Chapter Functions 4 pages 119–177 - Edl · Mathematics II Solutions Manual • Chapter 4, page...

“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 119

Chapter

4 Functions pages 119–177.....................................................................

INVESTIGATION 4A TABLES

4.01 Getting Started

For You to Explore1. There are a number of answers that would work for each

table. Here is a sample.

Table A

• Each output is 2 times the input, so A(n) = 2n.• Each output is 2 more than the previous output, so

A(n) = A(n − 1) + 2.

Table B

• Each output is equal to the input multiplied by the nextinput, so B(n) = n(n + 1).

• Each output is 2n more than the previous output, soB(n) = B(n − 1) + 2n.

Table C

• Each output is equal to the opposite of the input addedto 2, so C(n) = −n + 2.

• Each output is one less than the previous output,starting at 2, so C(n) = C(n− 1)− 1, where C(0) = 2.

Table D

• Each output is equal to the input multiplied by 2 morethan the input, so D(n) = n(n + 2).

• Each output is 2n + 1 more than the previous output,so D(n) = D(n − 1) + 2n + 1.

• Find any output by squaring the next input thensubtracting 1, so D(n) = (n + 1)2 − 1.

2. (a)

−2 0 2 4 6 8 10

−2

0

2

4

6

8

10

(4, 8)

(0, 0)

(1, 2)

(2, 4)

(3, 6)

(b) There are other ways to draw a function through the 5points. Here is one way to do it:

−2 0 2 4 6 8

−2

0

2

4

6

8

10

(4, 8)

(0, 0)

(1, 2)

(2, 4)

(3, 6)

Many other answers are possible.(c) There are many possible answers here. One is,

“double the input if the input is less than 5 andotherwise output 17.” Another is this rule:

A(n) = 2n + n(n − 1)(n − 2)(n − 3)(n − 4)

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The second part of this rule for A(n) will be zerowhen n = 0, 1, 2, 3, 4 but will be different for anyother input n. Similar rules can assign any arbitrarynumber for the next output(s).

On Your Own

3. Possible answers for Table E include:We can find any output by multiplying the input by 2,then adding 3, so

E(n) = 2n + 3

We can find any output by adding 2 to the previousoutput, starting at 3, so

E(n) = E(n − 1) + 2, where E(0) = 3

4. We can find any output by multiplying the input by 12 ,

then subtracting 2, so

F(n) = 1

2n − 2

We can find any output by adding 12 to the previous

output, starting at −2, so

F(n) = F(n − 1) + 1

2, where F(0) = −2

5. Possible answers for Table G include:We can find any output by multiplying the input by 3 thensubtracting 7, so

G(n) = 3n − 7

We can find any output by adding 3 to the previousoutput, starting at −7, so

G(n) = G(n − 1) + 3, where G(0) = −7

6. Possible answers for Table H include:We can find any output by multiplying the input by 5 thenadding 3, so

H(n) = 5n + 3

Each output is 5 more than the previous output, starting at3, so

H(n) = H(n − 1) + 5, where H(0) = 3

7. Possible answers for Table I include:Each output is the square of the input, so

I (a) = a2

8. Each output J (p) is equal to the square of the input p

multiplied by 2, so

J (p) = 2p2

9. We can find any output by taking the square of the inputthen adding 1, so

K(x) = x2 + 1

10. We can find any output by taking the square of the inputthen subtracting 25, so

L(x) = x2 − 25

Alternately, each answer is a product of two numbers.Finding rules for the two numbers that multiply to makethe output leads to

L(x) = (x − 5)(x + 5)

which is the same as L(x) = x2 − 25.11. We can find any output by multiplying the input by 6 then

adding 9, so

M(n) = 6n + 9

Each output is 6 more than the previous output, startingat 9, so

M(n) = M(n − 1) + 6, where M(0) = 9

12. The outputs N(t) are all perfect squares. Since eachoutput is the square of 3 more than the input t , we write

N(t) = (t + 3)2

Also, the outputs in Table N are the sum of the ones inTable I and Table M. This gives the alternate form

N(t) = t2 + 6t + 9

13. Each output O(n) is equal to the square of the input n

multiplied by 5, so

O(n) = 5n2

14. We can find any output by multiplying the input by 4 thenadding 1, so

P(b) = 4b + 1

Each output is 4 more than the previous output, startingat 1, so

P(b) = P(b − 1) + 4, where P(0) = 1

15. The outputs in Table Q are the sum of the outputs inTables O and P. So, their rules can be added togetherto get

Q(n) = 5n2 + 4n + 1

16. Each output R(w) is equal to the cube of the input w, so

R(w) = w3

17. Each output S(c) is equal to 3 more than the cube of theinput c, so

S(c) = c3 + 3

One way to notice this is that the outputs in Table S arethree larger than the outputs in Table R.

18. Each output T (n) is equal to 3 raised to the power of theinput n, so

T (n) = 3n

19. Each output U(x) is equal to 2 raised to the power of theinput x, so

U(x) = 2x


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20. Each output V (n) is equal to 1 less than 2 raised to thepower of the input n, so

V (n) = 2n − 1

Also, the outputs in Table V are each one less than theoutputs in Table U. So, any rule for Table U can be usedto build a rule for Table V.

Maintain Your Skills

21. (a)

x a(x)

0 01 12 43 94 165 25

(b)

x b(x)

0 −11 02 33 84 155 24

(c)

x c(x)

0 −41 −32 03 54 125 21

(d)

x d(x)

0 −91 −82 −53 04 75 16

22. Replace f (x) with 0 and x with 5:

0 = 52 − c

25 = c

23.

(x + 1)2 − x2 = 2x + 1

Multiply out the first expression in the identity.

x2 + 2x + 1 − x2 = 2x + 1

Then combine like terms to get

2x + 1 = 2x + 1

The expressions on both sides of this equation areidentical. This shows that the original identity is true forevery value of x.

4.02 Two Types of Definitions—Closed

Form & Recursive

Check Your Understanding

1.

Table BInput: n Output: B(n) �

0 0 21 2 42 6 63 12 84 20

2. (a) The outputs agree at B(0) = 0 and B(1) = 2, but thisrecursive rule gives b(2) = 4, which is incorrect.

(b) The output does not agree at B(1). The rule givesb(1) = 0, but b(1) should equal 2.

(c) This agrees with the table.(d) The output does not agree at B(0). The rule gives

b(0) = 2, but b(0) should equal 0.3. (a) Try rule (a) with the input 0 and output 0:

b(n) = 2n

0 = 2(0)

0 = 0

Rule (a) works with the input 0 and output 0. Now trythe input 1 and output 2:

b(n) = 2n

2 = 2(1)

2 = 2

Rule (a) works with the input 1 and output 2. Now trythe input 2 and output 6:

b(n) = 2n

6 = 2(2)

6 �= 4

Rule (a) does not work with the input 2 and output 6,so it does not match Table B.

(b) Rule (b) works with every input, so it fits Table B.(c) Rule (c) works with every input, so it fits Table B.(d) Rule (d) works for the first three inputs, but then fails.

This rule gives b(3) = 14, but the table saysB(3) = 12.

4.

Input Output �0 5 61 11 82 19 103 29 154 44


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5.

Input Output �0 6 31 9 32 12 33 15 34 18

6.

Input Output �0 5 −31 2 152 17 −133 4 −54 −1

7. (a) f (1) = f (0 + 1)

= f (0) · (0 + 1)

= 1 · 1

= 1

f (2) = f (1 + 1)

= f (1) · (1 + 1)

= 1 · 2

= 2

Continue this method to find f (3) through f (6). Youwill get f (3) = 6, f (4) = 24, f (5) = 120 andf (6) = 720.

(b) The factorial function on the calculator gives theseoutputs. Mathematically, the notation uses anexclamation mark. For example,

5! = 5 · 4 · 3 · 2 · 1 = 120

8. (a)

Input Output0 21 62 103 144 18

(b)

Input Output0 21 62 183 544 162

(c)

Input Output0 21 62 183 544 162

(d)

Input Output0 21 62 103 144 18

9. Here are some possible answers.

• Tables E, F, and G are all related because each has aconstant difference in the difference table.

• Tables I, J, and K are all related because they includean x2 term in the rule.

There are many more, such as the fact that some tablesare multiples of others, some are the sum of others, someare the reciprocals of others, and so forth.

On Your Own

10.

Side Length # Dots �

0 0 11 1 22 3 33 6 44 10 55 15

11. (a) Each output is equal to the previous output added tothe new input, so

T (n) = T (n − 1) + n

Here, n is the number of dots and T (n) is thetriangular number.

(b) You can think of the triangular number T (n) as thesum of the integers 1 + 2 + 3 + · · · + (n − 2)+(n − 1) + (n). (Note that this expression only makessense for n ≥ 6. If you use an argument like this tofind a closed form function, you will have to checkthat function for smaller values of n to be sure that itworks.) You could reverse the order of the addendswithout changing the sum.

T (n) = 1 + 2 + 3 + · · · + (n − 2)

+ (n − 1) + (n)

T (n) = (n) + (n − 1) + (n − 2) + · · · + 3 + 2 + 1

Add these two equations term by term to get

2 . T (n) = (n + 1) + (n + 1) + (n + 1)

+ · · · + (n + 1) + (n + 1) + (n + 1)

Because each of the previous equations had n

addends, their sum also has n addends.

2 . T (n) = n(n + 1)

T (n) = n(n + 1)

2


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12. (a)

Input: x Output: ax + b �

0 b a1 a + b a2 2a + b a3 3a + b a4 4a + b a5 5a + b

(b) f (x + 1) − f (x) = (a(x + 1) + b) − (ax + b)

= ax + a + b − ax − b = a

13.

Input: x Output: ax2 + bx + c �

0 c a + b1 a + b + c 3a + b2 4a + 2b + c 5a + b3 9a + 3b + c 7a + b4 16a + 4b + c 9a + b5 25a + 5b + c

f (x + 1) − f (x)

= (a(x + 1)2 + b(x + 1) + c) − (ax2 + bx + c)

= a((x + 1)2 − x2) + b(x + 1 − x) + (c − c)

= a(2x + 1) + b

= 2ax + a + b

14. You are given that g(1) = 1. From the definition,g(2) = g(1)+ 3 = 4. Then g(3) = g(2)+ 5 = 9. Finally,g(4) = g(3) + 7 = 16. The correct answer choice is D.


15. (a) Plug in two points in the equation to find slope. Herewe use the first two points:

m = �y

�x

= −4 − (−7)

1 − 0= 3

(b) m = �y

�x

= −11 − (−7)

1 − 0= −4

(c) m = �y

�x

= 1 12 − 2

1 − 0

= − 12

1

= −1

2

(d) A linear function has a constant difference for eachinput. The value of this difference is the slope. Thus,if you subtract any output from the next output, youfind the slope. Note that the inputs must beconsecutive integers to find the slope directly fromthe difference in outputs.

4.03 Constant Differences


1.

Table Gn G(n) �

0 −7 31 −4 32 −1 33 2 34 5

2. Each output is 3 more than the previous output, starting at−7, so:

f (n) ={−7 if n = 0

f (n − 1) + 3 if n > 0

3. The “hockey stick” property of difference tables lets youfind any output by adding all the differences up to thatoutput to the initial output of the function. Since thisfunction has a constant difference of 3, the sum of all thedifferences up to the n-th output will be 3n. This meansyou can find the n-th output directly by multiplying theinput by 3, then adding −7, so:

g(n) = 3n − 7

4. (a) • Using the closed form rule:

g(n) = 3n − 7

g(10) = 3(10) − 7

g(10) = 30 − 7

g(10) = 23

• Using the recursive rule:

f (10) = f (9) + 3

= (f (8) + 3) + 3

= (f (7) + 3) + 3 + 3

= · · ·= f (0) + 10 · 3

= −7 + 30

= 23

(b) Using the closed form rule:

g(n) = 3n − 7

g(10.1) = 3(10.1) − 7

= 30.3 − 7

= 23.3


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The recursive rule cannot be used here. The recursiverule starts with f (0), and could be used to find f (1),then f (2), then f (3), and so on. However, it can’t beused to find any values of f (n) when n is negative, orwhen n isn’t an integer.

5. The difference when the input n is 0 can be calculated bysubtracting f (0) from f (1), or 5 − 11. 5 − 11 is equal to−6, not 6. The flaw in Leslie’s logic is that some of thedifferences are −6, not 6.

6. The constant differences will cause the output to decreaseby 7 in four steps. The constant difference must be −7

4 .

n p(n) �0 3 − 7

4

1 54 − 7

4

2 − 12 − 7

4

3 − 94 − 7

4

4 −4

7. • Each output p(n) is 74 less than the previous output.

Subtract this constant difference 10 times to reach thevalue of p(10):

p(10) = p(0) − 10 · 7

4

= 3 − 70

4

= −29

2• Subtract the constant difference 100 times to reach the

value of p(100):

p(100) = p(0) − 100 · 7

4= 3 − 175

= −172

• Subtract the constant difference 263 times to reach thevalue of p(263):

p(263) = p(0) − 263 · 7

4

= 3 − 1841

4

= 12

4− 1841

4

= −1829

48. Since 2 yields 11 and 3 yields 14, the constant difference

is 3. From input 3 to input 7, you add that 4 times. Thatis, you add 12 to 14, to obtain 26. The correct answerchoice is D.

9. (a)n F(n) �

0 1 01 1 12 2 13 3 24 5 35 8 56 13

(b) The set of differences is the same as the set ofoutputs, moved one step down, with 0 for the firstdifference, F(1) − F(0). More formally, for n > 1,

F(n) − F(n − 1) = F(n − 2)

The reason you have to specify the first difference, isthat you can’t use this equation for n < 2, becauseF(−1) is not defined.

(c) Continue the pattern in the table until you reachF(10).

n F(n) �

0 1 01 1 12 2 13 3 24 5 35 8 56 13 87 21 138 34 219 55 34

10 89

F(10) = 89 if the pattern in the table continues.

On Your Own

10.

Table Mn M(n) �

0 9 61 15 62 21 63 27 64 33

11. The differences are constant, so each output is 6 morethan the last. This leads to the recursive definition

m(n) ={

9 if n = 0m(n − 1) + 6 if n > 0

12. The method presented in this lesson should lead directlyto m(n) = 6n + 9 as a rule that fits the table.

13. • Using the closed-form rule:

m(n) = 6n + 9

m(7) = 6(7) + 9

= 42 + 9

= 51

• Using the recursive rule:

m(7) = m(6) + 6

= (m(5) + 6) + 6

= (m(4) + 6) + 6 + 6

= · · ·= m(0) + 7 · 6

= 9 + 42

= 51


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Using the closed-form rule:

m(n) = 6n + 9

m(−7) = 6(−7) + 9

= −42 + 9

= −33

The recursive rule cannot be used, since its rule startsat m(0) and tells you how to get to the next output. It isnot possible to find m(−1) or any output from anegative input.The two models agree for any nonnegative integer

input. However, for negative numbers and nonintegers,the recursive form is undefined.

14. (a)

n D(n) �

0 1 11 2 22 4 43 8 84 16 165 32 326 64

(b) The differences are the same as the outputs. Moreformally,

D(n) − D(n − 1) = D(n − 1)

This can be rewritten as D(n) = 2 · D(n − 1), whichhelps to explain why the closed form rule D(n) = 2n

fits the table.(c) Continue the table until D(10) is found.

n D(n) �

0 1 11 2 22 4 43 8 84 16 165 32 326 64 647 128 1288 256 2569 512 512

10 1024

If the pattern continues, D(10) will be 1024.15. A closed form rule shows how to directly calculate the

output from the input. So, the output must equal someexpression that includes the input. To fit a table that hasconstant differences, the input must be multiplied by theconstant difference in the closed form definition of thefunction. This makes sense, because when the input isincreased by 1, which is the next entry in the table, theoutput changes by this constant difference. The output of0 is important, because it shows the value when noconstant difference has been added. Thus, at the end ofthe closed form rule the output of 0 is added.

Examples will vary.

16. One way is to look for tables with linear closed formrules. Another is to take common differences for eachtable and see what happens. Only tables A, C, E, F, G, H,M, and P have constant differences and can be matchedby linear rules.

17. The difference between the two inputs is 8 and thedifference between the two outputs is −12. To get theoutput of 11, you’ll add the constant difference eighttimes to the output of 3. The constant difference isone-eighth of the difference between the two outputs.This is 1

8 · −12 = − 32 .

18. (a) When the input increases by 6, the output increasesby 1. Thus, when the input increases by 1, the outputmust increase by 1

6 . To get the output for an input of0, subtract 1

6 nine times from the output of 9:

0 − 1

6(9) = −3

2

So, a closed-form rule for the Inverse of Table Mwould be

Output = 1

6· Input + −3

2

(b)Inverse of Table MInput Output �

0 − 32

16

1 − 43

16

2 − 76

16

3 −1 16

4 − 56

(c) The constant difference in this table, 16 , is the

reciprocal of the constant difference of 6 found inTable M.


19. The difference between the output of 2 and the output of5 is 27, while the difference between the inputs 2 and 5 is3. Since the output increases by 27 when the inputincreases by 3, the constant difference is 9. The rest of thenumbers in the table can be found using this difference.

Input Output �

0 −23 91 −14 92 −5 93 4 94 13 95 22 96 31

20. (a) m = �y

�x

= 22 − (−5)

5 − 2

= 27

3= 9


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(b)m = �y

�x

= 22 − (−5)

10 − 7

= 27

3= 9

(c) m = �y

�x

= 5 − 2

22 − (−5)

= 3

27= 1

9

4.04 Tables and Slope


1. (a) The constant difference is the change in the outputdivided by the change in the input.

m = 24 − 10

7 − 3

= 14

4

= 7

2(b) Trace the constant difference back three steps. The

output is 10 when the input is 3. Subtract the constantdifference 7

2 three times from the output:

10 − 3 · 7

2= 20

2− 21

2= −1

2

(c) Since the constant difference is 72 and the output is

−12 when the input is zero, the information in this

lesson leads to the rule

Output = 7

2· (Input) − 1

2

2. (a) m = �y

�x

= 24 − 10

7 − 3

= 7

2

(b) The equation is y−10x−3 = 7

2 . This can be simplified to2y = 7x − 1.

3. In order for this table to come from a linear rule, theslope between any two points must be the same. Since theslope between (0, −12) and (3, 5) is 17

3 and the slopebetween (3, 5) and (4, 10) is 5, this table could not havecome from a linear rule.

4. The slope between any two points on a line is the same,because a line has a constant slope. Since the slopebetween (0, −12) and (3, 5) is 17

3 and the slope between(3, 5) and (4, 10) is 5, there is not a line that passesthrough these three points.

5. Between the inputs 1 and 6 (a difference of 5) the outputchanges by −25. So the constant difference must be −5.Then, (2, a) is one constant difference below (1, 4), anda = −1. Similarly b = −26 is one constant differencebelow (6, −21).

6. The constant difference, or slope, is 12 and the output at 0

is −4. This means that one rule that agrees with thistable is

F(n) = 1

2n − 4

7. (a) Add the constant difference once to the output for 6 toget the output for 7:

−3 + 5 = 2

(b) Subtract the constant difference three times from theoutput for 6 to get the output for 3:

−3 − 3(5) = −18

(c) Subtract the constant difference three more times toget the output for 0:

−18 − 3(5) = −33

And since the constant difference is 5,

Output = 5 · Input − 33

You might also approach this problem by noting that thelinear function must be of the form

Output = 5 · Input − Output for Zero

Use any input-output pair to solve for the output for 0,and then use the linear function to compute any missingvalues.

8.

Table Kx K(x) �0 1 11 2 32 5 53 10 74 17

Tables that come from linear rules have a constantdifference, so the values in the � column in thedifference table are all the same if the rule is linear. InTable K, the values in the � column are not all the same,so this Table cannot come from a linear rule.

9.

x K(x) � �2

0 1 1 21 2 3 22 5 5 23 10 74 17


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On Your Own10. (a)

m = �y

�x

= −21 − 46 − 1

= −5

(b) Since the slope of the line is −5, the linear functionfor which this line is the graph must have a constantdifference equal to this slope. To find the output foran input of 2, you can just add the constant differenceto the output for 1. This gives you a = 4 − 5 = −1.

(c) To find the output for an input of 7, you can just addthe constant difference to the output for 6. This givesyou b = −21 − 5 = −26.

11. Yes, all the outputs are the same, so all the differencesare 0.

12. (a)

−10 −8 −6 −4 −2 0 2 4 6 8 10

−10

−8

−6

−4

−2

0

2

4

6

8

10

y = 7

(b) m = �y

�x

= 0−5

= 0

(c) Examine the graph of this line. Since for every valueof x the value of y is always 7, the equation is

y = 7

13. (a)

−10 −8 −6 −4 −2 0 2 4 6 8 10

−10

−8

−6

−4

−2

0

2

4

6

8

10

x = 7

(b) Answers may vary. Slope is defined as the change inthe value of y (�y) divided by the change in thevalue of x (�x). In this example, this fraction is 5

0 .There is no value for such a fraction—division byzero is undefined. Therefore, you can say there is “noslope,” or that the slope is undefined.

(c) Since the value of x is always 7, no matter what valuey takes on, an equation for this line is

x = 7

14.

Table On O(n) �

0 0 51 5 152 20 253 45 354 80

Table O cannot come from a linear rule because thedifferences are not constant.

15. There is a constant difference of 4 and P(0) = 1, so alinear rule is

P(b) = 4b + 1

16. (a) The outputs in Table Q are the sum of the outputs inTables O and P.

(b) The differences in Table Q are the sum of thedifferences in Tables O and P.


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17.w R(w) � �2 �3

0 0 1 6 61 1 7 12 62 8 19 18 63 27 37 24 64 64 61 30 65 125 91 366 216 1277 343

18. (a)x d(x) �0 3 21 5 22 7 23 9

(b) One answer is that the constant difference doesn’tcontinue. d(4) = 35, so the next difference in thetable would be 26.

d(x) acts just like x �→ 2x + 3 on 0, 1, 2 and 3, butnowhere else. In fact,

d(x) − (2x + 3) = x4 − 6x3 + 11x2 − 6x

= x(x3 − 6x2 + 11x − 6)

= x(x − 1)(x − 2)(x − 3)

So, d(x) − (2x + 3) will be 0 at 0, 1, 2, and 3, andnowhere else.

19. The slope is the difference of the outputs (21) divided bythe difference of the inputs (9). Thus it is 21

9 = 73 . The

correct answer choice is C.


20. (a)

x a(x) �

0 0 11 1 32 4 53 9 74 16

(b)

x b(x) �

0 0 21 2 62 8 103 18 144 32

(c)

x c(x) �

0 0 51 5 152 20 253 45 354 80

(d) x d(x) �

0 0 −101 −10 −302 −40 −503 −90 −704 −160

(e)(f) The function e(x) = x2 has all positive oddintegers in its difference column, including 25 ifyou work enough rows. As a result, e(x) = 2x2 has50 in its difference column because 2 · 25 = 50.Thus, with the function e(x) = kx2, 50 will be inthe difference column if 50

kis a positive odd

integer.Identify all odd factors of 50: 1, 5 and 25.1 · 50 = 50, so e(x) = 50x2 also has 50 in its

difference column.5 · 10 = 50, so e(x) = 10x2 also has 50 in its

difference column.

21. (a) x f (x) �

0 2 31 5 32 8 33 11 34 14

(b) x g(x) �

0 3 01 3 02 3 03 3 04 3

(c) x h(x) �

0 0 41 4 62 10 83 18 104 28

(d) x k(x) �

0 4 21 6 22 8 23 10 24 12

(e) x r(x) �

0 0 11 1 72 8 193 27 374 64

(f) x s(x) �

0 1 61 7 122 19 183 37 244 61


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4.05 Difference Tables for Polynomial

Functions


1.n y(n) � �2

0 −2 8 101 6 18 102 24 28 103 52 384 90

Since the second differences are constant, this table couldhave come from a quadratic rule. Continue the table tofind the matching quadratic

n y(n) − 5n2 �0 −2 31 1 32 4 33 7 34 10

Now, since y(n) − 5n2 has a constant difference of 3, youcan match it to the linear function y(n) − 5n2 = 3n − 2.This tells you that y(n) = 5n2 + 3n − 2, which matchesthe original table.

2. Start by calculating the constant second differences:

n y(n) � �2

0 10 12 −61 22 6 −62 28 0 −63 28 −64 22

The constant second difference is −6, so if the function isdefined by y(x) = an2 + bn + c, the value of a is equalto −6 divided by 2, or −3.Next, subtract −3n2 from the values of y(n). This cancelsout the an2 term, giving a table that can be used tocalculate the values of b and c:

Input: n Output: y(n) Output: −3n2 y(n) − (−3n2)

0 10 0 101 22 −3 252 28 −12 403 28 −27 554 22 −48 70

A rule for the y(n) − (−3n2) column is 15n + 10. Then b

is 15 (the constant difference for the y(n) − (−3n2)

column) and c is 10 (the first term in the y(n) − (−3n2)

column). A quadratic function that fits this table is

y(n) = −3n2 + 15n + 10

3.

w R(w) � �2 �3

0 0 1 6 61 1 7 12 62 8 19 18 63 27 37 24 64 64 61 30 65 125 91 366 216 1277 343

4.

x m(x) � �2 �3

0 4 −3 34 301 1 31 64 302 32 95 94 303 127 189 124 304 316 313 154 305 629 467 1846 1096 6517 1747

5. (a) Cubic functions appear to yield constant thirddifferences (�3). Testing other cubics confirms this.

(b) This constant third difference is equal to the leadingcoefficient multiplied by 6.

6. Try it with Exercise 4: To get the � of f (x), you need tocompute f (x + 1) − f (x). But

f (x + 1) − f (x)

= (5(x + 1)3 + 2(x + 1)2 − 10(x + 1) + 4)

− (5x3 + 2x2 − 10x + 4)

= 5((x + 1)3 − x3) + 2((x + 1)2 − x2)

− 10((x + 1) − x) + 4(1 − 1)

= 5(3x2 + 3x + 1) + 2(2x + 1) − 10(1) + 4(0)

and this is Sasha’s method.7. x f (x) � �2 �3

0 d a + b + c 6a + 2b 6a

1 a + b + c + d 7a + 3b + c 12a + 2b 6a

2 8a + 4b + 2c + d 19a + 5b + c 18a + 2b 6a

3 27a + 9b + 3c + d 37a + 7b + c 24a + 2b 6a

4 64a + 16b + 4c + d 61a + 9b + c 30a + 2b

5 125a + 25b + 5c + d 91a + 11b + c

6 216a + 36b + 6c + d

8. (a) Anything multiplied by 0 is equal to 0, so any valueof x that makes the factors (x − 3), (x − 5) or(x − 6) equal to 0 will satisfy the equation. If x = 3the factor (x − 3) turns into (3 − 3), which is equal to0. The whole equation looks like

(3 − 3)(3 − 5)(3 − 6) = 0

(0)(−2)(−3) = 0

0 = 0

So x = 3 satisfies this equation. Similarly, whenx = 5, the factor (x − 5) is equal to 0, and whenx = 6, the factor (x − 6) is equal to 0. x = 5 andx = 6 also satisfy this equation.


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(b) Multiply out the factors. First multiply two togetherand then multiply the result by the third factor:

(x − 3)(x − 5)(x − 6)

= (x2 − 5x − 3x + 15)(x − 6)

= (x2 − 8x + 15)(x − 6)

= x3 − 8x2 + 15x − 6x2 + 48x − 90

= x3 − 14x2 + 63x − 90

9.x v(x) � �2 �3

0 −90 50 −22 61 −40 28 −16 62 −12 12 −10 63 0 2 −4 64 2 −2 2 65 0 0 86 0 87 8

10. Start by calculating the constant second differences. Thissolution uses x for the input and y for the output:

x y � �2

0 7 −13 201 −6 7 202 1 27 203 28 474 75

The constant second difference is 20, so if the function isf (x) = ax2 + bx + c, the value of a is equal to 20divided by 2, or 10.Next, subtract the values of 10x2 from the values of y

(the output). This cancels out the ax2 term and gives atable that can be used to find b and c:

x y 10x2 y − 10x2

0 7 0 71 −6 10 −162 1 40 −393 28 90 −624 75 160 −85

A rule for the y − 10x2 column is −23x + 7. The value ofb is −23 and the value of c is 7. A quadratic function thatfits this table is

y = 10x2 − 23x + 7

11. (a)n F(n) � �2 �3

0 1 0 1 −11 1 1 0 12 2 1 1 03 3 2 1 14 5 3 25 8 56 13

Since the third differences are not constant, this tablecould not have come from a cubic function.

(b) The � value for any n is the same as the F(n) valuefor the previous n. Similarly, the �2 value for any n

is the same as the � value of the previous n and theF(n) value of n − 2. This diagonal down-and-rightpattern continues forever.

12. Begin by trying to find a constant difference. Thissolution uses x for the input and y for the output.

x y � �2 �3

0 −3 −2 2 121 −5 0 14 122 −5 14 26 123 9 40 384 49 785 127

The table has a constant third difference of 12, so a cubicfunction can generate this table. In a cubic functionax3 + bx2 + cx + d, the value of a is equal to theconstant difference divided by 6. So, in this table, thevalue of a is 2.

Subtract the outputs of 2x3 from the outputs of thetable. This effectively “cancels out” the ax3 term andgives a table of values from a quadratic function. Fromthis table, continue by using Tony’s method to find thevalues of b, c and d.

x y 2x3 y − 2x3

0 −3 0 −31 −5 2 −72 −5 16 −213 9 54 −454 49 128 −795 127 250 −123

The column y − 2x3 can be matched by a quadratic rule.The input is x and the output is bx2 + cx + d is theoutput. Find the constant second difference:

x bx2 + cx + d � �2

0 −3 −4 −101 −7 −142 −21

The constant second difference is −10, so if the rule isy = bx2 + cx + d, the value of b is equal to −10 dividedby 2, or −5.Then, subtract the values of −5x2 from the values ofbx2 + cx + d (the output). This cancels out the −5x2 termand gives a table that can be used to find c and d. Thenext table is a linear function, so two points are enough:

x −5x2 + cx + d −5x2 −5x2 + cx + d − (−5x2)

0 −3 0 −31 −7 −5 −2

A rule for the 5x2 + cx + d − (− 5x2) column is x − 3.The value of c is 1 and the value of d is −3.

The polynomial that fits the original table is

Output = 2(Input)3 − 5(Input)2 + Input − 3


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On Your Own

13.n y(n) � �2

0 −7 2 61 −5 8 122 3 20 183 23 384 61

Since the second differences are not constant, this tablecannot have come from a quadratic rule.

14.Input: a Output: b �

0 25 −141 11 −142 −3 −143 −17 −144 −31

Since the first difference is constant, this table comes froma linear rule. The constant difference is −14 and the outputfor 0 is 25, so a rule that agrees with this table is:

b = −14a + 25

15.n c(n) � �2

0 −8 8 −41 0 4 −42 4 0 −43 4 −44 0

The second difference is constant, so this table comesfrom a quadratic rule. Since the constant seconddifference is −4, the value of a in the rule an2 + bn + c

is −2. Subtract −2n2 from c(n) to get a table of a linearrule that can be used to find b and c. Two points areenough for the next table:

n c(n) −2n2 c(n) − −22

0 −8 0 −81 0 −2 2

The constant difference is 10 and the output for 0 is −8,so the linear rule is 10n − 8. Then the rule that fits theoriginal table is

c(n) = −2n2 + 10n − 8

16. (a) If this table comes from a cubic function, the thirddifferences will be constant. Here is a differencetable:

n D(n) � �2 �3

0 1 1 1 11 2 2 2 22 4 4 4 43 8 8 8 84 16 16 165 32 326 64

Since the third differences are not constant, this tabledoes not come from a cubic function.

(b) Each time, the numbers in the difference column areidentical to the numbers in the previous column. Thiswill continue for as many columns as you make.

17. Input: x Output: y � �2 �3

0 1 0 3 −31 1 3 0 22 4 3 23 7 54 12

The first differences are not constant, so this table cannotcome from a linear rule. The second differences are notconstant, so this table cannot come from a quadratic rule.And, the third differences are not constant, so this tablecannot come from a cubic rule.

18. Answers will vary. One way to do this is to draw thegraphs and the data points together. Another is to decidehow far away the data points are from the curves.Because you’ve used tables so much, you might make atable to compare the four alternatives.

x y 2x − 5 2. 8x − 0.6 x2 − x + 1x3 − 3x2

+2x + 10 1 −5 −0.6 1 11 1 −3 2.2 1 12 4 −1 5 3 13 7 1 7.8 7 74 12 3 10.6 13 25

Rule 3 seems to be the best fit.19. Follow the process of taking differences until a constant

appears, which in this case means you have to find all thedifferences you can—up to the fourth difference. Sinceyou haven’t yet done a problem with a fourth difference,you’ll also need to figure out what the constant differencetells you about the coefficient of the term raised to thefourth power in the polynomial. You can do this bymaking difference tables for quartic polynomials that youknow and looking for a relationship between the constantdifference you get and the known coefficient, or you canmake a difference table for the general quarticpolynomial—q(x) = ax4 + bx3 + cx2 + dx + e. Thenyou can subtract the term ax4 from the outputs and you’llhave a cubic polynomial, which you know how to dealwith.

y = 5

24x4 − 7

4x3 + 127

24x2 − 15

4x + 1

20. The constant third differences of a cubic will be theleading coefficient times 6. Since that coefficient is 1, thecorrect answer choice is C.


21. (a) x f (x)

0 141 62 03 −44 −65 −6


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(b)x g(x)

0 781 602 443 304 185 8

(c)x h(x)

0 01 −82 −143 −184 −205 −20

(d)x j (x)

0 91 02 −33 04 95 24

(e) When x is set to 5, k(x) must be equal to 0. The rulefor k(x) must then include (x − 5) as a factor,because (x − 5) is equal to 0 when x = 5, andanything multiplied by 0 is equal to 0. The same logicapplies when x is 6, so one possible rule for k(x) is:

k(x) = (x − 5)(x − 6)

The table for this rule would be

x k(x)

0 301 202 123 64 25 0

Many other rules are possible!(f) Because of the same logic applied to part (e), the rule

for m(x) must have x, (x − 1), and (x − 2) asfactors. The simplest rule for these three criteriawould be m(x) = x(x − 1)(x − 2). However, for thisrule, m(3) = 6. To get the output you want, multiplyyour rule by 4 to get m(x) = 4x(x − 1)(x − 2). Nowm(3) = 24, and all the other outputs you want arestill zero. This is only one of many possible solutions.Here is its input-output table:

x m(x)

0 01 02 03 244 965 240

22. (a)x f (x)

0 01 32 63 94 12

(b)x g(x)

0 01 42 83 124 16

(c)x h(x)

0 01 32 83 154 24

(d)x j (x)

0 01 32 63 154 36

(e) You can create a function that matches f (x) for theinputs 0 through 4, but not for other inputs, by addingon a term for k(x) that will equal zero at the inputswhose outputs you want to match, but won’t be zerofor other inputs. For example, x(x − 1)(x − 2) iszero when x = 0, 1, or 2, so the table forj (x) = 2x + x(x − 1)(x − 2) was identical to thetable for f (x) when x = 0, 1, or 2.

Extending the product you add to the function willallow it to match more entries in the table for f (x).So a function k(x) that matches all the entries forf (x) from 0 through 4 could be

k(x) = 3x + x(x − 1)(x − 2)(x − 3)(x − 4)

The input-output table for this function would beidentical to that for f (x), but f (5) = 15 andk(5) = 135. There are other possibilities for k(x).

4A MATHEMATICAL REFLECTIONS

1. Create a difference table for the function.

Input: n Output: B(n) �0 1 21 3 42 7 63 13 84 21

The difference you add to an output to get the n-th outputis equal to twice the value of n. For example, to get B(3),


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you add 2 · 3 to B(2). This leads to a recursive definitionfor g(n).

g(n) ={

1 if n = 0g(n − 1) + 2n if n > 0

2. Extend the difference table to show the second difference.

Input: n Output: B(n) � �2

0 1 2 21 3 4 22 7 6 23 13 84 21

Because the second difference is constant, you know thata quadratic rule will match this table. The coefficient ofthe n2 term in the quadratic rule will be half the constantdifference from the table or 1. Add a new column to yourtable showing B(n) − n2. You will be able to match thisnew column with a linear rule.

Input: n Output: B(n) � �2 B(n) − n2 �0 1 2 2 1 11 3 4 2 2 12 7 6 2 3 13 13 8 4 14 21 5

Look at the B(n) − n2 column. Its output for 0 is 1 and ithas a constant difference of 1. That means this columnmatches the linear rule n + 1. Since B(n) − n2 = n + 1,you know that a closed-form rule that matches the table isB(n) = n2 + n + 1.

3. The slope between any two points on a line is constant.That means the slope between any point (x, y) on the lineand (2, 5) will be equal to the slope between (2, 5) and(−4, 8). This relationship gives an equation for the linecontaining these two points.

y − 5

x − 2= 8 − 5

−4 − 2y − 5

x − 2= 3

−6y − 5

x − 2= −1

2

y − 5 = −1

2(x − 2)

y − 5 = −1

2x + 1

1

2x + y = 6

x + 2y = 12

4. Find the difference between an output and the previousoutput. If that difference is constant for every pair ofsequential outputs, the table can be matched by a linearrule a(Input) + b, where a is the constant difference and b

is the output for the input 0.5. If the first difference is constant, the function will be

linear. If the second difference is constant, the function

will be quadratic. If the third difference is constant, thefunction will be cubic.

6. Create a difference table, and compute differences untilyou find a constant difference.

Input Output � �2

0 1 4 21 5 6 22 11 8 23 19 10 24 29 125 41

Since the second difference is constant, this table will bematched by a quadratic rule. Half the constant difference,or 1, is the coefficient of the leading term.

Extend the table by including a column forOutput − (Input)2 and its differences.

Input Output Output − (Input)2 �0 1 1 31 5 4 32 11 7 33 19 10 34 29 13 35 41 16

The Output − (Input)2 column has a constant firstdifference, so it can be matched by a linear rule. Its outputfor input 0 is 1 and the constant difference is 3, so the ruleis 3(Input) + 1, and the quadratic rule for the originaltable is Output = (Input)2 + 3 (Input) + l.

INVESTIGATION 4B ABOUT FUNCTIONS


For You to Explore

1. (a) The product of 3 and 4 is 12.(b) The quotient of 3 and 4 is 3

4 .(c) The quotient of 6 and 8 is 6

8 , which reduces to 34 .

(d) The quotient of 0 and 23 is 023 , or 0.

(e) The quotient of 23 and 0 can be written as 230 , which

is not a real number.(f) Calculate SQRT (5) first, which is

√5. Now

calculate SQR (√

5), which is 5.(g) SQR (5) is 25, and SQRT(25) is 5.(h) Since SQRT (−5) is not defined as a real number,

SQR(SQRT (−5)) cannot be evaluated.(i) SQR (−5) = 25, then SQRT (25) = 5.(j) Calculate P(3, 4) and P(2, 6) first. Both are equal to

12. Q(12, 12) is equal to 1212 , or 1.

(k) Pierre de Fermat’s birthday is August 17.(l) 3 does not have a birthday, so BD(3) cannot be

calculated.(m) SUM(9, 16) is 25, then SQRT(25) is 5.(n) SQRT (9) = 3 and SQRT (16) = 4, then

SUM(3, 4) = 7.


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(o) BD(Isaac Newton) is January 4th, but SQRT

(January 4th) doesn’t make sense. Thus, SQRT

(BD(Isaac Newton)) cannot be evaluated.(p) SQRT (PR (−2, 1

3 )) is SQRT (−23 ). You can’t take

the square root of a negative number, so this isundefined.

(q) The reciprocal of 2 is 12 .

(r) The reciprocal of 2 is 12 , then the reciprocal of 1

2 is 2.(s) The reciprocal of 2 is 1

2 , then the reciprocal of 12 is 2,

and then the reciprocal of 2 is 12 again.

(t) Zero does not have a reciprocal, so this does not exist.2. The function m(x) might output the greatest integer that

is less than or equal to the input. Remember, though, thatthis is not the only possible rule that m(x) could follow.

3. (a) No, the statement is not always true. For example, ifa = 16 and b = 9, the statement is false.

(b) The statement is true whenever one of a or b is zero,and false otherwise.

4. (a) p can be any person.(b) Square roots exist for any positive number or 0, so

the domain of SQRT is any non-negative number.(c) Division by zero is invalid, so the domain of QUO

is any pair of real numbers (a, b) when b is notequal to 0.

(d) n can be any nonzero number.(e) x can be any nonnegative number, other than 2.

5. (a) BD(p) can be equal to any day of the year,depending on what person p is the input. Thus, therange of BD is any day of the year.

(b) SQRT outputs the positive square root of a number,except for SQRT (0) = 0. Thus, the range of SQRT

is any nonnegative number.(c) QUO can output any number, so the range of QUO

is all numbers.(d) Any positive number is the reciprocal of some other

positive number. Likewise for negative numbers. Therange of REC is all real numbers other than 0.

(e) You can try plugging in positive numbers bigger than2 and smaller than 2, and you will eventually see thatyou can get any output. The range is all real numbers.This can be shown with algebra as follows: For whichc is there a solution to

√x

x−2 = c? If c = 0, then x = 0works. If c �= 0, then square both sides, get rid of thefraction, multiply out and combine like terms toobtain: c2x2 − (4c2 + 1)x + 4c2 = 0. Thediscriminant of this quadratic equation is(4c2 + 1)2 − 4 · c2 · 4c2 = 8c2 + 1 > 0 for all c. So italways has two solutions; the larger solution worksfor c > 0 and the smaller solution works for c < 0.

On Your Own6. (a) This is not enough information. The input (a, b)

could have been 3 and 3, −1 and −9, 12 and 18, or

others.(b) Only one number has a square root of 9, and that’s 81.(c) Two numbers have a square of 9,−3 and 3. The input

could have been either, so this is not enough

information.7. (a) −3 and 3 both produce 9. Other examples work.

(b) There are many examples of people with the samebirthday—most twins, for example.

(c) (3, 6) and (4, 8) both produce 12 . Other examples

work.(d) Impossible. All numbers that have square roots have

unique square roots.(e) Impossible. No two numbers have the same

reciprocal.8. (a)

D(x, y) =√

x2 + y2

D(0, 0) =√

02 + 02

= √0

= 0

(b)D(3, 4) =

√32 + 42

= √9 + 16

= √25

= 5

(c)D(−3, 4) =

√(−3)2 + 42

= √9 + 16

= √25

= 5

(d)D(4, 3) =

√42 + 32

= √16 + 9

= √25

= 5

(e) Many answers are possible. One is the input (5, 0).There are many others, such as (−5, 0) and (−3, −4).

(f) Here is the graph:

−10 −8 −6 −4 −2 0 2 4 6 8 10

−10

−8

−6

−4

−2

0

2

4

6

8

10

(4, 3)

(0, −5)


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D(x, y) represents the distance from (0, 0) to (x, y).So, all the points that make D(x, y) = 5 are thepoints that are 5 units away from (0, 0). This is acircle with center (0, 0) and radius 5.

9. (a) B(x, y, z) = D(x, y) + z

B(0, 0, 0) = D(0, 0) + 0

= 0 + 0

= 0

(b) B(3, 4, 5) = D(3, 4) + 5

= 5 + 5

= 10

(c) B(4, 3, 5) = D(4, 3) + 5

= 5 + 5

= 10

(d) B(−3, 4, −5) = D(−3, 4) − 5

= 5 − 5

= 0

(e) B(5, 12, z) = 7

D(5, 12) + z = 7

13 + z = 7

z = −6

(f) The equation is the same as√

x2 + y2 + z = 0, soz = −√x2 + y2. The solutions are all triples(x, y,−√x2 + y2), where x and y are any realnumbers.

10. (a) D(x, y) is defined as√

x2 + y2.Since √ is defined to be nonnegative,

√x2 + y2 is

never negative, and D(x, y) is never negative.(b) B(x, y, z) can be negative if z is negative. One

example is B(3, 4, −12).


11. (a) f (x) = 3x − 7

f (10) = 3(10) − 7

= 23

(b)g(x) = x + 7

3

g(23) = 23 + 7

3= 10

(c) Compute f (0) first:

f (0) = 3(0) − 7

= −7

Now compute f (−7):

f (−7) = 3(−7) − 7

= −21 − 7

= −28

(d)g(g(−28)) = g

(−28 + 7

3

)

= g

(−21

3

)

= g(−7)

= −7 + 7

3= 0

(e)f (g(172)) = f

(172 + 7

3

)

= f

(179

3

)

= 3

(179

3

)− 7

= 179 − 7

= 172

(f) g(f (0.27)) = g(3(0.27) − 7)

= g(0.81 − 7)

= g(−6.19)

= −6.19 + 7

3

= 0.81

3

= 0.27

(g) Expanding this expression would take a lot of work.Instead, it is easier to use the pattern from theprevious examples. f (g(1000)) must equal 1000, sof (g(f (g(1000)))) = 1000.

12. (a) h(x) = (x − 1)2

h(4) = (4 − 1)2

= 9

(b) j (x) = √x + 1

j (9) = √9 + 1

= 4


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(c)h(h(3)) = h((3 − 1)2)

= h(4)

= (4 − 1)2

= 9

(d)j (j (16)) = j (

√16 + 1)

= j (5)

= √5 + 1

(e)j (h(5)) = j ((5 − 1)2)

= j (16)

= √16 + 1

= 5

(f) j (h(−5)) = j ((−5 − 1)2) = j (36) = √36 + 1 = 7

(g) Function j (x) has only nonnegative numbers in itsdomain, so j (−3) does not exist. Then, h(j (−3))

does not exist, either.13. (a) f (3) = 3 + 4 = 7

(b) f (f (3)) = f (7) = 7 + 4 = 11(c) f (f (f (3))) = f (11) = 11 + 4 = 15(d) f (f (f (f (3)))) = f (15) = 15 + 4 = 19(e) Each application of f adds four to the original input.

Since f has been applied twelve times, the effect willbe to add 4 twelve times.

f (f (. . . (f (3))))︸︷︷︸12 f ’s

= 51

14. (a) g(1) = 1+11 = 2

1 = 2(b) g(g(1)) = g(2) = 2+1

2 = 32

(c) g(g(g(1))) = g( 3

2

) = 32 +1

32

= 53

(d) g(g(g(g(1)))) = g( 5

3

) = 53 +1

53

= 85

(e) g(g(g(g(g(1))))) = g( 8

5

) = 85 +1

85

= 138

(f) You may have recognized the sequence of numeratorsand denominators in this problem as the Fibonaccinumbers, defined recursively as

F(n) =⎧⎨⎩

1 if n = 01 if n = 1F(n − 1) + F(n − 2) if n ≥ 2

In each case, if g is composed n times, the output willbe F(n+1)

F (n).

g(g(. . . (g(1))))︸︷︷︸10 g’s

= 14489

4.07 Getting Precise About Functions

Check Your Understanding1. (a) The graph of this function is a downward-facing

parabola. The vertex is at (36, 1296), so the range isall reals less than or equal to 1296.

(b) Here is the graph:

−20 0 20 40 60 80 100

−1500

−1000

−500

0

500

1000

1500

2. (a) If the base is x and the perimeter is 144, the height y

can be found using the perimeter. The formula for theperimeter of this rectangle is P = 2x + 2y, so144 = 2x + 2y. Then solve for y:

144 = 2x + 2y

144 − 2x = 2y

72 − x = y

The area is equal to xy. Since y = 72 − x, the area isA(x) = x(72 − x).

(b) Since the shape must be a rectangle, the length of thebase must be greater than 0. Also, the length must beless than 72, since the height of the rectangle mustalso be positive.

(c) The rectangle with the largest area is a square of sidelength 36, which has area 362 = 1296. A rectanglemust also have positive area. The graph shows thatany positive area less than or equal to 1296 ispossible.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 137

(d)

−20 0 20 40 60 80 100

−1500

−1000

−500

0

500

1000

1500

Note that this graph does not include the points (0, 0)

or (72, 0).3. No, these functions are not the same because they have

different domains and ranges. Even though the functionsagree in many places, they are not identical functions.

4. (a) Both tables are the same:

Input Output0 31 82 133 184 235 28

(b) The natural domain of H is all real numbers. Anynumber can be used as input to H . The naturaldomain of K is all nonnegative integers, by itsdefinition. Defining K(0) allows the definition ofK(1), which allows K(2), and so on, but only integervalues of n are allowed.

(c) H and K are not the same function, because theyhave different domains.

5. (a) Only positive numbers x and 0 can produce an outputfor h(x). If x is negative, then so is x3. Then

√x3 is

not a real number.(b) The range is all nonnegative numbers R

+. Anypositive number can be the output, as well as 0.

6. (a) The statement f : R2 → R indicates the domain and

target.(b) There are infinitely many pairs that make f

equal to 12. Some include (3, 2) and (0, 4). In fact,any number can be picked for a and b can befound.

(c) Since f (x, y) = 2x + 3y, any point that makes theequation true satisfies 2x + 3y = 12. This is thegraph of a straight line:

−10 −8 −6 −4 −2 0 2 4 6 8 10

−10

−8

−6

−4

−2

0

2

4

6

8

10

(0, 4)

(3, 2)

(−3, 6)

7. (a) 3(3)2 = 27 and 3(−3)2 = 27(b)

12 = 3x2

4 = x2

±√4 = x

2, −2 = x

(c) f (11) = 3(11)2 = 363 and f (10) = 3(10)2 = 300.363 − 300 = 63.

(d)

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

= 3(x2 + 2x + 1)

= 3x2 + 6x + 3

(e) f (x +1)−f (x) = (3x2 +6x +3)− (3x2) = 6x +3.(f) Since g(x) = 6x + 3, then g(10) = 6 · 10 + 3 = 63.

Alternately, g(10) = f (11) − f (10), which wascalculated earlier in this exercise.

8. (a) 5(b) 5(c) 10(d) A


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(e) g(x) = (x+1)2 −x2 = (x2 +2x+1)−(x2) = 2x+1(f) 6x + 3(g) 20x + 10(h) g(x) = (x+1)2 +10(x+1)−12−(x2 +10x−12) =

2x + 11(i) g(x) = 3(x+1)2+10(x+1)−12−(3x2+10x−12) =

6x + 139. (a) The natural domain for f is the set of inputs for

which it is defined. In this case, that’s all realnumbers such that x ≤ 6.

(b) Here is a sketch of the graph:

−5 5

−2

2

4

6

8

10

12

−4

On Your Own10. (a) Only positive numbers x and 0 can produce

an output for j (x). If x is negative,√

x is not a realnumber.

(b) The range is all nonnegative numbers R+. Any

positive number can be the output, as well as 0.11. (a) Answers will vary.

(b) Answers will vary. The answer will always be oneday beyond the answer to part (a).

(c) Yes, this is a function. As a rule, it has a consistentoutput for any input.

12. (a) The statement N : R2 → R indicates the domain and

target.(b) There are many possible answers. Some answers

include (4, 3) and (0, −5).

(c) The only pair of numbers that makes x2 + y2 = 0is (0, 0).

(d) It is the graph of x2 + y2 = 25, a circle with center(0, 0) and radius 5:

−10 −8 −6 −4 −2 0 2 4 6 8 10

−10

−8

−6

−4

−2

0

2

4

6

8

10

(4, 3)

(0, −5)

(e) The range is all actual outputs of the function. Sincesquares are never negative, the output of N cannot benegative. The output can be zero, and any outputk > 0 is the result of N(

√k, 0). Therefore the range

is R+, the nonnegative real numbers.

13. The two functions do not have the same domain. Thedomain of t (x) is R

+, nonnegative real numbers. Thedomain of f (x) is R, all real numbers. This happensbecause of the square root in the definition of t (x).

Specifically, t (−7) is undefined, since√−7 is not a

real number. However, f (−7) = −7 is fine. Thefunctions do not behave identically, so they are not thesame function.

14. (a) This function is a quadratic with vertex( 1

12 , −4924

).

Domain: R. Target: R. Range: f (x) ≥ −4924 .

(b) This function cannot accept numbers less than 3, andoutputs cannot be negative. Domain: x ≥ 3.Target: R. Range: R

+.(c) You cannot divide by 0, nor take the square root of a

negative number, so the domain is all x > 3. Thetarget is R and the range is all positive numbers.

(d) This function cannot accept 0 as input, and cannotoutput 1 (since 1

xcannot equal zero). Domain: x �= 0.

Target: R. Range: f (x) �= 1.(e) The input is any point (x, y). Any real number can be

the output. Domain: R2. Target: R. Range: R.

(f) The input is a value x and the output is apoint, the point on y = x3 that has that value of x.Domain: R. Target: R

2. Range: all points (x, y)

satisfying y = x3.(g) This function swaps x and y, so any possible input or

output is possible. Domain: R2. Target: R

2.Range: R

2.


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15. (a) f (g(2)) = f (−4) = 32(b) g(f (2)) = g(−4) = −10(c) f (g(z)) = f (z − 6) = (z − 6)2 − 4(z − 6) =

z2 − 16z + 60(d) g(f (z)) = g(z2 − 4z) = z2 − 4z − 6(e) f (f (z)) = f (z2 − 4z) = (z2 − 4z)2 − 4(z2 − 4z) =

z4 − 8z3 + 12z2 + 16z

(f) g(g(z)) = g(z − 6) = (z − 6) − 6 = z − 12(g) The equation to solve is a2 − 4a = a − 6. This

simplifies to the quadratic

a2 − 5a + 6 = 0

(a − 2)(a − 3) = 0

So a = 2 or a = 3.(h) Rules for f (g(b)) and g(f (b)) were found in

this exercise. The equation to solve isb2 − 16b + 60 = b2 − 4b − 6. This can besolved for b:

b2 − 16b + 60 = b2 − 4b − 6

−16b + 60 = −4b − 6

60 = 12b − 6

66 = 12b

5.5 = b

16. (a) 2f (a) + 3 = 5

2(3a + 2) + 3 = 5

6a + 4 + 3 = 5

6a + 7 = 5

6a = −2

a = −1

3

(b) f (2a + 3) = 5

3(2a + 3) + 2 = 5

6a + 9 + 2 = 5

6a + 11 = 5

6a = −6

a = −1

(c) f (2a + 3) = a + 3

3(2a + 3) + 2 = a + 3

6a + 9 + 2 = a + 3

6a + 11 = a + 3

5a = −8

a = −8

5

17. (a) 12290 = 3x + 2

12288 = 3x

4096 = x

(b) In this case, you go through the same process but endup with 4096 = x2. This has two solutions, x = 64and x = −64. That’s because when you raise a

negative number to an even power, you get a positiveoutput.

(c) Solve for the x3 term in the same manner, 4096 = x3.This has only one solution, x = 16.

(d) Solve to get 4096 = x6. Because x is raised to aneven power, there are two solutions, x = 4 andx = −4.

18. In Exercise 17, you worked with these same functionsand found that for the second and fourth functions, therewere some pairs of inputs that could result in the sameoutput. This would mean that Derman’s conjecture isfalse for those functions.

(a) For f (x) = 3x + 2, Derman’s conjecture is true.(b) For f (x) = 3x2 + 2, Derman’s conjecture is false.(c) For f (x) = 3x3 + 2, Derman’s conjecture is true.(d) For f (x) = 3x6 + 2, Derman’s conjecture is false.

19. The denominator cannot be 0, so x2 − 5x + 4 �= 0. Thisimplies that (x − 1)(x − 4) �= 0, so x �= 1 or 4. Thecorrect answer choice is D.

20. The “inverse Spiro trick” would be, “Divide by 4, add 3,multiply by 2, and subtract 3.” To undo the trick, youneed to work through each of Spiro’s functions in reverseorder and undo the calculations step by step.

21. (a) To find f (7), you first need to find a. You know that3a + 1 = 7, so a = 2. Then, f (7) = 2 − 2 = 0

(b) First, find a. 3a + 1 = 14, so a = 133 , and

f (14) = 133 − 2 = 7

3(c) 3a + 1 = 16, so a = 5. f (16) = 5 − 2 = 3(d) Go through the same process that you used for the

specific values. 3a + 1 = z, so a = z−13 .

f (z) = z−13 − 2 = z−1−6

3 = z−73

22. Answers will vary. Many answers are possible.Here are some:

f (x) = x + 5, g(x) = x − 3

f (x) = 5x, g(x) = 3x

f (x) = x2, g(x) = x3

Most of the time, the two functions should performrelated operations, such as adding or subtracting, forexample.


23. Graphing each function shows its behavior. Also, anyfunction with an even exponent, like f (x) = x2, cannotoutput negative numbers. Then f (x) has R

+ as its range,not R. Functions b, d, and e have a range of R.

24. (a)

g(x, y) = (3x + 5) − (3y + 5)

x − y

= 3x + 5 − 3y − 5

x − y

= 3x − 3y

x − y

= 3(x − y)

x − y

= 3


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(b)

g(x, y) = (5x + 5) − (5y + 5)

x − y

= 5x + 5 − 5y − 5

x − y

= 5x − 5y

x − y

= 5(x − y)

x − y

= 5

(c)

g(x, y) = (12x + 5) − (12y + 5)

x − y

= 12x + 5 − 12y − 5

x − y

= 12x − 12y

x − y

= 12(x − y)

x − y= 12

(d)

g(x, y) = (x2) − (y2)

x − y

= x2 − y2

x − y

= (x + y)(x − y)

x − y= x + y

(e)

g(x, y) = (x2 + x) − (y2 + y)

x − y

= x2 + x − y2 − y

x − y

= (x2 − y2) + (x − y)

x − y

= (x + y)(x − y) + (x − y)

x − y

= (x + y + 1)(x − y)

x − y= x + y + 1

(f)

g(x, y) = (3x2 + 2x − 1) − (3y2 + 2y − 1)

x − y

= 3x2 + 2x − 1 − 3y2 − 2y + 1

x − y

= (3x2 − 3y2) + (2x − 2y)

x − y

= 3(x + y)(x − y) + 2(x − y)

x − y

= (3x + 3y + 2)(x − y)

x − y

= 3x + 3y + 2

4.08 Algebra with Functions


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

f (3) = 2(3) + 3

= 9

(b) g(x) = 5x + 1

g(3) = 5(3) + 1

= 16

(c) f (g(3)) = f (5(3) + 1)

= f (16)

= 2(16) + 3

= 35

(d) f ◦ g(3) is notation that means exactly the samething as f (g(3)), so f ◦ g(3) = 35.

(e) g ◦ f (3) = g(9) = 5(9) + 1 = 46.(f) If f (3) is 9 and g(3) is 16, then it is

9 ·16 = 144.2. (a) g ◦ f (x) = g(2x + 3)

= 5(2x + 3) + 1

= (10x + 15) + 1

= 10x + 16

(b) f ◦ g(x) = f (5x + 1)

= 2(5x + 1) + 3

= (10x + 2) + 3

= 10x + 5

3. (a) The formula for the area of a circle is A = πr2,where A is the area and r is the radius. Here, A is insquare inches.

(b) Since r = 4t , substitute r in the previous formulawith 4t to get

A = π(4t)2

Another way to write this is A = 16πt2.4. (a) f ◦ g(x) = f (x2 + 3) = x2 + 3.

g ◦ f (x) = g(x) = x2 + 3 sincef (x) = x.

(b) f ◦ g(x) = f (2x − 7) = 2x − 7, andg ◦ f (x) = 2x − 7.

(c) f ◦ g(x) = f ((x − 4)3) = (x − 4)3, andg ◦ f (x) = (x − 4)3.

(d) An identity is something that doesn’t change the otherinput. For example, 0 is the identity for addition,since a + 0 = a for any number a. This is the samebehavior f (x) = x has under composition. So,f (x) = x is the identity function for composition.


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5. Use the method of Exercise 7. Since f (x) = 3x − 1, theresult of f ◦ g(x) is

3 · g(x) − 1

So, to find g(x), solve the equation 3 · g(x) − 1 = x:

3 · g(x) − 1 = x

3 · g(x) = x + 1

g(x) = x + 1

3

Note that if f and g describe actions, they reverse eachother: f multiplies by 3 and subtracts 1, while g adds 1and divides by 3.

6. Your goal is to find a function g that undoes f · f

multiplies a number by 2 and then adds 3 to the result. Toundo that process, you would first subtract 3 and thendivide by 2. Thus, g(x) = x−3

2 .Check to see that the composition g ◦ f (x) = x:

g ◦ f (x) = g(2x + 3)

= (2x + 3) − 3

2

= 2x

2= x

7. (a) If f (g(x)) = 4x2 + 1, then fill in the blanks for whatf (x) does:

2( ) + 5 = 4x2 + 1

The blank is filled by the expression for g(x). Thensolve for g(x):

2 · g(x) + 5 = 4x2 + 1

2 · g(x) = 4x2 − 4

g(x) = 2x2 − 2

Function g(x) must be g(x) = 2x2 − 2. Check it:

f (g(x)) = f (2x2 − 2) = 2(2x2 − 2) + 5

= 4x2 − 4 + 5 = 4x2 + 1

(b) You are given that g(2x + 5) = 4x2 + 1. Letu = 2x + 5. Then u − 5 = 2x, so x = u−5

2 .Substituting, you obtain

g(u) = 4

(u − 5

2

)2

+ 1 = 4

(u2 − 10u + 25

4

)+ 1

= u2 − 10u + 25 + 1 = u2 − 10u + 26

8. (a) f ◦ g(x) = f (cx + d) = a(cx + d) + b =acx + (ad + b)

g◦f (x) = g(ax+b) = c(ax+b)+d = acx+(bc+d)

(b) The two functions will be identical whenever(ad + b) = (bc + d). This can be simplified a bit bycollecting terms:

ad + b = bc + d

ad − d = bc − b

d(a − 1) = b(c − 1)

This last statement is a quick check to see iff ◦ g = g ◦ f . For example, the functionsf (x) = 3x + 4 and g(x) = 5x + 8 have thisproperty. Also, any functions where d = b = 0 ora = c = 1 have the property.

9. Start with f (x) = ax + b and find f ◦ f (x) =a(ax + b) + b = a2x + ab + b. You want this expressionto equal 4x + 9. The coefficients of the x terms of theseexpressions must be equal, so you know that a2 = 4. Thismeans that a = 2, −2.

You also want the constant terms in both expressions tobe equal, so ab + b = 9. If a = 2, this means that2b + b = 9 and b = 3. This would give youf (x) = 2x + 3, which works because2(2x + 3) + 3 = 4x + 6 + 3 = 4x + 9. If a = −2, then−2b + b = 9 and b = −9. This would give youf (x) = −2x − 9, which also works because −2(−2x − 9) − 9 = 4x + 18 − 9 = 4x + 9.

On Your Own

10. (a) f (4) = 42 − 1 = 15(b) g(4) = 3 · 4 + 1 = 13(c) f (4) · g(4) = 15 · 13 = 195(d) f (g(4)) = f (13) = 132 − 1 = 168(e) f ◦ g(4) has the same meaning as f (g(4)),

so it’s 168.(f) g ◦ f (4) = g(15) = 3 · 15 + 1 = 46

11. h(x) outputs the same values as the absolute valuefunction, and has the same domain of R. When x ispositive, h(x) = x. When x is negative, h(x) is thepositive opposite: h(x) = −x when x < 0. This is thesame as the definition of the absolute value function.

12. (a) f ◦ g(3) = f (1) = 2(b) g ◦ f (3) = g(0) = −2(c) f ◦ g(a) = f (a − 2) = (a − 2)2 − 5(a − 2) + 6 =

a2 − 9a + 20(d) g ◦ f (a) = g(a2 − 5a + 6) = (a2 − 5a + 6) − 2 =

a2 − 5a + 4(e) (f ◦ g) ◦ f (a) = (f ◦ g)(a2 − 5a + 6)

Since f ◦ g(a) = a2 − 9a + 20, this expression is

(a2 − 5a + 6)2 − 9(a2 − 5a + 6) + 20

This is a good answer, but the normal form (foundby expansion) is a4 − 10a3 + 28a2 − 15a + 2.

(f) f ◦ (g ◦ f ) = f (a2 − 5a + 4). Sincef (a) = a2 − 5a + 6, this expression is

(a2 − 5a + 4)2 − 5(a2 − 5a + 4) + 6

Remarkably, this expression also expands toa4 − 10a3 + 28a2 − 15a + 2.

(g) To find these, solve a2 − 9a + 20 = 0. The left sidefactors as (a − 4)(a − 5) = 0, giving two solutionsa = 4 and a = 5.

(h) Solve a2 − 5a + 4 = 0. The left side factors as(a − 1)(a − 4) = 0, giving two solutions a = 1and a = 4.


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13. (a) To find h ◦ (f ◦ g)(x), start by finding f ◦ g(x).

f ◦ g(x) = (x + 3)2 − 6(x + 3) + 8

= x2 + 6x + 9 − 6x − 18 + 8

= x2 − 1

Now you can compute h ◦ (f ◦ g)(x).

h ◦ (f ◦ g)(x) = (x2 − 1) + 1

= x2

(b) To find (h ◦ f ) ◦ g(x), start by finding h ◦ f (x).

h ◦ f (x) = (x2 − 6x + 8) + 1

= x2 − 6x + 9

Now compute (h ◦ f ) ◦ g(x).

f ◦ g(x) = (x + 3)2 − 6(x + 3) + 9

= x2 + 6x + 9 − 6x − 18 + 9

= x2

14. (a) You know that f (x) = x2 − 10x + 21, and you wantto find g(x) = ax + b so that f ◦ g(x) = x2 − 4.

f ◦ g(x) = x2 − 4

(ax + b)2 − 10(ax + b) + 21 = x2 − 4

(a2)x2 + (2ab)x + b2 − (10a)x − 10b + 21 = x2 − 4

(a2)x2 + (2ab − 10a)x + (b2 − 10b + 21) = x2 − 4

For these two expressions to be equal, they must havethe same coefficients and constant terms. This givesyou three equations in a and b:

a2 = 1

2ab − 10a = 0

b2 − 10b + 21 = −4

The first equation tells you that a = 1, −1. You canhandle these two cases separately.

If a = 1, then 2ab − 10a = 2b − 10. 2b − 10 = 0,so b = 5. If b = 5 is a solution for the equationb2 − 10b + 21 = −4, then you have found g.(5)2 − 10(5) + 21 = 25 − 50 + 21 = −4, sog(x) = x + 5.

You can check this by finding f ◦ g(x) for this g.

f ◦ g(x) = (x + 5)2 − 10(x + 5) + 21

= x2 + 10x + 25 − 10x − 50 + 21

= x2 − 4

So g(x) = x + 5 gives the result you want. Now lookat the case where a = −1. In this case,2ab − 10a = −2b + 10. Since −2b + 10 = 0, b = 5.You already know that b = 5 results in the correctconstant term, so this would give you g(x) =−x + 5. Check the composition.

f ◦ g(x) = (−x + 5)2 − 10(−x + 5) + 21

= x2 − 10x + 25 + 10x − 50 + 21

= x2 − 4

So g(x) = −x + 5 also gives the result you want.

(b) You know that f ◦ g(x) = x2 − 4, and you want tofind h(x) = ax + b so that h ◦ (f ◦ g(x)) = x2.

h ◦ (f ◦ g(x)) = x2

h(x2 − 4) = x2

a(x2 − 4) + b = x2

ax2 − 4a + b = x2

So a = 1 and −4a + b = 0. This implies that−4 + b = 0, so b = 4. Thus h(x) = x + 4.

15. Answers will vary. One possible answer is g(x) = x,since it satisfies f ◦ g = f = g ◦ f .

Other answers include g(x) = xk for any positiveinteger k.

16. You want g such that g(x − 5) = 3x2 − 11x − 20.Replace x by x + 5 everywhere. Then

g(x) = 3(x + 5)2 − 11(x + 5) − 20

= 3(x2 + 10x + 25) − 11x − 55 − 20 = 3x2 + 19x

The correct answer choice is A.17. (a) a ◦ b = a(x2 − 7) = 3(x2 − 7) + 1 = 3x2 − 20

(b) b ◦ c = b(x − 5) = (x − 5)2 − 7 = x2 − 10x + 18(c) (a ◦ b) ◦ c = (a ◦ b)(x − 5) = 3(x − 5)2 − 20 =

3x2 − 30x + 55(d) a ◦ (b ◦ c) = a(x2 − 10x + 18) =

3(x2 − 10x + 18) + 1 = 3x2 − 30x + 5518. This exercise suggests that function composition is

associative:

(a ◦ b) ◦ c = a ◦ (b ◦ c)

One of the best ways to do this proof is to consider apotato-and-arrow diagram of the functions involved. Nomatter which composition is performed first, an arrowchain leading through the three functions will bepreserved. Other methods are possible.

19. Start by graphing f (x) = 2x + 3 and g(x) = x2 and theline with equation y = x. Then to find a point(a, f ◦ g(a)) by following this process.

• Move vertically from the point (a, 0) to touch thegraph of y = g(x). The coordinates of this point are(a, g(a)) or (a, a2).

• Move horizontally from the point (a, a2) to the linewith equation y = x. This new point has coordinates(a2, a2).

• Move vertically from the point (a2, a2) to touch thegraph of y = f (x). This point has coordinates(a2, 2(a2) + 3) or (a2, f ◦ g(a)). This is they-coordinate you want to pair with an x-coordinateof a.

• Move horizontally from the point (a2, f ◦ g(a)) tothe vertical line x = a. Mark this point for yourgraph of f ◦ g(x).

Do this for several points until you can see the shape ofthe graph of the composition.


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Maintain Your Skills20. In each case, function g reverses function f . Use

backtracking to find a rule for g.

(a) g(x) = x − 3(b) g(x) = x + 3(c) g(x) = x−5

3(d) g(x) = x+5

3(e) g(x) = x−5

2(f) (g(x) = x−B

A

21. In each case, function g reverses function f . Usebacktracking to find a rule for g.

(a) g(x) = x − 3(b) g(x) = x + 3(c) g(x) = x−5

3(d) g(x) = x+5

3(e) g(x) = x−5

2(f) g(x) = x−B

A

4.09 Inverses: Doing and Undoing

Check Your Understanding1. (a) Yes, the inverse is f −1(x) = x.

(b) Yes, the inverse is g−1(x) = 1x

.(c) No, there are duplicate outputs. For example, the

output 16 corresponds to the inputs 4 and −4.(d) Yes, the inverse is k−1(x) = 3√x.(e) No, there are duplicate outputs. For example, the

output 0 corresponds to the inputs 0, 1, and −1.(f) Yes, the inverse is m−1(x) = x2 where x ≥ 0. The

domain x ≥ 0 is required, since the square rootreturns a positive result.

(g) No, there are duplicate outputs. For example, theoutput 5 corresponds to the inputs 5 and −5.

2. (a) If (1, 5) is on the graph of the original function, thenthe input 1 gives output 5. The inverse functionreverses this, so it must have the input 5 giving output1. Using notation:

f (1) = 5 → f −1(5) = 1

The point (5, 1) must be on the inverse function’sgraph.

(b) For every point (x, y) on the original graph, there isa point (y, x) on the graph of the inverse function.The graph has been reflected over the line withequation y = x.

(c) If the graph of f , consisting of points (x, y) isone-to-one, then the graph of f −1(x), consisting ofcorresponding points (y, x) must also be one-to-one.The operation of taking the inverse resulted in areflection of the graph over the line with equationy = x. If you do this reflection again, the new imagewill be the original graph of f .

3. (a) The natural domain for f as it’s defined is x ≤ 6.

(b)

−5 5

−2

2

4

6

8

10

12

−4

(c) Answers will vary. Here is a possible solution:

f (x) ={

x if x < 02x if x ≥ 0

(d) Answers will vary. Here is a possible solution:

f (x) =⎧⎨⎩

x if x < 02x if 0 ≤ x ≤ 6−x if x > 6

4. (a) If you restrict the domain to x ≥ 0, the function isone-to-one. You could also restrict the domain tox ≤ 0 to have a one-to-one function.

(b) If you restricted the domain to x ≥ 0, then the inversewill be x �→ √

x. If you restricted it to x ≤ 0, theinverse is x �→ −√

x.5. (a) The domain is {1, 5, 9, 13}.

(b)

1

5

9

13

3

7

11


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 144

(c) t (x) is a function because each input is matched withexactly one output. No inputs are duplicated.

(d) t (x) does not have an inverse function. In the table ofthe inverse function the input 11 would have twooutputs. The inverse is not a function.

Change the output for the input 13 to some number otherthan 3, 7, or 11.

6. Rewrite as

x = f −1(x)

f −1(x) − 1

then solve for f −1(x). To make it simpler, replacef −1(x) with y:

x = y

y − 1x(y − 1) = y

xy − x = y

xy − y = x

y(x − 1) = x

y = x

x − 1

Remarkably, the inverse function f −1(x) = xx−1 is the

same as f (x).7. Here are the graphs of h and j on the same axes, along

with the dotted line y = x:

−1 0 1 2 3 4 5 6

−1

0

1

2

3

4

5

6

The functions are reflections across the line y = x, andany point (a, b) on one corresponds to (b, a) on theother. They are inverse functions.

8. (a)g ◦ f (x) = 5(5x2 − 17x + 6)

= 25x2 − 85x + 30

(b) g−1(x) = x5 , so

h(x) = g ◦ f ◦ g−1(x)

= 25(x

5

)2 − 85(x

5

)+ 30

= x2 − 17x + 30

(c)

0 5 10 15x

h(x)

f(x)

y

−24

−30

−36

−42

−18

−12

−6

6

12

18

24

(d) The zeros of h are x = 2, 15, and the zeros of f arex = 2

5 , 3.9. The statement is false. A function can have an inverse

which returns the same value as the function for someinputs x, but that doesn’t mean that it is the identityfunction. For example, consider the functionf (x) = 2x + 3. Its inverse is f −1(x) = x−3

2 .f (x) = f −1(x) for x = −3, and f is not the identityfunction.

It is even possible for f (x) = f −1(x) for all inputs x

but still not be the identity function. Such functionsinclude

f (x) = 10 − x

f (x) = 1x

f (x) = x

x − 1

These functions are their own inverses, sof (x) = f −1(x) for all x, and the functions are still notthe identity.


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On Your Own10. Functions (b) and (c) are one-to-one. The others have

several places where an output corresponds to multipleinputs. Only in (b) and (c) are the outputs unique.

11. It is important to remember that 3 and 7 are the onlyknown inputs. Switch the inputs and outputs to findinformation about the inverse function. f −1(7) = 3 andf −1(5) = 7 are the only known pieces of informationabout the inverse function. The correct answer choiceis D.

12. To prove that if functions f, g : R → R are one-to-one,f ◦ g is also one-to-one, you can follow an input a

through the composition. f ◦ g(a) = f (g(a)). You knowthat g(a) is the only output that g produces for an inputof a and that no other input produces this output. f (g(a))

is the only output that f produces for an input of g(a),and no other input produces this output. That means thatwhen you input a into f ◦ g, you’ll get an output which isproduced for no other input, and that you’ll get the sameoutput every time. f ◦ g is one-to-one.

13. Take an example: if f is a doubling function and g isadding 3, then f ◦ g adds 3, then doubles. The inversefunction would divide by 2, then subtract 3, so it undoesf then g. The inverse function is

g−1 ◦ f −1(x)

The correct answer choice is B.14. Switch the variables as in the examples.

(a) x = 5m−1(x) + 3x − 3 = 5m−1(x)

m−1(x) = x − 35

(b) x = 2n−1(x) − 11x + 11 = 2n−1(x)

n−1(x) = x + 112

(c) x = −3p−1(x) + 4x − 4 = −3p−1(x)

p−1(x) = −x − 43

(d)x = q−1(x)

5− 0.6

x + 0.6 = q−1(x)

5q−1(x) = 5(x + 0.6)

q−1(x) = 5x + 3

15. (a) If the slope is nonzero, the line will always increase ordecrease. The line will never have two inputs give thesame output. If two inputs did give the same output,the slope between those two points would be zero!

(b) f −1(x) undoes the function f (x) = ax + b whichmultiplies by a and adds b. To undo that process,subtract b and divide by a, so f −1(x) = x−b

a

(c) The slope of f −1(x) is 1a

.(d) There are many possible answers, but you’re looking

for something where the slope of the function and itsinverse are equal. a = 1

a. This only happens if

a = 1, −1.If a = 1, and the function and its inverse are equal,

x + b = x − b. This is only true if b = 0, so thefunction f (x) = x is its own inverse.

If a = −1, and the function and its inverse areequal, −x + b = −(x − b). This equation is anidentity, so any linear function of the formf (x) = −x + b will work.

16. (a) g ◦ f (x) = 25(5x3 − 12x2 − 11x + 6)

= 125x3 − 300x2 − 275x + 150

(b) k−1(x) = x

5h(x) = g ◦ f ◦ k−1(x)

= 125(x

5

)3 − 300(x

5

)2 − 275(x

5

)+ 150

= x3 − 12x2 − 55x + 150

(c)

0 5−5 10 15x

h(x)

f(x)

y

−400

−500

−600

−300

−200

−100

100

(d) f has zeros at x = −1, 25 , and 3. h has zeros at

x = −5, 2, and 15.17. You’re looking for a linear function g(x) = ax + b so

that g ◦ f ◦ g−1 is a monic quadratic. If g(x) = ax + b,then g−1(x) = x−b

a.

g ◦ f (x) = a(7x2 − 15x + 2) + b

= 7ax2 − 15ax + 2a + b

g ◦ f ◦ g−1 = 7a

(x − b

a

)2

− 15a

(x − b

a

)+ 2a + b

= 7(x − b)2

a− 15(x − b) + 2a + b


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Since you just want this to be monic, you can chooseany value for b, but it’s most practical to choose b = 0. a,however, has to be 7. So, if g(x) = 7x, g ◦ f ◦ g−1 turnsout to be x2 − 15x + 14. However, any functiong(x) = 7x + b would work—it would just give you adifferent monic quadratic.

18. (a) Answers will vary. The graph must always go up andto the right. Graphs (b) and (c) from Exercise 10 areincreasing.

(b) The function is one-to-one unless there are two inputsa and b that give f (a) = f (b) (the same output). Thefact that the function is increasing means that ifb > a, then f (b) > f (a) automatically. And ifb < a, then f (b) < f (a) (why?). So, it’s impossibleto have a �= b and f (a) = f (b). Any increasingfunction must be one-to-one.

Maintain Your Skills19. (a) f (10) = 4(10) + 3 = 43

(b) f (10) is 43, so f −1(43) must be 10.(c) f (0) = 3 and f (3) = 15.(d) Since f (3) = 15, f −1(15) = 3. Next, f −1(3) = 0.(e) f (f −1(x)) is always equal to x, so

f (f −1(289)) = 289.(f) f −1(f (x)) is always equal to x, so

f −1(f (−162.3)) = −162.3.

(g) For the same reason as part (f), f −1(f (10)) = 10.Now compute f (10), which is 43.

20. (a) f (r) = 4r + 3(b) f (f (r)) = f (4r + 3) = 4(4r + 3) + 3 = 16r + 15(c) f (f (f (r))) = f (16r + 15) = 4(16r + 15) + 3 =

64r + 63(d) f (f (f (f (r)))) = f (64r + 63) =

4(64r + 63) + 3 = 256r + 255(e) f (f (f (f (f (r))))) = f (256r + 255) =

4(256r + 255) + 3 = 1024r + 1023. You can alsowrite this as (45) r + (45 − 1).

(f) f 12(r) = (412)r+(412−1) = 16777216r+16777215.

4.10 Graphing Functions


1. There is no consensus on how to round negative numbers.The most common method mathematicians use is toalways “round 1

2 up to the next greatest integer.” Somecalculators, including the TI-Nspire, will round “awayfrom 0”, so that, for instance, −3.5 rounds to −4; in thismethod, they are rounding the absolute value of thenumber, and then restoring its original sign. The answershere abide by the most common method.Answers may vary. Samples are given.

(a) f (x) = ⌊x + 1

2

⌋(b) f (x) =

⌊100x+ 1

2

⌋100

(c) f (x) = 5 · ⌊ x5 + 1

2

⌋

2. The domain is all real numbers, and the range is allintegers. The inverse of the floor function is not afunction; infinitely many numbers map to the sameinteger, so you cannot uniquely determine what inputcorresponds to a particular output. Ask students todemonstrate that this is so by providing examples; ask ifthey can find a counter-example.

3. In parts (a) and (c), answers may vary. Samples are given.(a)

f (x) =

⎧⎪⎪⎨⎪⎪⎩

23 − x

7, if −5 ≤ x ≤ 2

7 − 2x, if 2 < x ≤ 34 − x, if 3 < x ≤ 6

(b) Any horizontal line drawn will intersect the graph atmost once.

(c)

f −1(x) =

⎧⎪⎪⎨⎪⎪⎩

4 − x, if −2 ≤ x ≤ 17 − x

2, if 1 < x ≤ 3;

23 − 7x, if 3 < x ≤ 4

(d)

(−2, 6)

(4, −5)

−6 −2 2

2

4

6

0 4 6−2

−4

−6

−4

(1, 3)(3, 2)

x

y

(e) Each piece of the inverse of f is the inverse of thecorresponding piece of f, but they are written inreverse order. The range for each piece of f is theinterval for the corresponding piece of the inversefunction.

4. (a)

x

y

−2−2

−4

−4 2 4

2

4

0


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 147

(b)

642

2

4

6

0x

y

(c)

42

2

4

0x

y

(d)

42

2

4

0x

y

(e)

4 620

2

4

6

x

y

−2−4

(f)

420x

y

−4

−4

2

4

(g)

420x

y

−4

−4

2

4

(h)

20x

y

−2

−2

2

(i)

2x

y

−2

−2

2

5.

s(t) =⎧⎨⎩

10 if 0 ≤ t < 150 if 15 ≤ t < 22;

35 if 22 ≤ t < 57

605040302010

10

20

30

40

(15, 0) (22, 0)(15, 10)

(22, 35) (57, 35)

(0, 10)

0x

y

6. Yes; 4 is an integer, so it can be “pulled out” of the floorfunction. It is true that x + a� = x� + a for anyinteger a.

7. f is not a function, since it does not have a unique valuefor each x between −1 and 1.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 148

8. For parts (a)–(f), if the graph passes the horizontal linetest, it is one-to-one. If it does not, then it is notone-to-one.(a) no(b) yes(c) yes(d) yes(e) yes(f) no

On Your Own

9. (a)

20x

y

−2

2

(b) Answers may vary. Sample:�x = − −x�

10. (a)

f (x) =

⎧⎪⎪⎨⎪⎪⎩

x

2+ 1, if −6 ≤ x ≤ −2

x + 2, if −2 ≤ x ≤ 0x

8+ 2, if 0 < x ≤ 8

(b)

f1(x) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩

x − 32

, if −1 ≤ x ≤ 3x − 3 if 3 < x ≤ 5x + 11

8, if 5 < x ≤ 13

(c)

f2(x) =

⎧⎪⎪⎨⎪⎪⎩

x

2+ 4, if −6 ≤ x ≤ −2

x + 5, if −2 < x ≤ 0x

8+ 5, if 0 < x ≤ 8

11.

c(d) ={

60, if 0 ≤ d ≤ 560 + 5�d − 5 , if d > 5

1086Number of Gigabytes

42040

50

60

70

80

90

Cos

t ($)

12. No; for example, if x = 2.5, then 2 · 2.5� = 5� = 5,but 22.5� = 2 · 2 = 4.

13. No; for example, 2.52� = 6.25� = 6, but2.5�2 = 22 = 4.

14. Answers may vary. Samples are given.(a) f (x) = −6 + h3(x) · (7x + 7)

(b) f (x) = 2x + 4 − h−1(x) · (2x + 1)

(c) f (x) = 2 + h−3(x) · (x − 1) + h4(x) · (5 − x)

(d) f (x) = 1 − x2 + h1(x) · (x2) + h5(x) · (x2)

15. Answers may vary. Samples are given.

a (x) = −x + h0 (x) · (2x) or a (x) = −1h0(x)+1 · (x)

16. f (x) = −3+h−2 (x)+h−1 (x)+h0 (x)+h1 (x)+h2 (x)


17. This Exercise reviews evaluating expressions to preparestudents to find the roots of a polynomial using the FactorTheorem: if one factor evaluates to 0, then the entire termwill evaluate to 0 regardless of the values of the otherfactors.(a–e) 0

(f) They all equal 0. There is no other integer that willcause the expression to evaluate to 0.

18. Like Exercise 17, this Exercise reviews evaluatingexpressions to prepare students to find the roots of apolynomial using the Factor Theorem: if one factorevaluates to 0, then the entire term will evaluate to 0regardless of the values of the other factors.

(a) 3; 3(b) 69; 69(c) 4; 544(d) 24; 24(e) −9, −5, −3, 1, and 4. See Exercise 17, in which all

five x-values caused the last part of the secondexpression to evaluate to 0.

4B MATHEMATICAL REFLECTIONS

1. (a) f (3) = 3(3)2 − 6(3) + 1 = 27 − 18 + 1 = 10f (−3) = 3(−3)2 − 6(−3) + 1 = 27 + 18 + 1 = 46

(b)

f (x) = 123x2 − 6x + 1 = 12

3x2 − 6x − 11 = 0

Now you can use the quadratic formula to find thesolutions to this equation.

x = −(−6) ±√(−6)2 − 4(3)(−11)

2(3)

= 6 ± √36 + 1326

= 6 ± √168

6

= 3 ± √42

3


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 149

(c) f (11) − f (10) = 3(11)2 − 6(11) + 1

− (3(10)2 − 6(10) + 1)

= (363 − 66 + 1) − (300 − 60 + 1)

= (298) − (241)

= 57

(d) f (x + 1) = 3(x + 1)2 − 6(x + 1) + 1

= 3x2 + 6x + 3 − 6x − 6 + 1

= 3x2 − 2

(e) f (x + 1) − f (x) = 3x2 − 2 − (3x2 − 6x + 1)

= 3x2 − 2 − 3x2 + 6x − 1

= 6x − 3

(f) g(x) = f (x + 1) − f (x) = 6x − 3, sog(10) = 6(10) − 3 = 57.

2. (a) f (3) = 3(3) + 2 = 11(b) g(3) = 3 + 5 = 8(c) f (g(3)) = f (8) = 3(8) + 2 = 26(d) f ◦ g(3) = 26 (This is just different notation for the

calculation in the previous part.)(e) g ◦ f (3) = g(11) = 11 + 5 = 16(f) f (3) · g(3) = 11 · 8 = 88

3. (a) You want f ◦ g(x) = x, and this will be true if g isthe inverse of f . To find the inverse of f , think aboutwhat f does, and then create a g that will undo it.f takes an input, multiplies it by 3 and then adds 2.So the inverse of f will subtract 2 from an input andthen divide it by 3. g(x) = x−2

3(b) For g ◦ f (x) = x, the inverse function you found in

the previous part will also work. g(x) = x−23

(c) To find g(x), solve the equation for g(x).

f ◦ g(x) = x2

3(g(x)) + 2 = x2

3(g(x)) = x2 − 2

g(x) = x2 − 23

(d) You are given that g(3x + 2) = x2 and what you wantis g(input) = output. So replace 3x + 2 by a simplevariable, say u = 3x + 2, then rewrite the output.First you find that x = u−2

3 , so x2 = (u−2

3

)2.Therefore, g(u) = (

u−23

)2.(e) If g ◦ f (x) = f (x), then this g left its input

unchanged. A function that does that is g(x) = x, theidentity function.

(f) In the equation f ◦ g(x) = f (x), g behaves just likex, so the identity function would also work here.

g(x) = x

4. (a) g ◦ f (x) = 16(−3 + x + 8x2 + 4x3) =−48 + 16x + 128x2 + 64x3

(b) k−1(x) = x

4h(x) = g ◦ f ◦ k−1(x)

= −48 + 16(x

4

)+ 128

(x

4

)2 + 64(x

4

)3

= −48 + 4x + 8x2 + x3

(c)

0−5 1x

h(x)f(x)

y

−32

−40

−24

−16

−8

8

16

24

(d) h(x) has zeros at x = −6, −4, and 2. f (x) has zerosat x = − 3

2 , −1, and 12 .

5. Answers will vary, but should include the followingelements.

• A function accepts inputs and returns outputs. Eachinput produces the same output every time it isoperated on by the function. You can’t have morethan one output matched to any given input.

• The set of acceptable inputs for a function is calledits domain. The set of all possible outputs for afunction is called its range.

• A function is also an object in its own right, and canbe operated on with the Basic Rules of Algebra.

6. Answers will vary. Here is a sample response:The composition of two functions f and g can be

written as f ◦ g(x) or f (g(x)). Both types of notationmean the same thing—Take an input x and find its outputfor the function g; then use that output as the input for thefunction f and find the final output. If the functions f

and g are expressed algebraically, you can use the BasicRules of Algebra to write f ◦ g(x) as a new function. Justsubstitute the algebraic expression for g(x) into f andsimplify.

7. The function x �→ 3x + 7 takes an input and multiplies itby 3, and then adds 7. To undo that, you want a functionthat takes an input, subtracts 7, and then divides by 3. Insymbols, x �→ x−7

3 .


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 150

INVESTIGATION 4C EXPONENTIALFUNCTIONS


For You to Explore

1.

Input Output �

0 1 31 4 32 7 33 10 34 13 35 16 3

2.

Input Output �

0 1 01 1 22 3 43 7 64 13 85 21 10

3.

Input Output �

0 1 21 3 62 9 183 27 544 81 1625 243 486

4.

Input Output �

0 3 121 15 602 75 3003 375 15004 1875 75005 9375 37, 500

5.

Input Output �

0 1 − 12

1 12 − 1

4

2 14 − 1

8

3 18 − 1

16

4 116 − 1

32

5 132 − 1

64

6. In the last problem, f (x) = (12

)x = 2−x , the outputs arethe negatives of the inputs. So, a good guess might beg(x) = 2x . The table for g(x) is:

Input Output �0 1 11 2 22 4 43 8 84 16 165 32 32

The output column equals the difference column.g(x) = 0 also works.

7. (a)

Input Output ÷0 1 41 4 1.752 7 1.433 10 1.34 13 1.235 16 1.19

(b) The numbers in the ratio column appear to beapproaching 1. This makes sense because as theoutputs get larger and larger, adding three to theprevious output doesn’t change it as much as whenthe outputs are small.

8. (a) b(x) = x2 − x + 1

Input Output ÷0 1 11 1 32 3 2.333 7 1.864 13 1.625 21 1.48

(b) c(x) = 3x

Input Output ÷0 1 31 3 32 9 33 27 34 81 35 243 3

(c) d(x) = 3 · 5x

Input Output ÷0 3 51 15 52 75 53 375 54 1875 55 9375 5


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 151

(d) f (x) = ( 12

)xInput Output ÷

0 1 12

1 12

12

2 14

12

3 18

12

4 116

12

5 132

12

9. (a) If a table has constant differences, the function islinear and the rule will be in the formf (x) = (constant difference) · x + (a constant). Tofind the constant, look at the output for an input of 0.So, the function will bef (x) = (constant difference) · x + (f (0)).

(b) If the table has constant ratios, then the function isexponential and the rule will be in the formf (x) = (a constant) · (constant ratio)x . To find theconstant, look at the output for an input of 0. So, thefunction will be f (x) = f (0) · (common ratio)x .

10. If the common ratio is going to equal the output, yourtable will look like this:

Input Output ÷0 f (0) f (0)

1 f (1) f (1)

2 f (2) f (2)

3 f (3) f (3)

4 f (4) f (4)

5 f (5) f (5)

If the common ratios = the outputs,f (2)

f (1)= f (1),

f (3)

f (2)= f (2), and f (4)

f (3)= f (3). So,

f (2) = (f (1))2, f (3) = (f (2))2, and f (4) = (f (3))2.Now your table looks like this:

Input Output ÷0 f (0) f (0)

1 f (0)2 f (0)2

2 f (1)2 = f (0)4 f (0)4

3 f (2)2 = f (0)8 f (0)8

4 f (3)2 = f (0)16 f (0)16

5 f (0)32 f (0)32

You can let f (0) be some number, say 3, the exponentsfor the function are powers of 2. One possible solution isf (x) = 3(2x )

On Your Own11. (a) For x2 = 5, take the square root of both sides.

Remember that the solution could be the positive ornegative square root. x = ±√

5(b) Estimate the solution using your calculator. Since

22 = 4 and 23 = 8, a good guess to start would be alittle more than 2. x ≈ 2.322. Many students will

choose this equation as the most difficult for them tosolve.

(c) x = 52 = 25(d) Since 8 is a power of 2, rewrite the equation as

23 = 2x and x = 3.12.

O2 2

2

4

6

8

x

y

13.√

2 ≈ 1.414, and x = 1.414 is between x = 1 and x = 2.Since the graph of y = 3x is continuous, which means(informally) that you can trace it without picking up yourpencil, and increasing, you can say that 3

√2 must be

between 31 and 32, or between 3 and 9.14. One way you could make a more accurate estimate of

3√

2 is by “zooming” in on the graph to find a point whosex-coordinate is close to 1.414 . . . You could also use yourcalculator to evaluate 3

√2, either in the home screen or in

the graph screen. See the Technology Appendix for someideas about how to do this.

15. (a) f (a) = 3 · 2a

Input: a Output: f (a)

−2 34

−1 32

0 31 62 12

(b) g(a) = 30 · 2a

Input: a Output: g(a)

−2 304 = 15

2

−1 302 = 15

0 301 602 120

(c) h(a) = 15 · 5a

Input: a Output: h(a)

0 15

1 12 53 254 125


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 152

(d) j (a) = 27 · ( 13

)aInput: a Output: j (a)

−1 810 271 92 33 1

16. (a) k(a) = 4 · 3a

(b) L(h) = 100 · ( 12

)h(c) p(x) = 2 · ( 1

4

)x(d) Q(n) = 8 · ( 3

2

)n17.

x f (x) = 2x

0 11 21.4 2.6391.41 2.6571.414 2.6651.4142 2.6651.41421 2.6651.414213 2.665

18. (a) For an integer a > 0, 5b = 2a > 1, so b > 0.However, 2a is even and 5b is odd in this case, so theycannot be equal. For an integer a < 0, 5b = 2a < 1,so b < 0. Now 2a = 5b implies that 2−a = 5−b (bytaking reciprocals). Here the exponents are againpositive integers, and you get the same contradictionas in the first case. Therefore, you must have a = 0,so 5b = 20 = 1. Hence b = 0.

(b) Since 4 = 22 and 8 = 23, you can write

4c = 8d

(22)c = (23)d

22c = 23d

2c = 3d

There are many possible solutions. For example,c = 3 and d = 2, or c = 6 and d = 4. You can writec = 3

2 · d . If you choose even integers for d, you willget an integer value for c.


19. a, b, d, and e can all be found directly. There is nosolution for g. To answer c and f, you can estimate or usea calculator.

(a) f (0) = 20 = 1(b) f

(12

) = 212 = √

2(c) f (π) = 2π ≈ 8.825 (using a calculator)(d) f (f (2)) = f (22) = f (4) = 24 = 16

(e)

f (a) = 1

82a = 2−3

a = −3

(f)

f (a) = 7

2a = 7

a ≈ 2.807

This is found using a calculator.(g) 2a will never be a negative number, so there is no

solution.

4.12 Graphs of Exponential Functions


1. In the first graph, the y values are negative, so a < 0.Therefore, f (x) = −3 · 2x describes this graph. Thesecond graph is decreasing, so 0 < b < 1.f (x) = 3 · ( 1

2

)xdescribes this graph. The third graph

passes through (1, 6). f (x) = 3 · 2x describes this graph.The last graph passes through (1, 15). f (x) = 3 · 5x

describes this graph.2. (a) Given f (x) = a · bx and the points (0, 12) and (2, 3),

you can write the equations 12 = a · b0, so a = 12and 3 = 12 · b2, 1

4 = b2, so b = ± 12 . b �≤ 0, so

b = 12 . The function you are looking for is

f (x) = 12 · ( 12

)x(b) Write the equations 12 = a · b2 and 3 = a · b4. Solve

each equation for a, and you get a = 12b2 and a = 3

b4 .Therefore,

12

b2= 3

b4

12b4 = 3b2

12b4 − 3b2 = 0

3b2(4b2 − 1) = 0

3b2(2b + 1)(2b − 1) = 0

b = 0 or b = ± 12 Since b > 0, b = 1

2 . Substitute to

find a : 3 = a · ( 12

)4, 3 = a · 1

16 , and a = 3 · 16 = 48.The function you are looking for is f (x) = 48 · ( 1

2

)x3. (a) To find a possible solution, choose a convenient value

for a. A good choice would be 2 because72 ÷ 2 = 36 (a perfect square) or 8 because72 ÷ 8 = 9 (a perfect square). In the first case,72 = 2 · b2 and b = ±6, b �= −6, b = 6. Oneequation is f (x) = 2 · 6x . Then, let a = 8 so72 = 8 · b2 and b = ±3, b �= −3, b = 3. Anotherequation is f (x) = 8 · 3x .

(b) To find the general form, solve the equation72 = a · b2 for a in terms of b, or for b in terms of a.

a = 72b2 or b =

√72a

. The two resulting equations are

f (x) =(

72b2

)· bx and f (x) = a ·

(√72a

)x

.


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4. The proof of Lemma 4.2 is precisely the same as that ofLemma 4.1 with all the > signs reversed. Alternately, if0 < b < 1, then b = 1

cwith c > 1. Lemma 4.1 implies

that cx > 1. Now cx = ( 1b

)x = 1bx , so 1

bx > 1. Therefore,1 > bx . The proof of Theorem 4.3 mimics that ofTheorem 4.2 with the signs reversed, or again you canuse b = 1

cwith c > 1. Then f (x) = bx = ( 1

C)x = 1

cx .Since the denominator is strictly increasing, the fractionis strictly decreasing.

5. If two graphs are reflections over the y-axis, a point(x, y) on one graph will have an image (−x, y) on theother graph. We need to show that f (−x) = g(x).

f (−x) = 2−x

= 12x

=(

12

)x

= g(x)

6. x ≈ 2.8077.

2x = 17

· 4x

7 · 2x = 4x

7 = 4x

2x

7 =(

42

)x

7 = 2x

This equation is the same as the one for which youapproximated a solution in Exercise 6, so x ≈ 2.807.

8. (a)

O8 4 4 8

8

4

4

8

xg

fy

The graphs do not intersect in this window.(b) The graphs intersect once. The intersection is at

roughly x = 35.24 and y = 54.26. You can find thispoint by “zooming out” on your calculator and thenusing the “intersect” function. See the TechnologyAppendix for more about how to do this.

9. (a) 2.66514√

2 ≈ 3.999991207 · · · ≈ 4(b) Yes, (2

√2)

√2 = 2

√2.

√2 = 22 = 4

10. When you try decimal values to approximate (−2)√

2

with a calculator, you get an error: Non-real result. If youevaluate (−2)

√2, the result is 2

√2 · (−1)

√2. Exponential

functions bx are restricted to b > 0,because (−1)x cannot be calculated for many fractionalvalues. For example, (−1)

12 would be

√−1, which isundefined.

On Your Own11. The y-intercept is (0, a) because

f (0) = a · b0 = a · 1 = a

12. Test each point to see if it is on the graph.

A. f (0) = −3 · 20 = −3 · 1 = −3 �= 1. (0, 1) is NOT onthe graph.

B. f (−1) = −3 · 2−1 = −3 · 12 = − 3

2 �= 6. (−1, 6) isNOT on the graph.

C. f (−2) = −3 · 2−2 = −3 · 122 = −3 · 1

4 = − 34 =

−.75. (−2, −.75) IS on the graph.D. f (2) = −3 · 22 = −3 · 4 = −12 �= 36. (2, 36) is

NOT on the graph.

The correct choice is C.13. (a) Next year the cost will increase by 3 percent, so the

price will be 3.99 + .03 · 3.99 or

3.99 · (1 + .03) = 3.99 · 1.03 = $4.11.

The following year, the cost will increase by 3percent again 4.11 ·1.03 = $4.23. Notice that this is

4.11 · 1.03 = (3.99 · 1.03) · 1.03 = 3.99 · 1.032

= $4.23

(b) Multiply $ 3.99 by 1.03 ten times or by 1.0310.

3.99 · 1.0310 = $5.36

(c) Use your calculator. After 55 years the price, will be$ 20.28.

(d) C(n) = 3.99 · 1.03n. The cost will be the originalprice multiplied by ( 1+ percent increase/100) raisedto the power (number of years)

14. By evaluating 2x for some negative integer values, youcan find that 2−20 = 1

1048576 < 11000000 . Any value less

than −20 will also work.15. Since

√6 ≈ 2.449,

√6 is between 2 and 3. Therefore,

3√

6 must be between 32 = 9 and 33 = 27. You can reasonthis because we see that y = 3x is continuous andincreasing.

16. It is not possible for 2p = 7q if p and q are nonzerointegers. First consider positive integers p and q. If p is apositive integer, 2p is a power of 2, and therefore is aneven integer. If q is a positive integer, 7q is a power of 7,and is an odd integer. Since 2p is even and 7q is odd, theycannot be equal. If p and q are negative integers,consider 2p = 1

2−p and 7q = 17−q . Following the same

argument, the denominator of the first fraction is always


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 154

even and the denominator of the second fraction isalways odd, so they are never equal.

17. Since 8 is a multiple of 2, the expressions on each side ofthe equation will both be even numbers, so they could beequal. In particular, 2

pq = 8 = 23, so p

q= 3, and p = 3q.

There are many integers that satisfy this equation.18. (a)

Input Output−2 1

4

−1 − 12

0 11 −22 43 −8

(b) The points are alternately positive and negative, soyou cannot connect them with a smooth curvewithout crossing the x-axis. It is not possible for f (x)

to equal zero, so the curve cannot cross the x-axis.19. Suppose first that b and d are positive.

• If b �= d , there will be exactly one solution as long asa and c have the same sign.

• If a and c have opposite signs, there is no solution.Also, if b = d and a �= c, there are no solutions.

• If b = d and a = c, there will be infinitely manysolutions, because the two sides of the equation willbe identical.Other possibilities arise if you allow any of theparameters to be 0, or if you allow b and d to benegative.

20. (a) By definition, if x is a positive integer, then bx is justb · b · b · · ·︸︷︷︸

x copies

. If b > 0, then b · b · b · b · · · will be the

product of positive numbers and therefore positive. Ifx = 0, then bx = 1 > 0. If x < 0, then bx is thereciprocal of a positive number, so it is positive.

(b) If x is rational, then bx = bpq = q

√bp where p and q

are integers. If b > 0, you just showed that bp > 0and the q

√positive number will be positive.

Maintain Your Skills21. (a) (3

√2)

√2 = 3

√4 = 32 = 9

(b) (3√

2)2 = 32·√2

(c) (3−√2)−1 = 3

√2

(d) (3√

8)√

2 = 3√

16 = 34 = 81(e) 3

√2 · 3−√

2 = 3√

2+(−√2) = 30 = 1

(f) (3−√2)−

√2 = 3

√4 = 32 = 9

(g) 3√

2 · 5√

2 = (3 · 5)√

2 = 15√

2

(h) (33√2)

3√2 = 33√4

(i) (33√2)

3√4 = 33√8 = 32 = 9

22.

O 224x

yabcdf

4

These graphs are all translations of each other. Forexample, a(x) = 9 when x = 2, but b(x) = 9 whenx = 1. The graph of b(x) is the image of a(x) after atranslation of 1 unit to the left.

4.13 Tables of Exponential Functions

Check Your Understanding1. (a) A(n) = 18 · ( 1

3

)n(b) B(x) = −2 · 4x

(c) This is not an exponential function, because there isnot a constant ratio between successive terms. 6

4 �= 129

(d) D(z) = 13 · 6z

2. (a)

A(n) ={

18 if n = 013 · A(n − 1) if n > 0

(b)

B(x) ={−2 if x = 0

4. B(x − 1) if x > 0

(c) Not an exponential function.(d)

D(z) ={ 1

3 if z = 06. D(z − 1) if z > 0

3. q(x) = a · bx So you can set up two equations with thegiven points:

100 = a · b3

4 = a · b5

Then divide one equation by the other to get an equationin b alone, and solve for b:

1004

= a · b3

a · b5

25 = b−2

b2 = 125

b = 15


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 155

To find a, just substitute your value for b into either ofyour original equations:

100 = a ·(

1

5

)3

100 = a ·(

1

125

)

12, 500 = a

The equation is q(x) = 12, 500 · ( 15

)x. If you like, you

can check this equation with your second point.4. b = −1

5 is not a solution because negative bases are notallowed in the definition of an exponential function.Exercise 4 begins by telling you that q(x) is anexponential function.

5. You can find a linear function that solves the problem:f (x) = 2x + 16. Another polynomial function thatsolves the problem is f (x) = 2x + 16 +(x + 3)(x − 2) . . . You can also find an exponentialfunction that solves the problem:

10 = a · b−3

20 = a · b2

Divide:

1

2= b−5

1

2= 1

b5

2 = b5

215 = b

Find a:

10 = a ·(

215

)−3

10 = a · 2− 35

10 · 235 = a

The exponential equation we are looking for is

f (x) = 10 · 235 ·(

215

)x ≈ 15.157 · 1.149x

6. (a) You want to go from 100 to 300 in 5 steps, so b5 = 3and b = 3

15 . Since T (0) = 100 = a · b0 = a · 1, you

know that a = 100. The function is

T (x) = 100 ·(

315

)x

(b)x T (x) ÷0 100 3

15 ≈ 1.246

1 100 · 315 ≈ 124.57 3

15 ≈ 1.246

2 100 · 325 ≈ 155.18 3

15 ≈ 1.246

3 100 · 335 ≈ 193.32 3

15 ≈ 1.246

4 100 · 345 ≈ 240.82 3

15 ≈ 1.246

5 300 315 ≈ 1.246

7. (a) Since the car depreciates 20%, it retains 80% of itsvalue. Multiply by 80% or 0.8. After 1 year the valueis $20,000 · 0.8 = $16,000. After 2 yearsthe value is $16,000 · 0.8 = $12,800. After 3 yearsthe value is $12,800 · 0.8 = $10,240.

(b) This is an exponential function with a = 20,000 andb = 0.8.

V (n) = 20,000 · (0.8)n

(c) Since the function is decreasing, it will eventually beworth less than $1000. Using your calculator, youcan find that V (14) ≈ $880 which is less than $1000.

8.

y1 = a · bx1

y2 = a · bx2

Divide:y1

y2= bx1−x2

(y1

y2

) 1x1−x2 = b

Substitute to find a:

y1 = a ·((

y1

y2

) 1x1−x2

)x1

y1(y1y2

) x1x1−x2

= a

Simplify the expression for a:

a = y

−x2x1−x2

1 · yx1

x1−x22

The exponential function that passes through (x1, y1)

and (x2, y2) is

y = y

−x2x1−x2

1 · yx1

x1−x22 ·

((y1

y2

) 1x−x

)x

On Your Own

9. Substituting the given values into the function yields

a · b0 = 4

a · b2 = 25

Divide the second equation by the first to get b2 = 254 , so

b = 52 . The first equation simplifies to a = 4. The correct

answer choice is C.10.

n M(n) ÷0 16 1.51 24 1.52 36 1.53 54 1.54 81


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A closed-form rule would be M(n) = 16 · ( 32

)n, since

M(0) = 16 = a and each term is multiplied by 32 = b. To

find a recursive rule, use the fact that you multiply by 1.5to get from one term to the next.

M(n)

{16 if n = 032 · M(n − 1) if n > 0

11. (a)

n F(n) ÷0 1 11 1 22 2 33 6 44 24 55 120 66 720 7

(b) This is not an exponential function because the ratiois not constant.

(c) Since F(1) = 1, F (2) = 2 = 2 · 1, F (3) = 6 =3 · 2 · 1, and F(4) = 24 = 4 · 3 · 2 · 1, you cancalculateF(10) = 10 · 9 · 8 · 7 · · · 3 · 2 · 1 = 3,628,800

12. (a) y = 3 · 2x

(b) y = −514 ·(

514

)x

(c) y = 3 · 213 · ( 1

2

) x6

13. (a) Since the interest is added to Kara’s account, she willhave $1000 + 0.03 · $1000 = $1000 · (1 + 0.03) =$1000 · (1.03) = $1030 in 1 year. Each year youmultiply by 1.03 to get the new amount.$1030 · 1.03 = $1060.90 in 2 years.$1060.90 · 1.03 = $1092.73 in 3 years.

(b) To calculate the amount of money in the account after20 years, you would multiply $1000 by 1.03 twentytimes. $1000 · 1.0320 = $1806.11

14. (a) Multiply by 0.98 each time because if you pay 2%,there will be 100% −2% = 98% remaining. B(1) =$2000 · 0.98 = $1960; B(2) = $1960 · 0.98 =$1920.80; B(3) = $1920.80 · 0.98 = $1882.38

(b) B(n) = 2000 · (0.98)n

(c) The domain is 0, 1, 2 . . .12 since payment is madeeach month for 12 months.

15. Use the equation 1000 = 2000 · b12 so

b = (12

) 112 , b = 0.944. The payment should be 100%

–94.4% = 5.6% each month.16. The functions would not look the same because they have

different domains. f (x) is defined for all real numbers,and its graph would be a smooth, connected curve. g(x)

is only defined for nonnegative integers. There would beno points to the left of the y-axis, and points to the rightof the y-axis would only take on integer values of x andwould not be connected to each other.


17. (a) b(5) = 35 = 243(b) b(3) · b(2) = 33 · 32 = 27 · 9 = 243

(c) b(1) = 31 = 3(d) b(3)

b(2)= 33

32 = 33−2 = 31 = 3

(e) b(6) = 36 = 729(f) (b(2))3 = (32)3 = 93 = 729

18. (a) f (0) = b0 = 1(b) f (−3) = b−3 = 1

b3 = 1f (3)

= 1p

(c) f (8) = b8 = b3 · b5 = f (3) · f (5) = pq

(d) f (6) = b6 = (b3)2 = (f (3))2 = p2

(e) f (15) = b15 = (b3)5 = (f (3))5 = p5 orf (15) = b15 = (b5)3 = (f (5))3 = q3. This lastequation gives you a way to write p in terms of q.

4C MATHEMATICAL REFLECTIONS

1. An exponential function is a function of the formy = a · bx where a and b are real numbers,a �= 0, b > 0, and b �= 1. Examples will vary, but willhave the real numbers as the domain and if a is positive,the real numbers y > 0 as the range. (If a is negative, therange is the real numbers y < 0.) If the base is 0 < b < 1and a is positive, then the function is decreasing for all x.This is also the case if a is negative and b > 1. If b > 1and a is positive, the function is increasing for all x. Thisis also the case if a is negative and 0 < b < 1.

2. (a) The ratio between successive terms is 2, so to getfrom one term to the next, you multiply by 2. Tomake a complete recursive definition, though, youalso need to establish the first input-output pair.

g(x) ={−5 if x = 0

2 · g(x − 1) if x > 0

(b) g(x) = −5 · 2x

(c) No, because the recursive function has only theintegers 0, 1, 2, . . . as its domain, and the domain ofthe closed-form function is the real numbers.

3. (a)

x h(x) ÷0 9 2

3

1 6 23

2 4 23

3 83

23

4 169

(b) h(23) = 9 · ( 23

)23 ≈ 0.0008.(c) h(x) could be an exponential function because the

entries in the table show a constant ratio betweensuccessive terms. It does not have to be anexponential function, though, because throughLagrange Interpolation you could find a polynomialfunction that matches this table completely.

4. (a) 12 = 3x2, so x2 = 4 and x ± 2.(b) x = 5 · 4− 1

2 = 5 · (22)−12 = 5 · 22·− 1

2 = 5 · 2−1 = 52


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 157

(c) 16 = 4 · 32x so 4 = 32x and 22 = (25)x or 22 = 25x .Because exponential functions are one-to-one, youknow that 2 = 5x and x = 2

5 .(d) −5 = x · 27

23 , so −5 = x · (33)

23 = x · 33· 2

3 = x · 9.Since −5 = 9x, x = − 5

9 .5. Use the definition of the function to write an expression

for the left side of the equation.

f (m) · f (n) = bm · bn

Then use the Laws of Exponents.

bm · bn = bm+n

Use the definition of the function.

bm+n = f (m + n)

This gives you the expression on the right side of theequation.

6. Exponential functions are strictly increasing or strictlydecreasing. This implies that every exponential functionis one-to-one, and all one-to-one functions have inverses.

7. You get 6% of the amount in the account as interest at theend of the year, so in effect, each year you have 1.06times as much as the year before. If you start with$1,000, you’ll have 1000 · (1.06)n after n years. After 30years, you have 1000 · (1.06)30 ≈ 5743.49 or $5743.49.

INVESTIGATION 4D TRANSFORMING BASICGRAPHS


For You to Explore1.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

2. (a) This is the graph of y = x2 shifted 5 units down.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(b) This is the graph of y = x2 shifted 5 units to the left.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(c) This is the graph of y = x2 shifted 5 units up.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 158

(d) This is the graph of y = x2 shifted 3 units up and2 units to the left.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

3.

2

2

4

–2–2

–4

x

y

4. (a) This is the same as the graph of y = x3 shifted5 units down.

(b) This is the same as the graph of y = x3 shifted5 units left.

(c) This is the same as the graph of y = x3 shifted5 units up.

(d) This is the same as the graph of y = x3 shifted left2 units and up 3 units.

5. The graph of y = x2 has the y-axis as a line of symmetry,so if you were to fold the graph along the y-axis, the twohalves of the graph would match up. The graph of y = x3

has rotational symmetry. If you rotate the graph ofy = x3 180◦ around the origin, the rotated image willcoincide with the original graph.

The graph of y = x2 passes through quadrants I and II,and the graph of y = x3 passes through quadrants Iand III.

When you zoom in near the origin, the two graphs bothbegin to look the same for positive x-values. Away fromthe origin, it’s evident that the y-values of the graph ofy = x3 are negative for negative x-values, but that thegraph of y = x2 only has positive y-values. In quadrant I,you can see that the graph of y = x3 is steeper than thegraph of y = x2 away from the origin.

6.

2 4 6 8 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 159

7. (a) This is the graph of y = √x shifted 5 units down.

2 4 6 8 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(b) This is the graph of y = √x shifted 5 units to the left.

–4 –2 2 4 6 8 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(c) This is the graph of y = √x shifted 5 units up.

2 4 6 8 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(d) This is the graph of y = √x shifted 3 units up and

2 units to the left.

–2 2 4 6 8 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

8.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

9. (a) This is the graph of y = |x| shifted 5 units down.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 160

(b) This is the graph of y = |x| shifted 5 units to the left.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(c) This is the graph of y = |x| shifted 5 units up.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(d) This is the graph of y = |x| shifted 3 units up and 2units to the left.

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

10. (a)

�2

�2

2 4 6

2

4

6

O

yy �3x

y �log3 x

x

(b) The graphs are reflections of each other over the linewith equation y = x. If you were to fold the paperalong the line with equation y = x, the graph ofy = 3x would match up with the graph ofy = log3 x. This property is characteristic offunctions that are inverses of each other.

11. Only one linear equation passes through two points. Inthis example, the linear equation is y = 0. However,infinitely many quadratic equations, cubic equations andother polynomials are possible. One quadratic equationis y = (x + 3)(x − 6) and one cubic equation isy = x(x + 3)(x − 6). You can multiply the right sideof either equation by any polynomial and it will stillpass through these two points. For example,y = 2(x + 3)(x − 6) and y = (x4 + x) (x + 3)(x − 6)

still pass through these points.

On Your Own12. (a), (c)

(b) • Connect endpoints (0, 0) and (−3, −3).• Connect endpoints (−2, −2) and (−3, −1).• Connect endpoints (−3, −3) and (−5, −1).

(d) If you were to rotate F 180◦ around the origin,you’d get G. Or you might also see that if youwere to reflect F over the y-axis and G overthe x-axis, the two reflected images wouldcoincide.

13. It has 180◦ rotational symmetry. If you were to turn thepaper upside down, the graph of O would look the sameas it does when the paper is right side up.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 161

14. (a), (c)

(b) • Connect endpoints (0, 0) and (−3, 3).• Connect endpoints (−2, 2) and (−3, 1).• Connect endpoints (−3, 3) and (−5, 1).

(d) If you were to reflect F over the y-axis, you’dget H .

15. It has the y-axis as a line of symmetry. If you were to foldthe paper along the y-axis, the two halves of figure E

would coincide.16. (a)

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(b)

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(c)

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(d)

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

17. The value of C shifts the graph vertically. When C ispositive, the graph shifts up; when C is negative, thegraph shifts down. This makes sense because, since C isjust a constant, it adds the same number to every y-value.Thus, every y-value will be 3 more if C is 3 and 4 less ifC is −4.

18. (a)

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 162

(b)

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(c)

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

(d)

–10 –5 5 10

–10

–7.5

–5

–2.5

2.5

5

7.5

10

19. The parameter C has the same effect on the graph ofy = |x| as on the graph of y = x2. C simply shifts thegraph vertically. When C is positive, the graph shifts up.When C is negative, the graph shifts down.

20. (a) This is the graph of y = 3x shifted 5 units down.

2

�2

�2 Ox

y

(b) This is the graph of y = 3x shifted 5 units left.

x

y

�2�4�6

2

4

6

O

(c) This is the graph of y = 3x shifted 5 units up.

2

4

6

8

10

�2 2Ox

y

(d) This is the graph of y = 3x shifted 2 units left and3 units up.

21. (a)y

�2 2

2

O x

(b) The graph of y = 1x

can never intersect the y-axis,because division by 0 is undefined. To see why it cannever intersect the x-axis, assume that it does and seewhat happens. If the graph does intersect the x-axis,


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 163

then 1x

is equal to 0 for some value of x, say x = a.This a cannot be equal to 0, because division by zerois undefined. That means you can multiply both sidesof the equation by a in order to try to solve it. Thatwould give you 1 = 0, which is impossible. So thegraph cannot intersect the x-axis either.

22. (a) This is the graph of y = 1x

shifted 5 units down.

y�2

�2

�4

2Ox

(b) This is the graph of y = 1x

shifted 5 units left.

y

�2�4

�2

2

O x

(c) This is the graph of y = 1x

shifted 5 units up.

�2 2

4

6

x

y

(d) This is the graph of y = 1x

shifted 2 units left and3 units up.

y

�4

2

6

Ox

23.

O 2

2

�2

�2

x

y

Maintain Your Skills24. (a)

5x − 7 = 2x + 53x = 12x = 4

(b)

5 · x

3− 7 = 2 · x

3+ 5

5 · x

3− 2 · x

3= 12

3x

3= 12

x = 12

(c)

5 · x

10− 7 = 2 · x

10+ 5

5 · x

10− 2 · x

10= 12

3x

10= 12

3x = 120x = 40

(d)

5 · x

100− 7 = 2 · x

100+ 5

5 · x

100− 2 · x

100= 12

3x

100= 12

3x = 1200x = 400

25. If the solution to the original equation is x = s, thenwhen x is replaced by x

Cto make a new equation, the new

solution will be x = Cs.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 164

4.15 More Basic Graphs


1. (a) First, you’ll want to find the y-coordinate that goeswith the x-coordinate 4

5 .

(4

5

)2

+ y2 = 1

16

25+ y2 = 25

25

y2 = 9

25

y = ±3

5

This means that the points(

45 , 3

5

)and

(45 , − 3

5

)are

both on the unit circle.(b)

(− 45 , 3

5

),(− 4

5 , − 35

),(

35 , 4

5

),(

35 , − 4

5

),(− 3

5 , 45

), and(− 3

5 , − 45

)are all on the unit circle.

2. (a) Use the equation to find the x-coordinate that goeswith this y-coordinate.

(5

13

)2

+ x2 = 1

25

169+ x2 = 169

169

x2 = 144

169

x = ±12

13

So(

1213 , 5

13

)and

(− 1213 , 5

13

)are both on the unit circle.

(b) Use the equation to find the y-coordinate that goeswith this x-coordinate.

(8

17

)2

+ y2 = 1

64

289+ y2 = 289

289

y2 = 225

289

y = ±15

17

So(

817 , 15

17

)and

(817 , − 15

17

)are both on the unit circle.

3. Mariko noticed several patterns in the points she foundfor the unit circle. All the coordinates were fractions, andthe x and y-coordinates for any point had the samedenominator. That denominator was the largest of thethree numbers in a Pythagorean triple, and thenumerators were the other two numbers in the triple. Thereason this is a sensible way to find coordinates for theunit circle is that Pythagorean triples are numbers a, b,and c such that a2 + b2 = c2. If you divide each side ofthat equation by c2, you get

a2

c2+ b2

c2= 1.

That equation can be rewritten as

(a

c

)2 +(

b

c

)2

= 1.

That result shows you that the point(

ac, b

c

)satisfies the

equation for the unit circle, x2 + y2 = 1.4. (a) f (−x) = (−x)2 = x2 = f (x).

(b) The graph of y = x2 has the y-axis as a line ofsymmetry, so it is an even function.

5. (a)

f (−x) = (−x)3 + (−x)

= −x3 − x

= −(x3 + x)

= −f (x)

(b) The graph of f has 180◦ rotational symmetry. Thatmeans that if you were to replace each point (x, y)

by the point (−x, −y), you’d get the same graph.6. The following functions are even.

• y = x2—See Exercise 4.• y = |x|—If f (x) = |x|, then f (−x) = | − x| =

|x| = f (x).

The following functions are odd.

• y = x—If f (x) = x, then f (−x) = −x = −f (x)

• y = 1x

—If f (x) = 1x

, then f (−x) = 1−x

= − 1x= −f (x)

• y = x3—If f (x) = x3, then f (−x) = (−x)3 =−x3 = −f (x)

• y = x3 − x—If f (x) = x3 − x, then f (−x) =(−x)3 − (−x) = −x3 + x = −(x3 − x) = −f (x)

• y = x3 + x—See Exercise 5.

The following functions are neither even nor odd.

• y = √x—If f (x) = √

x, then f (−x) = √−x isonly defined for x ≤ 0. It isn’t equal to f (x) or−f (x).

• y = bx(b > 0, b �= 1)—If f (x) = bx , thenf (−x) = b−x = 1

bx . It isn’t equal to f (x) or −f (x).• y = logbx(b > 0, b �= 1)—If f (x) = logbx, then

f (−x) = logb(−x) is only defined for x < 0. It isn’tequal to f (x) or −f (x).

The graph of x2 + y2 = 1 is not the graph of a function,so it is neither even nor odd.

7. To find the x-intercepts for the graph, set y = 0. That’sbecause you want to find all of the points on the x-axis(which has equation y = 0) that satisfy the graph’sequation. You’ll get an equation that you can solve byfactoring.

0 = x3 + x

0 = x(x2 + 1)

x = 0 is the only real-number solution to the equation. (iand −i are two complex solutions.)


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 165

8. (a)x x − 3 (x − 3)2

−1 −4 160 −3 91 −2 42 −1 13 0 04 1 15 2 46 3 97 4 16

(b)

2

2

Ox

y

4

6

8

4 6

(c) The graph of y = (x − 3)2 is the same as the graph ofy = x2 shifted 3 units to the right

On Your Own9. (a)

g(−x) = (−x)3 − (−x)

= −x3 + x

= −(x3 − x)

= −g(x)

(b) The graph of y = x3 − x has 180◦ rotationalsymmetry. That means that if you were to replaceeach point (x, y) by the point (−x, −y), you’d getthe same graph.

10. (a) Near the origin, the x3-term in the equationy = x3 − x is much smaller (in absolute value) thanthe −x-term, so it’s negligible in comparison. It hasmuch less effect on the size of the y-coordinate.Look at x = 0.001. In that case, y = 0.000000001 −0.001 = −0.000999999 ≈ −0.001.

(b) Away from the origin, the x3-term in the equationy = x3 − x is much larger (in absolute value) thanthe −x-term, so the −x-term is negligible incomparison to the x3 term. Look at x = 1000. In thatcase, y = 1,000,000,000 − 1000 = 999,999,000 ≈1,000,000,000

11. To find the x-intercepts, set y = 0 and solve for x byfactoring.

0 = x3 − x

0 = x(x2 − 1)

0 = x(x − 1)(x + 1)

x = 0, 1, −1

There are three x-intercepts: (−1, 0), (0, 0), and (1, 0).

12. (a)

2

�2

�4

Ox

y

2

4

4

(b)

0 = (x − 3)3 − (x − 3)

= (x − 3)[(x − 3)2 − 1]= (x − 3)[(x − 3) − 1][(x − 3) + 1]= (x − 3)(x − 4)(x − 2)

x = 3, 4, 2

The x-intercepts are (2, 0), (3, 0), and (4, 0)

(c) The graph of y = (x − 3)3 − (x − 3) is the same asthe graph of y = x3 − x shifted 3 units to the right.

13. Start with the distance formula,√(x − 0)2 + (y − 0)2 = 5

Square both sides to get an equation for the circle.

x2 + y2 = 25

14. Start with the distance formula,√(x − 3)2 + (y − 0)2 = 5

Square both sides to get an equation for the circle.

(x − 3)2 + y2 = 25

15. (a) (x − 6)2 + y2 = 25(b) (x + 3)2 + y2 = 16(c) x2 + (y + 4)2 = 1(d) (x − 3)2 + (y + 4)2 = 1(e) (x + 2)2 + (y − 3)2 = 9(f) (x − a)2 + (y − b)2 = r2

16.y

�1

�1

1

1Ox

This is the same as the graph of x2 + y2 = 1 afterdilating by a factor of 1

3 in the x-direction. This kind ofgraph is called an ellipse.

17. Whenever you replace x by x − c, where c is a positiveconstant, the graph shifts to the right by c units. Thecorrect answer choice is D.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 166

Maintain Your Skills18. (a) You can solve x2 + 5x − 14 = 0 by factoring.

x2 + 5x − 14 = 0(x + 7)(x − 2) = 0

x = −7, 2

(b) If you substitute M = x − 3 into the equation(x − 3)2 + 5(x − 3) − 14 = 0, you will getM2 + 5M − 14 = 0. From the previous part, youknow then that M = −7, 2. This means thatx − 3 = −7, 2 and x = −4, 5.

(c) If you substitute N = x − 5 into the equation(x − 5)2 + 5(x − 5) − 14 = 0, you will getN2 + 5N − 14 = 0. From part (a), you know thenthat N = −7, 2. This means that x − 5 = −7, 2 andx = −2, 7.

(d) If you substitute P = x + 2 into the equation(x + 2)2 + 5(x + 2) − 14 = 0, you will getP 2 + 5P − 14 = 0. From part (a), you know then thatP = −7, 2. This means that x + 2 = −7, 2 andx = −9, 0.

19. If the solutions to the original equation are a, b, c, . . . ,the solutions to the equation where the variable x hasbeen replaced by x − C will be a + C, b + C, c + C, . . . .

4.16 Translating Graphs

Check Your Understanding1. (a)

x = M − 5 M y = √M

−5 0 0−4 1 1−1 4 2

4 9 311 16 420 25 5

(b) The graph of y = √x + 5 is the same as the graph of

y = √x shifted 5 units to the left.

(c) The domain is x ≥ −5 and the range is y ≥ 0.

2�2�4 Ox

y4

2. (a)

�2�4

3

O

y

x

(b)

2�2

�2

�4

�4�6 Ox

y

4

2

(c)

4 62Ox

y4

2

(d)

2�2x

2

4

6

3. (a)

�4

2

O

y

x

(b)

2�2�6x

y

8

6

4

2

(c)

8 104 62Ox

y4

2


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 167

(d)

4x

y

O

2

4

6

4. You should disagree with Walter. He did not replace everyinstance of x in the equation of the basic graph withx − 2. If he were to graph his equation, his graph wouldbe shifted two units below the graph he actually wanted.

5. (a) The graph of y = (x − 2)3 + 6(x − 2)2+11(x − 2) + 7 will be the same as the graph ofy = x3 + 6x2 + 11x + 7 shifted two units to the right.

(b)(x − 2)3 + 6(x − 2)2 + 11(x − 2) + 7

= x3 + 3(x2)(−2) + 3x(4) − 8+ 6(x2 − 4x + 4) + 11x − 22 + 7

= x3 − 6x2 + 12x − 8 + 6x2 − 24x + 24+ 11x − 22 + 7

= x3 − x + 1(c) Start with the basic graph for the equation

y = x3 − x. Shift that graph up 1 unit, and you havethe graph of y = x3 − x + 1 which is also the graphof y = (x − 2)3 + 6(x − 2)2 + 11(x − 2) + 7. Thegraph of y = x3 + 6x2 + 11x + 7 is the same as thisone, but shifted two units to the left. Here’s whatyour graph should look like.

�2

�2�4

�4

Ox

y

6. (a) The graph of y = (x + 3)2 is the same as thegraph of y = x2 shifted 3 units to the left, so itsvertex is (−3, 0). (That’s the same as the vertex ofthe parabola with equation y = x2 shifted 3 units tothe left.)

(b) The graph of y − 2 = (x + 3)2 is the same as thegraph of y = x2 shifted 3 units to the left and 2 unitsup, so its vertex is (−3, 2).

(c) The graph of y = (x + h)2 + k is the same as thegraph of y = x2 shifted h units to the left and k unitsup, so its vertex is (−h, k).

7. (a) To sketch the graph of y = x2 + 6x + 9, it will helpto notice that the quadratic can be factored, and thatthis equation can be rewritten as y = (x + 3)2. That

means that its graph will be the same as the graph ofy = x2 shifted 3 units to the left.

O�4 �2

2

4

x

y

(b) The graph of y = x2 + 6x + 7 must be the same asthe graph of y = x2 + 6x + 9 shifted 2 units down.

O�5 �3�2

2

x

y

On Your Own8. (a) To graph y = (x − 2)3 + (x − 2), translate the basic

graph of y = x3 + x 2 units to the right.

O 2 4�2

�4

4

2

x

y

(b) To graph y + 1 = (x − 2)3 + (x − 2), translate thegraph from part (a) 1 unit down. (You could alsothink of translating the basic graph of y = x3 + x

2 units to the right and 1 unit down.)

O 2 4�2

�4

2

x

y

(c) y = (x − 2)3 + x − 3 is just a simplified form ofy + 1 = (x − 2)3 + (x − 2), so its graph is the sameas in part (b).

O 2 4�2

�4

2

x

y


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 168

(d) y = x3 − 6x2 + 13x − 11 is the expanded form ofy = (x − 2)3 + x − 3, so its graph is the same as inparts (b) and (c).

O 2 4�2

�4

2

x

y

9. (a) The slope is 3.

O 2�2

2

x

y

(b) The slope is 3.

O 3

�2

2

x

y

(c) The slope is 3.

O 2

2

4

x

y

(d) All of the graphs are lines, and they’re all parallel toeach other with slope 3. The graph in part (b) is thesame as the graph in part (a) shifted 2 units to theright. The graph in part (c) is the same as the graph inpart (b) shifted 5 units up or the graph in part (a)shifted 2 units to the right and 5 units up.

10. (a)

2

2

�2

�2x

y

(b) y = 12 (x − 3)

(c) y = 12 (x − 3) + 1

11. (a) The line through the origin of slope 4 has equationy = 4x. If you translate that line 3 units to the right

and 1 unit up, it will pass through the point (3, 1).That means its equation must be y = 4(x − 3) + 1 ory = 4x − 11.

(b) The line through the origin of slope 23 has equation

y = 23x. If you translate that line 2 units to the left

and 1 unit down, it will pass through the point(−2, −1). That means its equation must bey = 2

3 (x + 2) − 1 or y = 23x + 1

3 (or even3y − 2x = 1).

(c) The line through the origin of slope m has equationy = mx. If you translate that line h units to the rightand k units up, it will pass through the point (h, k).That means its equation must bey = m(x − h) + k.

12. (a) The graph of y = (x − 3)2 + 6(x − 3) + 7 will be thesame as the graph of y = x2 + 6x + 7 shifted 3 unitsto the right. That’s because every instance of x in theequation y = x2 + 6x + 7 has been replaced byM = x − 3. So if you’re looking at a graph ofy = M2 + 6M + 7 and want to use it to graphy = (x − 3)2 + 6(x − 3) + 7, you’ll end up shiftingeach point 3 units to the right because x = M + 3.

(b) The graph of y − 2 = (x − 3)2 + 6(x − 3) + 7 will bethe same as the graph of y = (x − 3)2 + 6(x − 3) + 7shifted 2 units up. The effect of replacing y in theequation y = (x − 3)2 + 6(x − 3) + 7 by N = y − 2means that if you’re looking at a graph ofN = (x − 3)2 + 6(x − 3) + 7 and want to use it tograph y − 2 = (x − 3)2 + 6(x − 3) + 7, you’ll endup shifting each point 2 units up because y = N + 2.

(c)y − 2 = (x − 3)2 + 6(x − 3) + 7

y = x2 − 6x + 9 + 6x − 18 + 7 + 2y = x2

13. The two equations y − 1 = 4(x − 3) andy + 3 = 4(x − 2) are really the same. To see this, solveeach equation for y. They both simplify to y = 4x − 11,so they are the same, and Susan is incorrect in thinkingthat there are two different lines that meet therequirements.

14. In Derman’s way of thinking, the original basic graph ofthe parabola, in the picture below, is the graph of y = M2.The horizontal axis is the M-axis, and the line withequation M = 0 is the y-axis with respect to that M-axis.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 169

When he thinks about moving the coordinate axes, he’spicturing the relationship M = x − 3 in this way:

By looking at the corresponding values of M (above thenumber line) and x (below the number line), you can seethat the M-value is always 3 less than the x-value, soM = x − 3. You can see that the x-axis is the image ofthe M-axis after a translation of three units left.Now the y-axis with respect to the x-axis is in a differentplace. The y-axis should now be the line with equationx = 0. The old y-axis (with respect to the M-axis) nowhas equation x = 3. It’s still useful to know the equationof this line, because it’s the line of symmetry for theparabola in the new coordinate system.

15. Start with y = x2 and translate this 1 unit to the left and 2units down. The new equation corresponding to this isy + 2 = (x + 1)2. The correct answer choice is D.

Maintain Your Skills16. (a) The two graphs are reflections of each other over the

line with equation y = x. This is because thefunctions 3x and log3x are inverses of each other.

�2

�2

2 4 6

2

4

6

O

yy �3x

y �log3 x

x

(b) The graph of y = 3 · 3x = 3x+1 is a translation of thegraph of y = 3x one unit to the left. The graph ofy = log3

(x3

) = log3x − 1 is a translation of thegraph of y = log3x one unit down. The two newgraphs are reflections of each other over the line withequation y = x. This is because the functions 3 · 3x

and log3(

x3

)are inverses of each other.

�2

4 6

4

6

O

y =3.3x

y =log3 x3

x�2

(c) The graph of y = 9 · 3x = 3x+2 is a translation of thegraph of y = 3x two units to the left. The graph ofy = log3

(x9

) = log3x − 2 is a translation of the graphof y = log3x two units down. The two new graphs arereflections of each other over the line with equationy = x. This is because the functions y = 9 · 3x andy = log3

(x9

)are inverses of each other.

�2�4

�2

�4

2

2

O

y =9.3x

y =log3 x9

x

y

(d) The graph of y = 27 · 3x = 3x+3 is a translation ofthe graph of y = 3x three units to the left. The graphof y = log3

(x27

) = log3x − 3 is a translation of thegraph of y = log3x three units down. The two newgraphs are reflections of each other over the line withequation y = x. This is because the functionsy = 27 · 3x and y = log3

(x27

)are inverses of each

other.

�2�4

�2

�4

2 4 6

2

4

6

O

y �27.3x

y �log3

x

y

x27

(e) The graph of y = 3n · 3x = 3x+n is a translation ofthe graph of y = 3x n units to the left. The graph ofy = log3

(x3n

) = log3x − n is a translation of thegraph of y = log3x n units down. The two newgraphs are reflections of each other over the line withequation y = x. This is because the functionsy = 3n · 3x and y = log3

(x3n

)are inverses of each

other.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 170

4.17 Scaling and Reflecting Graphs

Check Your Understanding1. (a) Complete the following table.

x N = x2 + 1 y = N5

−2 5 1−1 2 2

5

0 1 15

1 2 25

2 5 13 10 2

(b) The y-values for the graph of 5y = x2 + 1 are smallerby a factor of 5 when compared to the correspondingy-values for the graph of y = x2 + 1. This means thatthe graph of 5y = x2 + 1 is the same as the graph ofy = x2 + 1 after that graph has been dilated verticallyby a factor of 1

5 . Both of the graphs will be parabolas,and they’ll have the y-axis as their line of symmetry.

(c)

Ox

2

�2 2

�2

y

2. (a)

Ox

4

�4 4

�4

y

(b)

O�4 4

2

y

x

(c)

O�4

4

2

y

x

(d)

O

�4

�4

4

4

y

x

(e)

O�4

�8

�12

�16

�4�8

4

8

12

16

4 8

y

x

3. (a) The domain is x ≥ 0, the range is y ≤ 0.

O

�2

2 4 6 8

yx

(b) The domain is x ≥ 0, the range is y ≤ 0.

O�2

�4

�6

�8

2 4 6 8y x

(c) The domain is x ≤ 0, the range is y ≥ 0.

O�2

2

x

y


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 171

(d) The domain is x ≤ 1, the range is y ≥ 0.

O�2�4�6

2

x

y

4. To dilate a graph by a factor of 2, replace every instanceof x in the equation for its basic graph with 1

2x. In thiscase, that would give you y = ( 1

2x)2.

Then to translate a graph 3 units up, replace everyinstance of y in the equation by y − 3. This gives youy − 3 = ( 1

2x)2, or (to graph it more easily on a graphing

calculator) y = ( 12x)2 + 3.

To translate a graph 4 units to the right, replace everyinstance of x in the equation by x − 4. This gives youy = ( 1

2 (x − 4))2 + 3, or y = ( 1

2x − 2)2 + 3.

5. (a)

O�2 2

2

4

x

y

(b)

�2 2

2

O

y

x

(c)

2

�2

�2 2

y

xO

6. Both Tony and Sasha are right. A function is even iff (−x) = f (x) for all x in its domain. However, it’s alsotrue that if you map every point (x, y) to the point(−x, y), the effect of such a mapping is to reflect thefigure over the y-axis. If a function is even, that reflectionwould map the graph of the function onto itself.

7. A function is even if f (−x) = f (x) for all x in thedomain of the function.

y = (−x)8 + 37(−x)6 − 71(−x)2 + 4= x8 + 37x6 − 71x2 + 4

8. Reflecting a function over the y-axis is the same asmapping each point (x, y) in a graph to the point(−x, y). If you do that mapping on the function g(x),you’ll get the same graph because g(−x) = f (−x)+f (− − x) = f (x) + f (−x) = g(x). In other words,g(x) is an even function.

9. (a) (b)

4

�2 4

y

xO

x3 � �

y7 1

x5 � �

y3 1

x � y � 1

(c) One way to find points is to set one of the variablesequal to 0 and solve for the other. If x = 0, then0 + y

12 = 1, and you have the point (0, 12). If y = 0,then x

−17 + 0 = 1, and you have thepoint (−17, 0).

(d) Because you can read the coordinates of thex-intercept and the y-intercept directly from theequation, (a, 0) is the x-intercept and (0, b) is they-intercept.

On Your Own10. (a) Notice that this is the basic graph of y = x3 + x after

the mapping (x, y) �→ (−x, y). The effect of such amapping is to reflect the graph over the y-axis.

�2

2

4

�4

�2

2

y

xO

(b) Notice that this is the basic graph of y = x3 − x

after the mapping (x, y) �→ (−x, y). The effect ofsuch a mapping is to reflect the graph over the y-axis.

�2

2

4

�4

�2

2

y

xO

(c) This is the same as the graph in part (a).

�2

2

4

�4

�2

2

y

xO


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 172

(d) This is the same as the graph in part (b).

�2

2

4

�4

�2

2

y

xO

(e) This is the basic graph of y = x3 + x after themapping (x, y) �→ (−x, −y). The effect of such amapping is to reflect the graph over the y-axis andthen over the x-axis. These two reflections, whencomposed, are equivalent to a rotation of 180◦ aroundthe origin. If you look at this graph upside down, itwill look the same as the basic graph does rightside up.

�2

2

4

�4

2

y

x

(f) This is the basic graph of y = x3 − x after themapping (x, y) �→ (−x, −y). As in part (e), theeffect of such a mapping is to reflect the graph overthe y-axis and then over the x-axis. These tworeflections, when composed, are equivalent to arotation of 180◦ around the origin. If you look at thisgraph upside down, it will look the same as the basicgraph does right side up.

�2

2

4

�4

�2

2

y

xO

11. (a) The graph of y = 1−x

is the same as the graph ofy = 1

xafter it’s reflected over the y-axis.

(b) The graph of y = 1−4x


xafter it’s reflected over the y-axis and has been

dilated horizontally by a factor of 14 .

(c) The graph of y = 4−x



dilated vertically by a factor of 4.(d) The graph of −y = 1

xis the same as the graph of

y = 1x

after it’s reflected over the x-axis. It looks thesame as the graph in part (a), because for this function,f (−x) = −f (x). (It’s an odd function.)

(e) The graph of −4y = 1x



dilated horizontally by a factor of 14 . It’s the same as

the graph in part (b).

(f) The graph of − y

4 = 1x



dilated vertically by a factor of 4. It’s the same as thegraph in part (c).

(g) The graph of −y = 1−x


x. Since the two reflections, shown separately in

parts (a) and (d) had the same effect, when youcompose the two reflections, they will undo eachother. (A reflection is its own inverse.)

(h) The graph of − y

4 = 1−4x


x. The transformations in parts (b) and (f) are

inverses of each other. The shrinking in part (b) isundone by the stretching in part (f), and reflection overthe y-axis is its own inverse.

12. All three of them are right. the mapping (x, y) �→(−x, −y) is what Tony describes. This is the mappingthat results from the composition of a reflection over they-axis (which maps x to −x) and a reflection over thex-axis (which maps y to −y). This composition of tworeflections is equivalent to a 180◦ rotation around theorigin.

13. The graph of (x + 3)2 + (y − 4)2 = 1 is a circle of radius1 with center (−3, 4). The graph of (−x + 3)2+(−y − 4)2 = 1 is a circle of radius 1 with center (3, −4).It’s also the same as the equation (x + 3)2 + (y − 4)2 = 1under the mapping (x, y) �→ (−x, −y), which isequivalent to a reflection over the x-axis and then overthe y-axis (or vice versa) or a rotation of 180◦ around theorigin. These interpretations are both equivalent, becausethe mapping (x, y) �→ (−x, −y) sends the center ofthe first circle to the center of the second, and doesn’thave an effect on the radius.

14. (a) This is an ellipse. You can think of it as a circle ofradius 6, centered at the origin, that has been dilatedby a factor of 1

3 horizontally and by a factor of 12

vertically.

O�2

�3

2

3

y

x

(b) This is the graph from part (a) after a translation of 1unit to the left and 4 units up. (The ellipse is nowcentered at (−1, 4).

O

2

4

�2

6

2

y

x


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 173

(c) This is the same as the graph from part (b). You canuse the algebraic simplification to transform one ofthese equations into the other.

O

2

4

�2

6

2

y

x

15. Reflecting over the y-axis corresponds to replacing x by−x. The correct answer choice is C.

16. Yes, the two ways of visualizing the situation areconsistent. If you think of the line as “more steep,” you’resaying that for every one-unit increase in an x-coordinateyou get a larger increase in the correspondingy-coordinates. Successive y-values in a table are fartherapart. This is exactly what happens in a verticalstretch—y-values are farther apart.

17. (a) The graph of y = (x − 1)3 + 3(x − 1)2 + 7(x − 1) +13 is the same as the graph of y = x3 +3x2 +7x +13translated one unit to the right.

(b)

(x − 1)3 + 3(x − 1)2 + 7(x − 1) + 13= x3 − 3x2 + 3x − 1 + 3x2

− 6x + 3 + 7x − 7 + 13= x3 + 4x + 8

(c) The graph of 18y = ( 1

2x)3 + ( 1

2x)+ 1 is the same as

the graph of y = x3 + x + 1 after it has been dilatedvertically by a factor of 8 and dilated horizontally bya factor of 2.

(d) This is the same as the basic graph of y = x3 + x

translated up one unit.

�2

2

4

�2

2

y

xO

(e) Think of going backwards along the trail establishedin parts (a) through (d) and undoing eachtransformation.

Start with a sketch of the basic graph of y =x3 + x. Translate it up one unit. Dilate it by a factorof 1

2 horizontally and a factor of 8 vertically. Then

translate it one unit to the left, and what you’ll have isthe graph of y = x3 + 3x2 + 7x + 13.

O

4

8

12

�3 2

y

x

18. In Derman’s way of thinking, the original basic graph ofthe parabola, drawn using the axes in the picture below, isthe graph of y = M2. The horizontal axis is the M-axis,and the line with equation M = 0 is the y-axis withrespect to that M-axis.

When he thinks about moving the coordinate axes, he’spicturing the relationship M = x

3 in this way:

By looking at the corresponding values of M (above thenumber line) and x (below the number line), you cansee that the M-value is always one-third of the x-value,so M = x

3 .The strange thing here is that the parabola does not

look any different, even though it is been dilatedhorizontally by a factor of 3.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 174

That is because in the new coordinate system, the x andy-axes are not on the same scale. This really affects thepicture. If you rescale the x-axis so that one unit in x isequal to one unit in y, the graph does look different.

Maintain Your Skills19. (a) Even

(b) Odd(c) Even(d) Odd(e) Even(f) Odd(g) f (x) = xn is even if n = 2k for some integer k and

odd if n = 2k + 1 for some integer k.

4D MATHEMATICAL REFLECTIONS

1. (a) A function is even if f (−x) = f (x) for all x in itsdomain. y = x2 and y = |x| are both even functions.

The graphs of even functions have the y-axis as aline of symmetry (left and right halves are reflectionsof each other).

(b) A function is odd if f (−x) = −f (x) for all x in itsdomain. y = x and y = x3 are both odd functions.

The graphs of odd functions have “rotationalsymmetry”. If you rotate it 180◦, you obtain the samegraph.

2. This graph is the same as the graph of y = |x| after atranslation of 2 units to the right and 3 units up.

O

2

4

2 4

y

x

3. y = (x + 3)3 + 14. The graph of x2 + y2 = 25 is a circle of radius 5 centered

at the origin. The graph of(

x2

)2 + (3y)2 = 25 is the sameas the graph of x2 + y2 = 25 after it has been dilated

horizontally by a factor of 2 and dilated vertically by afactor of 1

3 .

�3

3y

O 4�4x

5. y = 6x

6. The graph of y = (x − 3)2 is the same as the graph ofy = x2 after its been translated 3 units to the right.

7. The graph of (x + 1)2 + (y − 4)2 = 36 is a circle ofradius 6 centered at the point (−1, 4).

8. The graph of −2y = x3 − x is the same as the graph ofy = x3 − x after it has been reflected over the x-axis anddilated vertically by a factor of 1

2 .

CHAPTER REVIEW

1. Each entry is 5 times the previous one, so a recursiverule is

a(n) ={

1 if n = 0a(n − 1) · 5 if n ≥ 1

You can rewrite each entry as a power of 5, so a closedform is a(n) = 5n. Therefore,a(15) = 515 = 30, 517, 578, 125.

2.

0

1

2

3

4

5

1

3

8

16

27

41

2

5

8

11

14

3

3

3

3

x b(x) 2

Since the second differences are constant, a quadraticfunction will agree with the table.

3.

0 –3

1

2

3

4

4

15

30

49

7

11

15

19

4

4

4

x c(x) 2

The second differences are constant, so this can bematched by a quadratic function. Since the constant is 4,the leading coefficient is 2. The values of c(x) − 2x2 are−3, 2, 7, 12, 17. This is matched by the linear function5x − 3. Hence, c(x) = 2x2 + 5x − 3.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 175

4. (a) f (2) = 3(2) + 7 = 6 + 7 = 13(b) g(2) = 2+1

3 = 1(c) f (g(2)) = f (1) = 3(1) + 7 = 3 + 7 = 10(d) g(f (2)) = g(13) = 13+1

3 = 143

(e) f ◦ g(2) = f (g(2)) = 10(f) f (2) · g(2) = 13 · 1 = 13

5. (a) f ◦ g(x) = f (g(x)) = f (x2 + 4x + 1) =(x2 + 4x + 1) − 2 = x2 + 4x − 1

(b) g ◦ f (x) = g(x − 2) = (x − 2)2 + 4(x − 2) + 1 =x2 − 4x + 4 + 4x − 8 + 1 = x2 − 3

6. (a)

O4

4

4

y

x

O2 2

4

2

x

y

4

22

2

4

Ox

y

(b) f and h are one-to-one. g is not one-to-one.(c) To find f −1(x), rewrite the equation as

x = f −1(x)

2 + 3, and solve for f −1(x):

x = f −1(x)

2+ 3

x − 3 = f −1(x)

22(x − 3) = f −1(x)

2x − 6 = f −1(x)

Similarly, x = −3h−1(x) −→ h−1(x) = − x3 .

7. (a) You are looking for an equation in the formf (x) = a · bx . Write f (2) = a · b2 = 9

2 andf (3) = a · b3 = 27

2 . Solve each equation for a:

a =92b2

a =272b3

So,

92b2 =

272b3

92

· b3 = 272

· b2

9b3 = 27b2

9b3 − 27b2 = 09b2(b − 3) = 0

So b = 0 or b = 3. But b > 0, so b = 3. Substitute tofind a:

f (2) = a · 32 = 92

→ 9a = 92

→ a = 12

Therefore f (x) = 12 (3)x .

(b)

O 22

6

4

2

8

10

12

14

x

y

4

(0, 1 )

(2, 9 )

(3, 27)

2

2

2

8. (a)

x g(x) ÷0 −2 −20

−2 = 101 −20 −200

−20 = 102 −200 −2000

−200 = 103 −2000 10

To find the next term, you multiply by 10. So,

g(x) ={−2 if x = 0

10 · g(x − 1) if x > 0

(b) A closed-form definition will be in the formg(x) = a · bx. g(0) = −2 = a · b0. So, a = −2.Substitute to find b : g(1) = −20 = −2 · b1 → b =10. The closed-form definition is

g(x) = −2(10)x

(c) g(4) = −2(10)4 = −2(10,000) = −20,000 andg(6) = −2(10)6 = −2(1,000,000) = −2,000,000.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 176

9. (a)

O�4 �2 2

4

2x

y

The graph of y − 2 = |x + 1| is the same as thegraph of y = |x| after a translation 1 unit to the leftand 2 units up.

(b)

2 4 6

�2

�4

�6

�8

O

yx

The graph of (x − 4)2 + (y + 6)2 = 4 is the same asthe graph of x2 + y2 = 4 after a translation 4 units tothe right and 6 units down.

10. (a)y

x2

2

4

6

4 6O

The graph ofy

2= (x − 4)2 is the same as the graph

of y = x2 after a vertical stretch by a factor of 2 and atranslation 4 units to the right.

(b)

�2

2y

O 62�2�6x

The graph of(x

3

)2 + (2y)2 = 4 is the same as thegraph of x2 + y2 = 4 after a vertical shrink by a factorof 2 and a horizontal stretch by a factor of 3.

(c)

�2

2

4

�4

�2

2

y

xO

The graph of y = (−x)3 − (−x) is the same as thegraph of y = x3 − x after a reflection over the y axis.

11. Replace x withx

2for a horizontal stretch of factor 2:

y = | x2 |. To translate the graph 3 units left, replace x with

x + 3: y = | x+32 |. To translate the graph 7 units down,

replace y with y + 7. The final equation is

y + 7 = |x + 32

|

O�8 �6 �4 �2

�2

�4

2

yx

CHAPTER TEST

Multiple Choice1. For a translation of the graph 2 units to the right, replace

x with x − 2. For a translation of the graph 7 unitsdown, replace y with y + 7. The result is (x − 2)2

+(y + 7)2 = 4. The correct choice is C.2. For a horizontal stretch by a factor of 4, replace x

withx

4:

y = 1x4

= 4x

The correct choice is A.3. Choices A and C each give a correct amount for a(1),

but a(2) is incorrect for both. Choice D gives anincorrect amount for a(1). The correct answer choice isB.

4. Since it has constant first differences, it is linear. Theslope is 24−(−3)

4−1 = 273 = 9. Using the first given point,

b(n) − (−3) = 9(n − 1), or b(n) + 3 = 9n − 9, sob(n) = 9n − 12. Hence, b(20) = 180 − 12 = 168. Thecorrect answer choice is A.

5. f ◦ g(3) = f (g(3)) = f (4(3) + 1) = f (13) =2(13) − 3 = 26 − 3 = 23. So, the correct choice is C.

6. To find the inverse, rewrite the equation as:

f (f −1(x)) = x = 3√

f −1(x) + 1

To make the work simpler, replace f −1(x) with y. Solvefor y:

x = 3√y + 1x − 1 = 3√y

(x − 1)3 = ( 3√y)3

(x − 1)3 = y

(x − 1)3 = f −1(x)

So, the correct choice is B.


“000200010271723961_CH04_p119-177” 2013/2/27 14:0 page 177

Open Response7. (a)

O

y

x�2 2

4

The graph of y − 1 = (2x)2 is the same as the graphof y = x2 after a horizontal shrink by a factor of 2,and a translation 1 unit up.

(b)

�2

2

4

�4

�2 2O

y

x

The graph of y = (x2

)3 − x2 is the same as the graph of

y = x3 − x after a horizontal stretch by a factor of 2.(c)

O 42 6 8 10

2

y

x

The graph of 2y = |x − 5| is the same as the graph ofy = |x| after a vertical shrink by a factor of 2, and atranslation 5 units right.

(d)

2

4

6

�2�4 2O

y

x

The graph of y = 2x+1 is the same as the graph ofy = 2x after a translation 1 unit left.

8. The graph of (x + 1)2 + (y − 4)2 = 36 is a circle ofradius 6 centered at the point (−1, 4).

9. n m(n)

0 41 12 −23 −54 −85 −116 −147 −178 −209 −23

10 −26

The differences are constant; namely −3. Therefore, thefunction is linear. Since it starts at 4, m(n) = −3n + 4.Thus m(80) = −240 + 4 = −236.

10. The differences are the constant 3. So a recursivedefinition is

p(n) ={−8 if n = 0

p(n − 1) + 3 if n ≥ 1This is a linear function, starting at −8, and sop(n) = 3n − 8.

11.

0

1

2

3

4

4

7

8

7

4

−2

−2

x g(x)

3

1

−1

−3

2

−2

Since the second differences are constant, it isquadratic. This constant is −2, so the leading coefficientis −l. Now g(x) − (−x2) has the values 4, 8, 12, 16, 20.Since this starts at 4 and has 4 for a constant difference,the function 4x + 4 matches it. Henceg(x) = −x2 + 4x + 4.

12. (a)x h(x) ÷0 6 3

6 = 12

1 3323 = 1

2

2 32

3432

= 12

3 34

3834

= 12

4 38

(b)

h(x) ={

6 if x = 012 · h(x − 1) if x > 0

(c) Write a function in the form h(x) = a · bx . Sinceh(0) = 6, a · b0 = 6 and a = 6. Since h(1) = 3 andh(1) = 6 · b1, b = 3

6 = 12 . The closed-form

definition is

h(x) = 6(

12

)x

(d) h(8) = 6( 1

2

)8 = 6 · 1256 = 3

128


Chapter Functions 4 pages 119–177 - Edl · Mathematics II Solutions Manual • Chapter 4, page...

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Transcript of Chapter Functions 4 pages 119–177 - Edl · Mathematics II Solutions Manual • Chapter 4, page...