 2e loge (3x2 + 5) + 2 3x + 5 −3x 3x 8x + 12e − 2xe + 3 x 2 e −3 x − 3 4x + 9x2 + 16x − 18

+ 2 loge (4)

Exercise 7I — The quotient rule

11 −3.745 2 0 1 13 a i cos (x) esin (x)

ii

b i −sin (x) ecos (x)

ii

1 3e 3 2 e or 2 2 −

1 e 2

3 2

cos ( x ) or cot (x) ii 3 sin ( x ) − 1 − sin ( x ) d i or − tan (x) ii cos ( x ) 3 Exercise 7H — The product rule 1 a u = x + 3, v = 2x2 − 5x du dv b =1, = 4x − 5 dx dx c 6x2 + 2x − 15 2 a 4x2 + 12x2 loge (6x) or 4x2 (1 + 3 loge (6x)) 3x − 2 + 3 loge (2x) b x 3 D 4 a cos (x) − x sin (x) b 3 sin (x) + 3x cos (x) c 3ex + 5xex

Answers

 4x   4x  cos   cos (x) − sin   sin (x)  3  3

6 10 7 17.279 26 8 e 9 6 − 2π

p −2(x + 1) sin (x2 + 2x) + 3 cos (3x − 9)

706

4 3

e4 x − 3 x − − l −20e 5x sin (2 − x) − 4e 5x cos (2 − x) − cos (6 x ) 6 sin (6x ) − m x 2 x3

 x sin    4

c i

4

k 4e4x − 3 loge (6x) +

−

1 2

d e f g

j

1 sin (loge (x)) x h 4e4x cos (e4x) 1  1 i sin    x x2 2 j cos (loge (2x − 1)) 2x − 1 − 3x k 6e sin (2e3x) 3 l − sin (loge (10x)) x 4 x (3 x + 4) m cos2 ( x 3 + 2 x 2 ) 24 n  − 3x  5 cos2   5 

g

−5e3x

− 33xe3x 5x cos (3x + 1) − 3x5 sin (3x + 1) 6x2 loge (7x) + 2x2 − 2e 2 x − −2x 2e loge (2x − 5) + 2x − 5 40 log e (5 x ) 8 tan (5 x ) h + x cos2 (5 x ) i 5 cos (2x) cos (x) − 10 sin (2x) sin (x)

du dv = 1, =1 1 a u = x + 3, v = x + 7 b dx dx dy 4 c = dx ( x + 7)2 2 a u(x) = x2 + 2x, v(x) = 5 − x b u′(x) = 2x + 2, v′(x) = −1 − 2 10 x + x + 10 c f ′( x ) = (5 − x )2 2 3 sec (x) −2 3x 2 + 4 x − 4 b 4 a 2 ( x − 4) (3 x + 2)2 c

− (sin

( x ) + cos ( x )) ex

e 2 x (2 x − 1) x2 2 g cos2 (2 x ) e

d

33 (10 − x )2

f

3 x (2 log e (4 x ) − 1) (log e (4 x ))2

h

x 2 + 2 − 2 x ( x + 1) log e ( x + 1) ( x + 1)( x 2 + 2)2

i

e3 x + 2 (3 cos (2 x ) + 2 sin (2 x )) cos 2 (2 x )

j k l m n

−5x

3 a −e

− 4( x 2

+ 2) ( x 2 + 3 x − 2)2 − 10 x 3

b

+ 6 x 2 − 35 x + 7 e5 x

3x 2 + 5

d e

2 x3 − 9( x + 3)e − 3 x

(3 x

f

+ 8)2

4[ x − 2 − 2( x − 1) log e (8 x )] x 2 ( x − 2)2

x cos ( x ) − 2 sin ( x ) 2x2 4[ x sin (3 − 2 x) − cos (3 − 2 x )] p x3 − 3e 2 − 7 x (7 x + 22) q ( x + 3)2 4( x − 2)e 2 x r (2 x − 3)2 o

5 D 7 E

6 D 8 −3

9 Does not exist as f ′( x) = 10

4 (1 − log e (10)) 75

15 0

Exercise 7J — Mixed problems on differentiation 1 a C b P c Q d Q e P f Q g C h P i C j C k Q l C m C n Q 1 2 a x b 3 (sin (x) + x cos (x)) − 3 x − 14 c 4 x3 4 (cos ( x ) + x sin ( x )) d cos2 ( x ) e e5x (5 sin (x) + cos (x))

j 2 sin (x) cos (x)

l

3−x ex

g h

2 (3 x + 1)3 3 x +1 2(3 − x) sin (x2 − 6x) ex (cos (2x) − 2 sin (2x)) 2 cos2 (2 x ) − cos ( x ) sin 2 ( x ) 3 cos (3 x ) sin (3 x )

i 4(6 x − 5) e3 x − 5 x + 2 j 3x2 + 8x − 2 k ex (x2 + 2x + 3) (4 x 3 − 9 x 2 − 50)(2 x + 3)4 l ( x 3 − 5)2 − −4x m 4e [cos (4x − 3) + sin (4x − 3)] n −6 cos (3x) sin (3x) − 3 sin (3 x ) o cos (3 x ) 2

2(2 x 4 cos ( x 4 ) − sin ( x 4 )) x3 − 2 q sin 2 (2 x ) p

r

20 [log e (5 x − 1)]3 5x − 1

1 − 2 log e ( x ) 2x2 −5 x+3 t cos x−2 ( x − 2)2 u 3x4 [5 cos (2x + 1) − 2x sin (2x + 1)] s

( )

3 − 4 log e x 32 2x3 w (sin (x) + x cos (x)) ex sin (x) − x −6 cos (x) sin (x) − 7e 7x − 3x2

v

2 x, x < −1 or x > 1 4 a i  − −  2 x, 1 < x < 1

y

ii f(x) {x2 1{

1

2x log e ( x )

m −sin (x) ecos (x) 1 or sec2 (x) n cos2 ( x )

x

2 − 2e 2 x z −3 sin (x) cos2 (x) y 18 cos (6x ) +

6 5 4 3 2 1

5 4 3 2 1 10

2

3

4

f '(x) 2x, x < 1

f '(x) 2x, x > 1

1

2

3

4

5

x

f '(x) 2x, 1< x < 1

Answers

Answers 7H ➜ 7j

x (2 x + 9) log e ( x ) − x 2 − 9 x + 8 x (log e ( x ))2 g 2(2 − x) sin (x2 − 4x) − 1 h e x ( − loge (5x)) x cos ( x ) i or cot (x) sin ( x ) f

k

c

[5 cos (4x − 7) + 4 sin (4x − 7)] 3x + 8

707

2 x + 2, x < − 2, x > 0 b i  −2 < x < 0  − 2 x − 2, ii

2

y

7 6 f '(x) = 2x 2, 5 f (x) {x2 2x{ x>0 4 3 2 1 x 1 2 3

5 4 3 2 1 10

2 f '(x) 2x 2,

3 2 < x < 0

4

5 f '(x) 2x 2, x < 2

π  2 cos (2 x ), 0 < x < 2 5 a i   − 2 cos (2 x ), π < x < π  2

ii

4x or 4x sec2 (2x2 − 3) cos2 (2 x 2 − 3) 15 3x + 6x loge (6x) − 2 x 2 sin ( x 2 ) − cos ( x 2 ) 16 x2 17 2 cos (2x) esin (2x) y 18 a 14 f ′(x) =

y 2 1 0

1

2

b

8 2e2x − 1 − 9 a − 2 xe x b x = 0 2 3x 10 3 x −2 11 a = 1 12 f ′(x) = 6 cos (2x), f ′(2π) = 6 5 or 5 sec2 (5x) 13 f ′(x) = cos2 (5 x )

f(x) {sin (2x){

3 P 4

P 2

3P 4

P x

f '(x) 2 cos (2x), 0 x P 2 f '(x) 2 cos (2x), P x P 2

b R \{0, 3}

π 3π −  sin ( x ), 0 < x < 2 , 2 < x < π i  π 3π sin ( x ), <x<  2 2 ii

1

0

1

P 2

P

f '(x) sin (x), 0 < x < P 2

Chapter review Short answer y 1

1 0

2 x − 2, x < 0, x > 2 19 f ′( x ) =  −  2 x + 2, 0 < x < 2

f '(x) sin (x), 3P < x 2P 2 f(x) {cos (x){

y

2

x

x

3

3P 2

2P

x

f '(x) sin (x), P < x 3P 2 2

π  cos ( x ), 0 < x < 2 20 f ′( x ) =   − cos ( x ), − π < x < 0  2 Multiple choice 1 B 2 D 5 A 6 C 9 A 10 D 13 C 14 C 17 B 18 E 21 D 22 B Extended response 1 a i x = −20 and x = 0 ii (−28, −20) ∪ (0, 12) iii (−20, 0) y b

3 7 11 15 19 23

E D E A C B

4 8 12 16 20 24

C B A A D C

dy — dx

f '(x)

2 3 4 5

4 a 3x2 + 2 a x2 − 4 a 6x3 + 43 x2 − 3

25

28 20

b 5 b 5

b i −8 14 ii 48 6 a 18 x 7 x2 + 4

708

Answers

b 6 5

0 1012 x

dy 3x = ( x + 20) dx 200 e i 1.875 ii −1.5 iii 4.5 f Yes, the largest absolute value of the gradient is 4.5, that is, the steepest section. g 30.24 m (at x = 12) 2 a Domain f (x) = [0, ∞), range f (x) = [0, ∞) Domain g(x) = R, range g(x) = [1, ∞) c x = 10

b i 2 x 2 + 1

d

ii 4x + 1

c Domain f (g(x)) = R, range f (g(x)) = [2, ∞) Domain g(f(x)) = [0, ∞), range g(f (x)) = [1, ∞) 2x d i ii 4 x2 + 1 4 e i ii 4 5 y 3 a 7 6 5 4 3 2 1 0

3 2 1

1

1 2 3 4 5 6

b x = 5

x

c x = 2 and x = 5

 − 2 x, x < 2  2< x<5 d f ′( x ) =  2, x>5 1, e y

3x + y = 3 + loge (3) − x + a2 + 32 11 a ytangent = 2ax − + 1, ynormal = 2a x a + b ytangent = , ynormal = − 2 ax + a (2a + 1) 2 a 2 10

a2

2

2

c ytangent = 4 ae 4 a x + e 4 a (1 − 8a 2 ), −

x 1 − 4 a2 2 + e4 a 2 + e 2 4 ae 4 a 12 y = 8x + 2(1 – loge (2)) 13 b = −124 Exercise 8B — Sketching curves 1 a (0, 8) a local max. b (−1, 2) a local max., (1, −2) a local min. c (2, −8) a local min.

ynormal =

d (−2, −8) a local min., ( 23 , 1 13 27 ) a local max.

e f g h

(0, 0) a positive point of inflection, (1, 1) a local max. (−2, 4) a local max., (0, 0) a local min. (3, −4) a local min. (0, 8) a positive point of inflection −

7 6 5 4 3 2 1 0

3 2 1

1

9 x + y = π

i ( 12 , 6 14 ) a local max. j (0, 5) a local min., (1, 6) a positive point of inflection k (−3, 54) a local max., (3, −54) a local min. l (−3, 16) a local max., ( 13 , −214 27 ) a local min. 1 2 3 4 5 6

x

2

3

4

m (0, 12) a negative point of inflection n (0, 0) a negative point of inflection, (3, −27) a local min. y y 2 a b f(x) (8, 0)

( 1, 2)

CHAPTER 8


2

y 8 x2

c

d

y

(1, 2)

y —) ( 2–3 , 1 13 27

g(x) 0

x

( 2, 8)

f(x)

x

0 (2, 8)

e

x

1

f

y

y ( 2, 4)

(1, 1) (0, 0)

x

0

y x2 (x 3)

0 (0, 0)

x

g(x)

g

y

y 5 6x x2

h

y

f(x)

(0, 8) x

0

0

x

Answers 8a ➜ 8b

Exercise 8A — Equations of tangents and normals 1 y = 5x − 4 2 At (−6, 0) y + 7x + 42 = 0 and at (1, 0) y = 7x − 7 3 x + y = 3 4 x + 3y + 21 = 0 5 a i y = 2x ii x + 2y = 5 b i y = 6x + 16 ii x + 6y = 22 c i x + 4y = 4 ii 2y = 8x − 15 d i y = 7x + 1 ii x + 7y + 43 = 0 e i 4y = x + 4 ii 4x + y = 18 f i 3y = x + 6 ii y + 3x = 12 g i x + y = 1 ii y = x + 3 h i y = 4x ii x + 4y = 0 i i y = 2x − 3 ii x + 2y + 1 = 0 j i y = 2x + 1 ii x + 2y = 2 − − − k i y = 3e 1x + 4e 1 ii ex + 3y = 3e 1 − e l i 2y = x − 2 + 2 loge (2) ii y + 2x = 4 + loge (2) m i 3y = 2x + 3 loge (3) ii 3x + 2y = 2 loge (3) π 3 3 π ii y = x + − n i x + y = + 3 3 2 2 o i 3x + 2y = 3π ii 3y = 2x − 2π 2 p i y = 2 x + ii x + 2 y = 1 2 6 a B b C 7 y = 2x 8 y = −1

1 0

x

0

2

(3, 4)

Answers

709

i

j

y

(– 1–2, 6 1–4)

y y 3x4 8x3 6x2 5 +ve point of inflection (1, 6)

x

0

y

g(x)

(0, 5) Minimum

y x2 x 6

k

5 a (− 2 , −4) a local min., (0, 0) a local max., ( 2 , −4) a local min. b (−2, 0), (0, 0) and (2, 0) c y

0

(0, 0) x

2

g(x)

2 x

0

( 2, 4)

( 2, –4)

( 3, 54)

a (−2, −27) a local min., (1, 0) a positive point of inflection

x

0

6 c (1, 0), (−3, 0) and (0, −3) d y y x4 6x2 8x 3

(3, 54) y

l

(1, 0)

x

0 3

3 y x3 4x2 3x 2

( 3, 16)

( 2, 27) x

0 —) ( 1–3 , 2 14 27

m

7 a (−2, −18) a local min., (0, −6) a local max., (1 14 , −9 107 256 ) a local min. b (0, −6) c y y x4 x3 5x2 6

y

x

0

y

(0, 0)

x

0

( 1– , 1–4)

c

( 1– , 1–4)

y ( 1, 0)

1 0

( 1–4,

f(x) 4 x

(0, 0) 0

0

3

2

y

x3

x2

5

x

—) (3 2–3 , 14 22 27

g

( 2, 36) y

y

16x 16

f(x)

1 (0, 1)

16

4

0

1 4

x

—) (2 2–3, 14 22 27

710

Answers

1

g(x)

0

f h(x)

4 a (−2, 36) a local max. and (2 23 , −14 22 27 ) a local min. b (−4, 0), (1, 0), (4, 0) and (0, 16) c

x

y ( 1, 36)

—) (2 1–3 , 18 14 27

y 5 ( 2–3 , 1 — ) 27

1

—– ) 256

e

4

d

g(x)

1–3

0

1

( 1, 0) (1, 0)

x

3

(2, 4)

2

y

3 a (−1, 0) a local max., (2 13 , −1814 27 ) a local min. − − c ( 1, 0), (0, 4)

f(x)

0

1 x

2

(3, 27)

y (0, 0)

1 0

b

y f(x)

g(x) (0, 0)

—–) (1 1–4 , 9107 256

( 2, 18)

8 a

h(x)

n

x

0 (0, 6)

(0, 12)

x

x

(2, 0)

y ( 12–3 ,

—) 1422 27

3

0

h(x)

5

x

(3, 36)

h

No stationary points

y

f(x)

2

(1, 2)

Not a zero gradient 0 1

1

ii −2 < x < −1 and x > 0 iii x < −2 and −1 < x < 0 y 16 f(x) x

( 1, 2) 2

i

y

g(x)

j

2

y

h(x)

20

( 4, 4)

5

( 1, 0) 0

( 2, 0)

(1, 0)

1

x

(0, 1) –1

x

0

2

1

9 B 10 A 11 C 12 E 13 a x = −3 a local min., x = 0 a local max. b x = −2 a local max., x = 1 a local min., x = 4 a local max. c x = −2 a negative point of inflection, x = 3 a local min. d x = −5 a local min., x = 2 a positive point of inflection e x = −3 a local max., x = 0 a local min., x = 2 a local max. f x = 1 a local max., x = 5 a local min. y 14 f '(x) 2x 4 0

3

x

An example of the answer. 17 a a = −1 and b = −8 − b At x = 43 a local max. and at x = 2 a local min. 18 a a = −2 and b = 5 b (−1, 4) and (0, 5) c (−1, 4) a local min., (0, 5) a local max. and (1, 4) a local min. 19 a −(x + 2)(3x − 2b + 2)  2(b − 1) 4(b + 2)3  b  , and (−2, 0) 27   3 c (−2, 0) is a minimum turning point.

 2(b − 1) 4(b + 2)3   3 ,  is a maximum turning point. 3

d y = 8(b − 4)x + 32 e b = 7 20 a f ′(x) = 2(x + b)(3x + b − 1)

 1 − b − (2b + 1)3  − b  ,  and ( b, 0) 27  3

x

2

0

c y = 2(b + 2)(b + 5)x − (b + 2)(b + 14)

4

f ′( x) < 0 if x < 2 and f ′( x) > 0 if x > 2 15 a i y f '(x) x2 4x

0

–4

x

ii x < −4 and x > 0 iii −4 < x < 0 y b i g'(x) 3x2 4x 7

0

1

x

ii x < −2 13 and x > 1 iii −2 13 < x < 1 c i y h'(x) 4x3 12x2 8x

2 1

0

x

c

dV = h 3 − 2h 2 dh

d

dh − = 3t 2 − 2e t dt

e

dM = 1 + log e (3k ) dk

f

dL 4 = 2t − 2 + 2π sin (π t ) dt t

2 D 3 C 4 a 8 workers b $1112 5 20 metres 6 x = 1 hour for max. and x = 7 hours for min. 7 a x = −5 ºC for min. and x = 15 ºC for max. b Minimum is N = 291 rabbits (round down) and maximum is N = 1625 rabbits. 3π 8 a Minimum v = −0.8 cm/s when t = 4 π b Maximum v = 0.8 cm/s when t = 4 9 a i 12 cm ii 132 cm  πt  π b R = 6 + 2 cos  4  c Max. =

π +12 12 − π , min. = 2 2

Answers

Answers 8c ➜ 8c

2 1–3

3 2 −1 as b > 0 2 Exercise 8C — Maximum and minimum problems when the function is known dC dP 5 = 5 − 4x 1 a b = 3n 2 + 4 n − dx dn 2 n

d b =

711

10 a b c 11 a b 12 a 13 a b

1 January 1998 941 cheetahs i 824 ii 507 (round down) 232 elephants Approx. 1 January 2009 75 b $50 c $3750 90 kg Approx. 77.3 kg after 3 months

Exercise 8D — Maximum and minimum problems when the function is unknown 1 Both numbers are 5. 2 2 and 6 3 b A = 60x − x2 c Length and width are each 30 cm. d Max. area is 900 cm2. 4 10 000 m2 40 5 Radius = cm (or 12.73 cm), length = 40 cm, π circumference = 80 cm 6 a L = 4.5 − 3x b V = 9x2 − 6x3 c Edges are 1 m, 1.5 m and 2 m, for max. d V = 3 m3 7 b V = 50x − 12 x3

c 192 cm3

8 6301.1 cm2 9 Approx. 6564 cm3 10 x = 1.5 km 800 hours v c 223.61 km/h 12 Approx. 13.86 cm 13 1239 cm3

−

b C = 40 000 v 1 + 0.8 v

11 a

c t = 60 min

712

Answers

60

90

t

3t

=

1 (3t 2 + 4) 2

3

b a =

3t 2 + 4

–

3t 3t 2 + 4 9t 2

( 3t 2 + 4 )3

12

=

3

(3t 2 + 4) 2

c v = 1.5, a = 163 8 9

a b c a b

c d 10 a 11 a 12 a

v = 3t 2 − 24t + 36 t = 6 x = 0 and t = 2 x = 32 a = 12 and −12 3000 − 371.4 people/day dN − 4000 = dt 8t + 1 − 97.56 people/day D b A B b C dH − π πt = sin dt 6 12

c E

π 2 π π 3 m/h ii m/h iii m/h 12 6 12 c Min. H = 8 m at t = 12 hours and max. H = 12 m at t = 24 hours 13 a i $2000 ii $2378.22 b 126.07 $/year c 68.65 $/year 14 0.2 mL/h 15 a 150 cents b (8, 13) b i

−

−

2 d −15 cents/month 3 e 175 cents f t = 13 16 a 7 m/s b 0.4 m/s2 c (0, 6) and (9, 16) d 0.2 m/s2 Exercise 8F — Related rates 5 cm/s 1 4624π −3 2 223 3 0.5 cm/s 4 320π cm3/min 1 5 45 cm/s 6 60π cm2/s 7 B 8 0.26 m/min

c

4 5 units 1 15 4.06 units Exercise 8E — Rates of change dV dS dA b c 1 a dr dh dt dC dI dv d e f dx dp dt 2 a dA = 2π r b 20π m2/m dr b −100π cm3/cm 3 a dV = − 4π r 2 dr 4 6 cm2/cm 5 a i 10 m/s ii −10 m/s b When t = 1 the projectile is rising but when t = 3 it is falling. dv 1 6 a = (90t2 − t3) dt 100 b V

0

7 a v =

9 2 3 m/s 1 m/min 10 24π h 11 a r = 3

b V =

π h3 9

Exercise 8G — Linear approximation 1 a 1.03 b 0.998 c 4.9 d 1.0005 2 2π cm3 or about 6 cm3 3 4% 4 a 43.74 cm2 b 0.44 cm2

c

75 cm /s cm/s 512π

7 8

5 E −k ; 12.5% decrease in volume 6 16 7 h ≈

1 4

8 5% increase in perimeter, 10% increase in area Chapter review Short answer 1 (−3, 86) a local max. and (2, −39) a local min. 2 y ( 2, 4)

2

(0, 0) 0

b c d 10 a

2 x

y = x2 (4 x2)

3 4 5 6 7

2x + y = 6 y = 3x − 6 + 3 loge (4) 8x + 6y + 31 = 0 and 8x + 6y = 31 1430 bees a x = 12 units b 16 sq. units

8

2 3

b

cm/min

9 P(4, 2); 2 5 10 18 Multiple choice 1 C 2 5 D 6 9 C 10 13 A 14 17 B 18

3 7 11 15 19

E D C D D

A A B A A

4 8 12 16 20

A is (0.5, 1) and B is (−0.5, 1) y = 4x − 1 c (0, −1) 1 cm e 1.125 cm 7 weeks i 98.25 rabbits/week ii 44.15 rabbits/week P0 = 267 (round down) 782 rabbits i 33 rabbits/week ii 44 rabbits/week 39 weeks

9 a y = 14 400 − h 2

( 2, 4)

4

a b d a b c d e f

D E B E D

5h dy = cm/s 14 400 − h 2 dt i 4.47 cm/s ii 7.54 cm/s 5.67 cm/s f ′(x) = −(x + 2a)(3x + 2a − 6)  − 2(a − 3) 4(2a + 3)3  − ,   and ( 2a, 0) 3 27

 4 500  c (−2, 0) is a local minimum turning point.  ,  3 27  is a local maximum turning point. Therefore (−2a, 0) is a local minimum turning point and  − 2(a − 3) 4(a + 3)3  ,  3 27   is a local maximum turning point. d y = −4(a − 6)(a − 1)x + 4(a − 1)(3a + 7)  3a + 7  e  ,0 , a ≠ 1  a − 6  dy = 8(4 − a) dx a +12 c x = 8

b y = 8(4 − a)x − (4 − a)( a + 12)

11 a

d a = −8

e y = 96x − 48

Extended response 1 a 10 2 − r 2 = h b

1 πr2 3

× 100 − r 2 c

2000π 3 3 cm 27

Integration Exercise 9A — Antidifferentiation 1 a 12 x2 + c b 15 x5 + c d

1 6 2x

+ c

e

g

2 + c x3

h

j

1 4 8x

k

+ c

5

m 35 x 3 + c − 5 + c p 2x2

n

s 16 x + c

t

q

2 a + 5x + c b x3 + 2x2 − 10x + c x2

−5

x

+ c

c

1 8 8x

f

−2 5 5x

+c +c

3

+c

4 3

x 2 + c

i

1 5 25 x

−

1 + c 9x3

l

2 3

7 16 4 7 x −9

x 12 x

+ c

+ c

7 4

3

x2 + c 4

x7 + c 2 r 5 + c x o

+c

c 2x5 + 32 x4 + 2x + c d

−2 6 3x

e

1 4 4x

+ 14 x4 − 2x3 + x2 + c

Answers 8D ➜ 9a

2 a 3000 b Approx. −10.11 bacteria/min 3 a 1.4 m b 3.4 m c 0.4 m/s 2π 3 d m/s 15 4 a 300 m b 40 m/min c 700 m d −20 m/min 100 5 a A = 2πr 2 + b r = 2 cm r 2 c A = 75.1 cm d $3005 27000 6 a l = w2 108000 b A = 2w2 + w c w = l = 30 cm (that is, a cube with side length of 30 cm) 54 000 d L = 10w + w2 e w = 22.10 cm; l = 55.26 cm 1080 27000 f C = 0.02w2 + + 5w + w w2 g w = 23.74 cm; l = 47.91 cm h 223 cents

CHAPTER 9

+ 12x − 13 x3 + c

Answers

713

f

1 x3 3

− 2x2 − 21x + c

g

5 3 3x

+ 5x2 − 5x + c

h

1 4 4x

− 73 x3 + 2x2 − 28x + c

i 14 x4 + x3 − 2x2 + c 3 B 1 3

(x + 3)3 + c

b

c

1 5

(2x + 1)5 + c

d

e

1 5 30 (6x + 5) + c −1 4 4 (4 − x) + c −4 5 15 (8 − 3x) + c −1 −1 2 (2x + 3) + c −1 −3 2 (4x − 7) + c −2 1 10 (6 − 5x) + c

i k m o

f h j l n p

6 C 7 a 3 loge |x| + c

1 4 4 (x − 5) + c −1 6 9 (3x − 4) +

e

e

loge |x| + c

f loge |x + 3| + c

8 5 loge |5x −5 2 loge |3

m o

−

j

4 3

+ 6| + c

l

+ 2x| + c

n

3 2 loge |2x − 5| + c −2 7 loge |6 + 7x| + c −3 11 loge |6 − 11x| +

p

(0, 0)

g

loge |3x + 2| + c

1 5 5x

+ x2 + loge |x| + c b −1

c 3x + 2 loge |x| + x + c d e

1 2

g

−3 8

loge |6 − 10x| + c f −

(4x + 1) 2 + c h −

−

2 i loge | x | + 2x 1 + 15 2 x + c j 2

5

k 2 x 2 − x2 + 92 x 3 + c l 5

3

1

1 18

loge |2x + 1| + c

1 3

(2x − 5)6 + c

1 2 2x

+ 4x + 8 loge |x| + c − 2 loge |3 − x| + c + 14 x4 + c

−

−

m 25 x 2 + 43 x 2 − 2 x 2 + c n −5x 2 + x 1 + x2 + c 11 a f (x) = 2x2 + x + 2 b f (x) = 5x − x2 − 5 3 1 d f (x) = 12 x2 + 23 x 2 − 103 c f (x) = 3x + 2 − x

c

1 2x 2e − −x

e + c

e e5x + c

k

1 e6x + c 12 − 1 6x 2 e + c x 3e 3 + c

m

x 6e 2

g i

7 4x 4e

+c

h

2 3x 9e

+c

l

+c

x 0.4 e 4 −x

+c

n − 9e 3 + c e x + e− x +c p 2

3 13 e3x − 32 e2x + 3ex − x

1 7

e

1 − 2 cos ( 2x) −2 9 cos (6x)

Answers

f

+c

2 x + 2ex + 12 e2x

c

h f (x) = (1 − 2x) + 2 i f (x) = loge | x + 5 | + 2 j f (x) = −4 loge | 7 − 2x | + 7

− 1 −3x 3e

j 4e

o ex − e−x + c

f f (x) = 2 x − x 2 − 6 (x + 4)4 + 1

d

−2x

+ c

−1 3

1 4

b 14 e4x + c

+ c

6 a

−4

714

1 a

4 14 x4 − x3 + 2e3x 5 a D

g f (x) =

(0, 0) (1.44, 0) x 1 2

Exercise 9B — Integration of e x , sin (x) and cos (x)

e f (x) = 43 x 3 − x3 + 50x

4

x

f (x)

2 1

3 2

3 2 2 3x

(1, 1) (0, 0)

c

(3x + 1)6 + c

1 2 4x

x

1

f (x)

x

( 2.77, 0)

q 23 loge |4 − 3x| + c r 4 loge |5 − 2x| + c 8 A 9 2x + 7 loge |x| + c 10 a

f

f (x)

loge |x| + c

h −2 loge |x + 4| + c

loge |5 − x| + c

0

1 2

(1, 3)

−

5x) 3 + c

4 7

k

f (x)

x

0

−

x

0

− 8) 5 + c

d

loge |x + 5| + c

d

−

+ 5) 2 + c

loge |x| + c

g 3 loge |x + 3| + c

f (x)

9x)11 + c

6 5

−6

c

b 8 loge |x| + c 7 3

x

1

2

1 3 4 (4x − 1) + c −1 5 5 (7 − x) + c 1 33 (8 − −1 12 (6x −1 15 (3x −2 3 (7 –

0

1

c

c

i

f (x)

4 E

5 a

g

12 a k = −8 b y = 41 13 a k = −1 2 b g( x) = 4 − − loge | x | x c 3 − loge (4) f (x) b 14 a

g

b C

cos (3x) + c

sin (7x) + c

b

−1 4

d

1 6 sin (2x) + −1 sin (−3x) 3

cos (4x) + c c

+ c

f

+c

+ c

h 2 sin (4x) + c

j 2 sin (−x) + c l 2 sin  x  + c  2

i 2 cos (3x) + c k −3 cos  x  + c  3  − x m 12 cos   + c  4

n 10 cos  x  + c  5  − x p 12 sin +c  2 

o 16 sin  x  + c  4 q s u

−6 4 5

 2x  cos   + c  3

 3x  r 8 sin   + c  4

cos  5 x  + c  2

−5

t

cos (π x) + c

π

v

sin  7 x  + c  4

− 12 7

 πx 6 sin   + c  2 π

−π − sin  π x  + c x cos 4x  + c  π  4 π  3 7 a sin (x) − cos (x) + c b − 12 cos (2x) − sin (x) + c

w

c

−6

1 4

1 2

sin (4x) − cos (2x) + c

g 6  sin  π x  − cos  π x   + c  2   π   3 

8

1 4x 4e

− cos (2x) +

1 4 4x

b

1 4 1 4x − 2 1 3 3x + 2

+c

sin (2x) + e + c x

c d 12 a b c

x 2 + x2 +

+ 1)5 + c

2

d i

4x + 3

5

+c

ii

2x − 5 + c

ii

4x + 3 + c

e i 4(2x + 3)(x2 + 3x − 7)3 ii −

2x ( x 2 − 1)2

ii

2 a A 3 E 4 a i 4e4x − 5

b B

f

i

6 − 5x

ii

1 (x2 4 −

+ 3x − 7)4 + c

2 +c x2 − 1

1 4x − 5 2e −

+c

6 − 5x

ii 2e

1 x2 2e

+c

+c

c i 2 xe x

ii

d i (1 − 2 x )e x − x 5 a i cos (x − 5) b i 3 cos (3x + 2)

ii e x − x + c ii sin (x − 5) + c ii 2 sin (3x + 2) + c

c i −4 sin (4x − 7)

ii

i 6 a b c d e 7 a b 8 a 9 a 10 a 11 a

i 4 sin (3 − 4x) 5 i 5x + 2 2x i x2 + 3 2x − 4 i x2 − 4 x i 3 cos (x) − x sin (x) x cos ( x ) − sin ( x ) i x2 i [ex (sin (x) + cos (x)] i sin (x) + x cos (x) i ex + xex

2

−1 4 −1 2 −

cos (4x − 7) + c

−1 2

cos (3 − 4x) + c

ii cos (6x − 3) + c ii 2 sin (2 − 5x) + c ii

ii 4 loge (5x + 2) + c ii 6 loge (x2 + 3) + c ii

1 2

loge (x2 − 4x) + c

ii sin (x) − x cos (x) 2 sin ( x ) ii x ii 3ex sin (x) ii x sin (x) + cos (x) ii xex − ex x6 i 12x5(1 − 3x)(2 − 3x)5 ii (2 − 3x)6 + c 2 3x 2 + 2 i ii 2 x 3 + 2 x + c 2 x3 + 2x 1 b 3x + loge |x − 1| + c 3+ x −1 2 b 5x − 2 loge |x + 2| + c 5− x−2 5 4+ b 4x + 25 loge |2x − 3| + c 2x − 3 4 −3+ b −3x − 2 loge |3 − 2x| + c 3 − 2x

Answers

Answers 9b ➜ 9c

1 a 1 b

3

−x

 πx 6 cos   + 5x + c  3 π f (x) = 4 + sin (x) f (x) = 1 − 2 cos (2x)  x f (x) = 3 2 + 12 sin    4  x  x f (x) = 4 sin   + 2 cos   − 4  4  2 k = −1 −6  πx y= cos   − x + 7  6 π 6 y=1+ π 2 3

1 2 2 (x

h

1 (3x − 1)5 + c c 3 cos  x  + 2e 2 − 15  3 d 13 loge |3x − 2| + 14 e4x + 5 sin  x  + c  5

f

ii

g

−x

 x  x e −6 cos   − 6 sin   + 5e  2  3

b i 10x(x2 + 1)4 1 c i 2x − 5

f

loge |2x + 3| + 12 e2x + c

−

− 2)8 + c

d i 6 sin (6x − 3) e i −5 cos (2 − 5x)

9 x3 − sin (2x) + 2e3x 10 a

1 2 (3x

−

+ 12 cos (8x) + 7x + c

1 2

ii

2

− 2 cos (x) + c

1 6x 2e

1 a i 24(3x − 2)7

2

e sin (4x) + 16 cos (2x) + c

h

Exercise 9C — Integration by recognition

b i 5e

 x d 2 cos   − 12 sin (2x) + c  2 5 2 2x

c f  π  = −2.02  6 

−

−

f

13 a k = −4 b f (x) = 2 sin (2x) − 4ex + 3

715

dy − sin ( x ) − = = tan ( x ) dx cos ( x ) b −loge |cos (x)| + c − cos ( x ) 13 sin ( x )

1

12 a

14

1 6

e 5 loge (4) + 2e2 − 2e 2 f

15 1 sin (ax + b) a −1 16 cos (ax + b) a 1 17 e ax + b a − 1 18 a cos (3π x + 1) + c 3π − 1 sin (1 − 4π x) + c b 4π

p π (2π − 2 2 + 4) ≈ 23.4

c 33 12

6 a 26

b 41

7 a 28

b e 1 + 1 + e c loge (24) d 6

e 23 13 8 a 10 9 a 6.875 10 10.5

f 7.25 b 7 b 5.375

−

g 100 c 8.5 c 6.125

Exercise 9E — The fundamental theorem of integral calculus

2 a

1 8 4 (e − −2

20 14

716

Answers

(e − e )

c 12

d 40

3

4

e

∫1 ( x3 − 9 x 2 + 20 x) dx f ∫−2 (− x3 − 4 x 2 − 4 x) dx

g

∫−1 e x dx

i

∫02 2 sin (2 x) dx

2

h − 74 k 6313.2

i 38 l 5949

b

∫0 (4 − x) dx

d

∫−3 3x 2 dx

−1

3

0

1

π

4 a 8

5 a

1 13

b 8

c

g e − e y

−1

7 3

4 −

h

∫1 e 2 x dx

j

∫0 2 cos  3  dx

3π

 x

d 26

−1 2

h

−8

e 22

−2

(e − e ) i 2

b

y 4x

j 3

y

12

0

1 2

f 2 loge (5) 3

∫1 x 2 dx

f 56

d

− 4 −3

1 3

c

e 8

b 3(1 − e

c 3

∫1 2 x dx

f

c

q 35 loge (5) or 0.966 s 3.65

c 2(e − e2)

e

π 4e 4

3 a

5 13

−

1)

−2 3

Exercise 9F — Signed areas The answers to all area questions are in square units. 1 a 8 b 8 2 a 4.5 b 4.5

m −1360 n 60 49 o 24.8 − 3 loge (4) or 20.64 p 37 13 r 2

b

g 3 h 6 k = 1 or −4 7 k = 2 8 k = 3 9 a = 2 loge (3) 10 k = π 6 11 a 16 b −8 12 D 13 A

Exercise 9D — Approximating areas enclosed by functions All answers for areas in questions 1 to 15 are in square units. 1 10 2 a 8 b 42 3 a A b A c B 4 a 30 b 22 5 a 32 b 32 c 39 d 16 e 29 f 12 g 26

b

−4 3

−

6 πx  sin  + 5 + c  2  π

g 12 23 j −1

n 0

o

d 10

f esin (x) + c

d

m 9 3 − 9

5 a −9

1 π x + 3 e +c π −3 πx  cos  2 +  + c d π  3

1 a

h 0 j −6 l 0

q 14 23 + 2 cos (9c) − 2 cos (3c) r 6 + log (π) − 6 sin (0.5c) e 3 18 4 a B b A

c

1 3 1 3

−

(e 2 + e2 − e 6 − e6)

g 1 i 20 − 10 2 k 0

loge |3x2 − 4|

e

−

1 2

c

3

y

x 0 1 2 y6 x

x

y x2

d

y

−2

3)

(e12 + 15 − e6)

0

1

3

x

1

x

01 y 4 x2

y

e

0 1

g

h

y loge ( –2x ) 2

4

y

y 3 sin (2x)

3 x

0

3

6 a 2

c 5 13

b 1

g e − e

i 1 7 a E

–P 2

d −1

f 2 23

e 4

x

0

3

y

−

h

4 3 1 2

5

3

1

8 a

∫0 f ( x) dx − ∫2 f ( x) dx

b

∫1 g( x) dx − ∫−3 g( x) dx

c

∫−1 h( x) dx − ∫−3 h( x) dx

f 49 13

x

P

j 4 b A

2

5 8 + e3 − e 1 1 Exercise 9G — Further areas All answers for areas are exact and in square units unless stated otherwise. 1 a 4 12 b 36 64 2 1 c d 21 13 e 312 2 a 85 13 b 57 16 3

y 2ex

2

x

4

0

y

f

y x

−

3 A 6 a i

1 0

−

(e 1 − e 2) b i

9 a B b D y 10

c E

0

x

y

d i

y x3 4x

ii

f

y

i

x

1

a 2 23

b

ii 1 − e

1 c 312

y 1 3 cos (2x)

2

a 13

632

P – 2

y

y 2 sin (x)

x

3P P — 4

π 3 + 4 2

0

1

x

0 –P –P –P P

Answers 9d ➜ 9g

4

3 −1

ii

2 1

4 1 0 P–

x

2

y

g i

−2

y e x 0

5 12

−

(1 − e 4)

x

0

2 y x3 x2 2x

1 2

y e2x

1

1

y

x

y

e i

10 23

12

ii 5 43

x

2

a 5 13

2 0

1 2

1 y — x2

2 0 1 2

4

b c 16 y 11

ii

12

y x2 4

2

x

3

0 −2

ii 2 loge (3)

y –2x

y

c i

2

3

x

2

0 1

∫2 g( x) dx − ∫−2 g( x) dx + ∫−3 g( x) dx

e

5 C

ii 6

y 3 3x2

∫−4 f ( x) dx − ∫−5 f ( x) dx

d

1

y 2

−4

−2

4 D

y 3

−1

2

h 40 12

g 8

h i b

π + 4

y cos ( –x) 2 2P

1

3 2

ii 4 − 2

y

0 P P –

x

2

y x 1 1

3

x

4 2 Exact area is 2 3 − 1 − or approx. 0.578 3 14 2 loge (2)

i

y

i

P–

1

6

P– –P 2

3

1

y sin (3x) 0 –P

ii

2 3

x

6

Answers

717

j

y yx x 8 6 4 2 1 x 0 1234

i

ii 12 45

2 a 16

d i x = 1

ii 1 85

() 3 4

(or approx. 2.15)

ii 2 2

h 43 3 C 4 D 5 E y 6 a i x = 0 and 4 ii 2

ii

b i x = −3 and 1

ii

iii 10 23 y 2x 0 1

3

c i x = −1 and 1

ii

x y 3 x2

y

y x2 1

y 1 x2

1 y ——— (x 3)2

b

d i x = −2 and 2

x

ii

11 + e 13 x-intercept is loge (3) (or 0.55), area is 2 14 No x-intercepts, area is 6π + 1. 15 a 1 + loge (x) b x loge (x) − x c 4 loge (4) − 3 (or approx. 2.55) 2x 16 a 2 x +2 1 2

−4

e i x = 0 and −1

∫0 (4 − x 2 − 3x) dx

1

∫0 2

∫1

3

i x = 0 and 1

ii

Answers

∫1 ( x − 1) dx

d

∫−2 (8 − 2 x 2 ) dx

dx h

b i x = 0, 1 and 2

ii

y

y x2 y x

∫−1 1

∫−1

(9 − x 2 − e x ) dx (1 −

x2 )

dx

iii

1 3

iii

1 2

iii

1 2

x

1 y

y x3 yx 0

x

1

y

y 3x2

2

1

1 3

x

1

y x3 2x

2

b

(3 x − x 3 ) dx f

(x + ex )

0

1

π 2 Exercise 9H — Areas between two curves All answers are in square units unless stated otherwise.

iii

y 1 x2

f

7 a i x = −1, 0 and 1 ii

20 a loge (4) b

c

y y (x 1)2

1

(3e 4 + 1) (or approx. 82.40) 2 9 2π − 4 (or approx. 2.28) 1

1

iii 21 13

1

18

∫0 ( x − x 2 ) dx

x

2

0

c loge (1.5) (or approx. 0.41) 17 a 2 23 b 5 13

g

0

ii

loge (x2 + 2)

1 a

y x2 4

y 4 x2

4 9

1 2

e

y

2

1 4

12 a x = −2 and y = 0 b

b

iii 2 23

x

0 1

1

x3

1 0 1

x

y

1 4

y

iii 10 23

4

2 5

3− 3 9 2

718

0

y 4x

(e6 − e3)

11 a

e 2 14

ii e3 + 1 − 3e (or approx. 12.93) −

10

d 21 13

yx

e i No x-intercepts ii 2e − 2e 1 (or approx. 4.7) f i x = 2 and −2 ii 139.2 1 3

c 2 16

g 1 12 + e2 − e (or approx. 6.17) ii 1 − 4 loge

8

1 2 −1

f 17 13 − e + e (approx. 14.98)

7 a i x = −2 and 2 π b i x = 4 c i x = 1

g i x = 2

b

0

8 a

2 3

1 d 2312

b 2 2 e 1 18

c 4

1

2

x

−

f e − e 1 − 2 (or approx. 1.09) π2 (or approx. 3.23) g 2 + 8 − h 2e − 2e 2 (or approx. 5.17) 3 9 e − e − 4 (or approx. 13.37) 10 3 12

11

π 3 12 (or approx. 0.45) 12 14 a x = −4 and 4 c 9 13

6 a 4 L b V = 5t +

c 200 L d 4.79 L/min e 5 h 20 min 7 a dv dt

1 2

11 10 9

13 3 loge (2) − 2 b x = −3 and 3

5 2 16 m2 1 16 a x = −20 and 20 c 93 13 m2

40  πt  sin    40  π

0

4

12

8

Hours

t

b 4 h c i 60 L ii 50.6 L 8 a 56 250 m3 b 5 9 1.26 m − 11 a y = 83 x2 + 6 b 38 m2

b 40 m d 840 m3

300 ) m2 π 60 000 b (30 000 − ) m3 π 17 a (150 −

12 b 3 loge (2) − 2

Exercise 9I — Average value of a function

Chapter review Short answer 1 f (x) = x3 − x2 − 3 2 a k = 0 4 4  πx b sin   + 1 −  4 π π

3(2 − 3 ) c 149 π 1 2 a 12 [50 loge (5) − 8 loge (2) − 21] 3 log e (2) b π c 14 (3e4 + 1) 1 a 8

b

d

302 2 45

3 a

1 2

b

−1 2

d

e6 − 1 6

(e2 − 1)  e 2 − 1 log e   2 

dt

64

0

8 16

t

Seconds

d 466 23 m e The distance travelled in 10 seconds.

8 π

dy = 2(x + 1) cos (x2 + 2x) and antiderivative is dx 12 sin (x2 + 2x) 4 a −2 b 2ex − 2x + 1 5 (5.5 − e2 + e) sq. units − 3 6 a −1 49 b 2 7 k = 12 or 2 3

8 a

y

1 y —— x 2

0

x

2

b loge (4) 9 166 23 sq. units π2  + 2 sq. units (or approx. 3.23) 10   8  1 17 sq. units 1 12 a y = −2x + 2 b (2 e − 2) sq. units Multiple choice 1 B 2 5 D 6 9 D 10 13 E 14 17 D 18 21 E 22 25 B

E E A B A D

3 7 11 15 19 23

C C C E B B

4 8 12 16 20 24

A B C C C A

Answers

Answers 9h ➜ 9j

4 C 5 E Exercise 9J — Further applications of integration − ( 2 − x )3 1 f ( x ) = +4 3 π 2 1 + 12 3 a 0 m b y = −0.01(x + 1)3 + 0.03x + 0.01 or y = −0.01x3 − 0.03x2 c 54 cm 4 a $34.56 b C = 40n − 200e0.01n + 200 c $3656.34 d i $36.56 ii $30.66 5 a i 0 m/s ii 48 m/s b i t = 8 ii 64 m/s c dx (m/s)

c 1 −

c 380 m3

719

Extended response 1 a $0 b C = 25t3 + 25t2 + 800t c $7750 d $4450 e $1050  πt  2 a T = 10 − 5 sin   b No  12  c 15 °C at 6 pm d 5 °C at 6 am e i 7.5 °C ii Approx. 13.5 °C f 4 pm − − 3 a (2, e) b y = (−2e 1)x + e + 4e 1 −1 c (0, e + 4e ) d (0, 1) − e (2 + 4e 1) sq. units 4 a 1 + loge (x) b x loge (x) − x c 3 m − d (e − e 2) m2 (or approx. 2.58 m2) − e 20(e − e 2) m3 (or approx. 51.66 m3) − 5 a 5 2 °C/cm b T = 180 − 40 x c i 82.02 °C ii 66.86 °C

Exam practice 3 Short answer 1 a x f (x) 1 3.4 2 2.7 3 1.5 4 0.7 5 0.3 Total

Area 3.4 2.7 1.5 0.7 8.3

dy − = ( x + 1)e − x + e − x = − xe − x dx c -20e-1(2e-4 - 1) ex 2 a ∫ y dx = x 4 − 2 sin ( x ) + 2 2x x 3 b ∫ y dx = − 2 3 2x 3 a The coordinates of P are ( 3 , 3 ) and ( 2 4 a =

2

7 2 2

dx = 0.12 x dt 6 a y = 4(x - 1) 1 b 3

b V = π r 2 h = 250 ⇒ h =

720

Answers

y 2r P 2 y 500 r 1 2 3 4 5 6 7 8 9 10 r

d r ≈ 3.4 125 π f i r ≈ 3.414 cm ii Define g(x) = the gradient function on a CAS calculator and determine sign of gradient close to 3.414 (for example, use r = 3.4 and r = 3.5). The gradient sequence in the vicinity of r = 3.414 is negative–zero–positive, which corresponds to a local minimum. iii 220 cm2 g r ≈ 2.710 cm h 1.47 e r =

3

Discrete random variables Exercise 10A — Probability revision 1 a e

b

1 4

c

f

1 36 1 2

g

1 6 5 12

ii

5 12

iii

1 3

2 3

iii

1 4

iv 1

b 1 4

2 a i

ii

3 a 18

b 83

c

3 8

d

1 8

1 4

c

3 8

d

5 8

17 40

d

3 20

1 2

4 a 83

b

5 0.7 6 0.1 7 a 0.2

b 0.8

8 a

2 5 3 8 1 3

b

13 40

c

b

1 2

c No 11

d Yes

Yes

Column 1 Column 2 Column 3 B

B′

Row 1

A

0.22

0.10

0.32

Row 2

A′

0.08

0.60

0.68

0.30

0.70

1

Row 3

500π r 500 = 2π r 2 + r πr2

5 6

+ 125 + 13 = 1

12 a

250 , πr2

d

c i 23 e

− 3 3 , ). 2 2

1 18 7 12

10 No

5

Multiple choice 1 A 2 E 3 B 4 B 5 C Extended response a S = 2πr2 + 2πrh

y 2Pr2 500 r

0

e 9 a

3

so S = 2π r 2 +

y 700 600 500 400 300 200 100

CHAPTER 10

b

b

c

b i Pr(A ∩ B) represents an unfit smoker, Pr(A ∩ B) = 0.22. ii Pr(A′ ∩ B′) represents a fit non-smoker, Pr(A′ ∩ B′) = 0.60. c i 0.10 ii 0.08 iii 0.32 iv 0.70

1 3 a e 14 a 15 a 16 a

2 3

b

1 2

Yes 0.7 0.105 0.7

b 0.4 b 0.39 b 0.3

17 a

25 64

b

c

9 64

1 3

c 0.5 c 0.495 c 27 35

d

15 32

d

c

2 5

b

1 15

25 a

25 72

b

19 66

26 A 28 a

17 32

8 15

d

1 4

7 15

b

2 9

b

22 31

Exercise 10B — Discrete 1 a Discrete c Continuous e Continuous g Continuous 2 a x 0 1 4

Pr(X = x) b

random variables b Continuous d Discrete f Discrete h Discrete 1

2

1 2

1 4

3– 4 1– 2 1– 4

4

5

Pr(X = x)

1 13

1 13

7 13

2 5

11 a

5 6

1

x

2

1 8

c

5 18

g

5 9

3 8

b

Pr(X x) 0.5 0.4 0.3 0.2 0.1

1 8

7 8

Pr(X x) 0.5 0.4 0.3 0.2 0.1

d

Pr(X x) 0.4 0.3 0.2 0.1 2 4 6 8 10 x

d

Pr(X x) 0.4 0.3 0.2 0.1

1 18

1 12

7 36

e

1

2

3

4

5

Pr(X = x)

1 5

1 5

1 5

1 5

1 5

3 5

c

x

1

4

9

16

25

36

Pr(X = x)

1 6

1 6

1 6

1 6

1 6

1 6

1 2

c

x

2

3

4

5

6

Pr(X = x)

1 6

1 6

1 6

1 3

1 6

2 5

c

0

1

2

3

4

Pr(X = x)

1 45

1 9

1 5

13 45

17 45

14 a

1 36

29 36

x

1

2

3

4

3 160

1 10

9 32

3 5

0

1

2

3

0.1715

0.4115

0.3292

0.0878

x Pr(X = x)

b 0.1060 5 a (1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (1, 2), (2, 2), (3, 2), 1 (4, 2), (5, 2), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5) 3 5

ii Pr(B) =

2 5

iii Pr(C) = 12 25

iv Pr(D) = 19 25

c i Pr(A | B) = 35

ii Pr(B | C) =

1 2

iii Pr(C | D) = 10 19 16 a

1 2 3 4 x

1 9

5 36

x

b i Pr(A) =

5 10 15 20 x

x

1 6

5 36

b

x

0

1

2

Pr(X = x)

4 25

12 25

9 25

4 25

+ 12 25 +

9 25

Answers 10a ➜ 10B

3 8

1 9

1 1 1 36 18 12

Pr(X = x)

Pr(X = x)

c

3

1 13

13 x 0

4 a

b

12

3 a HHH, HHT, HTH, HTT, THH, THT, TTH, TTT b x = 0, 1, 2, 3 c x 0 1 2 3

d

2

b 65

Pr(X x)

d 0.1

3 13

35 36

10 a

3 a 0.1035 (4 d.p.) 3 b 0.3929 (4 d.p.) c 0.2746 (4 d.p.)

1 11

1

b 16

9 a

31

c

x

Pr(X = x)

6 7

b 0.64

b 0.6

b k = −5 was discarded as it would result in negative probabilities. 8 a x 2 3 4 5 6 7 8 9 10 11 12

f

27 B

55 64

29 a 0.3277 0 0.5 3 32

c

7 a

7 33

8 C 19 D 20 D 21 B 22 1 23 a 0.4 b 0.5 c 0.74 d 0.26 3 g h 12 e 0.76 f 18 25 5 15 7 i 20 24 a

5 a, d 6 a 0.2

d No

=1

Answers

721

17 a 0.54 e

31 55

b 0.55

c 0.54

f 4

g 4

d 0.24

18 a 2.24 b 2.96 9 $1452 1 20 a i 3.5 b i 9 c i 0.5 Exercise 10D — Measures random distributions 1 2.3, 0.81

8 C 1 19 C 20 E 21 D 22 B 23 D 24 a x = 0, 1, 2, 3, 4 b

x

0

1

2

3

4

Pr(X = x)

1 16

1 4

3 8

1 4

1 16

Exercise 10C — Measures of centre of discrete random distributions 1 7.11 2

1 4 a = 18 , E(X) = 5 13

5 b = 0.15, E(X) = 2.39 6 k = 0.05, E(X) = 13 7 a x 1

2

3

4

5

6

1 6

1 6

1 6

1 6

1 6

1 6

Pr(X = x)

b $0.02 b 12.9456

c 29.1276

b 79.1091

c 878.99

b 5.41

c 2.33

15 a

x

2 3 4 5 6 7 8 9 10 11 12

Pr(X = x)

1 1 1 36 18 12

1 9

5 36

1 6

5 36

1 12

1 9

1 18

1 36

x

0

1

2

3

4

Pr(X = x)

1 16

1 4

3 8

1 4

1 16

x Pr(X = x)

b 1 45 2 13

Answers

b

9 13

0

1

2

3

8 125

36 125

54 125

27 125

c 2 c

5 23

d

1 615

y

2

1.5

1

0

0.5

0.3

0.15

0.05

b $1.60 c $0.51 14 0.95

b 2 0 a = 0.15, b = 0.05 1 11 a = 0.1, b = 0.2 12 a $3.75 b No, because although his expected gain is $3.75 per game, he must pay $5 to play each game. Therefore his loss per game will be $1.25. c No, because the expected gain is not equal to the initial cost of the game. 13 a $0 b No, because even though her expected gain (that is, her average gain per game in the long run) is $0, she could still lose $40 in one single game. c Yes, because E(X) = 0. 14 11 15 17

722

3 a $1.42 4 4.96 5 a 3.2364 d 80.91 6 a 8.7899 d 219.7475 7 10 8 5

Pr(Y = y)

b 7

17 a

55 b 2 64

13 a

b 3 12

16 a

ii 5 ii 8 ii –2 of variability of discrete

2 a 5 83

10 a 8.3 11 14 12 11.36

3 a = 0.09, E(X) = 5.42

9 a

d 18.48

9 2 13 , 1.0556, 1.0274

13 18

8 a

c 7.16

x

1

3

5

Pr(X = x)

1 35

9 35

5 7

b 4 13 25 16 a

c 1.0443

d

34 35

x

2

3

4

5

Pr(X = x)

3 50

4 25

3 10

12 25

b 4 15 21 (≈ 0.9165) 5 d Pr(2.3669 ≤ X ≤ 6.0330) = Pr(3 ≤ X ≤ 6) = 0.94 17 a 0.2 b 1 c 2.2 d 1.36 e 1.1662 f 1 18 a 0 ≤ X ≤ 8 b 4 ≤ X ≤ 16 c 21 ≤ X ≤ 49 d 11.2 ≤ X ≤ 32 c

f 12 65 ≤ X ≤ 22 16

e 8.3 ≤ X ≤ 11.1 19 a 7 20 a

c 0.94

x

1

2

3

4

Pr(X = x)

1 4

3 8

1 4

1 8

b 2.25 1 A 2 22 E 23 C

b 5.83

c 0.97

d

1 6

CHAPTER 11

Chapter review Short answer 1 a

1 12

2 a 0.15 3 a

2 5

9 24

b

d

1 8

iv

5 8


b 0.25 b

4 15

4 a i 0.3 ii 0.5 b 2.7 min c 10 5 a Teacher to check 6 a

5 12

c

iii 0.5

(0.625)

1 10

b

x

x

1

2

3

4

Pr(X = x)

1 30

2 15

3 10

8 15

x

2

3

4

5

6

7

Pr(X = x)

7 36

7 36

7 36

7 36

7 36

1 36

1 b 412

c

7 12

8 No, it is not a fair game. 9 No 10 a i 4 ii 20 b

1 243

1 a 1 b 12 a 13 7 14 a 15 a 16 a

i 4 i 7 $8.10

b

ii 4 ii 4

40 cents 10 cents 4 x Pr(X = x)

b $510

c 2

b $2000 b $50

c 20% c 10%

0

1

2

3

4

0

1 20

3 20

3 10

1 2

c 3 14 Multiple choice 1 D 2 4 D 5 7 D 8 10 B 11 3 D 14 16 C 17 19 B

2

3

4

5

Pr(X = x) 0.16807 0.36015 0.3087 0.1323 0.02835 0.00243

4

4 625

7 a

1 16

b

8 27

c

b

1 16

c

65 81 3 8

8 a 0.096 b 0.0064 c 0.008 d 0.2 9 a 0.0518 b 0.2592 10 a 0.2627 b 0.0084 c 0.2568 d 0.2568 11 0.0381 12 0.0924 13 a 0.1023 b 0.2001 14 0.2281 15 0.0653 16 a 0.0528 b 0.6676 17 a 0.2734 b 0.2965 c 0.1678 18 C 19 E 20 B 21 D 22 C 23 a i 0.0102 ii 0.9898 b 2 24 18 tickets 25 22 26 a 5 645 726 b 352 858 c $1 446 717.80 27 a 2 b Positively skewed Pr(X x) 28 a i 0.2

C A B A C B

3 6 9 12 15 18

0.1

E D C A E D

0.0

0 1 2 3 4 5 6 7 8 910 x

ii Positively skewed

b i

Pr(X x) 0.3 0.2

b $2.14 d 70 cents Loss of 54 cents Loss of 54 cents Loss of 54 cents b

0.1 0.0

iii No iii No iii No

0 1 2 3 4 5 6 7 8 910 x

ii Symmetrical or normally distributed

c i

Pr(X x) 0.3

19 32

0.2

d 2.5625 packages

e Var(X) = 2.3086, SD(X) = 1.5194 f 1 g $126.56 h 10 19

Answers 10c ➜ 11a

c

1

0.3

Extended response 1 a $14.70 c 61.25 cents e $2.10 2 a i $720 ii b i $40 ii c i $60 ii d 2.7% 3 a 18 7 8

0

5 0.0036 6 a 32 81

b 3 13 7 a

Exercise 11A — The binomial distribution 1 b, d and f constitute a Bernoulli trial. 2 a 0.2613 b 0.0446 c 0.2461 d 0.0092 e 0.0073 f 0.1969 3 a 5 b 0.3 c

0.1 0.0

0 1 2 3 4 5 6 7 8 910 x

ii Negatively skewed

Answers

723

29 a i Positively skewed ii Symmetrical or normally distributed iii Negatively skewed b Controls skewness 30 a Symmetrical or normally distributed b The curve would resemble more of a bell shape that is tall and narrow. 31 a Positively skewed b Negatively skewed 32 a Symmetrical or normally distributed b Positively skewed c Symmetrical or normally distributed d Negatively skewed Exercise 11B — Problems involving the binomial distribution for multiple probabilities 1 a 0.1792 b 0.6826 2 a 0.23 b 0.85 c 0.92 d 0.45 3 a 0.4718 b 0.9692 4 a 0.3370 b 0.3438 c 0.0333 31 6 5 a 32 b 163 c 31 6 a

1 9

b

0.65 Thursday 0.35

Poor

724

Answers

0.30

Best Match

b 0.25 b 3500 (train), 7000 (bus) b 3933 (Pete), 1967 (Sam) b 35 b 0.652 b 0.3014

c 0.652

Exercise 11D — Expected value, variance and standard deviation of the binomial distribution 1 a 6 b 1.6 b 1.35

c b b b

0.0178 0.0464 0.0170 0.9967

c 6

d 3 43

b b 16 19

0.1045 0.0930 A D

3 a c 4 a 5 a 6 a 7 a 8 a 9 a 10 a 11 a 12 a

1.26 3.24 5 4.62 12 1.67 7.5 5.4 28 96 1 2

b 2.74 d 4.16 c 1.58 c 1.88 c 2.19 b 0.5155 b 0.5 b 0.4613 b 0.4488 b 24 b 20

13 a

3 4

b b d b

0.1878 0.9216 0.5327 0.2276

b 0.4256 b b b b

14 a 13 15 a

0.0973 0.2741 0.0760 0.9573

1 5

6 C 1 19 E

0.65

Good

0.35

Poor

0.55

Good

0.45

Poor 0.65

0.65

Good

0.35

Poor

0.55

Good

0.45

Poor 0.55 0.35

0.55 0.45

0.45

2 a 0.6525 b 0.6652

Kingfisher

2 a 4.8

Poor

Good

Best Match

0.70

d 37 12

Good

b 0.6115

b 0.777 4 a 0.64 5 0.49 6 0.525 7 a 3500 8 a 3661 9 a 265 10 C 11 a 25 12 a 0.1479

Kingfisher

0.20

c 50

Thursday 0.35

Best Match

0.80

9 65

22 a

Exercise 11C — Markov chains and transition matrices 1 a Saturday Friday 0.65

Kingfisher

c

48 81

7 0.028 8 a 0.0878 b 0.9822 9 a 0.3669 10 a 0.9830 11 a 0.5000 c 0.9645 12 a 0.2553 13 a 0.1751 14 A 15 D 17 D 18 C 20 0.1509 21 a 0.9392 22 a 0.9106 c 0.6767 23 a 0.0173 c 0.8960 24 a 0.5981 c 0.2486 25 a 0.6778 26 a 0.8516 27 a 0.3438 28 a 0.9672

3 a 0.3

0.35 0.65

Good Poor Good Poor Good Poor Good Poor

b 2.5 b 3.55 b 4.8

b 16 b 27 b 15 17 E 20 D

4 5

b 3.2

c 0.1450 c 0.0001

d 0.3980

18 C 21 D c 1.79

3 a 0.0020 2 b 0.7939 24 0.2642 25 a 0.2335 b 0.3828 26 400 27 a 0.5000 b 0.6535 c 0.3465 28 a 5000 b 2500 c 50 29 a 6 b The company has an extremely popular product. 30 a 0.5455 b 0.3340 c 0.6 d 0.8795 e 0.2769 31 a 160 students b SD(X) = 12, µ − 2σ = 136, µ + 2σ = 184 c There is a probability of 0.95 that between 136 to 184 students (inclusive) will have a reading level that is inadequate to cope with high school. 32 a 1260 patients b SD(X) = 19.44, µ − 2σ = 1221.12, µ + 2σ = 1298.88 c There is a probability of 0.95 that between 1222 to 1298 patients (inclusive) will be cured.

Chapter review Short answer 1 a 165 2 a 0.243

31 b 32 b 0.001

1 c 32 c 0.081

3 a

27 128

b

54 255

4 a

256 625

b

1 9

5 a 4C1 p1 (1 − p)3 6

e Tellya has 45.4% of the market, Yodacall has 54.6% of the market. f Tellya has 41.2% of the market, Yodacall has 58.8% of the market. g Tellya has 40% of the market, Yodacall has 60% of the market. Both companies will be viable, but in the long term Tellya’s market share will drop by half, while Yodacall’s share will increase (in fact, will triple).

b p =

1 4

CHAPTER 12

5 16

7 a 0.14 8 a 0.45 9 0.234 10 a i 50 b i 40

b 0.23 b 0.2

11 a p = 15

b n = 50

12 a 8

b 6.4

Multiple choice 1 C 2 B 3 5 B 6 C 7 9 A 10 D 11 13 B 14 B 15 17 E 18 B 19

B 4 C 8 E 12 B 16 A

Continuous distributions Exercise 12A — Continuous random variables y 1 a A pdf

ii 25 ii 8

( 1– , 2) 2

(0, 2)

0

D A D D

y

0

Not a pdf

x

( 1– , 0) 2

(0, 2)

c

y

Not a pdf

( 1, 2) 1

1

0

x

(1, 0)

y

d ( 1, 3–2 )

1

A pdf (1, 3–2 )

3– 2

0

x

1

y

e

A pdf

1

P – 4

f

x

P– 4

Not a pdf

y

P , 2) ( 3— 2

2 P

g

0

P P

3— 2

Answers 11B ➜ 12a

Extended response 1 a 135 b 0.1 c $60 2 a 1 b 0.0015 c 0.2639 d 10 3 a 1 b 0.2642 c 0.0003 4 a i 0.0018 ii 0.8826 iii 0.2761 b 3 5 a 0.0719 b 0.8432 c 0.3378 6 a 1440 b σ = 24, µ - 2σ = 1392, µ + 2σ = 1488 c This means that there is a probability of about 0.95 that between 1392 and 1488 of the 2400 people selected will be cured by the drug. 7 X ~ Bi(80, 15 ) a 0.0164 b 0.9713 c 0.7550 8 a X ~ Bi(8, 0.8) i 0.1678 ii 0.2936 iii 0.214 b i 0.2951 ii 0.2849 9 a 0.018 b 0.8336 c 0.3600 d 0.36 10 a In the first week of August, 229 customers will visit Kaz’s Kitchen and 171 people will visit Al’s fine dining. b In the long run, 229 people will dine at Kaz’s Kitchen each week and Al will get 171 customers each week. 11 a Let ti = the number of Tellya customers at the end of January month i. Let yi = the number of Yodacall customers at the end of January month i. ti + 1 = 0.85ti + 0.1yi and yi + 1 = 0.15ti + 0.9yi b Tellya has 400 customers or 80% of the market; Yodacall has 100 customers or 20% of the market. c Tellya has 350 customers or 70% of the market, Yodacall has 150 customers or 30% of the market. d Tellya has 52.6% of the market, Yodacall has 47.4% of the market.

b

x

1– 2

2

P– 2

x

0

y

A pdf

1 1– e 0

(e, –1e) 1

e

x

Answers

725

y

h

2

(2, 2)

0

(1, 0) 2

b

c 1 e 1

f

3 38

h

6 53

3

2 5

4

2 13

5

6 103

6

7 2

4 15

( P – , 0) 2

11 a

b c 12 a

3 2

 1 ,a≤x≤b  10 a f (x) =  b − a 0, elsewhere  b p = 5

0.5 a

a 0

Q

a=2 i 0.25 ii 0.75 iii 49 y (6,

2

b i 18 13 a

1– 4

)

6

(0, –1 ) 8

0

x

(4, 0)

b Teacher to check c i 0.5475 ii 0.1179 iv 0.9583 v 0.5182 y 14 a

iii 0.8118 vi 0.3583

(0, 1) x

0

c i 0.3935 ii 0.7788 y 6 a

0 –P ( – , 0) 4

iii 0.4378

(1, 1)

c i 43 15 a

0

27 32

y

( 10 –2 , 0) 3

y

1

x

10

ii

−3 4

i 0.1563 ii 0.1040 iii 0.7398 iv 0.8767 0.5371 0.6916

x

1– 2

(0, 1– ) 4

1

ii

( P– , 0) 4

x

2− 3 2− 3 iii 2 4

1 2

y

b

1– 2

x

0

c i 0.5 ii 23 iii 16 iv 13 v 13 7 a = 1.4 8 b = 3

726

( P– , 0) 2

y

1– 4

Exercise 12B — Using a probability density function to find probabilities of continuous random variables 1 a 0.5625 b 0.543 125 c 0.6696 2 a 0.9630 b 0.2963 c 0.2593 d 0.2693 e 0.2693

9 a 3 − 7 , or 0.3542

c i

ii

6 − 19 , or 0.8206 2

2− 2 2

iii

3 −1 2

iv

b

π 6

(0, –1 ) 2

1– 2

9 k =

b 4 a b 5 a

b

f (Q )

a– 2

7 3 11 8 k = π

3 a

6 − 22 , or 0.6548 2

1 18

1 d e −1

g

c 10 a

x

1 2

2 a

Not a pdf

Answers

3 2 2− 2 2

( P , –1 ) 2

P

x

16 a

8 a 1 b

y

y (0,1)

a1 a2

a3

ii 0.5

2 3

iii

y (0,

(P – , 0) 4

x

c Mean = median = mode = 0 d The distribution is symmetrical.

x

0

b i 0.75 17 a

( –P , 0) 0 4

1 ) 2P

9 a

1 4

b 4 8 − 2

d

4

e 4 12 − 4 4

12 − 2

10 a

c

4

4−2

y P– 2

1

−

e x ,x∈R π ii 0.9214 iv 0.6987

b 0.5

c

d i 0.2398 iii 0.6746

2

Exercise 12C — Measures of central tendency and spread 1 a Mean −1.75, median −1.75, no mode b Mean

−2 3,

median 2 − 2, mode 0

b Mean 3.5, median 2 − 3 − 4 , mode 4 c Mean 1.7, median 1.746, mode 2

b Variance

11 36 ,

c Variance

11 225 ,

2 3

standard deviation standard deviation

1 4 a Variance 18 , standard deviation

11 m = 127 n =

11 6 11 15

9 32

a 0.7769 b 0.1733 a i 0.5132 ii 0.2413 b 1111 g c 0.6321 a i 0.0952 ii 0.1647 b 0.9512 c 105.36 h a Between 30 and 40 minutes b i 0.3 ii 0.2 c 35 minutes d 31 minutes a i 0.9682 ii 0.9873 b 0.0045 c 0.0032 a 0.1110 b 0.2815 a y

2 6

d b 2.3863

0

d 0.5

d 0.9720

1 100

c 0.6623

d 0.68

Exercise 12E — The normal distribution 1 a b

y ( –P , 1– ) 2 2

1 – 2

c Mean =

c 0.2796

c 0.1790

10 (20, 0) x

0

b

100 9

iii 0.8187

–P 2

(P , 0)

4

6

8 10 12 14 16

5 10 15 20 25 30 35

c x

π , variance = 0.4674 2 e 0.4142

3 2 1

0

1

2

3

2 a i 15 b i 1 23

ii 15 ii 1.66

c i and ii

d i and iii

iii 45 iii 5

Answers

Answers 12B ➜ 12E

c 0 e 103 a 2 a 0.5 b

c 0.0111 iii 0.1653

(10, 10k)

10k

b Variance 0.0742, standard deviation 0.2725 c Variance 0.4674, standard deviation 0.6837 5 a 103 b 103 loge (2) 6 7

−6 7

Exercise 12D — Applications to problem solving 1 a 2 hours b µ = 1.29 hours c 0.4375 d 0.2857 ii 43 iii 0.617 2 a i 18 3 4 5 6 7 8 9

2 a Mean −2.37, median − 6, mode − 11

x

b µ = 0.5 c As the graph is symmetrical, the median is 0.5. d The mode is equal to the maximum of the graph. In this case we have two points where the maximum is the same. Therefore the mode = 0, 1.

b

c Mean 0.75, median 3 12 , mode 1

3 a Variance 43 , standard deviation

1

1– 2

x

0

727

3 a

i ii

iii

0 10 20 30 40 50 60 70 80 90100110120

b It widens the graph and increases the range. c It moves the graph to the right. 4 a b

c

3 2 1

0

1

2

3

5 10 15 20 25 30 35

20 30 40 50 60 70 80

5 6

a b c a b

i The mean increases as time goes on. The standard deviation also increases as time goes on. µ = 65, σ = 5 (65, 1 )

4 5 6

a d a d a

b e b e b

0.9235 0.7462 0.6554 0.3260 0.75

7 a 0.46 8 0.14 9 0.72

b

8 31

10 a m =

b m = −1.75

Probability

50 55 60 65 70 75 80 x

7 a Mean: translation of 20 units in the positive x direction; standard deviation: dilation factor of 12 from the x-axis, dilation factor of 2 from the y-axis b Mean: translation of 9 units in the positive x direction; standard deviation: dilation factor of 2 from the x-axis, dilation factor of 12 from the y-axis 8 a 68% b 95% c 99.7% d 16% e 2.5% f 0.15% 9 a 16% b 2.5% c 0.15% 10 a 68% b 95% c 99.7% d 16% e 16% f 0.15% 11 a 28 ≤ X ≤ 52 b 16 ≤ X ≤ 64 c 4 ≤ X ≤ 76 12 a 25.8 ≤ X ≤ 28.6 b 24.4 ≤ X ≤ 30 c 23 ≤ X ≤ 31.4 d 28.6 e 30 f 31.4 13 a 16 ≤ X ≤ 17.2 b 15.4 ≤ X ≤ 17.8 c 14.8 ≤ X ≤ 18.4 d 16 e 15.4 f 14.8 14 a 1 (102.3, – 21.4 2P

M 3S MS 2 MS M MS MS 2 MS 3 38.1 59.5 80.9 102.3 123.7 145.1 166.5

b i 34% ii 47.5% iii 49.85% iv 81.5% 15 a 81.5% b 2.35% 16 a 2.4 kg ≤ X ≤ 3.6 kg b 1.8 kg ≤ X ≤ 4.2 kg c 1.2 kg ≤ X ≤ 4.8 kg 17 B 18 D 19 C 20 C 21 B 22 B 23 a 10.59 ≤ X ≤ 13.41 b 9.17 ≤ X ≤ 14.83 c 7.76 ≤ X ≤ 16.24 Exercise 12F — The standard normal distribution 1 a 0.8413 b 0.0668 c 0.0401 d 0.9966 e 0.2721 f 0.8644 −5 3

2 a 3

b

3 a 13

b −0.5

Answers

−3 2

d 2 a 2 23 a 24 a b 25 a b

c 0.0082 f 0.5486 c 0.25

0.0082

0.3364

c 0.9022 c 0.6826 16 E

1.80

2.00

0.6006

0.0548

$1.74 0.9804 b 980 0.1056 b 17 i 0.7309 ii 0.6218 iii 0.0914 0.1679 i 0.1870 ii 0.9665 0.6449

Exercise 12G — The inverse cumulative normal distribution 1 a 1.282 b −0.842 c −0.100 d 0.126 2 a 1 b 0.675 c 0.253 d 0.496 3 a −0.675 b −0.253 c 0.583 d 2.576 4 a 11.166 b 9.493 c 9.336 d 11.683 5 a 35.984 b 36.698 c 30.774 d 33.497 6 a 5.182 b 1.527 c 17.525 7 E 8 A 9 C 10 B 11 A 12 D 13 B 14 a 173.16 cm b 153.27 cm 15 a 44.58 mm b 45.42 mm 16 48.65 seconds 17 1.907 18 7.896 19 11.704 grams 20 26.305 21 25.844 22 3 minutes 44.27 seconds 23 a > 1.7 kg b Between 1.3 kg and 1.7 kg 24 a 64 b 44 Chapter review Short answer 1 a y

b Yes it could.

(0, 4) ( 1– , 0) 2

c 1.2 c 1.133

c 0.7623

1 a 0.0863 1 b 0.0115 12 a 0.1587 b 0.3874 d 0.8302 e 0.2391 13 A 14 B 15 E 17 A 18 A 19 E 20 0.8676 21 a 0.6006 b 0.0548 c Cost ($) 1.40 1.60

5 2P

728

0.3874 0.0317 0.0808 0.0501 0.75

0

x

2 a

A = 163

y

4 5 6 7

(4, 2)

2

0

x

4

2

b

3 16

3 a

1 4

4 a

3 8

c m =

( 132 ) 3

b

1 3

b

7 8

c 2

d 2

1 8

8 3

5 a 6 7

b

c 8 a 0.25 c 5 min 48 s a 5

6

7

c 0.191 e 88 minutes

b 1

9 10 11

9 17 25 33 41 49

c

y

0

3

6

b

2

6 14 22 30 38 46 54

1

2

t

Exam practice 4 Short answer 1 34% 1 2 a 15

b

4 45

c 1

d

1 8

3 a 74%

b 37

28

 0.75 0.20  4 a T =    0.30 0.60 

9

8 a

(1.42, 1.54)

d 0.41

6

9 6 3

4

6

8 10 12 14

c

b  Initial proportion of coffee purchasers   0 . 9  S= =   In itial proportion of biscotti purchasers   0 . 4  c 0.755 4

 0 . 75 0 . 20  0.9  d   ×    0 . 30 0 . 60  0.4 

70 80 90 100110120130

9 Mean: translation of 22 units in the positive x direction; standard deviation: dilation factor of 13 from the x-axis, dilation factor of 3 from the y-axis 10 a 68% b 95% c 99.7% 11 a 16.5 ≤ X ≤ 22.7 b 13.4 ≤ X ≤ 25.8 c 10.3 ≤ X ≤ 28.9 d 22.7 e 25.8 f 28.9 2 3

0.28 0.37 m = 2 0.16

b 0.5 b 0.28 b 0.40 b m = 1.5 b 0.34

c 0.51

y 1

b

2.5 2

c 0.32

3 7 11 15 19

Extended response 1 a i 0.4066 b 0.8946 2 a 1.82 cm c 5% 3 a 163.2 seconds c 0.7357

ii 0.1653 iii 0.0672 c 6:26 pm d $1.06 b 5% d 0.9474 b 0.0096 d 137.5 seconds

4 8 12 16 20

Extended response a i f (-2) = 0.054 ii f (1) = 0.242

0.5

( 2, 0.054)

Multiple choice 1 A 2 D 5 C 6 A 9 E 10 D 13 D 14 D 17 D 18 C

C B A C C

Multiple choice 1 E 2 B 3 C 4 B 5 D

C D E A C

5 1 .5

0

(1, 0.242) 0.5

1

1.5

2

2.5 x

.5

c 0.3401 d Pr(-1 < X < 1) ≈ 0.7194 (left rectangular) ≈ 0.6410 (right rectangular) e 0.683 y 1

g

(0, 0.3989)

( 1, 0.2421) ( 2, 0.1330)

2.5 2

0.5

5 1 .5

0

(1, 0.2421) 0.5

1

1.5

(2, 0.1330) 2

2.5 x

.5

Answers 12F ➜ 12g

2 a 1 13 a 14 a 15 a 16 a

0.0062 0.8413 0.1535 2.34 cm 0.1963 26.77 cm 0.0625 3.4 3 hours, 31 minutes 11.9 b

0

d 4 b 0.325

8

a b c a b c a b c a

h 0.6827

Answers

729

Index absolute value function 92–5, 212–15 graphing 304–5 hybrid functions, as 94–5 addition of ordinates 206–10, 221 addition rule of probability 488–9, 498, 529 angles complementary 269–70 measurement of 258–9 negative 267 antidifferentiation 416–23, 474 see also integration areas approximation methods 435–8, 475 between graph and x-axis 435–8 between two curves 460–4, 475 bound by a curve and the x-axis 455–6 enclosed by functions, approximating 435–8 finding without sketch graphs 456–8 further 455–8 signed 448 average rates 325 Bernoulli trials and sequences 537–46 binomial distribution 537–48, 581 expected value, variance and standard deviation of 573–7, 582 graphs of 546–7, 581 multiple probabilities, problems involving 552–6, 581 binomial expansions 1 binomial theorem 1–5, 47 chain rule 341–3, 366 change-of-base rule 138 change, rates of 396–8, 408 combinations 496–7, 498, 529 complementary angles 269–70 completing the square 23 composite functions 108–10, 122 graphing 307–9 conditional probability 492–3, 498, 529 constant rates 324 continuous distributions 591–3 continuous random variables 589–93, 643 applications to problem solving 613–16, 644 normal distribution 619–23 probability density functions and 595–601 standard normal distribution 626–33 cosine 261 graphs 282–8, 315 special cases 262 cubic function gradient function of 327 power form, in 65–9 cubic graphs 29–34, 48 curves, sketching 376–84 definite integrals 442–6, 474

730

Index

degrees converting radians to 259–60 converting to radians 260–1 finding number of degrees in one radian 259 derivatives e x 344, 366 loge (x) 347–9, 366 sin (x), cos (x) and tan (x) 351–4, 366–7 x n 338–9, 366 difference functions 103–4, 122 graphing 303–4 differentiation 325 chain rule 341–3, 366 first principles, using 334–6, 366 mixed problems on 360–4 product rule 355–7, 367 quotient rule 357–9, 367 dilation 57–8, 60, 65, 71, 79, 87, 97–8, 173–5, 183–5, 283, 291 discrete random distributions measures of centre of 510–16, 529–30 measures of variability 518–26, 530 discrete random variables 501–6, 529 expectation theorems 512–14 discriminant 22 division of polynomials 11–14 domain and range of functions 18–20 restricted 18, 33 equations absolute value 92–5, 121 base e 151–3, 167 composite 108–10, 122 exponential 142–6, 151–3, 167 finding, for graphs of exponential and logarithmic functions 202–4 functional 108–10, 122 literal 159–61, 167 logarithmic 147–9 natural (base e) logarithms, with 154–5, 167 tangents and normals 373–5, 408 trigonometric 273–81, 295–7, 315 Euler’s number 151 events 484, 497 independent 489 mutually exclusive 489 exact values 264 expectation theorems 512–14 exponential and logarithmic modelling 162–4 graphs, using 216–17 exponential equations 142–6, 167 base e 151–3 exponential functions 220 absolute values, with 212–15 ex, integration of 426–7 graphs of, with base e 191

finding areas without sketch graphs 456–8 functional equations 108–10, 122 functions 49 absolute value 92–5, 121, 212–15, 304–5 average value of 466–8 composite 108–10, 122, 307–9 continuous 325 cubic 29–31, 65–9, 327 difference 103–4, 122 domain and range 18–20 exponential 172, 191, 212–15, 220, 426–7 gradient 325–7 graphing 307–9 hybrid 94–5 inverse 240–3, 252 limits 332–6, 366 logarithmic 212–15, 220–1 polynomial, characteristics of 376–84 power 71–5, 78–83 probability density functions 595–601 product 105–6, 306–7 quadratic 22, 59–60, 326–7 range of 18–20 restricting 245–9, 252 smooth 325 square root function in power form 86–90 sum 103, 122, 303–4 tangent, graphs of 291–5 their inverses and 234–9, 252 trigonometric 311–12, 316, 427–8 fundamental theorem of integral calculus 441–6

sine and cosine functions, of 282–8, 315 stationary points 377–8, 408 straight line 15–17 sum function 303–4 tangent function, of 291–5, 315–16 transformations 57–62 trigonometric 282–8, 315 truncus 78–83 types of 111–17 hybrid functions 94–5 hyperbola 71–5 index laws 132–6, 167 instantaneous rates 325 integral calculus, fundamental theorem of 441–6 integrals, properties of 417–21 definite 442–6, 474 integration ex, sin (x) and cos (x) 426–8 further applications of 468–71 recognition, by 430–3 interquartile range 610–11 intersection 485, 529 interval notation 18 inverse cumulative normal distribution 636–40 inverse functions 240–43, 252 inverses 156–8, 167 functions and 234–9, 252 relations, of 229–33 Karnaugh maps and probability tables 491–2, 498, 529

gradient rates of change and 324–8, 366 gradient function cubic functions, of 327 graphing from the graph of a function 325 quadratic functions, of 326–7 straight lines, of 326 graphs 221, 303–9, 316 absolute value 212–15, 221, 304–5 addition of ordinates 206–10, 221 antiderivative function and original function, relationship 421–3, 474 binomial distribution 546–7, 581 composite functions 307–9 cosine 282–8, 315 cubic 29–34, 48 exponential and logarithmic modelling, using 216–17 exponential functions with any base 172–80 exponential functions with base e 191–6 hyperbola 71–5 inverse, of 229–33 linear 15–20, 47–8 logarithmic, to any base 182–90 logarithmic, to base e 197–200 maximum and minimum problems, solving 386–94, 408 polynomial functions, characteristics of 376–84 power functions, of 120 product functions 306–7 properties of straight line 15–17 quadratic 22–7, 48 quartic 38–44, 48–9

linear approximation 404–6 linear graphs 15–20, 47–8 literal equations 159–61, 167 logarithm laws 138–41, 167 logarithmic equations using any base 147–9 logarithmic functions with absolute values 212–15, 221 logarithmic graphs to any base 182–90, 220–1 Markov chains 559–70, 581 transition matrices 565–70, 581 mathematical model 111 matrices transformations, with 96–101, 121–2 transition 565–70, 581 using to describe a reflection in the line y=x 229–30 maximum and minimum problems where function known 386–9, 408 where function unknown 390–4, 408 mean 573–7, 582, 605 measures of central tendency and spread 605–11, 643–4 median 515–16, 605 mode 515–16, 605 modelling 111–17, 122 exponential and logarithmic 162–4 trigonometric 299–301 modulus functions, graphing 304–5 see also absolute value function negative angles 267 negative quartics 39

Index

731

normal distribution 619–23, 644–5 common probabilities associated with 619–20 graph 619 inverse cumulative normal distribution 636–40, 645 properties 619 standard 626–33, 645 symmetry properties 630–3 ordinates, addition of 206–10, 221 parabola and transformations 57–62 Pascal’s triangle 2, 5, 47 percentiles 601, 638 polynomials 6–10, 47 division of 11–14 evaluating 8–10 positive quartics 38 power functions hyperbola 71–5 truncus 78–83 probability 485–98 addition rule of 488–9, 529 combinations 496–7, 498, 529 conditional 492–3, 498, 529 independent events 489 Karnaugh maps and probability tables 491–2, 498, 529 mutually exclusive events 489 outcomes 484, 497, 529 terminology 484–5, 497, 529 tree diagrams 494–6, 498, 529 Venn diagrams 485, 497, 529 probability density function 590–3, 643 using to find probabilities of continuous random variables 595–601, 643 product functions 105–6 graphing 306–7 product rule 355–7, 367 quadratic function in power form 59–60 quadratic graphs 22–7, 48 quantiles 601, 638 quartic graphs 38–44, 48–9 negative 39 positive 38 quotient rule 357–9, 367 radians 258–63, 314 converting degrees to 260–1 converting to degrees 259–60 finding number of degrees in one 259 random variables, discrete 501–6 discrete probability distributions 502–6 rates of change 396–8, 408 rates, related 401–3, 409

732

Index

reflection 58, 60, 65, 71–2, 79, 87, 97–8, 178–80, 187–90, 285–6, 291 related rates 401–3, 409 relations and their inverses 229–33 restricting function 245–9, 252 sample point 484, 497, 529 sample space 484, 497, 529 sigma notation 445 signed areas 448–51 sine 261 graphs 282–8, 315 special cases 262 sketching curves 376–84 square root function in power form 86–90 standard deviation 521–3, 573–7, 582, 607 interpreting 523–6 stationary points 377–8, 408 types of 377–8 straight line graphs, properties of 15–17 sum functions 103, 122 graphing 303–4 symmetry exact values and 264–71, 314 tangent 262 graphs 291–5, 315–16 special cases 262 tangents and normals, equations of 373–5, 408 transformations 57–62 combination of 59, 61, 66, 72, 88 matrices, with 96–101 transition matrices 565–70, 581 translation 59, 98–9, 286–7, 292 horizontal 60, 66, 72, 79–80, 87, 176–7, 186–7 vertical 60, 66, 72, 80, 87, 176, 185–6 tree diagrams 494–6, 498, 529 trigonometric equations 273–81, 295–7, 315 trigonometric functions, integration of 427–8 trigonometric functions with an increasing trend 311–12, 316 trigonometric graphs 282–8, 315 finding equations of 295–7 trigonometric modelling 299–301 truncus 78–83 turning point form 23 union (probability) 484, 497, 529 unit circle 258–63, 314 symmetry properties 264–5 variability of discrete random distributions, measures of 518–26, 529–30 variance 518–21, 573–7, 582, 607 Venn diagrams 485, 497, 529

Maths Quest Methods Year 12 2v2b6d

Overview 26281t

More details 6y5l6z

Related Documents 3h463d

Maths Quest Methods Year 12 2v2b6d

Mathematical Methods Cas - Maths Quest 11 1z2y60

Maths Quest 11 Mathematical Methods Cas Prelims 61l6r

Maths Talent Quest 2gm1h

Maths Quest 8.pdf 5y342p

Cambridge Maths Year 12 3-unit e6w4m

More Documents from "William Nguyen" 514hs

Maths Quest Methods Year 12 2v2b6d

Clark Chan x453b

270760144-detective-conan.pdf 1e4p2d

Mqmon 4s4e22

374371453-powermill-2018.pdf 5a14

Introduction To Sap Bridge 5h5823