Methods TI-nspire CAS calculater companion

8/10/2019 Methods TI-nspire CAS calculater companion

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MATHS QUEST 12TI-NSPIRE CAS CALCULATOR COMPANION

MathematicalMethods CAS

VC E M AT H E M ATI CS U N I T S 3 & 4

RAYMOND ROZEN | BRIAN HODGSON | NICOLAOS KARANIKOLASBEVERLY LANGSFORD-WILLING | MARK DUNCAN | TRACY H ERFT

LIBBY KEMPTON | JENNIFER NOLAN | GEOFF PHILLIPS


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First published 2013 byJohn Wiley & Sons Australia, Ltd42 McDougall Street, Milton, Qld 4064

Typeset in 10/12 pt Times LT Std

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Introduction

This booklet is designed as a companion to Maths Quest 12 Mathematical Methods CAS Second Edition .

It contains worked examples from the student text that have been re-worked using the TI-Nspire CXCAS calculator with Operating System v3.

The content of this booklet will be updated online as new operating systems are released by TexasInstruments.

The companion is designed to assist students and teachers in making decisions about the judicious use ofCAS technology in answering mathematical questions.

The calculator companion booklet is also available as a PDF le on the eBookPLUS under thepreliminary section of Maths Quest 12 Mathematical Methods CAS Second Edition .

iv Introduction


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CHAPTER 1 Graphs and polynomials 1

CHAPTER 1

Graphs and polynomialsWORKED EXAMPLE 1

Use the binomial theorem to expand x(2 3) 4 . TH IN K WR IT E

1 On a Calculator page, press: MENU b 3: Algebra 3 3: Expand 3

Complete the entry line as:expand((2 x 3)4)Then press ENTER .

2 Write the result. (2 x 3)4 = 16 x 4 96 x 3 + 216 x 2 216 x + 81


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2 Maths Quest 12 Mathematical Methods CAS

WORKED EXAMPLE 8

Given that = + + = + = P x x x x Q x x x x x R x x( ) 6 2 3 , ( ) 2 5 2 and ( ) 4,2 4 5 4 2 2 nd:a + P x Q x( ) ( )b P x R x( ) ( ).

TH IN K WR IT E/ DI SP LAY

a 1 On a Calculator page, dene the

polynomials P ( x ), Q( x ) and R( x ). Todo this, press: MENU b 1: Actions 1 1: Dene 1Complete the entry lines as:Dene p( x ) = 6 2 x + 3 x 2 + x 4 Dene q( x ) = x 5 2 x 4 + x 2 5 x 2Dene r ( x ) = x 2 4Press ENTER after each entry.

a

2 To calculate P ( x ) + Q( x ), completethe entry line as:

p( x ) + q( x )Press ENTER

b To calculate P ( x ) R( x ), complete the entryline as:

p( x ) r ( x )Press ENTER .

b


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WORKED EXAMPLE 10

If = + + = = p x ax x x bx p p( ) 3 5, ( 1) 5 and (2) 655 4 3 , nd the values of a and b. TH IN K WR IT E/ DI SP LAY

1 On a Calculator page, dene thepolynomial P ( x ) by completing the entryline as:

Dene p( x ) = a x 5 + x 4 3 x 3 + b x 5Then press ENTER .To calculate the values of a and b,complete the entry line as:solve( p(1) = 5 and p(2) = 65, a )Then press ENTER .

2 Write the answer. Given p( x ) = ax 5 + x 4 3 x 3 + bx 5 and solving p(1) = 5 and p(2) = 65 gives a = 2 and b = 6.

WORKED EXAMPLE 11

Find the quotient, Q( x), and the remainder, R( x), when + x x x3 2 84 3 2 is divided by the linearexpression + x 2.

TH IN K WR IT E

1 On a Calculator page, press: MENU b 2: Number 2

7: Fraction Tools 7 1: Proper Fraction 1Complete the entry line as:

propFrac + +

x x x x

3 2 8

2

4 3 2

Then press ENTER .

2 Write the answer. Dividing x 4 3 x 3 + 2 x 2 8 by x + 2 givesa quotient, Q( x ), of + x x x 5 12 243 2 and a

remainder, R( x ), of 40.


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WORKED EXAMPLE 12

Determine whether or not = D x x( ) ( 3) is a factor of P ( x) = 2 x3 4 x2 3 x 8. TH IN K WR IT E

1 On a Calculator page, dene P ( x ) bycompleting the entry lines as:Dene p( x ) = 2 x 3 4 x 2 3 x 8

p(3)Press ENTER after each entry.

2 Since p(3) = 1, ( x 3) is not a factorof P ( x ).

p x (3) 0 so ( 3) is not a factor of P ( x ).


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WORKED EXAMPLE 13

a Factorise P ( x) = 2 x3 x2 13 x 6. b Solve 2 x3 x2 13 x 6 = 0. TH IN K WR IT E

a 1 On a Calculator page, dene thepolynomial P ( x ) by completing theentry line as:

Dene p( x ) = 2 x 3 x

2 13 x 6Then press ENTER .

To factorise P ( x ), complete the entryline as:factor ( p( x ))Then press ENTER .

a

2 Write the answer. Factorising p( x ) = 2 x 3 x 2 13 x 6 givesP ( x ) = ( x 3)( x + 2)(2 x + 1)

b 1 To solve P ( x ) = 0, complete the entryline as:solve p x x ( ( ) 0, )=Then press ENTER .

b

2 Write the answer. Solving 2 x 3 x 2 13 x 6 = 0 gives= = =

x x x 2, 1

2, or 3


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2 Sketch the graph. y

x 021

(1, 3)

(2, 5)

3

5

f ( x ) = 1 2 x , x ( , 1)

3 State the domain and range. The domain is ( , 1).The range is (3, ).

WORKED EXAMPLE 22

Sketch the graph of = + y x x3 8 2 2, showing the turning point and all intercepts, roundinganswers to 2 decimal places where appropriate.

TH IN K WR IT E/ DR AW

1 On a Graphs page, complete the functionentry line as:

f ( x ) = 3 + 8 x 2 x 2Then press ENTER .To label the coordinates of the interceptsand turning point, press: MENU b 6: Analyze Graph 6Select the appropriate action.

2 Sketch a parabola through these points. y

x 0 41 5

(2, 11)

(0.35, 0)

f ( x ) = 3 + 8 x 2 x 2

(4.35, 0)3

6

9

12

(0, 3)


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WORKED EXAMPLE 25

Sketch the graph of = y x x x10 83 2 , showing all intercepts. TH IN K WR IT E/ DR AW

1 Use a CAS calculator to solve for x .

2 State the x -intercepts. The x -intercepts are 2, 1 and 4.3 Sketch the graph of the cubic. y

x 0

y =

x

3

x

2

10

x 8

8

4 2 1


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WORKED EXAMPLE 27

Sketch the graph of = + + y x x x x7 5 104 3 2 , showing all intercepts. TH IN K WR IT E/ DR AW

1 On a Calculator page, dene thepolynomial P ( x ) by completing the entryline as:

p x x x x x Define ( ) 7 5 104 3 2

= + +

Then press ENTER .To nd the y-intercept, complete the entryline as: p(0)Then press ENTER .To nd the x -intercepts, complete theentry line as: solve = p x x ( ( ) 0, )Then press ENTER .

2 To sketch the graph of P ( x ), open aGraphs page.

Complete the function entry line as: f x p x 1( ) ( )=Then press ENTER .

3 Sketch the graph of the quartic. y

x 05, 0)( 5, 0)(

1 2 3

(1, 0)(2, 0)

(0, 10)

3 2 1


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WORKED EXAMPLE 28

Sketch the graphs of each of the following equations, showing the coordinates of all intercepts.Use a CAS calculator to nd the coordinates of the turning points, rounding to 2 decimal places asappropriate.a y = x2 ( x 1)( x + 2) b y = ( x + 3) 2( x 1) 2


a1

To nd the x and y intercepts, on aCalculator page, complete the entrylines as:solve( y = x 2 ( x 1) ( x + 2), y) | x = 0solve( y = x 2 ( x 1) ( x + 2), x ) | y = 0Press ENTER after each entry.

a

2 State the intercepts and their nature. The graph touches the x -axis at x = 0 and crossesthe x -axis at x = 2 and x = 1.The y-intercept is at y = 0.

3 To nd the coordinates of the turningpoints, open a Graphs page.Complete the function entry line as:

f 1( x ) = x 2( x 1)( x + 2)Then press ENTER .

Then press: MENU 6: Analyze graph 6 2: Minimum 2Move the upper and lower boundsinto the correct position to locate thestationary points. Similarly for thelocal maximum at (0, 0).

4 Sketch the graph of the quartic.

0 x

y

(0, 0) (1, 0)

(0.69, 0.40)

(2, 0)

(1.44, 2.83)


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b 1 To nd the x and y intercepts, on aCalculator page, complete the entrylines as:solve( y = ( x + 3) 2 ( x 1)2, y) | x = 0solve( y = ( x + 3) 2 ( x 1)2, x ) | y = 0Press ENTER after each entry.

b

2 State the intercepts and their nature. The graph touches the x -axis at x = 3 and at x = 1.The y-intercept is at y = 9.

3 To nd the coordinates of the turningpoints, open a Graphs page.Complete the function entry line as:

f 1( x ) = ( x + 3)2( x 1) 2Then press ENTER .

Then press: MENU 6: Analyze graph 6 2: Minimum 2Move the upper and lower boundsinto the correct position to locate thestationary points. Similarly for thelocal maximums.

4 Sketch the graph of the quartic.

0 x

y

(1, 16)

(0, 9)

(1, 0)(3, 0)


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WORKED EXAMPLE 29

Determine the equation of the graph shown.


1 On a Calculator page, complete the entryline as:a x x x x y( 3) ( 1) ( 1) ( 2) + +

Then press ENTER .To calculate the value of a , complete theentry line as:

= = y a x solve( 3, ) | 0

Then press ENTER .To nd the equation of y, complete theentry line as:

y | a =1

2

Then press ENTER .

2 Write the equation. y x x x x ( 2)( 1)( 1)( 3)

2=

+ +

x

y

03 1

3

21


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WORKED EXAMPLE 31

ax 7 y = 02 x + ( a 9) y = 0

Find the value(s) of a , where a is a real constant. Consider a set of simultaneous equations thathave a unique solution.

TH IN K WR IT E

1 On a Calculator page, to create the matrixof coefcients, press:Press: MENU b 7: Matrix & Vector 7 1: Create 1 1: Matrix 1Choose 2 and 2 as the number of rowsand columns, complete as shown, andpress ENTER .

2 Press: MENU b 7: Matrix & Vector 7 3: Determinant 3to nd the determinant in terms of a .Then complete the entry line assolve( a 2 9a + 14 = 0, a ), and pressENTER .

3 Write the answer. There is a unique solution when the determinantis non-zero, so for a unique solution a R/{2, 7}


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WORKED EXAMPLE 32

For the linear simultaneous equations give below:a determine the values of t for which there are i innitely many solutions and ii no solutionsb determine the unique solution for the equations in terms of t, specifying the restrictions on t.

tx 3 y = 62 x + ( t 5) y = 3 t

TH IN K WR IT E

1 On a Calculator page, complete the entryline as:

solve

=

t t

t det 3

2 50,

Press: MENU b 3: Algebra 3 7: Solve System of Equations 7 2: Solve System of Linear Equations 2Choose 2 as the number of equations, and x ,and y as the variables.

2 Complete the entry lines as:

linsolve =+ =

=

tx y

x t y t x y t

3 6

2 ( 5) 3 ,{ , } | 3

linsolve =+ =

=

tx y

x t y t x y t

3 6

2 ( 5) 3 ,{ , } | 2

and press ENTER after each entry.

3 State the solution. There is a unique solution

=

x t

15

3 and =

+

= y

t

t

3 6

3 when t R /{2, 3}

There is no solution when t = 3.There is an innite number of solutions whent = 2.


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WORKED EXAMPLE 33

Solve the following simultaneous linear equations. x + 2 y 3 z = 11 4 x 3 y + z = 12 3 x y z = 14

TH IN K WR IT E

1On a Calculator page, press: MENU b 3: Algebra 3 7: Solve System of Equations 7 2: Solve System of Linear Equations 2Choose 3 as the number of equations, and x ,

y and z as the variables. Complete the entryline as:

linsolve

+ = + = =

x y z

x y z x y z

x y z

2 3 11

4 3 12,{ , , }

3 14

and press ENTER .

2 State the solution. There is a unique solution x = 2, y = 3 and z = 5


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WORKED EXAMPLE 34

Consider the following system of simultaneous equations.

kx y + z = 83 x + ky + 2 z = 2

x + 3 y + z = 6For what values of k, is there: i a unique solution? ii no solution?iii an innite number of solutions?

TH IN K WR IT E

1 On a Calculator page, to create the matrix ofcoefcients, press:Press: MENU b 7: Matrix & Vector 7 1: Create 1 1: Matrix 1Choose 3 and 3 as the number of rows andcolumns, complete as shown, and pressENTER .

2 Press: MENU b 7: Matrix & Vector 7 3: Determinant 3to nd the determinant in terms of k . Thencomplete the entry line assolve( k 2 7k + 10 = 0, k ),and press ENTER .

3 Now press: MENU b 3: Algebra 3 7: Solve System of Equations 7 2: Solve System of Linear Equations 2Choose 3 as the number of equations, and

x , y and z as the variables. Complete theentry lines as:

linsolve + =+ + =

+ =

k y z

x ky z x y z

x y z

8

3 2 2,{ , , }

3 6

and press ENTER .


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4 Then complete the entry lines as:

linsolve

+ =+ + =

+ + =

=

k y z

x ky z x y z

x y z

k

8

3 2 2,{ , , }

3 6

| 2

linsolve

+ =+ + =

+ + =

=

k y z

x ky z x y z

x y z

k

8

3 2 2,{ , , }

3 6

| 5

5 State the solution. i There is a unique solution when k 5 or k 2. ii There is no solution when k = 5.iii There is an innite number of solutions when

k = 2.

WORKED EXAMPLE 35

The cubic function with the general equation y = ax 3 + bx 2 + cx + 8 passes through the points (1, 2)(2, 4) and (4, 8). Find the values of a , b and c.

TH IN K WR IT E

1 On a Calculator page,Dene f ( x ) = ax 3 + bx 2 + cx + 8Now press: MENU b 3: Algebra 3 7: Solve System of Equations 7

2: Solve System of Linear Equations 2Choose 3 as the number of equations, anda , b and c as the variables. Complete theentry line as:

linsolve

==

=

f

f a b c

f

(1) 2

( 2) 4,{ , , }

(4) 8

and press ENTER .2 State the solution. a = 1, b = 3 and c = 4, the cubic is

y = x 3 3 x 2 4 x + 8


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WORKED EXAMPLE 36

Solve these ve linear simultaneous equations using matrices and a CAS calculator.

2v + w 3 x + 2 y z = 12v + 3w + 4 x y + 2 z = 13v 2w + 5 x 2 y 3 z = 32

3v w + 2 x y 3 z = 183v + 3w 4 x + 3 y 2 z = 9

TH IN K WR IT E

1 On a Calculator page, create the matrix ofcoefcients, press: MENU b 7: Matrix & Vector 7 1: Create 1 1: Matrix 1Choose 5 and 5 as the number of rows andcolumns, complete as shown, and store thematrix as a , then press ENTER .

2 Repeat, choosing 5 and 1 as the number ofrows and columns, and store the matrix as b,then press ENTER .

3 To nd the solution, complete as showna1b

4 State the solution. v = 2, w = 4, x = 1, y = 3 and z = 5.


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CHAPTER 2 Functions and transformations 19

CHAPTER 2

Functions and transformationsWORKED EXAMPLE 3

Given the equation y = kx 2, determine the effect on the graph y = x2, when k = {2, 3, 4}. Sketch thegraphs.



f 1( x ) = x 2

Then press ENTER .

2 Complete the function entry lines as: f 2( x ) = 2 x 2

f 3( x ) = 3 x 2

f 4( x ) = 4 x 2Press ENTER after each entry.

3 Answer the question by describing the

changes in words.

As the value of k increases the graph becomes

thinner and stretches away from the x -axis.


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WORKED EXAMPLE 6

Find the equation of the curve, if it is of the form y = a ( x b)3 + c.


1 Write the general equation of the cubicfunction.

y = a ( x b)3 + c

2 Write the coordinates of the stationarypoint of inection ( b, c) and hence statethe values of b and c.

The stationary point of inection is (1, 3).So b = 1, c = 3.

3 Substitute the values of b and c into thegeneral formula.

y = a ( x 1) 3 + 3

4 The graph passes through the point (0, 5)( y-intercept). Substitute the coordinatesof this point into the equation.

Using (0, 5):5 = a (0 1) 3 + 3

5 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve(5 = a (0 1) 3 + 3,a )Then press ENTER .

6 Write the solution for the equation. a = 27 Substitute the value of a into

y = a ( x 1)3 + 3. y = 2( x 1)3 + 3

x 0

3

5


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CHAPTER 2 Functions and transformations 21

WORKED EXAMPLE 18

Given f : [0, ) R , where = f ) x and g( x) = af ( x) + b, where a and b are positive realconstants, consider the effect on g( x) as a and b increase individually.

TH IN K WR IT E

1 On a Graphs page, complete thefunction entry line as:

+ x = 2Insert a slider by pressing: MENU b 1: Actions 1 A: Insert Slider AThen press ENTER .Repeat to insert a second slider.Grab the slider for each variable andmove it back and forth taking note of theeffect of each variable increasing.

2 Write your description in words. As a increases, the graph is dilated away from

the x -axis, with the graph stretched further fromthe x -axis. As b increases, the graph is translated upparallel to the y-axis.


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WORKED EXAMPLE 21

Express f ( x ) = |5 x 4 | as a hybrid function, dening the domain of each part and graphing thefunction.


1 Break the function into two parts: anegative and positive part.

) x 5 x 4 x 0(5 4 0

4

2 Simplify the domain and function foreach.

First function: 5 x 4

First domain: 5 x 4 04

Second function: (5 x 4) = 5 x + 4Second domain: 5

4

3 Rewrite the function in hybrid formwith the two rules with their respectivedomains.

=

5 b 0.5 x 1.4


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

2 Write the answers in exact form.The calculator defaults to base e whensolving exponential equations in exactform.

Note: ln ( A) log e ( A). This will bediscussed later in this chapter.

Note: In part b , the inequality has been

changed to because g 0.

a Solving 2 x > 5 for x gives

> x log (5)

log (2)

e

e

.

b Solving 0.5 x 1.4 for x gives

x

log 7

5log (2)

.e

e

3 a Solving 2 x > 5 for x gives x > 2.322,correct to 3 decimal places.

b Solving 0.5 x 1.4 for x gives x 0.485,correct to 3 decimal places.


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CHAPTER 3 Exponential and logarithmic equations 31

WORKED EXAMPLE 18

Solve for x in each of the following:

a log x (4) = 2 b )( = log 3 x 1125 . TH IN K WR IT E/ DI SP LAY

a

&b

1 On a Calculator page, press:

MENUb

3: Algebra 3 1: Solve 1Complete the entry lines as:solve(log x (4) = 2, x )

solve =

x (log 3, ) x

1125

Then press ENTER .

2

Write the solutions.a

Solving log x (4)=

2 for x gives x =

2.b Solving =

log 3 x

1125

for x gives x = 5.

WORKED EXAMPLE 22

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


a

&b

1 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry lines as:solve(e x = 3, x )solve(e x 3e

x = 2, x )Press ENTER after each entry.

2 Write the answers in exact form. a Solving e x = 3 for x gives x = log e (3).b Solving e x 3e

x = 2 for x gives x = log e (3).3 Write the answers in approximate

form, correct to 3 decimal places.a Solving e x = 3 for x gives x = 1.099, correct to

3 decimal places.b Solving e x 3e

x = 2 for x gives x = 1.099,correct to 3 decimal places.


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WORKED EXAMPLE 23

Solve for x, giving your answer both in exact form and correct to 3 decimal places, given thatlog e ( x) = 3.


1 Rewrite using e x = y log e ( y) = x . log e ( x ) = 3e3 = x

2 Write the answer in exact form. x = e33 On a Calculator page, press:

MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve(ln ( x ) = 3, x )Then press ENTER .

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

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

WORKED EXAMPLE 25

Calculate the inverse of y = 3e x + 1. TH IN K WR IT E/ DI SP LAY

1 Interchange x and y to write the inverseequation.

x = 3e y+

1

2 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve( x = 3e y

+ 1, y)

Then press ENTER .

3 Write the equation of the inverse. The inverse of y = 3e x + 1 is

= y

x log

31e where x > 0


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CHAPTER 3 Exponential and logarithmic equations 33

WORKED EXAMPLE 26

Calculate the inverse of f ( x) = 2 log 10 ( x 1) + 1. TH IN K WR IT E

1 Interchange x and y to write the inverseequation.

x = 2 log 10 ( y 1) + 1

2 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve( x = 2 log 10 ( y 1) + 1, y)Then press ENTER .

3 Write the equation of the inverse. = ++

y10

101

x

2

1

2

4 The solution from the CAS can be furthersimplied. Rewrite the answer with anegative power.

= + +

y 10 10 1 x

1 2

1

2

5 Add the powers when multiplying indicesof the same base.

y 10 1 x 12

1

2= +

+ +

6 Simplify the powers and write the answer. = +

y 10 1 x 1

2

f x ( ) 10 1 x

11

2= +

WORKED EXAMPLE 27

Solve e kx = 5 + 2e kx for x , where k R \{0}.


1 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:

solve( ek x

= 5 + 2e k x

,x ) Note: You must put in a multiplicationsign between the k and the x .Then press ENTER .

2 Write the answer. = + 1 lo 33 52

e , k R \{0}.


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WORKED EXAMPLE 28

Solve for x, given that12

og log (62 where p > 0.


1 On a Calculator page, press: MENU b

3: Algebra 3 1: Solve 1Complete the entry line as:

solve =

x p x 12

log ( ) 5 log ( ) log (6),2 2 2

Then press ENTER .

2 Write the answer. Note that the CAS

gives the incorrect restriction p

0. Thecorrect restriction is p > 0 for log 2 ( p) tobe dened.

x = 36 p10

WORKED EXAMPLE 29

Solve the following equations using a CAS calculator. Give your answers correct to 3 decimalplaces.a e x = x3 b log e ( x) = x 2


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

2 Write the solutions correct to 3 decimalplaces.

a Solving e x = x 3 for x gives x = 1.857or x = 4.536.

b Solving log e ( x ) = x 2 for x gives x = 0.159or x = 3.146.


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CHAPTER 4 Exponential and logarithmic graphs 35

CHAPTER 4

Exponential and

logarithmic graphsWORKED EXAMPLE 3

Sketch the graph of f ( x) = 2 2 x 1, showing intercepts and asymptotes, and stating the domainand the range


1 To graph y = 2 2 x 1 on a Graphs page,complete the function entry line as:

f 1( x ) = 2 2 x 1

Press ENTER . Note: The horizontal asymptote at y = 2 isnot displayed.

2 To locate the x-intercept on a Calculatorpage, press:

MENU b 3: Algebra 3 1: Solve 1Complete 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: f 1(0) and press ENTER .Write the y-intercept in coordinate form.

0, 32 are the coordinates of the y-intercept.


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5 Sketch the graph.

x 0 1 2

2

(2, 0)

(0, )

1

y

1

2 1

y = 223

Asymptote

f ( x ) = 2 2 x 1

6 State the domain and the range. The domain is R and the range is ( , 2).


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WORKED EXAMPLE 6

Sketch the graph of f ( x) = 2 log 10 (3 x) 2, showing intercepts and asymptotes, and stating thedomain, range and transformations. Give exact values or round to 3 decimal places.


1 To graph y = 2 log 10 (3 x ) 2 on aGraphs page, complete the function entry

line as: f 1( x ) = 2 log 10 (3 x ) 2Press ENTER .

Note: The vertical asymptote at x = 3 isnot displayed.

2 To locate the x-intercept on a Calculator

page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve( f 1( x ) = 0, x )Press ENTER .

3

Write the x -intercept in coordinate form. The coordinates of the x -intercept are (

7, 0).4 To calculate the y-intercept, complete the

entry line as: f 1(0) and press .Write the y-intercept in coordinate form.

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

5 Sketch the graph.

x 0 2 2 4 6

y

2(0, 2 log 10 (3) 2)

( 7, 0)

x = 3Asymptote

f ( x ) = 2 log 10 (3 x ) 26 State the domain and the range. The domain is ( , 3) and the range is R.

7 State the transformations . Reection in the y-axis, dilation 2 units fromthe x -axis, vertical translation 2 units down,horizontal translation 3 units to the right


42/112


WORKED EXAMPLE 11

Sketch the graph of f ( x) = 2 e x, marking the asymptote and intercepts. State the

transformations, domain and range. Give exact answers. Check using a CAS calculator. TH IN K WR IT E/ DR AW

1 State the rule. f ( x ) = 2 e x

2 State the basic shape. Exponential curve3 State the transformations. A reection in the x -axis and a reection in the

y-axis. The vertical translation is 2 units up.4 Find the horizontal

asymptote by translatingthe asymptote of f ( x ) = e x up 2 units.

The horizontal asymptote is y = 2.

5 Find the y-intercept by making x = 0or by reecting (0, 1) in the x -axis andtranslating it up 2 units.

If x = 0, y = 2 e 0

= 2 1 = 1or (0, 1) (0, 1) (0, 1)The y-intercept is 1.

6 Find the x -intercept by making y = 0and solving the equation.

e x =

e

1 x

log e (e x ) = x log e (e) = x 1

If y = 0, 2 e x = 0

e x = 2

1

= 2

e x = 1

log e (e x ) = log e1

2

x = log e1

2

so the x -intercept is log e1

2

.

7 Sketch the graph.

x

y = 2

0

y

2

2

1 1

(0, 1)(log e (

), 0)21

Asymptote

f ( x ) = 2 e x

8 State the domain and the range. 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:

f 1( x ) = 2 e x

Then press ENTER . Note: The horizontal asymptote at y = 2 isnot displayed.


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WORKED EXAMPLE 15

Sketch the graph of f ( x) = 2 3 log e(1 x), marking the asymptote and intercepts.State the domain and range.


1 State the rule. f ( x ) = 2 3 log e (1 x )2 Find the vertical asymptote by translating

the line x = 0 one unit to the right or bymaking 1 x = 0.

Vertical asymptote is x = 1.

3 Find the y-intercept by making x equal to0 and solving the equation.

If x = 0, y = 2 3 log e (1)= 2

4 Find the x -intercept by making y equalto 0.

If y = 0, 2 3 log e (1 x ) = 0 3 log e (1 x ) = 2 log e (1 x ) =

23

= 1 x

x = 1 x 0.95 (to 2 decimal places)

5 To graph f ( x ) = 2 3 log e (1 x ) on aGraphs page, complete the function entryline as:

f 1( x ) = 2 3 ln (1 x )Then press ENTER .

Note: The vertical asymptote at x = 1 isnot displayed.

6 Sketch the graph, remembering thatthere is a reection in both the x- and the

y-axes.

x

x = 1

0

y

4

2 1 1

(0, 2)(1 e , 0)

23

2Asymptote

f ( x ) = 2 3 log e

(1 x )

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


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WORKED EXAMPLE 16

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

TH IN K WR IT E

1 Use the point on the y-axis and substitutevalues into the given equation.

For (0, 2):2 = Ae 0 + B2 = A + B [1]

2 Substitute the coordinates of anotherpoint into the given equation.

For ( 2.44, 0):0 = Ae

2.44 + B [2]3 On a Calculator page, press:

MENU b 3: Algebra 3 1: Solve 1Press t to obtain the expressiontemplate and choose the simultaneousequation template for two unknowns.Complete the entry line as:

a

aasolve

2

0,

=+e =

Press ENTER .

4 Write down the solutions correct to2 decimal places.

A = 2.19 and B = 0.19

WORKED EXAMPLE 23

Sketch the graph of y = x2e x using a CAS calculator. Show all axis intercepts and any asymptotes. TH IN K WR IT E/ DI SP LAY

1 To graph y = x 2e x on a Graphs page,complete the function entry line as:

f 1( x ) = x 2e x

Then press ENTER . Note: y = x 2e x is asymptotic to the negative x -axis, as when x , x 2e x 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.

x 0

y

4

(0, 2)

1

(2.44, 0)34 2 1


45/112


WORKED EXAMPLE 24

For the function = 2+ 3 y :a sketch the graph of = 2+ y , showing any asymptotesb calculate all axis intercepts both in exact form and correct to 2 decimal placesc state the domain and range.


a 1 To graph y x log ( 2) 3e= 2 + with aCAS calculator, open a Graphs pageand complete the function entry lineas:

f 1( x ) = 2 ln 3

Then press ENTER . Press t andselect the absolute value template.

a


x

3

2

1

12

2 3 4 5 6 7 1 2 30 4 5 1

x = 2

Asymptote

b 1 To locate the x -intercepts, express y3 as a hybrid

function.

b y x

2 l ( 2

2 log ( ( 2 x e

2 Substitute y = 0 and solve eachequation.

2 log e ( x + 2) 3 = 0

lo ( 3

2+ =

3 Write the equation in exponentialform and solve for x .

23

x = 2.481 694 Repeat for the other x -intercept. 2 log e (

( x + 2)) 3 = 0

og ( ( x 3

2

=

x

(

23

x = 6.481 69

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

e 2, 0

and e 2, 0

(2.48, 0) and ( 6.48, 0)


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

y 2 3o

y 2 3= o

3

3

y = 1.613 717 Write down the coordinates of the

y-intercept in exact form and correctto 2 decimal places.

(0, 2 log e (2) 3)(0, 1.61)

c1

State the domain. c x R \{

2}2 State the range. y R


47/112


WORKED EXAMPLE 26

The population of wombats in Snubnose Gully is increasing according to the equation:W = 100 e0.03 t

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 rst 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.


a 1 State the rule. a W = 100 e0.03 t

2 Find W when t = 0. When t = 0, W = 1003 Write the answer in a sentence. The initial size of the population is 100 wombats.

b 1 Find W when t = 2. b When t = 2, W = 100 e0.03 2 = 100 1.0618 106 (nearest whole number)

2 Write the answer in a sentence. After 2 years there are 106 wombats.3 Find W when t = 10. When t = 10, W = 100 e0.03 10

= 100 1.3499 135 (nearest whole number)

4 Write the answer in a sentence. After 10 years there are 135 wombats.c 1 To graph f ( x ) = 100 e0.03 x on a Graphs

page, complete the function entryline as:

f 1( x ) = 100 e0.03 x

Then press ENTER .Note the viewing window settings:Xmin:0, Xmax:30, Ymin:0 andYmax:250When drawing your graph, label theaxes with the given variables.

c

d 1 Convert the year 2020 to the correctnumber of years.

d t = 2020 1998= 22 years

2 Find W by substituting t = 22 into theequation.

When t = 22, W = 100 e0.03 22= 193.479= 193 (nearest whole number)

3 Write the answer as a sentence. In the year 2020, there are approximately

193 wombats.


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e 1 Let W = 250. e 250 = 100 e0.03 t

2 Divide both sides by 100. 2.5 = e0.03 t

3 Take natural logs of both sides. log e (2.5) = log e (e0.03 t )

0.03 t = log e (2.5)

4 Divide both sides by 0.03. 1

.03log e ( )2.5=

5 Evaluate the answer using a CAS

calculator.t = 30.543 t = 31 (nearest year)

6 Express t = 31 as a year and write theanswer as a sentence.

There will be 250 wombats in the year 2029.


49/112

CHAPTER 5 Inverse functions 45

CHAPTER 5

Inverse functionsWORKED EXAMPLE 6

If f ( x) = ln( x + 1) + 1,a nd f

1( x) b draw the graph of f ( x) and its inverse f 1( x).

TH IN K WR IT E

a 1 Let y = f ( x ). a y = ln( x + 1) + 1

2 Interchange x and y. x = ln( y + 1) + 1

3 To make y the subject, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve(ln( y + 1) + 1 = x , y)(This means to solve the equationwith respect to y).Then press ENTER .

4 Write the answer in the form f 1( x ). f

1( x ) = e x 1 1

b 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) + 1Then press ENTER .Complete the function entry line as:

f 2( x ) = e x 1 1Then press ENTER .

b


50/112


WORKED EXAMPLE 12

a Sketch the graph of f ( x) = x2 3 x + 3, showing the turning point and relevantintercept(s).

b Find the rule of the inverse by an algebraic method and sketch this graphon the same set of axes together with the line y = x.

c Is the inverse a function?d The inverse is a reection in the line y = x of the original function f ( x). Use this information to

nd 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 3 x + 3 so that f

1( x) exists. TH IN K WR IT E/ DR AW

a Use a CAS calculator to help in drawing agraph of f ( x ) including the relevant points.

a y

x

1

34

2

0 2 31 4

(1.5, 0.75)

f ( x ) = x 2 3 x + 3

b 1 To nd the equation of f 1( x ), let

y = x 2 3 x + 3. Interchange x and y.

b x = y2 3 y + 3

2 To make y the subject, on aCalculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve( y2 3 y + 3 = x , y)Then press ENTER .

3 Write the answer in the formof f

1( x ). ) x 4 x 3

2

+

4 Use the calculator to draw the graphsof f ( x ), f

1( x ) and y = x. y

x

1

34

2

0 2 31 4 5 6 7

f ( x ) = x 2 3 x + 3 y = x

c The inverse is a one-to-many relation andtherefore is not a function.

c The graph of the inverse does not pass thevertical line test, as it is a one-to-many relation,and therefore it is not a function.


51/112

CHAPTER 5 Inverse functions 47

d The points of intersection between f ( x ) and f

1( x ) will occur at the intersections of thegraph 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( x 2 3 x + 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 y

x

1

34

2

0 2 31 4

f ( x ) = x 2 3 x + 3

y = x

(1.5, 0.75)

The points of intersection are at (1, 1) and (3, 3).

e For the left-hand branch of the parabola, weneed f ( x ) = x 2 3 x + 3, since the domain is( , a ). It is a one-to-one function. Locate the turning point of the graph andthe x -coordinate will be the value of a . To locate the minimum of

f ( x ) = x 2 3 x + 3 on a Calculator page,press: MENU b 4: Calculus 4 7: Function Minimum 7Complete the entry line as:

f Min( x 2 3 x + 3, x )Then press ENTER .The x -value of the turning point is

3

2, which

implies that3

2= .

e

f R x x , 3+

=


52/112


53/112

CHAPTER 6 Circular (trigonometric) functions 49

CHAPTER 6

Circular (trigonometric)

functionsWORKED EXAMPLE 1

Convert the following to degrees, giving the answer correct to 2 decimal places.

a 2c b 6.3 c c 9

10

TH IN K WR IT E

a ,b

&c

1 On a Calculator page, type: 2 Note: The radian sign does not need

to be entered if the calculatoris already in radian mode.

To convert 2 to degrees, pressCatalog k , press D to get to theitems beginning with the letter D.Then select DD.Then press ENTER .

2 Repeat this process for parts b and c .

3 Write the answers. a 2c 114.59 (correct to 2 decimal places)b 6.3 c 360.96 (correct to 2 decimal places)

cc9

10

= 162 .


54/112


55/112


WORKED EXAMPLE 7

Find the exact value of:

a in 4

3

b tan 5

6

.

TH IN K WR IT E

1 On a Calculator page, complete the entrylines as:

a sin 4

3

b tan 5

6

Press ENTER after each entry.

2 Write the answers.sin

4

3

3

2

=

tan5

6

3

3

=

WORKED EXAMPLE 9

Find all solutions to the equation xcos( ) 2

2=

in the domain [0, 2 ].

TH IN K WR IT E

1 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:

solve x x x (cos( ) 2

2, ) 0 = | 2

Then press ENTER .

2 Write the answer. x 3

4,5

4

=


56/112


WORKED EXAMPLE 10

Find all solutions to the equation sin ( ) = 0.7 in the domain [0, 4 ]. Give your answers correct to4 decimal places.

TH IN K WR IT E


3: Algebra 3 1: Solve 1Complete the entry line as:solve(sin ( ) = 0.7, ) | 0 4 Then press ENTER .

2 Write the answer. If sin ( ) = 0.7, 0 4 , then = 0.7754 or 2.3662 or 7.0586 or 8.6494,correct to 4 decimal places.


57/112


WORKED EXAMPLE 14

Find the general solution of the equation (( x2 sin ) 32 =

and hence nd all solutions for x in thedomain 0 x 2 .

TH IN K WR IT E

1 On a Calculator page, press: MENU b 3: Algebra 3 1: Solve 1Complete the entry line as:solve x x (2 sin (2 ) 3, )=

Then press ENTER .

2 Write the two separate solutions andspecify n Z .

Solving 2 sin (2 x )=

3

gives

x n

x n

n Z (6 1)

6and

(3 2)

3, .

=

=

+

3 Substitute n = 0 and n = 1 and n = 2 intoeach of the general solutions, to nd thesolutions in the domain 0 x 2 .

n = 0: x 6

=

and x 2

3

=

n = 1: x 5

6

= and x

5

3

=

n = 2: x x 11

6and

8

3

= =

For 0 x 2 , x 2

3,5

3,5

6,11

6

=

4 This answer can be veried using a CAScalculator.Complete the entry line as:solve x x x (2sin (2 ) 3, ) 0 2 = |

Then press ENTER .


58/112


59/112


WORKED EXAMPLE 25

While out in his trawler John North, a sherman, notes that the height of the tide in the harbourcan be found by using the equation:

h t5 2 cos6

== ++

,

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 rst 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? TH IN K WR IT E

a 1 On a Graphs page, complete thefunction entry lines as:

f x x

f x

1( ) 5 2 cos6

2( ) 6

= +

=Press ENTER after each entry.

Select an appropriate window.

a

2 Find when high tide occurs. t = 0, 12, 24, . . .A high tide of height 7 m occurs at midnight, noonthe next day, and midnight the next night.

b 1 Find the minimum value of h. b For minimum h,

t cos6

1 =

So h = 5 + 2 1 = 3Alternatively, min. value =median amplitude so h = 5 2 = 3.

2 Find the difference between high andlow tides.

The difference between high and lowtides is 7 3 = 4 metres.

c Use the information from the previous pageto sketch the graph.

c h

t

2

4

6

0 2 4 6 8 1012141618202224


60/112


d 1 Press: MENU b 6: Analyze Graph 6 4: Intersection 4Move the cursor to the left of therst point of intersection, pressENTER , then move the cursorto the right of the rst point of

intersection, press ENTER ,the coordinates of the rst point ofintersection are displayed. Repeat forthe other points of intersection.

d

t = 2, 10, 14, 22, . . .2 Write the answer in words. 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 and2 am the next morning.

WORKED EXAMPLE 26

Using addition of ordinates, sketch the graph of y = sin ( x) + cos ( x) for the domain [0, 2 ]. TH IN K DR AW

1 On a Graphs page, complete the functionentry 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.

2 Sketch the graphs. y

x 4

2

34

74

54

1

1

2

2

0 232

3 Erase y = sin ( x ) and ( y) = cos ( x ) to see thenal graph.

y

x 4

2

34

32

74

54

1

2

0 21

2


61/112


WORKED EXAMPLE 27

Sketch the graph of y = |3 cos (2 x)| over the domain [0, 2 ]. TH IN K DR AW


f 1( x ) = |3 cos (2 x )|

Then press ENTER .Select an appropriate window.


x 4

2

34

32

74

54

1

1

2

3

2

3

0 2


62/112


WORKED EXAMPLE 28

Find the domain and sketch the graph of the product function y = x sin ( x). Use a CAS calculatorfor assistance.


1 The required function can be viewed asa product of two functions: f ( x ) = x and

g( x ) = sin ( x ). The domain of the productfunction is equal to the intersection of thedomains of the two individual functions.

Let f ( x ) = x , dom f ( x ) = RLet 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.2 On a Graphs page, complete the function

entry line as: f 1( x ) = x sin ( x )Then press ENTER .Select an appropriate window.

3 To locate the maximums and minimums,press:Trace along the graph to nd a localmaximum.Press ENTER to lock in the point.This method can also be used to locateminimums and intercepts.


63/112


WORKED EXAMPLE 30

For the pair of functions f ( x) = cos ( x) and g x x( ) :==a show that f ( g( x)) is denedb nd f ( g( x)) and state its domainc sketch the graph of f ( g( x)), using a CAS calculator for assistance.


a For f (g( x )) to exist, the range of g must be asubset of the domain of f . So nd both therange of g and the domain of f to show thatthis condition is observed.

a f ( x ) = cos ( x ); domain of f ( x ) = Rg x x ( ) := range of g( x ) = R+ {0}Range of g( x ) domain of f ( x ) f (g( x )) is dened

b 1 Form the composite function f (g( x ))by substituting g( x ) into f ( x ).

b f g x x ( ( )) cos( )=

2 The domain of f (g( x )) must be thesame as the domain of g( x ). Since thedomain of g( x ) is R+ {0}, so is thedomain of f (g( x )).

Domain of f (g( x )) = R+ {0}

c On a Graphs page, complete the functionentry line as: f x x 1( ) cos( )=

Then press ENTER .Select an appropriate window.

c


64/112


WORKED EXAMPLE 32

Consider a remote island where global warming has caused the temperature to increase by0.1 degree each month. The mean daily temperature is modelled by the function

T m m m( ) 16 0.1 6 cos6

= + + ,

where T is the temperature in degrees Celsius and m is the number of months after January 2008.a Sketch a graph of the function for a ve year period from January 2008, using a CAS calculator

for assistance.b Find the mean daily temperature for March 2009c When will the mean daily temperature rst reach 23 degrees?

TH IN K WR IT E

a On a Graphs page, complete the functionentry line as:

f x x x 1( ) 16 0.1 6 cos6

= + +

Then press ENTER .Select an appropriate window:XMin: 0

XMax: 60YMin: 2.5YMax: 30

a

b March 2009 occurs when m = 14.On a Calculator page, complete the entryline as:

f 1(14)Then press ENTER .

b

The mean daily temperature in March 2009 wouldbe 20.4 degrees Celsius.

c 1 To solve m m23 16 0.1 6 cos6

, = + +

for m, return to the Graphs page.Complete the function entry line as: f 2( x ) = 23Then press ENTER .

c


65/112


2 Press: MENU b 6: Analyze Graph 6 4: Intersection 4Move the cursor to the left of the rstpoint of intersection, pressENTER , then move the cursorto the right of the rst point of

intersection, press ENTER ,the coordinates of the rst point ofintersection are displayed.

m = 11.5631Hence, the rst time the temperature reaches23 degrees Celsius will be during the 12th monthafter January 2008. That is, during January 2009.


66/112


67/112

CHAPTER 7 Differentiation 63

CHAPTER 7

DifferentiationWORKED EXAMPLE 12

If f x

) x ,2 + use a CAS calculator to nd:

a f ( x) b f (2). TH IN K WR IT E


4: Calculus 4 1: Derivative 1Complete the entry lines as:

+

d

dx x x

x 2

83 2

x x

8| 2 x +

Press ENTER after each entry.Alternatively, press t to obtain theexpression template and choose thederivative template.

2 Write the answers using the correctnotation for the derivative.

a f ( x ) 8

2

b f (2) = 2


68/112


WORKED EXAMPLE 14

If f x

) x 1

2 x , nd f ( x). Check your answer using a CAS calculator.

TH IN K WR IT E

1 Write the equation. =) x 1

2

2 Express f ( x ) in index form. y = 3 ) x 2

12

3 Express y as a function of u. Let 1

2 where u = 2 x 2 3 x

4 Differentiate y with respect to u.

ydu

1

5 Express u as a function of x . u = 2 x 2 3 x

6 Differentiate u with respect to x .

u

x 3

7 Find f ( x ) using the chain rule.

=

y x

) x 33

8 Replace u as a function of x and simplify.

=

3 )12

232

=

(4 3)

2(2

=

x 3 4

2 (2 x 3

9 On a Calculator page, press: MENU b 4: Calculus 4 1: Derivative 1

Complete the entry line as:

x

1

2 Press ENTER .

10 Note the answer from the CAS is notexpressed in the same format obtainedpreviously.

=

) x (4 3)

2 x 3 x ) x 3 x )


69/112


WORKED EXAMPLE 24

Find the derivative of y = sin (5 x). TH IN K WR IT E

1 Write the equation. y = sin (5 x )

2 Express u as a function of x and ndu

x . Let u = 5 x so

u

x 5

3 Express y as a function of u and nd yu

. y = sin ( u) so = yu

os ( )u

4 Find y x

using the chain rule. =dy x

5cos ( )

5 Replace u with 5 x . = 5 cos (5 x )

6 It is important to note what happens whena CAS calculator is used to answer thisquestion. On a calculator page, completethe entry line as:

( ) x Press ENTER .

Note: The calculator gives an incorrectanswer as it is set in Degree mode.

In differentiating circular functions, thecalculator needs to be set in Radian modeas the angles are given in radians.

Write the answer, using the correctnotation for the derivative.

= y

dx 5cos (5 )


70/112


WORKED EXAMPLE 33

For the function: f ( x) = | x2 4 x|:a nd the derivativeb sketch the graphs of y = f ( x) and y = f ( x) on the same set of axes.

TH IN K WR IT E

a 1 As f ( x ) = | x 2 4 x | is a composite

function, apply the chain rule to ndthe derivative,

f ( x ) where g( x ) = x 2 4 x andh( x ) = | x |.

a f ( x ) = h(g( x ))

f ( x ) = g ( x ) h (g( x ))=

0if x < 0 or x > 4 and x 2 4 x < 0 if

0 < x < 4.

y

x 0

1

2

3

4

21

34

1 2 3 4 51

3 Write the derivative with the correctdomain.

= ) x 2 or 4

2 x + 4< x

4 To differentiate f ( x ) = | x 2 4 x | with aCAS calculator, on a Calculator page,press: MENU b 4: Calculus 4 1: Derivative 1Alternatively press t to obtainthe expression template and choose thederivative template.

Complete the entry line as: ( ) x Press ENTER .

Note: The answer from the CAS is notexpressed as a hybrid function with thecorrect domain and hence cannot beused.

| x|.sign ( x 4) + sign ( x ). | x 4 | is not anacceptable answer.


71/112


b 1 Sketch the graph of f ( x ) = | x 2 4 x |. b

x

y

12 2

468

3 0 1 2 3 4 5 6 7

2468

10

f ' ( x ) = 2 x 4, x < 0

f ' ( x ) = 2 x + 4, 0 < x < 4

f ' ( x ) = 2 x 4, x > 4

f ( x ) = | x 2 4 x |

2 Sketch the graph of the derivative, 0.8, n)Then press ENTER .

b

2 Write the solution. Solving 1 0.7 n > 0.8 for n impliesn > 4.512 34

3 Interpret the results and answer thequestion.

Grant would need to take 5 shots to ensure aprobability of 0.8 of scoring at least one bullseye.


94/112


WORKED EXAMPLE 8

So Jung has a bag containing 4 red and 3 blue marbles. She selects a marble at random and thenreplaces it. She does this 7 times. Find the probability, correct to 4 decimal places, that:a at least 5 marbles are redb greater than 3 are redc no more than 2 are red.

TH IN K WR IT E

a 1 State the probability distribution anddene and assign values to variables.

a Let X = number of red marbles selectedn = 7

p = 4

7 X ~ Bi(7,

4

7 )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 EEnter values of n and p , usingTab e to move between elds.Enter lower bound as 5 and upperbound as 7, as the number of trialsis 7.

Press ENTER .

3 Write the solution. Pr( X 5) = binomCdf(7,4

7, 5, 7)

= 0.359 3454

Answer the question and round to4 decimal places. The probability that at least 5 red marbles arechosen is 0.3593.

b 1 Repeat as above, using the CAScalculator to nd Pr( X > 3).

Note: Only inclusive values can beentered into the CAS calculator, Pr( X > 3) = Pr( X 4)Remember that only discretevalues are possible for a binomialdistribution.

b Pr( X > 3) = binomCdf(7,4

7 , 4, 7)= 0.653 100 08


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WORKED EXAMPLE 13

Using the above data for attending the gym or aerobics class, nd:a the proportion of people attending the gym and aerobics class on the 5th dayb the number of people attending the gym or aerobics class in the long term.

TH IN K WR IT E

a 1 Write down the transition matrix. a T 0.8 0.7

0.2 0.3=

2 Write down a suitable initial statematrix. In this case, it is the initialnumbers of people attending the gymand aerobics class.

S 150

500 =

3 Identify which state matrix isrequired.

As S 0 corresponds to day 1, therefore day 5corresponds to the state matrix S 4.

4 On a Calculator page, complete theentry lines as:

t

s

0.8 0.7

0.2 0.3

150

500

Press ENTER after each entry.

5 Calculate the proportion of peopleattending the gym or aerobics on the5th day.That is, S 4 = T 4 S 0.

6 Write the solution. S 4 = T 4 S 0

= 0.8 0.70.2 0.3

15050

4

= 155.555

44.445

7 Answer the question (rounding to thenearest whole number).

On the 5th day, there will be 156 people at thegym and 44 people attending the aerobics class.


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CHAPTER 12 Continuous distributions 95

CHAPTER 12

Continuous distributionsWORKED EXAMPLE 4

A random variable, X , has its frequency curve dened as ) xlo ( .5

, 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.

TH IN K WR IT E

a 1 The required probability can beobtained by evaluating the deniteintegral of f ( x ) over the interval[2, 4].

Note: Remember Pr( X < 4) = Pr( X 4) because Pr( X = 4) = 0On a Calculator page, press: MENU b 4: Calculus 4 3: Integral 3Complete the entry line as:

x x ln(0.5 )2

4

Press ENTER .

a

2 Write the solution, rounding to4 decimal places.

Pr( X < 4) = Pr(2 X < 4)

= x

dx lo ( .5 )

2

= 0.3863

b Find the required probability as shownabove, this time using the interval[2.5, 3.5].Write the solution, rounding to 4 decimal

places.

b Pr(2.5 X 3.5) = x

x log (0.5

22.3.5

= 0.2004


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WORKED EXAMPLE 5

A random variable, X , has its frequency curve dened as f ( x) =

e1

20 x

0 elsewhere

x1

.a Draw the graph of f ( x).b Show that f ( 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. TH IN K WR IT E

a Draw the graph of f ( x ). It is a decreasingfunction with a starting point (0,

1) and a

horizontal asymptote y = 0.

a y

0

12(0, )

x

b 1 A pdf must be greater than or equalto 0 for all values of x . Checkwhether this condition is observed byinspecting the graph of f ( x ).

b f ( x ) 0 for all x

2 Find the total area under the curveby evaluating the denite integral of

f ( x ).Note that the interval over which theintegral needs to be evaluated is[0, ). So, in this case, evaluate

x im ) x .

A = elim 1

2

x 1

=

eim 1

2

1

2

1

2

= lim x 1

20

=

e

lim 1

x 12

0

3 Substitute the terminals in andevaluate the limit. Remember that

lim 1

= 0.

=

e

lim 1 1

0

= 0 + 11

= 1

4 Both conditions required for thefunction to be a pdf are observed.State your conclusion.

Since f ( x ) 0 for all x and the total area under thecurve is 1, f ( x ) is a pdf.

c i The required probability can beobtained by evaluating the deniteintegral of f ( x ) over the interval[0, 3]. Use the expression for theantiderivative found in part b tospeed up your calculations. Giveyour answer correct to 4 decimalplaces.

c i Pr( X < 3) =

e1

2

1

= 1

0

3

1

= e

1 1

20

= 0.7769


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ii 1 To nd Pr( X > 2.5),evaluate the integral of f ( x )over the interval [2.5, ].Alternatively, nd Pr( X < 2.5)and subtract it from 1. Thismethod avoids nding limits,as the interval over which theintegral needs to be evaluatedis [0, 2.5], that is, it does notinvolve .

ii Pr( X > 2.5) = 1 Pr( X < 2.5)

= 1

e1

2

12.

= 1 e

1

x 1

0

.

= 1

e1 1.2

0

= 1 0.7135= 0.2865

2 The required probability canalso be obtained by using theCAS over the interval [2.5, )On a Calculator page, press: MENU b 4: Calculus 4 3: Integral 3

Complete the entry line as:

e1

2

x 1

.5

Press ENTER

3 Write the solution, roundingto 4 decimal places.

Note: The answer is the sameas above.

Pr( X > 2.5) = e x 121

.5

= 0.2865

iii 1 Write the appropriate

statement for the conditionalprobability.

iii = X X > X X >

Pr(Pr[( 3)]

Pr( 3 X )Pr(Pr(2.5 3)

Pr( 3 X )

= 0.06337

0.7769

= 0.0816


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WORKED EXAMPLE 8

Find the variance and standard deviation for the following probability density function.

f ( x) = x

12

e sewhere.

TH IN K WR IT E

1 On a Calculator age, press:Using CAS, press MENU b 1: Actions 1 1: Dene 1Complete the entry line

Dene f x x ( ) 1

2=

Then press: MENU b 4: Calculus 4 3: Integral 3Complete the entry line as:

xf x dx ( )1

2

Then press ENTER .Repeat for E ( X 2)

x f x dx ( )2

1

2

2 Calculate the variance and the standard

deviation as shown

X

X

Var( ) 11

144

SD( ) 11

12

=

=


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WORKED EXAMPLE 9

The probability density function for X is given by x e

) xlo (

0, elsewher=

< x.

Calculate, correct to 3 decimal places:a the mean b the medianc the standard deviation d Pr( 2 X + 2 ).

TH IN K WR IT E

a 1 Write the formula for nding themean of the continuous randomvariable.

a = = X x x dx E( ) log ( )ee0

2 We need to use the CAS calculator tointegrate and nd the probability.On a Calculator page, complete theentry line as:

x ( ))1

Then press ENTER .

3 Write the solution and round to3 decimal places.

l x og ( x

= 2.097

b 1 Write the formula for nding themedian.On a Calculator page, press: MENU b

3: Algebra 3 1: Solve 1Complete the entry line as:

=solve 121

Then press ENTER .

b x x log (m

1

1

2 Write the solution. Solving x log (112

for m implies

m = 0.1866 or m = 2.1555.3 As 1 m e, so m must be the

bigger of the two possible solutions.Answer the question and round to3 decimal places.

The median is 2.156.


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c 1 We rst need to nd the variance.On a Calculator page, complete theentry line as:

( (2.097261 2Then press ENTER .

c Var( X ) = x )21 2.

2 Calculate the standard deviation,correct to 3 decimal places.

SD( X ) = Var( )

= .176047371282

= 0.420d 1 Find the intervals

2 and

+ 2 . d

2

= 2.097

2 0.420

= 1.2581 + 2 = 2.097 + 2 0.420

= 2.93642 State the interval

2 X + 2 . 2 X + 2

= 1.2581 X 2.9364= 1.2581 X e, since 2.9364 > e (the upper

value).

3 Calculate Pr( 2 X + 2 )using the CAS calculator.

Pr( 2 X + 2 ) = x og (e

1.2581

= 0.969Note that in this example, 96.9% of the data lieswithin 2 standard deviations of the mean, which isclose to the estimated value of 95%.


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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)

TH IN K WR IT E

a1

Draw a diagram and shade the regionrequired.a

0 2 z

2 On a Calculator page, press: MENU b 5: Probability 5 5: Distributions 5 2: Normal Cdf 2Enter the values as shown.

Note: As we are dealing with a Z variable, = 0 and = 1.The lower limit is as there is noend point for the function.

Press ENTER .

3 Write the solution and round theanswer to 4 decimal places.

Pr( Z < 2) = normCdf ( ,2,0,1) = 0.9772.

b 1 Draw a diagram and shade the regionrequired.

b

0 z0.728


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2 For Pr( Z 0.728), repeat as aboveon the CAS calculator, this time thelower limit is 0.728 and the upperlimit is .

Remember: Pr( X = a ) = 0 Pr( X > a ) = Pr( X a )


Pr( Z 0.728) = normCdf ( 0.728, ,0,1) = 0.7667.

c 1 Draw a diagram and shade the regionrequired.

c

0 z2.02 1.59

2 For Pr( 2.02 < Z < 1.59), repeat asabove on the CAS calculator. Thistime the lower limit is 2.02 and theupper limit is 1.59.


Pr( 2.02 < Z < 1.59) = normCdf ( 2.02,1.59,0,1) = 0.9224.


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WORKED EXAMPLE 16

If X is normally distributed with = 50 and = 8, calculate, correct to 4 decimal places:a Pr( X > 55) b Pr(28 < X < 65) c Pr( X < 40 | X < 70).

TH IN K WR IT E

a 1 Draw a diagram and shade the regionrequired.

a

x 50 55

2 For Pr( X > 55) repeat as shown inWorked example 15.

Note: As we are dealing with an X variable, the mean and standarddeviation have changed and are now50 and 8, respectively.

Press ENTER .


Pr( X > 55) = normCdf (55, ,50,8) = 0.2660


b

x 50 6528

2 For Pr(28 < X < 65), repeat as aboveon the CAS. This time the lowerlimit is 28 and the upper limit is 65.Write the solution and round theanswer to 4 decimal places.

Pr(28 < X < 65) = normCdf (28,65,50,8) = 0.9666.


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c 1 Write the rule for conditionalprobability.

Note: Pr( X < 40 | X < 70) = Pr( X < 40). This is given by theoverlapping region in the diagrambelow.

X

Pr ( X < 40)Pr ( X < 70)Region required

40 7050

c =Pr( X 7< )r[( 70 ]

r( 70)

Pr( 4 X 0)

Pr( 7 X 0)

2 Find the individual probabilities ofthe fraction using the CAS, as seenpreviously in the example.

Pr( X < 7< 0)r( 40)

r( 70)

normCdf( ,40,50,8)

normCdf( ,70,50,8)=

=.105650

.993790


r( 70)Pr( 4< 0)

Pr( 7< X 0)

mCdf( ,40,5 ,8)

mCdf( , 0,5 ,8)=

= 0.1063


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WORKED EXAMPLE 20

Find the value of c, correct to 3 decimal places, in the following.a Pr( Z < c) = 0.57b Pr( Z c) = 0.91

TH IN K WR IT E

a 1 To calculate Pr( Z < c) = 0.57, draw

a diagram and shade the regionrequired.

a

z0 c

57%

2 As the shaded area is to the leftof the unknown value, we can useinvNorm straight away.On a Calculator page, press: MENU b 5: Probability 5 5: Distributions 5

3: Inverse Normal 3Enter the values as shown.

Note: As we are dealing with a Z variable, = 0 and = 1.

Press ENTER .

3 Write the solution and round to 3decimal places.

c = invNorm(0.57,0,1)= 0.176

b 1 To calculate Pr( Z c) = 0.91, drawa diagram and shade the regionrequired.

b

z0c

91%

2 For this example, the shaded area isto the right of the unknown value, sowe subtract the given area from 1.

Pr( Z < c) = 1 Pr( Z > c)= 1 0.91= 0.09


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3 Find the value of c using the CAS asshown above.


c = invNorm(0.09, 0, 1) = 1.341


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WORKED EXAMPLE 23

X is normally distributed with a mean of 10 and a standard deviation of 2. Calculate x1, correct to3 decimal places, if:a Pr( X x1) = 0.65b Pr( X > x1) = 0.85.

TH IN K WR IT E

a1

Draw a diagram and shade the regionrequired.a

x 10

0.65

x 1

2 As the shaded area is to the leftof the unknown value, we can useinvNorm straight away.

Note: This time we are dealingwith an X variable, so the mean andstandard deviation have changedfrom 0 and 1, respectively. From theinformation,

= 10 and = 2.Enter the values as shown.

Press ENTER .


x 1 = invNorm(0.65,10,2) = 10.771


b

x 10

0.85

x 1

2 For this example, the shaded area isto the right of the unknown value, sowe subtract the given area from 1.

Pr( Z < x 1) = 1 Pr( Z > x 1) = 1 0.85 = 0.15

3 Find the value of c using the CAS asshown previously. Write the solutionand round to 3 decimal places.

x 1 = invNorm(0.15,10,2) = 7.927


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Methods TI-nspire CAS calculater companion