Essentials of PreCalculus, 1e, AN SSM Chapter...

Chapter 1 Functions and Graphs

Section 1.1 1. 2 10 40

2 10 402 40 12 30

15

xx

xxx

+ =+ =

= −==

0

3. 10225 −=+ xx

4123

21025

−=−=

−−=−

xxxx

5. 2( 3) 5 4( 5)2 6 5 4 20

2 11 4 202 4 20 11

2 992

x xx x

x xx x

x

x

− − = −− − = −

− = −− = − +− = −

=

7.

32

21

43

=+x

9229

689869

32 12

21

43 12

=

=−=

=+

⎟⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛ +

x

xx

x

x

9. 3

215

32

−=− xx

12301834

183304

321 65

32 6

=+−=−

−=−

⎟⎠

⎞⎜⎝

⎛ −=⎟⎠

⎞⎜⎝

⎛ −

xxx

xx

xx

11. 6.34.02.0 =+x

16

2.32.0==

xx

13. ( ) ( )

( ) ( )

( ) ( )

3 35 11 05 4

3 320 5 11 20 05 412 5 15 11 0

12 60 15 165 03 225

75

n n

n n

n nn n

nn

+ − − =

⎡ ⎤+ − − = −⎢ ⎥⎣ ⎦+ − − =+ − + =

− = −=

15. 2 2

3( 5)( 1) (3 4)( 2)

3 12 15 3 2 814 7

12

x x x x

x x x xx

x

+ − = + −

+ − = − −=

=

17. 0.08 0.12(4000 ) 4320.08 480 0.12 432

0.04 481200

x xx x

xx

+ − =+ − =

− = −=

19. 01522 =−− xx

326

282

or 52

102

822

822

6422

6042

)1(2)15)(1(4)2()2(

15 2 1

2

−=−

=−

=

==+

=

±=

±=

+±=

−−−±−−=

−=−==

x

x

x

x

x

cba

21. 0721898 2 =−+ yy

24 83

024or 0380)24)(38(

−==

=+=−=+−

yy

yyyy

23. 073 2 =− xx

37

073or 00)73(

=

=−==−

x

xxxx

Copyright © Houghton Mifflin Company. All rights reserved.

2 Chapter 1: Functions and Graphs

25. 09)5( 2 =−−x [( 5) 3][( 5) 3] 0

8 0 or 2 0 8 2

x xx x

x x

− − − + =− = − =

= =

27. 2

2

2

2 15 0

2 1 15 1

( 1) 161 1

1 4

x x

x x

xx

x

− − =

6

− + = +

− =

− =±= ±

x = 1 + 4 or x = 1 – 4 x = 5 x = −3

29. 012 =−+ xx 21 1 4(1)( 1)2(1)

1 1 4 1 52 2

x

x

− ± − −=

− ± + − ±= =

1 5 1 5 or 2 2

x x− + − −= =

31. 0142 2 =++ xx

222 or ,

222

2

224

2244

844

8164

)2(2)1)(2(444 2

−−=

+−=

±−=

±−=

±−=

−±−=

−±−=

xx

x

x

x

33. 23 5 3x x 0− − = 2( 5) ( 5) 4(3)( 3)

2(3)5 25 36 5 61

6 65 61

661 615 5, or

6 6 6 6

x

x

x

x x

− − ± − − −=

± + ±= =

±=

= + = −

35. 2

2

1 3 1 02 4

1 34 1 4(0)2 4

x x

x x

+ − =

⎛ ⎞+ − =⎜ ⎟⎝ ⎠

4413 or ,

4413

4413

43293

)2(2)4)(2(433

0432

2

2

−−=

+−=

±−=

+±−=

−−±−=

=−+

xx

x

x

x

xx

37. 0232 2 =++ xx

2

2 3 2

3 3 4 2 22 2

3 9 82 2

3 1 2 2 or22 2 2 2

3 1 4 22 2 2 2

a b c

x

x

x

x

= = =

− ± − ⋅ ⋅=

− ± −=

− + −= = = −

− − −= = = −

39. 532 += xx

22932

2093

)1(2)5)(1(4)3()3(

5 3 1053

2

2

±=

+±=

−−±−−=

−=−===−−

x

x

x

cbaxx

41. 1132 <+x

482

<<

xx

( , 4)−∞

43. 1634 +>+ xx

6

122−<

>−xx

( , 6)−∞ −


Section 1.1 3

45. 1916 ≥+− x

3

186−≤

≥−xx

( , 3]−∞ −

47. 75)2(3 +≤+− xx

813

1387563

−≥

≤−+≤−−

x

xx

13 , 8

⎡ ⎞− ∞⎟⎢⎣ ⎠

49. )4(2)53(4 −>−− xx

22814

822012

<−>−

−>+−

xx

xx

( , 2)−∞

49. )4(2)53(4 −>−− xx

22814

822012

<−>−

−>+−

xx

xx

( , 2)−∞

51. 2 7 0( 7) 0

x xx x

+ >+ >

The product is positive. The critical values are 0 and –7.

( 7)x x −

) ,0()7 ,( ∞∪−−∞ 53. 2 7 10 0

( 5)( 2) 0x xx x

+ + <+ + <

The product is negative. The critical values are –5 and –2.

( 5)( 2)x x+ +

)2 ,5( −−

55. 2832 ≥− xx2 3 28 0

( 4)( 7) 0x xx x

− − ≥+ − ≥

The product is positive or zero. The critical values are –4 and 7.

( 4)( 7)x x+ −

( , 4] [7, )−∞ − ∪ ∞ 57. 2

26 4 5

6 5 4 0(3 4)(2 1) 0

x x

x xx x

− ≤

− − ≤− + ≤

The product is negative or zero. The critical values are 4 1

3 2 and .−

(3 4)(2 1)x x− +

⎥⎦

⎤⎢⎣

⎡−34 ,

21

59. 4<x

)4 ,4(

44

−

<<− x



61. 91 <−x

9 1 9 8 10

xx

− < − <− < <

( 8, 10)−

63. 303 >+x

27 33303or 303

>−<>+−<+

xxxx

) ,27( )33 ,( ∞∪−−∞

65. 412 >−x

2 1 4 or 2 1 42 3 2 5

3 5 2 2

x xx x

x x

− < − − >< − >

< − >

⎟⎠

⎞⎜⎝

⎛ ∞∪⎟⎠

⎞⎜⎝

⎛ −∞− ,25

23 ,

67. 53 ≥+x

2 853or 53

≥−≤≥+−≤+

xxxx

) ,2[ ]8 ,( ∞∪−−∞

69. 14103 ≤−x

14 3 10 14 4 3 24

4 83

xx

x

− ≤ − ≤− ≤ ≤

− ≤ ≤

⎥⎦

⎤⎢⎣

⎡− 8 ,34

71. 4 5 24x− ≥

4 5 245 2

285

xx

x

8− ≤− ≤ −

≥

or 4 5 245 20

4

xxx

− ≥− ≥

≤ −

] 28( , 4 , 5

⎡ ⎞−∞ − ∪ ∞⎟⎢ ⎠⎣

73. 05 ≥−x

Because an absolute value is al-ways nonnegative, the inequality is always true. The solution set consists of all real numbers.

( , )−∞ ∞

75. 04 ≤−x

Because an absolute value is always nonnegative, the inequality 4 0x − < has no solution. Thus the only solution

of the inequality 04 ≤−x is the solution of the equation x – 4 = 0.

4=x 77. 35

3535

AA LW

LW

LW

===

=

2

2

272 2

2 2 27 352 2 27

70 2 27

2 27 70 0 (2 7)( 10) 0

PP L W

L W

WW

W W

W WW W

== +

+ =

⎛ ⎞ + =⎜ ⎟⎝ ⎠

+ =

− + =− − =

72

7352

70 7 10

W

L

LL

=

=

==

or 10

35 35 10 3.5

W

LWL

L

=

===

The rectangle measures 3.5 cm by 10 cm.

79. A = 1500 = lw

wl

wlP15000

32600

=

+−=

2

2

2

2 3 600150002 3 600

30,000 3 600

3 600 30,000 0

3( 200 10,000) 03( 100)( 100) 0

l w

ww

w w

w w

w ww w

+ =

⎛ ⎞ + =⎜ ⎟⎝ ⎠

+ =

− + =

− + =− − =

w = 100 ft 15000 150 ft

100l = =

The dimensions are 100 feet by 150 feet.


Section 1.1 5

81. Plan A: 5 + 0.01x

Plan B: 1 + 0.08x

5 + 0.01x < 1 + 0.08x 4 < .07x 57.1 < x

Plan A is less expensive if you use at least 58 checks.

83. Plan A: 100 + 8x

Plan B: 250 + 3.5x

100 + 8x > 250 + 3.5x 4.5x > 150 x > 33.3

Plan A pays better if at least 34 sales are made.

85. 10468 ≤≤ F

4020

725936

104325968

≤≤

≤≤

≤+≤

C

C

C

....................................................... Connecting Concepts 87.

)1(21253 += nn

220)22)(23(05062

==−+=−+

nnnnn

So 1 . 2 3 21 22 253+ + + + + =L

89. 22420 xxR −=

0)210(202420 2

>−>−

xxxx

The product is positive. The critical values are 0 and 210.

2 (210 )x x−

(0, 210)

91. 0 ,64 ,48 16 0000

2 ==>++−= svsstvts

0)1)(3(160)34(16

0486416

486416

2

2

>−−−>+−−

>−+−

>+−

tttt

tt

tt

The product is positive. The critical values are t = 3 and t = 1.

16( 3)( 1)t t− − −

1 second < t < 3 seconds The ball is higher than 48 ft between 1 and 3 seconds.

93. a. 01.025.4 ≤−s b. 4.25 0.01, or 4.25 0.01

4.26 4.24 critical valuess s

s s− = − = −

= =

26.424.4 ≤≤ s

....................................................... Prepare for Section 1.2 94. 4 ( 7) 3

2 2+ − = −

95. 50 25 2 5 2= =

96.

?3 2

5 3( 1) 25 5

y x= −

= − −≠ − No, the equation is not true.

97. 20 3 20 ( 2)( 1)

x xx x

= − += − −

x = 2 or x = 1



98. 3 ( 1) 3 1 2 2− − − = − + = − = 99. 2 2( 3) (4) 9 16 25 5− + = + = =

Section 1.2

1.

3. a.

b. 10

)8490()7590()8196()6993()96141()90108()90129()87111()7299()6384( average −+−+−+−+−+−+−+−+−+−=

21 27 24 39 18 45 24 15 15 6 234 23.410 10

+ + + + + + + + += = =

The average increase in heart rate is 23.4 beats per minute. 5. 2 2

2 2

( 8 6) (11 4)

( 14) (7)

196 49245

7 5

d = − − + −

= − +

= +

=

=

7. 2 2

2 2

( 10 ( 4)) (15 ( 20))

( 6) (35)

36 12251261

d = − − − + − −

= − +

= +

=

9. 2 2

2 2

(0 5) (0 ( 8))

( 5) (8)

25 64

89

d = − + − −

= − +

= +

=

11. 2 2

2 2

2 2

( 12 3) ( 27 8)

(2 3 3) (3 3 2 2)

( 3) (3 3 2 2)

3 (27 12 6 8)

3 27 12 6 8

38 12 6

d = − + −

= − + −

= + −

= + − +

= + − +

= −

13. 2 2

2 2

2 2

2 2

2 2

( ) ( )

( 2 ) ( 2 )

4 4

4( )

2

d a a b b

a b

a b

a b

a b

= − − + − −

= − + −

= +

= +

= +

15. 2 2

2 2

2 2

2

2

( 2 ) (3 4 ) with 0

( 3 ) ( )

9

10

10 (Note: since 0, )

d x x x x x

x x

x x

x

x x x

= − − + − <

= − + −

= +

=

x= − < = −


Section 1.2 7

17. 2 2

22 2

2

2

(4 ) (6 0) 10

(4 ) (6 0) 10

16 8 36 100

8 48 0( 12)( 4) 0

x

x

x x

x xx x

− + − =

⎛ ⎞− + − =⎜ ⎟⎝ ⎠

− + + =

− − =− + =

2

−

12 or 4x x= = The points are (12, 0), ( 4, 0).−

19. 1 2 1 2, 2 2

1 5 1 5, 2 2

6 4, 2 2

(3, 2)

x x y yM

+ +⎛=⎜ ⎟⎝ ⎠⎛ ⎞+ − +=⎜ ⎟⎝ ⎠⎛ ⎞=⎜ ⎟⎝ ⎠

=

⎞

21. 6 6 3 11, 2 2

12 8, 2 2

(6, 4)

M ⎛ ⎞+ − +=⎜ ⎟⎝ ⎠⎛ ⎞=⎜ ⎟⎝ ⎠

=

23. 1.75 ( 3.5) 2.25 5.57,

2 21.75 7.82,

2 2( 0.875, 3.91)

M + − +⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞= −⎜ ⎟⎝ ⎠

= −

25.

27.

29.

31.

33.

35.

37.

39. Intercepts: ( )0 6, ,

512,0 ⎟

⎠

⎞⎜⎝

⎛

41. ( ) ( ) 0) (5, ,5 0, ,5 ,0 −

52 +−= yx 43. (0, 4), (0, −4), (−4, 0)

4|| −= yx

45. )0 ,2(),2 ,0( ±±

422 =+ yx

47. )0 ,4(),4 ,0( ±±

4 |||| =+ yx

49. center (0, 0), radius 6 51. center (1, 3), radius 7 53. center (−2, −5), radius 5 55. center (8, 0), radius 1

2



57. 222 2)1()4( =−+− yx 59. ( )2225

41

21

=⎟⎠

⎞⎜⎝

⎛ −+⎟⎠

⎞⎜⎝

⎛ − yx 61. 222 )0()0( ryx =−+−

2 2 2

2 2 2

2

2 2

2 2 2

( 3 0) (4 0)

( 3) 4

9 16

25 5

( 0) ( 0) 5

r

r

r

r

x y

− − + − =

− + =

+ =

= =

− + − =

63. 2 2 2

2 2 2

2 2 2

2 2 2

2

2 2

2 2 2

( 2) ( 5)

( 1) ( 3)

(4 1) ( 1 3)

3 ( 4)

9 16

25 5

( 1) ( 3) 5

x y r

x y r

r

r

r

r

x y

+ + − =

− + − =

− + − − =

+ − =

+ =

= =

− + − =

65. 2 2

2 2

2 2 2

6 5

6 9 5 9

( 3) 2

x x y

x x y

x y

− + = −

− + + = − +

− + =

center (3, 0), radius 2

67. 2 2

2 2

2 2 2

14 8 56

14 49 8 16 56 49 16

( 7) ( 4) 3

x x y y

x x y y

x y

− + + = −

− + + + + = − + +

− + + =

center (7, −4), radius 3

69. 2 2

2 2

2 2

22

22 2

4 4 4 6363 4

1 6 4 4

1 162

1 ( 0) 42

x x y

x x y

x x y

x y

x y

+ + =

+ + =

3 14

+ + + = +

⎛ ⎞+ + =⎜ ⎟⎝ ⎠

⎛ ⎞+ + − =⎜ ⎟⎝ ⎠

center 1 ,02

⎛ −⎜⎝ ⎠

⎞⎟ , radius 4

71. 2 2

2 2

2 2 2

3 15 2 4

1 3 9 15 1 4 2 4 4 4

1 3 52 2 2

x x y y

x x y y

x y

− + + =

− + + + + = + +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + + =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

94

center 1 3 5, , radius 2 2 2

⎛ ⎞−⎜ ⎟⎝ ⎠

73. 2 2( 4 2) (11 3)

36 64100

10

d = − − + −

= +

==

Since the diameter is 10, the radius is 5. The center is the midpoint of the line segment from (2,3) to (-4,11).

222 5)7()1(

center )7,1(2113,

2(-4)2

=−++

−=⎟⎠⎞

⎜⎝⎛ ++

yx

75. Since it is tangent to the x-axis, its radius is 11.

222 11)11()7( =−+− yx


Section 1.2 9


79.

81.

83.

85.

87. )3 ,9(

21 ,

25

=⎟⎠

⎞⎜⎝

⎛ ++ yx

5 1361 185

32

1 and 92

5 therefore

===+=+

=+

=+

yxyx

yx

Thus (13, 5) is the other endpoint.

89.

)7 ,2(2

)8( ,2

)3(−=⎟

⎠

⎞⎜⎝

⎛ −+−+ yx

therefore 3 22

3 47

x

xx

−=

− ==

and 8 72

8 16

y

yy

−= −

− = −= −

4

Thus (7, −6) is the other endpoint.

91. 2 2

2 2

2 2

2 2

(3 ) (4 ) 5

(3 ) (4 ) 5

9 6 16 18 25

6 8

x y

x y

x x y y

x x y y

2

0

− + − =

− + − =

− + + − + =

− + − =

93. 2 2 2 2

2 2 2 2 2

2 2 2 2 2 2

2 2

2 2

2 2 2

(4 ) (0 ) ( 4 ) (0 ) 10

(4 ) (0 ) 100 20 ( 4 ) (0 ) ( 4 ) ( )

16 8 100 20 ( 4 ) ( ) 16 8

16 100 20 ( 4 ) ( )

4 25 5 ( 4 ) ( )

16 200 625 25 ( 4 ) ( )

x y x y

2x y x y x

x x y x y x x y

x x y

x x y

x x x y

− + − + − − + − =

− + − = − − − + − + − − + −

− + + = − − − + − + + + +

− − = − − − + −

+ = − − + −

⎡ ⎤+ + = − − + −⎢⎣2 2 2

2 2 2

16 200 625 25 16 8

16 200 625 400 200 25 25

x x x x y

x x x x y

⎥⎦

⎡ ⎤+ + = + + +⎢ ⎥⎣ ⎦

+ + = + + +

y

Simplifying yields . 225259 22 =+ yx 95. The center is (-3,3). The radius is 3.

222 3)3()3( =−++ yx

....................................................... Prepare for Section 1.3 97. 2

23 4

( 3) 3( 3) 4 9 9 4 4

x x+ −

− + − − = − − =−

98. { 3, 2, 1, 0, 2}{1, 2, 4, 5}

DR

= − − −=



99. 2 2(3 ( 4)) ( 2 1) 49 9 58d = − − + − − = + = 100. 2 6 0

2 63

xxx

− ≥≥≥

101. 2 6 0( 2)( 3) 0

x xx x

− − =+ − =

2 0 3 02 3

x xx x

+ = − ==− =

–2, 3

102. 3 4, 6 5a x a x= + = − 3 4 6 5

9 33

x xx

x

+ = −==

3(3) 4 13a = + =

Section 1.3

1. Given ,13)( −= xxf

a. (2) 3(2) 16 15

f = −= −=

b. ( 1) 3( 1) 13 14

f − = − −= − −= −

c. (0) 3(0) 10 1

1

f = −= −= −

d. 2 23 13 3

2 11

f ⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= −=

e. ( ) 3( ) 13 1

f k kk

= −= −

f. ( 2) 3( 2) 13 6 13 5

f k kkk

+ = + −= + −= +

3. Given ,5)( 2 += wwA

a. 2(0) (0) 55

A = +=

b. 2(2) (2) 59

3

A = +==

c. 2( 2) ( 2) 59

3

A − = − +==

d. 2(4) 4 521

A = +

=

e. 2

2

2

( 1) ( 1) 5

2 1 5

2 6

A r r

r r

r r

+ = + +

= + + +

= + +

f. 2

2( ) ( ) 5

5

A c c

c

− = − +

= +

5. Given ,1)(

xxf =

a. 21

21)2( ==f

b. 21

21)2( =

−=−f

c.

35

3 1355

1

3 5 51 15 3 3

f ⎛ ⎞− =⎜ ⎟⎝ ⎠ −

=

= ÷ = ⋅ =

d. 121

21)2()2( =+=−+ ff

e. 4

1

4

1)4( 222

+=

+=+

cccf

f. h

hf+

=+2

1)2(

7. Given ,)(

xxxs =

a. 144

44)4( ===s

b. 155

55)5( ===s

c. 122

22)2( −=

−=

−−

=−s

d. 133

33)3( −=

−=

−−

=−s

e. . ,0 Since ttt =>

1)( ===tt

ttts

f. . ,0 Since ttt −=<

1)( −=−

==t

tttts

9. a. Since .13)( use ,24 +=<−= xxPx 111121)4(3)4( −=+−=+−=−P

b. Since .11)( use ,25 2 +−=≥= xxPx

( ) ( ) 611511552

=+−=+−=P c. Since .13)( use ,2 +=<= xxPcx 13)( += ccPd. Since ,21 then ,1 ≥+=≥ kxk

. 11)( use so 2 +−= xxP

102

1112

11)12(11)1()1(

2

2

22

+−−=

+−−−=

+++−=++−=+

kk

kk

kkkkP


Section 1.3 11

11. 2 3 73 2 7

2 7 , is a function of .3 3

x yy x

y x y

+ ==− +

=− + x

13. 2

22

2

2, is a not function of .

x y

y x

y x y

− + =

= +

= ± + x

15. since offunction anot is ,4 xyxy ±=

. of values twoare there0each for xx >

17. . offunction a is ,3 xyxy =

19. 2 2

2 , is a not function of .

y x

y x y

=

=± x

21. Function; each x is paired with exactly one y.

23. Function; each x is paired with exactly one y. 25. Function; each x is paired with exactly one y. 27. numbers. real all ofset theisDomain 43)( −= xxf 29. numbers. real all ofset theisDomain 2)( 2 += xxf 31. { }.2 isDomain

24)( −≠+

= xxx

xf 33. { }.7 isDomain 7)( −≥+= xxxxf

35. { }.22 isDomain 4)( 2 ≤≤−−= xxxxf

37. { }.4 isDomain 4

1)( −>+

= xxx

xf

39.

Domain: the set of all real numbers

41.

Domain: the set of all real numbers

43.

Domain: { } 6 6x x− ≤ ≤

45.

Domain: { } 3 3x x− ≤ ≤ 47. a. (2.8) 0.37 0.34int(1 2.8)

0.37 0.34int( 1.8)0.37 0.34( 2)0.37 0.68$1.05

C = − −= − −= − −= +=

b. c(w)

49. a. Yes; every vertical line intersects the graph in one point. b. Yes; every vertical line intersects the graph in one point. c. No; some vertical lines intersect the graph at more than one point. d. Yes; every vertical line intersects the graph in at most one point.

51. Decreasing on ( ; increasing on [ , 0]−∞ 0, )∞ 53. Increasing on ) ,( ∞−∞



55. Decreasing on ( ; increasing on [ ; decreasing on [ ; increasing on [3, 3]−∞ − 3, 0]− 0, 3] , )∞ 57. Constant on ; increasing on [ 0] ,(−∞ 0, )∞ 59. Decreasing on ( ; constant on [0, 1]; increasing on [, 0]−∞ 1, )∞ 61. g and F are one-to-one since every horizontal line intersects the graph at one point.

f, V, and p are not one-to-one since some horizontal lines intersect the graph at more than one point. 63. a. 2 2 50

2 50 225

l ww lw l

+ == −= −

b.

2(25 )

25

A lwA l l

A l l

== −

= −

65. 100 ,6500000,80)( ≤≤−= tttv

67. a. ( ) 5(400) 22.80

2000 22.80C x x

x= += +

b. xxR 00.37)( = c. ( ) 37.00 ( )

37.00 [2000 22.80 ]37.00 2000 22.8014.20 2000

P x x C xx

x xx

= −= − += − −= −

Note x is a natural number.

69. 15 153

155

5 1515 5

( ) 15 5

hr

hr

r hh r

h r r

−=

−=

= −= −= −

71. 22 )50()3( += td

600 meters, 25009 2 ≤≤+= ttd

73. 22 )6()845( ttd +−= miles where t is the number of hours after 12:00 noon

75. a.

22

2

Circle22

2

Area2

4

C rx r

xr

xr

x

ππ

π

π ππ

π

==

=

⎛ ⎞= = ⎜ ⎟⎝ ⎠

=

22

2

Square4

20 4

54

Area 54

5252 16

C sx s

xs

xs

xx

=− =

= −

⎛ ⎞= = −⎜ ⎟⎝ ⎠

= − +

2 2

2

5Total Area 254 2 16

1 1 5 254 16 2

x xx

x x

π

π

= + − +

⎛ ⎞= + − +⎜ ⎟⎝ ⎠

b.

x 0 4 8 12 16 20

Total Area 25 17.27 14.09 15.46 21.37 31.83

c. Domain: [0, 20].

77. a. 2 2 2

2

Left side triangle

20 (40 )

400 (40 )

c x

c x

= + −

= + −

2 2 2

2

Right side triangle

30

900

c x

c x

= +

= +

2 2Total length 900 400 (40 )x x= + + + − b.

x 0 10 20 30 40

Total Length 74.72 67.68 64.34 64.79 70

c. Domain: (0, 40).


Section 1.3 13

79. x 5 10 12.5 15 20

Y(x) 275 375 385 390 394

answers accurate to nearest apple

81. 2

2( ) 5 1

6 0 ( 3)( 2) 0

3 0 or 2 0 3 2

f c c c

c cc c

c cc c

= − − =

− − =− + =

− = + == =−

83. 1 is not in the range of f(x), since

.11or 11 ifonly 111 −=−=+

+−

= xxxx

85. Set the graphing utility to “dot” mode.

WINDOW FORMAT

Xmin=-4.7

Xmax=4.7

Xscl=1

Ymin=-5

Ymax=2

Yscl=1

87.

WINDOW FORMAT

Xmin=-4.7

Xmax=4.7

Xscl=1

Ymin=-5

Ymax=1

Yscl=1

89.

....................................................... Connecting Concepts 91. 426)24()39()( 3

2 =−=−−−=xf 93. 20)21216()( 20 =−−−=xf

95. a. 3623532)7(5)1(3)7,1( =−+=−+=fb. 132)3(5)0(3)3,0( =−+=fc. 122)4(5)2(3)4,2( =−+−=−fd. 302)4(5)4(3)4,4( =−+=fe. 2132)2(5)(3)2,( −=−+= kkkkkff. 1182155632)3(5)2(3)3,2( −=−−++=−−++=−+ kkkkkkkf

97.

122

1185=

++=s

214336)1)(4)(7(12

)1112)(812)(512(12)11,8,5(

===

−−−=A

99. 2

23 3

2 3 0( 1)( 3) 0

a a

a aa a

a+ − =

+ − =− + =

1=a or 3−=a



101.

....................................................... Prepare for Section 1.4 103. 5 ( 2) 7d = − − = 104. The product of any number and its negative reciprocal is –1.

17 17

− ⋅ =−

105. 4 4 8

2 ( 3) 5− − −=− −

106. 3 2( 3)3 2 6

2 9

y xy x

y x

− =− −− =− +

=− +

107. 3 5 15

5 3 13 35

x yy x

y x

− =− =− +

= −

5 108. 3 2(5 )

0 3 2(5 )0 3 10 2

10 52

y x xx xx xx

x

= − −= − −= − +==

Section 1.4

1.

23

23

3147

1212 −=

−=

−−

=−−

=xxyym

3. 21

4002

−=−−

=m

5. The line does not have a slope since .012 == xx 7.

616

)3(442

=−−

=−−−

−−=m

9.

199

1933

31926

)4(37

21

27

=⋅==−−

−=m

11. h

fhfh

fhfm )3()3(33

)3()3( −+=

−+−+

=

13.

hfhf

hfhfm )0()(0

)0()( −=

−−

=

15. m = 2

y-intercept (0, –4)

17. m = 13

−

y-intercept (0, 4)

19. m = 0 y-intercept (0, 3)

21. m = 2 y-intercept (0, 0)


Section 1.4 15

23. m = –2 y-intercept (0, 5)

25. m = 34

−

y-intercept (0, 4)

27. Use with bmxy += 1, 3.m b= =

3y x= + 29.

Use bmxy += with 3 1, .4 2

m b= =

3 1 4 2

y x= +

31. Use with bmxy += 0, 4.m b= =

4y = 33. 2 4( ( 3))

2 4 124 10

y xy x

y x

− =− − −− =− −

=− −

35. 4 1

1 33 34 4

m −=

− −

= = −−

413

43

44

49

43

)3(431

+−=

++−=

−−=−

xy

xy

xy

37. 1 112 712 125 5

m − −=

−−

= =−

1211 ( 7)5

12 84115 5

12 84 555 5 5

12 295 5

y x

y x

y x

x

− = −

− = −

= − +

= −

39. ( ) 2 3 12 4

2

f x xxx

= + =−=−=−

41. ( ) 1 4 3

4 212

f x xx

x

= − =− =

=−

43. ( ) 3 5

2

22

2( 2)4

xf x

x

xx

= − =

− =

= −=−

45. ( ) 3 123 12 0

3 124

f x xx

xx

= −− =

==

The x-intercept of the graph of f(x) is (4,0).

Xmin = − 4, Xmax = 6, Xscl=2, Ymin = −12.2, Ymax = 2, Yscl = 2



47. 1( ) 54

1 5 04

1 54

20

f x x

x

x

x

= +

+ =

=−

=−

The x-intercept of the graph of f(x) is ).0,20(−

Xmin Xmax = 30, Xscl = 10, ,30−=Ymin Ymax = 10, Yscl = 1 ,10−=

49. Algebraic method: 1 2( ) ( )4 5 6

3 113

f x f xx x

x

x

=+ = +

=

=

Graphical method: Graph and 4 56

y xy x

= += +

They intersect at .316 ,

31

== yx

Xmin = −7.8, Xmax = 7.8, Xscl = 2, Ymin = −2 ,Ymax = 10, Yscl = 2

51. Algebraic method: 1 2( ) ( )

2 4 123 16

163

f x f xx x

x

x

=− =− +

=

=

Graphical method: Graph and 2 412

y xy x

= −=− +

They intersect at .326 ,

315 == yx

Xmin = − 4, Xmax = 10, Xscl = 2, Ymin = −2, Ymax = 10, Yscl = 2

53. 1505 1482 2.87528 20

m −= =−

The value of the slope indicates that the speed of sound in water increases 2.875 ft per s for a one-degree increase in temperature.

55. a. 29 13 1.45

20 9( ) 13 1.45( 9)( ) 1.45

m

H c cH c c

−= ≈−

− = −=

57. a. 63,000 38,000 25002010 2000

( ) 63,000 2500( 2010)( ) 2500 4,962,000

m

N t tN t t

−= =−

− = −= −

b. (18) 1.45(18) 26 mpgH = ≈ b. 60,000 2500 4,962,0005,022,000 2500

2008.8

tt

t

= −==

The number of jobs will exceed 60,000 in 2008.


Section 1.4 17

59. a. 240 180 3018 16

( ) 180 30( 16)( ) 30 300

m

B d dB d d

−= =−

− = −= −

b. The value of the slope means that a 1-inch increase in the diameter of a log 32 ft long results in an increase of 30 board-feet of lumber that can be obtained from the log. c. board feet (19) 30(19) 300 270B = − =

61. Line A represents Michelle Line B represents Amanda Line C represents the distance between Michelle and Amanda.

63. a. Find the slope of the line.

842.13870

70108110180

≈=−−

=m

Use the point-slope formula to find the equation.

94.18842.194.128842.1110

)70(842.1110)( 11

−=−=−−=−

−=−

xyxyxyxxmyy

b. 94.18)90(842.1 −=y

14784.146

94.1878.165≈=−=

yy

65. ( ) 92.50 (52 1782)( ) 92.50 52 1782( ) 40.50 1782

P x x xP x x xP x x

= − += − −= −

40.50 1782 040.50 1782

178240.5044, the break-even point

xx

x

x

− ==

=

=

67. ( ) 259 (180 10,270)

( ) 259 180 10,270( ) 79 10,270

P x x xP x x xP x x

= − += − −= −

79 10,270 079 10.270

10.27079

130, the break-even point

xx

x

x

− ==

=

=

69. a. 275$2750275)0(8)0( =+=+=C b. 283$2758275)1(8)1( =+=+=C c. 355$27580275)10(8)10( =+=+=C d. The marginal cost is the slope of ,2758)( += xxC which is 8 (dollars).

71. a. ttC 75.600.500,19)( +=

b. ttR 00.55)( =c. ( ) ( ) ( )

( ) 55.00 (19,500.00 6.75 )( ) 55.00 19,500.00 6.75( ) 48.25 19,500.00

P t R t C tP t t tP t t tP t t

= −= − += − −= −

d. 48.25 19,500.0019,500.00

48.25404.1451 days 405 days

t

t

t

=

=

= ≈

73. The graph of .

43 has 243 −=−=+ myx

415

43

343

43

)1(433

+−=

++−=

−−=−

xy

xy

xy



75. The graph of .1 has 4 −==+ myxThus we use a slope of 1.

2 1( 1)1 21

y xy xy x

− = −= − += +

77. The equation of the line through (0,0) and P(3,4) has

slope 4 .3

The path of the rock is on the line through P(3,4) with

slope 3 ,4

− so 34 ( 3)43 944 43 9 44 43 254 4

y x

y x

y x

y x

− =− −

− =− +

.

=− + +

=− +

The point where the rock hits the wall at y = 10 is the point

of intersection of 425

43

+−= xy and y = 10.

3 25 104 43 25 40

3 155 feet

x

xxx

− + =

− + =− =

= −

Therefore the rock hits the wall at ).10 ,5(−The x-coordinate is –5.

79. a. [ ] )10,3()13,3()12,2( so 1 22 QQhhQh =+=+++=

515

23510

==−−

=m

b. [ ] )41.5,1.2()11.2,1.2()12,2( so 1.0 22 QQhhQh =+=+++=

1.41.041.0

21.2541.5

==−−

=m

c. [ ] )0401.5,01.2()101.2,01.2()12,2( so 01.0 22 QQhhQh =+=+++=

01.401.0

0401.0201.250401.5

==−

−=m

d. As h approaches 0, the slope of PQ seems to be approaching 4. e. [ ] 12 ,2 5, ,2 2

2211 ++=+=== hyhxyx

[ ]2 2 22 12 1

2 1 5 (4 4 ) 1 5 4 4(2 ) 2

hy y h h h hm hx x h h h

+ + −− + + + − += = = = =

− + −+

81.

hxh

hxhh

hxhh

xhxhxxhxxhxm +=

+=

+=

−++=

−+−+

= 2)2(22)( 222222


Section 1.4 19


Substitute1212

xxyy

−− for m in the point-slope form )( 11 xxmyy −=− to yield )( 1

1212

1 xxxxyyyy −

−−

=− , the two-point form.

85. 3 11 (

4 521 ( 5)12( 5)

1 2 102 10 12 11

y x

y x

y xy x

y xy x

−− = −−

− = −−

=− −− =− +

=− + +=− +

5)

87. .5 and 3 with 1 Use ===+ ba

by

ax

1535

)1(1553

15

153

=+

=⎟⎠

⎞⎜⎝

⎛ +

=+

yx

yx

yx

89.

.3 with 1 Use abby

ax

==+

1 Since (5, 2) is on the line, 3

5 2 13

5 23 3 (1)3

15 2 317 3173

yxa a

a a

a aa a

aa

a

+ =

+ =

⎛ ⎞+ =⎜ ⎟⎝ ⎠

+ ==

=

173

11717

3

1

3173

317

Thus

=+

=+

=⎟⎠⎞

⎜⎝⎛

+⎟⎠⎞

⎜⎝⎛

yx

yx

yx

91. 3 2

2 3

2 3

2

2

3(1 ) 3 3(1 3 3 ) 31 1

3 9 9 3 3

9 9 3

(9 9 3 )

9 9 3

h h h hh h

h h hh

h h hh

h h hh

h h

3+ − + + + −=+ −

+ + + −=

+ +=

+ +=

= + +

93.

The slope of the line through (3, 9) and (x, y) is .2

1539 so ,

215

=−−

xy

Therefore

2

2

2( 9) 15( 3)2 18 15 45

2 15 27 0 Substitute into this equation.

2x 15 27 0(2 9)( 3) 0

9 or 32

y xy x

y x y x

xx x

x x

− = −− = −

− + = =

− + =− − =

= =

( ) ( )

.2

15 is 481 ,

29 and 9) (3, containing line theof slope theand , ofgraph on the is

481 ,

29point The

itself.point theis but this ,9 ,393 ,3 If

.481 ,

29

481

29 ,

29 If

2

22

22

⎟⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛

⇒====

⎟⎠

⎞⎜⎝

⎛⇒=⎟⎠

⎞⎜⎝

⎛===

xy

xyx

xyx



....................................................... Prepare for Section 1.5 95. 23 10 8 (3 2)( 4x x x x+ − = − + ) 2)

0

96. 2 28 8 16 ( 4x x x x x− = − + = − 97. 2( 3) 2( 3) 5( 3) 7

18 15 726

f − = − − − −= + −=

98. 22 1(2 1)( 1) 0

x xx x

− − =+ − =

12

2 1 0 1 0

1

x x

x x

+ = − =

=− =

99. 2 3 2 0x x+ − =

23 (3) 4(1)( 2)2(1)

3 172

x− ± − −

=

− ±=

100. 2

2

2

53 16 64 5

16 64 48 0

4 3 0( 1)( 3) 0

t t

t t

t tt t

=− + +

− + =

− + =− − =

1, 3t =

Section 1.5

1. d 3. b 5. g 7. c 9. 2

2

2

( ) ( 4 ) 1

( 4 4) 1 4

( 2) 3 standard form,

f x x x

x x

x

= + +

= + + + −

= + −

vertex (−2, −3), axis of symmetry x = −2

11. 2

2

2

( ) ( 8 ) 5

( 8 16) 5 16


f x x x

x x

x

= − +

= − + + −

= − −

vertex (4, −11), axis of symmetry x = 4

13. 2

2

2

2

( ) ( 3 ) 19 93 14 4

3 4 92 4 4

3 5 standard form,2 4

f x x x

x x

x

x

= + +

⎛ ⎞= + + + −⎜ ⎟⎝ ⎠

⎛ ⎞= + + −⎜ ⎟⎝ ⎠

⎛ ⎞= + −⎜ ⎟⎝ ⎠

vertex ⎟⎠

⎞⎜⎝

⎛ −−45 ,

23 , axis of symmetry x =

23

−

15. 2

2

2

2

( ) 4 2

( 4 ) 2

( 4 4) 2 4


f x x x

x x

x x

x

=− + +

=− − +

=− − + + +

=− − +

vertex (2, 6), axis of symmetry x = 2


Section 1.5 21

17. 2

2

2

2

2

( ) 3 3 7

3( 1 ) 71 33 1 74 4

1 28 332 4 4

1 313 standard form,2 4

f x x x

x x

x x

x

x

=− + +

=− − +

⎛ ⎞=− − + + +⎜ ⎟⎝ ⎠

⎛ ⎞=− − + +⎜ ⎟⎝ ⎠

⎛ ⎞=− − +⎜ ⎟⎝ ⎠

vertex ⎟⎠

⎞⎜⎝

⎛431 ,

21 , axis of symmetry x =

21

19. 5

)1(210

2==

−=

abx

255025)5(10)5()5( 2

−=−=−== fy

25)5()(

)25 (5,vertex 2 −−=

−

xxf

21.

0)1(2

02

==−

=abx

1010)0()0( 2 −=−== fy

10)(

)10 (0,vertex 2 −=

−

xxf

23. 3

26

)1(26

2=

−−

=−

−=

−=

abx

101189

1)3(6)3()3( 2

=++−

++−== fy

10)3()(

)10 (3,vertex 2 +−−= xxf

25.

43

)2(23

2==

−=

abx

847

856

818

89

749

89

749

1692

7433

432

43 2

=

+−=

+−=

+−⎟⎠⎞

⎜⎝⎛=

+⎟⎠

⎞⎜⎝

⎛−⎟⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛= fy

847

432)

847 ,

43vertex

2+⎟

⎠

⎞⎜⎝

⎛ −=

⎟⎠⎞

⎜⎝⎛

xf(x

27. 81

)4(21

2=

−−

=−

=abx

1617

1616

162

161

181

161

181

6414

181

814

81 2

=

++−=

++−=

++⎟⎠⎞

⎜⎝⎛−=

+⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛−=⎟⎠

⎞⎜⎝

⎛= fy

1617

814)

1617 ,

81vertex

2+⎟

⎠

⎞⎜⎝

⎛ −−=

⎟⎠⎞

⎜⎝⎛

xf(x



29. 2

2

2

2

( ) 2 1

( 2 ) 1

( 2 1) 1

( 1) 2

f x x x

x x

x x

x

= − −

= − −

= − + − −

= − −

1

vertex (1, −2) The y-value of the vertex is −2. The parabola opens up since a =1> 0. Thus the range is { }.2−≥yy

2

2( ) 2 2 1

0 2 30 ( 3)( 1)

f x x x

x xx x

= = − −

= − −= − +

1 301or 03−==

=+=−xx

xx

31. 2

2

2

2

2

( ) 2 5 152 125 25 252 12 16 16

5 8 2524 8 8

5 1724 8

f x x x

x x

x x

x

x

=− + −

⎛ ⎞=− − −⎜ ⎟⎝ ⎠⎛ ⎞=− − + − +⎜ ⎟⎝ ⎠

⎛ ⎞=− − − +⎜ ⎟⎝ ⎠

⎛ ⎞=− − +⎜ ⎟⎝ ⎠

2⎛ ⎞⎜ ⎟⎝ ⎠

vertex ⎟⎠

⎞⎜⎝

⎛8

17 ,45

The y-value of the vertex is .8

17

The parabola opens down since a = −2 < 0.

Thus the range is .8

17

⎭⎬⎫

⎩⎨⎧

≤yy

2

2( ) 2 2 5 1

2 5 3 0(2 3)( 1) 0

f x x x

x xx x

= =− + −

− + =− − =

1 23

01or 032

==

=−=−

xx

xx

33. 2

2

2

2

2

2

( ) 3 6

( 3 ) 69 93 64 4

3 962 4

3 24 92 4 4

3 152 4

f x x x

x x

x x

x

x

x

= + +

= + +

⎛ ⎞= + + + −⎜ ⎟⎝ ⎠

⎛ ⎞= + + −⎜ ⎟⎝ ⎠

⎛ ⎞= + + −⎜ ⎟⎝ ⎠

⎛ ⎞= + +⎜ ⎟⎝ ⎠

vertex ⎟⎠

⎞⎜⎝

⎛−451 ,

23

The y-value of the vertex is 4

15 .

The parabola opens up since a =1 > 0.

Thus the range is .4

15

⎭⎬⎫

⎩⎨⎧

≥yy

No, 3 .4

15

⎭⎬⎫

⎩⎨⎧

≥∉ yy

35. 2

2

2

( ) 8

( 8 16) 16

( 4) 16

f x x x

x x

x

= +

= + + −

= + −

minimum value of –16 when x = −4


Section 1.5 23

37. 2

2

2

2

( ) 6 2

( 6 ) 2

( 6 9) 2

( 3) 11

f x x x

x x

x x

x

=− + +

=− − +

=− − + + +

=− − +

9

maximum value of 11 when x = 3

39. 2

2

2

2

2

( ) 2 3 132 123 9 92 12 16 16

3 8 924 8 83 124 8

f x x x

x x

x x

x

x

= + +⎛ ⎞= + +⎜ ⎟⎝ ⎠⎛ ⎞= + + + −⎜ ⎟⎝ ⎠

⎛ ⎞= + + −⎜ ⎟⎝ ⎠

⎛ ⎞= + −⎜ ⎟⎝ ⎠

2⎛ ⎞⎜ ⎟⎝ ⎠

minimum value of 81

− when x = 43

−

41. 2

2

2

( ) 5 11

5( ) 11

5( 0) 11

f x x

x

x

= −

= −

= − −

minimum value of –11 when x = 0

43. 2

2

2

2

1( ) 6 1721 ( 12 ) 1721 ( 12 36) 17 1821 ( 6) 352

f x x x

x x

x x

x

=− + +

=− − +

=− − + + +

=− − +

maximum value of 35 when x = 6 45.

27)0(64327

643)( 22 +−−=+−= xxxh

a. The maximum height of the arch is 27 feet.

b. 23(10) (10) 27643 (100) 27

6475 271675 43216 16

357 522 feet16 16

h =− +

=− +

=− +

=− +

= =

c. 276438)( 2 +−== xxh

2

2

2

2

38 2764319

6464( 19) 364( 19)

364( 19)

31983

8 19 33

8 57320.1

x

x

x

x

x

x

x

x

x

− =−

− =−

− =−− =−− =−

=

=

=

≈

feet 1.20 when 8)( ≈= xxh

47. a. 3 2 6003 600 2

600 23

w lw l

lw

+ == −

−=

b.

2

600 23

22003

A w llA l

l l

= ⋅

⎛ ⎞−=⎜ ⎟⎝ ⎠

= −

c. 2

2 2

2

2 ( 300 )32 ( 300 150 ) 15,0003

In standard form,2 ( 150) 15,0003

A l l

A l l

A l

=− −

=− − + +

=− − +

The maximum area of is produced when

215,000 ft

150 ftl = and the width 600 2(150) 100 ft3

w −= = .



49. a. 3.594.97.0)( 2 ++−= tttT

917567.0

857.907477.0

7477.03.59

747

7947.0

3.597947.0

3.597.04.97.0

2

2

222

2

2

+⎟⎠⎞

⎜⎝⎛ −−≈

+⎟⎠

⎞⎜⎝

⎛ −−≈

⎥⎦⎤

⎢⎣⎡++⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛⎥⎦⎤

⎢⎣⎡+−−=

+⎟⎠

⎞⎜⎝

⎛ −−=

+⎟⎠

⎞⎜⎝

⎛ −−=

t

t

tt

tt

tt

The temperature is a maximum when

756

747

==t hours after 6:00 A.M.

Note 75 (60 minutes) 43 minutes. ≈

Thus the temperature is a maximum at 12:43 P.M. b. The maximum temperature is approximately 91°F.

51. 2

2

2

( ) 1.43 11.44 47.68

( ) 1.43( 8 ) 47.68

( ) 1.43( 4) 24.8

N t t t

N t t t

N t t

= − +

= − +

= − +

minimum at t = 4, or 1993 for 2500 homes

53. 2

2( ) 0.002 0.03 8

(39) 0.002(39) 0.03(39) 8 3.788 3

h x x x

h

=− − +

=− − + = >

Solve for x using quadratic formula. 2

20.002 0.03 8 0

15 4000 0

x x

x x

− − +

+ − =

=

215 (15) 4(1)( 4000)2(1)

15 16,225 , use positive value of 2

56.2

x

x

x

− ± − −=

− ±=

≈ Yes, the conditions are satisfied.

55. a.

( )

2

2

2

2 2

2

( ) 0.018 1.476 3.41.4760.018 3.40.018

0.018 ( 82 ) 3.4

0.018 82 41 3.4 0.018(41)

0.018 ( 41) 33.658

E v v v

v v

v v

v v

v

=− + +

⎛ ⎞=− − +⎜ ⎟⎝ ⎠

=− − +

=− − + + +

=− − +

2

The maximum fuel efficiency is obtained at a speed of 41 mph. b. The maximum fuel efficiency for this car, to the nearest mile per gallon, is 34 mpg.

57. Let y = 0, then xx 60 2 +=

)6(0 += xx x = 0 or x + 6 = 0 x = −6 The x-intercepts are (0, 0) and (−6, 0).

Let x = 0, then 0)0(60)( 2 =+=xfThe y-intercept is (0, 0).

59. Let y = 0, then 2

2

0 3 5 6

5 5 4( 3)( 6)2( 3)

x x

x

=− + −

− ± − − −=

−

Since the discriminant is negative, there are no x-intercepts.

47)6)(3(452 −=−−−

Let x = 0, then 66)0(5)0(3)( 2 −=−+−=xfThe y-intercept is (0, −6).

61.

740)2.0(2

2962

=−

−=−a

b

520,109)740(2.0)740(296)740( 2 =−=R Thus, 740 units yield a maximum revenue of $109,520.

63. 85

)01.0(27.1

2=

−−=−

ab

25.2448)85(7.1)85(01.0)85( 2 =−+−=P Thus, 85 units yield a maximum profit of $24.25.


Section 1.5 25

65.

2

( ) ( ) ( )(102.50 0.1 ) (52.50 1840)

0.1 50 1840

P x R x C xx x x

x x

= −= − − +

=− + −

The break-even points occur when or P(x) = 0. )()( xCxR =

Thus, 1840501.00 2 −+−= xx250 50 4( 0.1)( 1840)2( 0.1)

50 17640.2

50 420.2

x− ± − − −

=−

− ±=

−− ±

=−

40 or 460x x= = The break-even points occur when x = 40 or x = 460.

67. Let x = the number of people that take the tour. a.

2

( ) (15.00 0.25(60 ))(15.00 15 0.25 )

0.25 30.00

R x x xx x

x x

= + −= + −

=− +

b. 2

2

( ) ( ) ( )

( 0.25 30.00 ) (180 2.50 )

0.25 27.50 180

P x R x C x

x x x

x x

= −

= − + − +

=− + −

c. 55)25.0(2

50.272

=−

−=−a

b

25.576$

180)55(50.27)55(25.0)55( 2

=−+−=P

d. The maximum profit occurs when x = 55.

69. ttth 12816)( 2 +−=

a. seconds 4)16(2

1282

=−

−=−a

b

b. feet 256)4(128)4(16)4( 2 =+−=h

c. 20 16 1280 16 ( 8)

16 0 or 8 00 8

t tt t

t tt t

=− +=− −

− = − == =

The projectile hits the ground at t = 8 seconds.

71. 519.1014.0)( 2 ++−= xxxy

feet 302875.305)5.42(19.1)5.42(014.0)5.42(

5.42)014.0(2

19.12

2

≈=++−=

=−

−=−

y

ab



73.

The perimeter is 48 = . hrhr +++ 2 πSolve for h.

( ) hrr

hrr

=−−

=−−

2 π4821

22 π48

Area = semicircle + rectangle

rr

rr

rrrr

rrrr

rrrr

rhrA

48 2π21

48 2ππ21

2 π48 π21

)2 π48( π21

)2 π48(212 π

21

2 π21

2

2

222

2

2

2

+⎟⎠

⎞⎜⎝

⎛ −−=

+⎟⎠

⎞⎜⎝

⎛ −−=

−−+=

−−+=

−−⎟⎠

⎞⎜⎝

⎛+=

+=

Graph the function A to find that its maximum occurs when r ≈ 6.72 feet. Xmin = 0, Xmax = 14, Xscl = 1 Ymin = −50, Ymax = 200, Yscl = 50

1 (48 π 2 )21 (48 π(6.72) 2(6.72))26.72 feet

h r r= − −

≈ − −

≈

Hence the optimal window has its semicircular radius equal to its height.

Note: Using calculus it can be shown that the exact

value of r = h = .4π

48+


Section 1.5 27

....................................................... Connecting Concepts 75. abxbaxxf ++−= )()( 2

a. x-intercepts occur when y = 0.

bxaxbxaxbxax

abxbax

===−=−

−−=++−=

0or 0

))((0)(0 2

Thus the x-intercepts are (a, 0) and (b, 0).

b. 2)1(2

)(2

babaa

b +=

+=− which is the x-coordinate of

the midpoint of the segment joining (a, 0) and (b, 0).

77. know We.)(Let 2 cbxaxxf ++=

)2( 224or 1424have we(1)Equation from and 4 implies This

4)0()0(0

(1) 1)2()2()2(2

2

−=+=++=

=++=

=++=

babac

cba)f(

cbaf

The x-value of the vertex is 2, and by the vertex formula we

have 2 = a

b

2

− , which implies b = −4a.

Substituting –4a for b in Equation (2) gives us

43343843)4(24

=

−=−−=−−=−+

a

aaa

aa

Substituting 34

for a in Equation (2) gives us

362323

32434

−=−=−=+

−=+⎟⎠

⎞⎜⎝

⎛

bbb

b

Thus the desired quadratic function is

.4343)( 2 +−= xxxf

79. 32 2 2

16P x

x w= = += +

w

a. xw −= 16b. Area

2(16 )

16

A xwA x x

A x x

== −

= −

81. The discriminant is which is always positive. Thus the equation has two real zeros for all values of b.

,4)1)(1(4 22 +=−− bb

83. Increasing the constant c increases the height of each point on the graph by c units. 85.

Let x = one number. Then 8 – x = the other number. .4at vertex ,8)8(2

8

22 ===−=−=

−

−−

a

bxxxxxP

4 and 4 are numbers The .48 and 4 Thus, =−= xx . 87. 3

223

11 )( , , , hxyhxxxyxx +=+===

22223223322333

1212 33)33(3333)( hhxx

hhhxxh

hhxhhx

hxhxhhxx

xhxxhx

xxyym ++=

++=

++=

−+++=

−+−+

−−−

=



....................................................... Prepare for Section 1.6 89. 2( ) 4 6

4 22 2(1)

2

f x x xba

x

= + −

− =− =−

=−

90. 4

2

4

2

3(3) 243(3) 24.310(3) 1

3( 3) 243( 3) 24.310( 3) 1

(3) ( 3)

f

f

f f

= = =+

−− = = =− +

= −

91. 3

3( 2) 2( 2) 5( 2) 16 10 6

(2) [2(2) 5(2)] [16 10] 6( 2) (2)

f

ff f

− = − − − =− + =−

− =− − =− − =−− =−

92. 2

2

2

2

2

( 2) ( 2) ( 2) [ 2 3] 4 1 3

( 1) ( 1) ( 1) [ 1 3] 1 2 1

(0) (0) (0) [0 3] 0 3 3

(1) (1) (1) [1 3] 1 4 3

(2) (2) (2) [2 3] 4 5 1

f g

f g

f g

f g

f g

− − − = − − − + = − =

− − − = − − − + = − =−

− = − + = − =−

− = − + = − =−

− = − + = − =−

93. 0,

2 2a a b b b− + += =

midpoint is (0, b)

94. 0, 02 2

a a b b− + − += =

midpoint is (0, 0)

Section 1.6

1.

3.

5.

7.

9.

11.

13. Replacing leaves the equation unaltered. Thus the graph is symmetric with respect to the y-axis. xx −by 15. Not symmetric with respect to either axis. (neither) 17. Symmetric with respect to both the x- and the y-axes. 19. Symmetric with respect to both the x- and the y-axes. 21. Symmetric with respect to both the x- and the y-axes. 23. No, since simplifies to (−y) = −3x – 2, which is not equivalent to the original equation 2)(3)( −−=− xy .23 −= xy 25. Yes, since implies which is the original equation. 3)()( xy −−=− ,or 33 xyxy −==− 27. Yes, since simplifies to the original equation. 10)()( 22 =−+− yx 29.

Yes, since xxy

−−

=− simplifies to the original equation.


Section 1.6 29

31.

symmetric with respect to the y-axis

33.

symmetric with respect to the origin

35.

symmetric with respect to the origin

37.

symmetric with respect to the line x = 4

39.

symmetric with respect to the line x = 2

41.

no symmetry

43. Even since ).(77)()( 22 xgxxxg =−=−−=− 45. Odd, since 5 3

5 3( ) ( ) ( )

( ).

F x x x

x xF x

− = − + −

=− −=−

47. Even 49. Even 51. Even 53. Even 55. Neither 57.

59. a. ( 2)f x+

b. ( ) 2f x +



61. a. ( 3)( 2 3,5) ( 5,5)(0 3, 2) ( 3, 2)(1 3,0) ( 2,0)

f x+− − = −

− − = − −− = −

63. a. ( )f x−

b. ( ) 1

( 2,5 1) ( 2,6)(0, 2 1) (0, 1)(1,0 1) (1,1)

f x +− + = −

− + = −+ =

b. ( )f x−

65. a. ( )

( 1,3) (1,3)( 2, 4)

f x−−− =− −

67.

b. ( )

( 1, 3)(2, 4) (2,4)

f x−− −

−− =

69.

71. a.

b.


Section 1.6 31

73. a.

75.

b.

77.

79.

81.

83. a.

b.

c.

....................................................... Connecting Concepts 85. a.

11)1(

2)( 2 +++

=x

xf

b.

1)2(

2)( 2 +−−=

xxf



....................................................... Prepare for Section 1.7 87. 2 2(2 3 4) ( 3 5) 1x x x x x+ − − + − = +2 2 88. 2 3 2

3 2(3 2)(2 3) 6 2 4 9 3 6

6 11 7 6

x x x x x x x x

x x x

− + − = − + − + −

= − + −

89. 2

2(3 ) 2(3 ) 5(3 ) 2

18 15 2

f a a a

a a

= −

= − +

+

2

90. 2

2

2

(2 ) 2(2 ) 5(2 ) 2

2 8 8 5 10

2 3

f h h h

h h h

h h

+ = + − + +

= + + − − +

= +

91. Domain: all real numbers except x = 1 92. 2 8 0

4x

x− =

=

Domain: x > 4

Section 1.7

1. 2

2( ) ( ) ( 2 15) ( 3)

12 Domain all real numbers

f x g x x x x

x x

+ = − − + +

= − −

2

2( ) ( ) ( 2 15) ( 3)

3 18 Domain all real numbers

f x g x x x x

x x

− = − − − +

= − −

2

3 2( ) ( ) ( 2 15)( 3)


f x g x x x x

x x x

= − − +

= + − −

{ }2( ) / ( ) ( 2 15) /( 3)5 Domain | 3

f x g x x x xx x x

= − − += − ≠ −

3. 2( ) ( ) (2 8) ( 4)3 12 Domain all real numbers

f x g x x xx

+ = + + += +

( ) ( ) (2 8) ( 4)4 Domain all real numbers

f x g x x xx

− = + − += +

2( ) ( ) (2 8)( 4)

2 16 32 Domain all real numbers

f x g x x x

x x

= + +

= + +

[ ]{ }

( ) / ( ) (2 8) /( 4)2( 4) /( 4)

2 Domain | 4

f x g x x xx x

x x

= + += + +

= ≠ −

5. 3 2

3 2( ) ( ) ( 2 7 )


f x g x x x x x

x x x

+ = − + +

= − +

3 2

3 2( ) ( ) ( 2 7 )


f x g x x x x x

x x x

− = − + −

= − +

3 2

4 3 2( ) ( ) ( 2 7 )


f x g x x x x x

x x x

= − +

= − +

{ }

3 2

2( ) / ( ) ( 2 7 ) /

2 7 Domain | 0

f x g x x x x x

x x x x

= − +

= − + ≠

7. 2 2

2( ) ( ) (2 4 7) (2 3 5)

4 7 12 Domain all real numbers

f x g x x x x x

x x

+ = + − + + −

= + −

2 2( ) ( ) (2 4 7) (2 3 5)2 Domain all real numbers

f x g x x x x xx

− = + − − + −= −

2 2

4 3 2 3 2 2

4 3 2

( ) ( ) (2 4 7)(2 3 5)

4 6 10 8 12 20 14 21 35

4 14 12 41 35 Domain all real numbers

f x g x x x x x

x x x x x x x x

x x x x

= + − + −

= + − + + − − − +

= + − − +

2 2

2

( ) / ( ) (2 4 7) /(2 3 5)2 51 Domain | 1,

22 3 5

f x g x x x x xx x x x

x x

= + − + −

⎧ ⎫−= + ≠ ≠ −⎨ ⎬⎩ ⎭+ −


Section 1.7 33

9. =+ )()( xgxf { }3 Domain | 3x x x x− + ≥

=− )()( xgxf { }3 Domain | 3x x x x− − ≥

=)()( xgxf { }3 Domain | 3x x x x− ≥

=)(/)( xgxf { }3 Domain | 3x x xx−

+ ≥x

11. =+ )()( xgxf { }22|Domain 24 2 ≤≤−++− xxxx

=− )()( xgxf { }22|Domain 24 2 ≤≤−−−− xxxx

=)()( xgxf ( ) { 22|Domain 24 2 ≤≤−+⎟⎠

⎞⎜⎝

⎛ − xxxx }

=)(/)( xgxf { }22|Domain 24 2

≤≤−+− xx

xx

13. 2))(( 2 −−=+ xxxgf

182525

2)5()5()5)(( 2

=−−=

−−=+ gf

15. 2

2( )( ) 2

1 1 1( )2 2 2

1 1 24 2

94

f g x x x

f g

+ = − −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 2+ = − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − −

= −

17. 2

2( )( ) 5 6

( )( 3) ( 3) 5( 3) 69 15 630

f g x x x

f g

− = − +

− − = − − − += + +=

19. 2

2( )( ) 5 6

( )( 1) ( 1) 5( 1)1 5 612

f g x x x

f g

− = − +

− − = − − − += + +=

6

21. ( )( )2

3 2 2

3 2

3 2

( )( ) 3 2 2 4

2 6 4 4 12

2 10 16 8

( )(7) 2(7) 10(7) 16(7) 8686 490 112 8300

fg x x x x

x x x x x

x x x

fg

= − + −

8= − + − + −

= − + −

= − + −= − + −=

23. 3 2

3 2( )( ) 2 10 16 8

2 2 2 2( ) 2 10 16 85 5 5 5

16 40 32 8125 25 5

384 3.072125

fg x x x x

fg

= − + −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

= − + −

−= = −

−

25.

( )

2 3 2( )2 4

1 1( )2 2

1 1( 4) 42 2

122

1 52 or 2 2

f x xxg x

f x xg

fg

⎛ ⎞ − +=⎜ ⎟ −⎝ ⎠

⎛ ⎞= −⎜ ⎟

⎝ ⎠⎛ ⎞

− = − −⎜ ⎟⎝ ⎠

= − −

= − −

27. 1 1( )

2 2

1 1 12 2 2

1 14 2

14

f x xg

fg

⎛ ⎞= −⎜ ⎟

⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠

= −

= −

12

29. [ ]2( ) 4 (2 4)( ) ( )

2 2( ) 4 2 4

2

2

x h xf x h f xh h

x h xh

hh

+ + − ++ − =

+ + − −=

=

=



31. [ ] 2

2 2

2

( ) 6 ( 6)( ) ( )

2 ( ) ( ) 6 6

2 ( )

2

x h xf x h f xh h

x x h h xh

x h hh

x h

+ − − −+ − =

+ + − − +=

+=

= +

2

33. 2 2

2 2 2

2

( ) ( ) 2( ) 4( ) 3 (2 4 3)

2 4 2 4 4 3 2 4

4 2 4

4 2 4

f x h f x x h x h x xh h

3x xh h x h x xh

xh h hh

x h

+ − + + + − − + −=

+ + + + − − − +=

+ +=

= + +

35. 2 2

2 2 2

2

( ) ( ) 4( ) 6 ( 4 6)

4 8 4 6 4

8 4

8 4

f x h f x x h xh h

x xh h xh

xh hh

x h

+ − − + + − − +=

− − − + + −=

− −=

=− −

6

37. [ ]

[ ][ ]

( )( ) ( )3 5

2 3 56 10 76 3

g f x g f xg x

xxx

=

= +

= += + −= +

o [ ][ ]

[ ]

( )( ) ( )2 7

3 2 7 56 21 56 16

f g x f g xf x

xxx

=

= −

= − += − += −

o

39. 2

2

2

( )( ) 4 1

4 1 2

4 1

g f x g x x

x x

x x

⎡ ⎤= + −⎢ ⎥⎣ ⎦⎡ ⎤= + − +⎢ ⎥⎣ ⎦

= + +

o [ ][ ] [ ]2

2

2

( )( ) 2

2 4 2 1

4 4 4 8 1

8 11

f g x f x

x x

x x x

x x

= +

= + + + −

= + + + + −

= + +

o

41. [ ]

3

3

3

( )( ) ( )

2

5 2

5 10

g f x g f x

g x x

x x

x x

=

⎡ ⎤= +⎢ ⎥⎣ ⎦⎡ ⎤=− +⎢ ⎥⎣ ⎦

=− −

o [ ][ ]

[ ] [ ]3

3

( )( ) ( )5

5 2 5

125 10

f g x f g xf x

x x

x x

=

= −

= − + −

=− −

o

43. [ ]( )( ) ( )

21

23 515( 1)6

1 16 5 5

11 5

1

g f x g f x

gx

xx

x xx

xx

x

=

⎡ ⎤= ⎢ ⎥+⎣ ⎦⎡ ⎤= −⎢ ⎥+⎣ ⎦

+= −+ +− −=

+−=+

o [ ][ ]

[ ]

( )( ) ( )3 5

23 5 1

23 4

f g x f g xf x

x

x

=

= −

=− +

=−

o


Section 1.7 35

45. [ ]

2

2

2

2

2

( )( ) ( )

1

1 1

1

1| |

g f x g f x

gx

x

xx

xx

=

⎡ ⎤= ⎢ ⎥

⎣ ⎦

⎡ ⎤= −⎢ ⎥

⎣ ⎦

−=

−=

o [ ]

2

( )( ) ( )

1

1

1

11

f g x f g x

f x

x

x

=

⎡ ⎤= −⎣ ⎦

=⎡ ⎤−⎣ ⎦

=−

o

47.

3( )( )5

2

35

2 53

g f x gx

x

x

⎡ ⎤= ⎢ ⎥

−⎢ ⎥⎣ ⎦

=−⎡ ⎤⎢ ⎥

−⎢ ⎥⎣ ⎦− −

=

o

[ ]

2( )( )

325

325

35x 2

3

5 2

f g x fx

x

x

xx

x

⎡ ⎤= −⎢ ⎥⎣ ⎦

=⎡ ⎤− −⎢ ⎥⎣ ⎦

=+

=+

=+

o

Use the results to work Exercises 49 to 63. 49. 2

2( )( ) 4 2 6

( )(4) 4(4) 2(4) 664 8 666

g f x x x

g f

= + −

= + −= + −=

o

o

51. 2

2( )( ) 2 10 3

( )( 3) 2( 3) 10( 3) 318 30 351

f g x x x

f g

= − +

− = − − − += + +=

o

o

53. 4 2

4 2( )( ) 9 9 4

( )(0) 9(0) 9(0) 44

g h x x x

g h

= − −

= − −=−

o

o

55. ( )( ) 4 9( )(8) 4(8) 9

41

f f x xf f

= += +=

o

o

57. 4 3 2

4 3 2( )( ) 3 30 75 4

2 2 2 2( ) 3 30 755 5 5 5

48 240 300 4625 125 2548 1200 7500 2500

6253848625

h g x x x x

h g

=− + − +

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞=− + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

=− + − +

− + − +=

=−

o

o 4

59. 2

2( )( ) 4 2 6

( )( 3) 4( 3) 2( 3) 6

12 2 3 6

6 2 3

g f x x x

g f

= + −

= + −

= + −

= +

o

o

61. 2

2

2

( )( ) 4 2 6

( )(2 ) 4(2 ) 2(2 ) 6

16 4 6

g f x x x

g f c c c

c c

= + −

= +

= + −

o

o −

3

63. 4 2

4 2

4 3 2 2

4 3 2 2

4 3 2

( )( ) 9 9 4( )( 1) 9( 1) 9( 1) 4

9( 4 6 4 1) 9 18 9 49 36 54 36 9 9 18 19 36 45 18 4

g h x x xg h k k k

k k k k k kk k k k k kk k k k

= − −

+ = + − + −

= + + + + − − − −

= + + + + − − −

= + + + −

o

o



65. a. 2 and 5.1 rAtr π==

[ ]22

2

so ( ) ( )

(1.5 )

2.25 (2)9 square feet28.27 square feet

A t r t

t

π

π

ππ

=

=

==≈

b. 5.1 tr =

[ ]

2

2

3

2 2(1.5 ) 3 and1 so31( ) (1.5 ) 3t 3

2.25

h r t t

V r h

V t t

t

π

π

π

= = =

=

=

=

Note:

( )

2 2

2 3

3

1 1 1 ( )3 3 31 3 (2.25 ) 2.253

(3) 2.25 (3)60.75 cubic feet190.85 cubic feet

V r h r hA

t t

V

π π

tπ π

ππ

= = =

= =

==≈

67. a. Since 2 2 2

2 2

2

2

2

2

4 ,

16

16

(48 ) 16 48

2304 96 16

96 2288

d s

d s

d s

d t s

t t

t t

+ =

= −

= −

t= − − ⋅ = −

= − + −

= − +

b.

2

(35) 48 35 13

(35) 35 96(35) 2288

153 12.37 ft

s

d

= − +

= − +

= ≈

69. ))(())(( xFYxFY =o converts x inches to yards.

F takes x inches to feet, and then Y takes feet to yards.

71. a. On [ ]0, 1 , 0a =

0)0()(8.99)1()(

101

====∆+

=−=∆

CaCCtaC

t

Average rate of change 8.9908.991

)0()1(=−=

−=

CC

This is identical to the slope of the line through

(0, C(0)) and (1, C(1)) since (1) (0) (1) (0)1 0

C Cm C−= = −

−C

b. On [ ]0, 0.5 , 0a = 5.0=∆t

Average rate of change (0.5) (0) 78.1 0 156.20.5 0.5

C C− −= = =

c. On [ ]1, 2 , 1a = 112 =−=∆t

Average rate of change (2) (1) 50.1 99.8 49.71 1

C C− −= = = −

d. On [ ]1, 1.5 , 1a = 5.015.1 =−=∆t

Average rate of change (1.5) (1) 84.4 99.8 15.4 30.80.5 0.5 0.5

C C− − −= = = = −


Section 1.7 37

71. e. On [ ]1, 1.25 , 1a = 25.0125.1 =−=∆t

Average rate of change (1.25) (1) 95.7 99.8 4.1 16.40.25 0.25 0.25

C C− − −= = = = −

f. On [ ]1, 1 , t+ ∆ Con ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( ) ( )

( ) ( )

3 2

3 2

2 3 2

2 3

1 25 1 150 1 225 1

25(1 3( ) 3 ) 150(1 2 ) 225(1 )

25 75( ) 75 ) 25 150 300 150 225 225

100 75 25(1) 100

t t t t

t t t t t

t t t t t

t tCon

+∆ = +∆ − +∆ + +∆

= + ∆ + ∆ − + ∆ + ∆ + +∆

= + ∆ + ∆ + ∆ − − ∆ − ∆ + + ∆

= − ∆ + ∆=

t

Average rate of change ( )32

2 3

2

100 75( ) 25 100(1 ) (1)

75( ) 25( )

75( ) 25( )

t tCon t Cont t

t tt

t t

− ∆ + ∆ −+∆ −= =∆ ∆

− ∆ + ∆=∆

= − ∆ + ∆

As approaches 0, the average rate of change over t∆ [ ]1, 1 t+ ∆ seems to approach 0.

....................................................... Connecting Concepts 73. [ ]

[ ]( )( ) ( )

2 35(2 3) 1210 15 1210 27

g f x g f xg x

xxx

=

= += + += + += +

o [ ][ ]

( )( ) ( )5 12

2(5 12) 310 24 310 27

f g x f g xf x

xxx

=

= += + += + += +

o

( )( ) ( )( )g f x f g x=o o 75. [ ]

61

6130 30

1 16 2 2 4 2

1 1

( )( ) ( )

61

5

2

30 11 2(2 1)

152 1

xx

xx

x xx x

x x xx x

g f x g f x

xgx

x xx x

xx

⎛ ⎞⎜ ⎟

−⎝ ⎠

−

− −− + +

− −

=

⎡ ⎤= ⎢ ⎥−⎣ ⎦

=−

= =

−= ⋅− +

=+

o

[ ]

52

52

5 2 42 2

( )( ) ( )

52

6

1

30 302 2

30 22 2(2 1)

152 1

xxx

x

2x x xx x

f g x f g x

xfx

x xx x

x xx x

xx

−

−

− + +− −

=

⎡ ⎤= ⎢ ⎥−⎣ ⎦⎛ ⎞⎜ ⎟⎝ ⎠=

−

− −= =

−= ⋅− +

=+

o

( )( ) ( )( )g f x f g x=o o 77. [ ]

[ ][ ]

( )( ) ( )2 3

2 3 32

22

g f x g f xg x

x

x

x

=

= +

+ −=

=

=

o [ ]( )( ) ( )

3232 3

23 3

f g x f g x

xf

x

xx

=

⎡ ⎤−= ⎢ ⎥⎣ ⎦⎡ ⎤−= +⎢ ⎥⎣ ⎦

= − +=

o



79. [ ]( )( ) ( )

4144

14

14 4 4

14

14

1 4

g f x g f x

gx

x

xxx

xx x

xx

=

⎡ ⎤= ⎢ ⎥+⎣ ⎦⎡ ⎤−⎢ ⎥+⎣ ⎦=

⎡ ⎤⎢ ⎥+⎣ ⎦

+ −+=

++= ⋅

+=

o

1

[ ]( )( ) ( )

4

44 1

44

44

44

f g x f g x

xfx

xx

x xx

xx

x

=

⎡ ⎤−= ⎢ ⎥⎣ ⎦

=⎡ ⎤− +⎢ ⎥⎣ ⎦

=− +

=

= ⋅

=

o

81. [ ]

3

33

3 3

( )( ) ( )

1

1 1

g f x g f x

g x

x

xx

=

⎡ ⎤= −⎢ ⎥⎣ ⎦

⎡ ⎤= − +⎢ ⎥⎣ ⎦

==

o [ ]3

33

( )( ) ( )

1

1 1

1 1

f g x f g x

f x

x

xx

=

⎡ ⎤= +⎣ ⎦

⎡ ⎤= + −⎣ ⎦= + −=

o

....................................................... Chapter 1 True/False Exercises 1. False. .33but ,9)3()3(Then .)(Let 2 −≠=−== ffxxf 2. False. .)( and 2)(Let 2xxgxxf == .4)2()2())((but ,2x)())((Then 2222 xxxgxfgxfxgf ===== oo

3. True 4. True 5. False . .9)3(3)3()]([ whereas,9]3[)]([ .3)(Let 222 xxxfxffxxxfxxf ====== 6.

False . .124

14

12

)1()2(Then .)(Let 2

22 ≠====

ffxxf

7. True 8. False . .431)3()1( whereas,2)2()31(Then .)(Let =+=+−==+−= ffffxxf 9. True 10. True 11. True 12. True 13. True

....................................................... Chapter Review 1. 1243 =− z [1.1]

4 9

94

z

z

− =

= −

2. 5634 +=− yy [1.1]

4

82−=

=−yy

3. xxx 14)32(32 =−− [1.1]

xxxxx

=−=−=+−

23614962


Chapter Review 39

4. )1(3)23(25 mm −=+− [1.1]

32

23336133465

−=

=−−=−−=−−

m

mmmmm

5. [1.1] 01832 =−− yy

3or 6

03or 06

0)3)(6(

−==

=+=−

=+−

yy

yy

yy

6. [1.1] 0492 2 =+− zz0)4)(12( =−− zz

21

4 1204or 012

=

===−=−

z

zzzz

7. [1.1] 13 2 =+ vv

013 2 =−+ vv

613 1

)3(2)1)(3(41 1 2

±−=

−−±−=

v

v

8. [1.1] 2243 ss −=

4413

43293

)2(2)4)(2(43 3

0432

2

2

±−=

+±−=

−−±−=

=−+

s

s

s

ss

9. 7553 +≤− cc [1.1]

6

122−≥

≤−cc

10. [1.1] )43(257 −−> aa

11313

8657

>>

+−>

aa

aa

11. [1.1] 0122 ≥−− xx

] [ ) ,4 3 ,(

.3 and 4 are valuesCritical

0)3)(4(

∞∪−−∞

−

≥+− xx

12. [1.1] 12 2 <− xx

0)1)(12(012 2

<−+<−−

xxxx

Critical values are 21

− and 1.

121

<<− x

13. 352 >−x [1.1]

1 422 82352or 352

<><>

−<−>−

xxxx

xx

) ,4( )1 ,( ∞∪−∞

14. 431 ≤− x [1.1]

135

335 4314

−≥≥

≤−≤−≤−≤−

x

xx

15. 2 2

2 2

(7 ( 3)) (11 2)

10 9 100 81 181

d = − − + −

= + = + =

[1.2] 16. 2 2

2 2

( 3 5) ( 8 ( 4))

( 8) ( 4) 64 16 80 4 5

d = − − + − − −

= − + − = + = =

[1.2]

17.

⎟⎠

⎞⎜⎝

⎛−=⎟⎠

⎞⎜⎝

⎛ −=⎟

⎠

⎞⎜⎝

⎛ +−+ 10 ,21

220 ,

21

2128 ,

2)3(2 [1.2]

18. ( 2 ,224 ,

24

2)11(7 ,

284

−=⎟⎠

⎞⎜⎝

⎛ −=⎟

⎠

⎞⎜⎝

⎛ −++− ) [1.2]

19. center (3, −4), radius 9 [1.2] 20. 2 2

2 2

2 2

10 4 20

10 25 4 4 20 25 4

( 5) ( 2) 9

x x y y

x x y y

x y

+ + + =−

+ + + + + =− + +

+ + + =

[1.2]

center (−5, −2), radius 3



21. 222 5)3()2( =++− yx [1.2] 22. 2 2

2 2

2 2 2

2 2

2 2

( 5) ( 1)

(3 5) (1 1)

8 0

8

( 5) ( 1) 8

2

2

2

x y r

r

r

r

x y

+ + − =

+ + − =

+ =

=

+ + − =

[1.2]

23. a. [1.3] 2(1) 3(1) 4(1) 5

3(1) 4 53 4 52

f = + −= + −= + −=

b. 2( 3) 3( 3) 4( 3) 53(9) 12 527 12 510

f − = − + − −= − −= − −=

c. 543)( 2 −+= tttf

d. 2

2 2

2 2

( ) 3( ) 4( ) 5

3( 2 ) 4 4 5

3 6 3 4 4 5

f x h x h x h

x xh h x h

x xh h x h

+ = + + + −

= + + + + −

= + + + + −

e. 2

23 ( ) 3(3 4 5)

9 12 15

f t t t

t t

= + −

= + −

f. 2

2

2

(3 ) 3(3 ) 4(3 ) 5

3(9 ) 12 5

27 12 5

f t t t

t t

t t

= + −

= + −

= + −

24. a. 2(3) 64 3

64 955

g = −

= −

=

[1.3]

b. 2( 5) 64 ( 5)

64 2539

g − = − −

= −

=

c. 2(8) 64 (8)

64 640

0

g = −

= −

==

d. 2

2

( ) 64 ( )

64

g x x

x

− = − −

= −

e. 2642)(2 ttg −=

f. 2

2

2

2

(2 ) 64 (2 )

64 4

4(16 )

2 16

g t t

t

t

t

= −

= −

= −

= −


Chapter Review 41

25. a. [1.7]

2

( )(3) [ (3)][3 8][ 5]

( 5) 4( 5)25 205

f g f gff

== −= −

= − + −= −=

o

b. 2

( )( 3) [ ( 3)]

[( 3) 4( 3)][9 12][ 3]

[ 3] 811

g f g f

ggg

− = −

= − + −= −= −= − −= −

o

c.

2

2

2

( )( ) [ ( )][ 8]

( 8) 4( 8)

16 64 4 32

12 32

f g x f g xf x

x x

x x x

x x

== −

= − + −

= − + + −

= − +

o

d. 2

2

2

( )( ) [ ( )]

[ 4 )]

[ 4 ]

4 8

g f x g f x

8

g x x

x x

x x

=

= +

= + −

= + −

o

26. a. [ ]

2

( )( 5) [ ( 5)]| ( 5) 1 |

[6]

2(6) 772 779

f g f gff

− = −

= − −

=

= += +=

o [1.7]

b. 2

( )( 5) [ ( 5)]

[2( 5) 7][57]

| 57 1 |56

g f g f

gg

− = −

= − +== −=

o

c. [ ]

2

2

2

2

2

( )( ) [ ( )]| 1 |

2(| 1 |) 7

2( 1) 7

2( 2 1) 7

2 4 2 7

2 4 9

f g x f g xf x

x

x

x x

x x

x x

=

= −

= − +

= − +

= − + +

= − + +

= − +

o

d. 2

2

2

2

( )( ) [ ( )]

[2 7]

| 2 7 1 |

| 2 6 |

2 6

g f x g f x

g x

x

x

x

=

= +

= + −

= +

= +

o

27. 2 2

2 2 2

2 2 2

2

( ) ( ) 4( ) 3( ) 1 (4 3 1)

4( 2 ) 3 3 1 4 3 1

4 8 4 3 3 1 4 3 1

8 4 3

(8 4 3)

8 4 3

f x h f x x h x h x xh h

x xh h x h x xh

x xh h x h x xh

xh h hh

h x hh

x h

+ − + − + − − − −=

+ + − − − − + +=

+ + − − − − + +=

+ −=

+ −=

= + −

[1.7]

28. 3 3

3 2 2 3 3

2 2 3

2 2

2 2

( ) ( ) ( ) ( ) ( )

3 3

3 3

(3 3 1)

3 3 1

g x h g x x h x h x xh h

x x h xh h x h x xh

x h xh h hh

h x xh hh

x xh h

+ − + − + − −=

+ + + − − − +=

+ + −=

+ + −=

= + + −

[1.7]



29.

f is increasing on [3, ∞) f is decreasing on (−∞, 3] [1.3]

30.

f is increasing on [0, ∞) f is decreasing on (−∞, 0] [1.3]

31.

f is increasing on [−2, 2] f is constant on (−∞, −2] ∪ [2, ∞) [1.3]

32.

f is constant on . . . , [−6, −5), [−5, −4), [−4, −3), [−3, −2), [−2, −1), [−1, 0), [0, 1), . . . [1.3]

33.

f is increasing on (−∞, ∞) [1.3]

34.

f is increasing on (−∞, ∞) [1.3]

35. Domain { }number real a is xx [1.3] 36. Domain { }6≤xx [1.3] 37. Domain { }55 ≤≤− xx [1.3] 38. Domain { }5 ,3 ≠−≠ xxx [1.3] 39.

2510

)1(437

−=−

=−−−−

=m [1.4]

12223

form slope-point )1(23

+−=−−=−+−=−

xyxyxy

40. 711

07011

=−−

=m [1.4]

110 ( 07

117

y x

y x

)− = −

=

41. 3 4 8

4 33 24

x yy x

y x

− =− =− +

= −

8 [1.4]

Slope of parallel line is .43

219

43

222

23

43

1123

43

23

4311

)2(4311

+=

+−=

+−=

−=−

−=−

xy

xy

xy

xy

xy

42. 2 5 15 2 1

2 25

x yy x

y x

00

=− +=− +

=− +

[1.4]

Slope of perpendicular line is .25

5( 7) [ ( 3)]257 ( 3)25 1572 25 15 72 25 15 142 2 25 12 2

y x

y x

y x

y x

y x

y x

− − = − −

+ = +

+ = +

= + −

= + −

= +


Chapter Review 43

43. 2

2

2

( ) ( 6 ) 10

( ) ( 6 9) 10 9

( ) ( 3) 1

f x x x

f x x x

f x x

= + +

= + + + −

= + +

[1.5] 44. 5)42()( 2 ++= xxxf [1.5]

3)1(2)(

25)12(2)(

5)2(2)(

2

2

2

++=

−+++=

++=

xxf

xxxf

xxxf

45. 38)( 2 +−−= xxxf [1.5]

19)4()(

163)168()(

3)8()(

2

2

2

++−=

++++−=

++−=

xxf

xxxf

xxxf

46. 1)64()( 2 +−= xxxf [1.5]

45

434)(

49

44

434)(

491

169

234)(

1234)(

2

2

2

2

−⎟⎠⎞

⎜⎝⎛ −=

−+⎟⎠

⎞⎜⎝

⎛ −=

−+⎟⎠

⎞⎜⎝

⎛ +−=

+⎟⎠⎞

⎜⎝⎛ −=

xxf

xxf

xxxf

xxxf

47. 543)( 2 −+−= xxxf [1.5]

311

323)(

34

315

323)(

345

94

343)(

5343)(

2

2

2

2

−⎟⎠⎞

⎜⎝⎛ −−=

+−⎟⎠

⎞⎜⎝

⎛ −−=

+−⎟⎠

⎞⎜⎝

⎛ +−−=

−⎟⎠⎞

⎜⎝⎛ −−=

xxf

xxf

xxxf

xxxf

48. 96)( 2 +−= xxxf [1.5]

0)3()(

99)96()(

9)6()(

2

2

2

+−=

−++−=

+−=

xxf

xxxf

xxxf

49.

166

)3(2)6(

2==

−−=

−ab [1.5]

81163

116)1(311)1(6)1(3)1( 2

=+−=

+−=+−=f

Thus the vertex is (1, 8).

50. 0

)4(20

2==

−ab [1.5]

101010

10)0(4)0( 2

−=−=

−=f

Thus the vertex is (0, −10).

51.

51260

)6(2)60(

2=

−−

=−

−=

−ab [1.5]

16111300150

11300)25(611)5(60)5(6)5( 2

=++−=

++−=++−=f

Thus the vertex is (5, 161).

52. 4

28

)1(2)8(

2−=

−=

−−−

=−

ab [1.5]

30163214

)4()4(814)4( 2

=−+=

−−−−=−f

Thus the vertex is (−4, 30).



53. )3 ,1() ,( ,32 ,

1112

11 =−=+

−+= yxxy

m

ybmxd [1.4]

554

54

5

441

33221

3)3()1(22

==

−=

+

−−=

+

−−+=

d

d

d

d

54. a. Revenue = 13x b. Profit = Revenue − Cost

13 (0.5 1050)13 0.5 105012.5 1050

P x xP x xP x

= − += − −= −

c. Break even ⇒ Revenue Cost13 0.5 1050

12.5 105084

x xxx

== +==

The company must ship 84 parcels. [1.4]

55.

[1.6]

56.

[1.6]

57. The graph of is symmetric with respect to the y-axis. [1.6] 72 −= xy 58. The graph of is symmetric with respect to the x-axis. [1.6] 32 += yx 59. The graph of is symmetric with respect to the origin. [1.6] xxy 43 −= 60. The graph of is symmetric with respect to the x-axis, y-axis, and the origin. [1.6] 422 += xy 61.

The graph of 143 2

2

2

2=+

yx is symmetric with respect to the x-axis, y-axis, and the origin. [1.6]

62. The graph of is symmetric with respect to the origin. [1.6] 8=xy 63. The graph of xy = is symmetric with respect to the x-axis, y-axis, and the origin. [1.6] 64. The graph of 4=+ yx is symmetric with respect to the origin. [1.6] 65.

a. Domain all real numbers Range { }4≤yy b. g is an even function [1.6]

66.

a. Domain all real numbers Range all real numbers b. g is neither even nor odd [1.6]


Chapter Review 45

67.

a. Domain all real numbers Range { }4≥yy b. g is an even function [1.6]

68.

a. Domain { }44 ≤≤− xx

Range { }40 ≤≤ yy b. g is an even function [1.6]

69.

a. Domain all real numbers Range all real numbers b. g is an odd function [1.6]

70.

a. Domain all real numbers Range { }integereven an is yy b. g is neither even nor odd [1.6]

71. 74)( 2 −+= xxxF [1.6]

11)2()(

47)44()(

7)4()(

2

2

2

−+=

−−++=

−+=

xxF

xxxF

xxxF

72. 56)( 2 −−= xxxA [1.6]

14)3()(

95)96()(

5)6()(

2

2

2

−−=

−−+−=

−−=

xxA

xxxA

xxxA

73. 43)( 2 −= xxP [1.6]

4)0(3)( 2 −−= xxP

74. 382)( 2 +−= xxxG [1.6]

5)2(2)(

83)44(2)(

3)4(2)(

2

2

2

−−=

−++−=

+−=

xxG

xxxG

xxxG



75. 664)( 2 +−−= xxxW [1.6]

433

434)(

49

424

434)(

496

169

234)(

6234)(

2

2

2

2

+⎟⎠⎞

⎜⎝⎛ +−=

++⎟⎠

⎞⎜⎝

⎛ +−=

++⎟⎠

⎞⎜⎝

⎛ ++−=

+⎟⎠⎞

⎜⎝⎛ +−=

xxW

xxW

xxxW

xxxW

76. xxxT 102)( 2 −−= [1.6]

225

252)(

225

42552)(

)5(2)(

2

2

2

+⎟⎠

⎞⎜⎝

⎛ +−=

+⎟⎠

⎞⎜⎝

⎛ ++−=

+−=

xxT

xxxT

xxxT

77.

[1.6]

78.

[1.6]

79.

[1.6]

80.

[1.3]

81.

[1.3]

82.

[1.3]


Chapter Review 47

83. 2

2( )( ) 9

6

3f g x x x

x x

+ = − + +

= + −

[1.7]

The domain is all real numbers.

2

2

2

( )( ) 9 ( 3

9 3

12

f g x x x

x x

x x

− = − − +

= − − −

= − −

)

7


2

3 2( )( ) ( 9)( 3)

3 9 2

fg x x x

x x x

= − +

= + − −


2 9( )3

( 3)( 3)3

3

f xxg x

x xx

x

⎛ ⎞ −=⎜ ⎟⎜ ⎟ +⎝ ⎠− +=

+= −

The domain is{ }3 .x x ≠ −

84. 3 2

3 2( )( ) 8 2

2 12

4f g x x x x

x x x

+ = + + − +

= + − +

[1.7]


3 2

3 2

3 2

( )( ) 8 ( 2 4

8 2 4

2 4

f g x x x x

x x x

x x x

)− = + − − +

= + − + −

= − + +


3 2

5 4 3 2( )( ) ( 8)( 2 4)

2 4 8 16 3

fg x x x x

x x x x x

= + − +

2= − + + − +


3

2

2

2

8( )2 4

( 2)( 2 4)2 4

2

f xxg x x

x x xx x

x

⎛ ⎞ +=⎜ ⎟⎜ ⎟ − +⎝ ⎠

+ − +=− +

= +

The domain is restricted when . 0422 =+− xx

number real anot is which 2

1222

1642

)1(2)4)(1(4)2()2( 2

−±=

−±=

−−±−−=

x

x

x

Therefore the domain is all real numbers. 85. Let x = one of the numbers and 50 – x = the other number.

Their product is given by 2 2(50 ) 50 50 .y x x x x x x= − = − = − +

Now y takes on its maximum value when

50 50 25.2 2( 1) 2

bxa

− − −= = = =

− −

Thus the two numbers are 25 and (50 – 25) = 25. That is, both numbers are 25. [1.5]

86. Let x = the smaller number. Let x + 10 equal the larger number. The sum of their squares y is given by

2 2

2 2

2

( 10)

20 100

2 20 100

y x x

x x x

x x

= + +

= + + +

= + +

Now y takes on its minimum value when

5420

)2(220

2−=

−=

−=

−=

abx

Thus the numbers are –5 and (−5 + 10) = 5. [1.5]



87. 23)( tts = [1.5]

a. Average velocity 2 23(4) 3(2)4 2

3(16) 3(4)2

48 122

36 18 ft/sec2

−=−−=

−=

= =

b. Average velocity 2 23(3) 3(2)3 2

3(9) 3(4)1

27 12 15 ft/sec1

−=−

−=

−= =

c. Average velocity 2 23(2.5) 3(2)

2.5 23(6.25) 3(4)

0.518.75 12

0.56.75 13.5 ft/sec0.5

−=−−=

−=

= =

d. Average velocity 2 23(2.01) 3(2)

2.01 23(4.0401) 3(4)

0.0112.1203 12

0.010.1203 12.03 ft/sec

0.01

−=−−=

−=

= =

e. It appears that the average velocity of the ball approaches 12 ft/sec.

88. ttts += 22)( [1.5]

a. Average velocity 2 22(5) 5 [2(3) 3]

5 32(25) 5 [2(9) 3]

250 5 [18 3]

250 5 18 3

234 17 ft/sec2

+ − +=−

+ − +=

+ − +=

+ − −=

= =

b. Average velocity 2 22(4) 4 [2(3) 3]

4 32(16) 4 [2(9) 3]

132 4 [18 3]

132 4 18 3

115 15 ft/sec1

+ − +=−

+ − +=

+ − +=

+ − −=

= =

c. Average velocity 2 22(3.5) 3.5 [2(3) 3]

3.5 32(12.25) 3.5 [2(9) 3]

0.524.5 3.5 [18 3]

0.524.5 3.5 18 3

0.57 14 ft/sec

0.5

+ − +=−

+ − +=

+ − +=

+ − −=

= =

d. Average velocity = 2 22(3.01) 3.01 [2(3) 3]

3.01 32(9.0601) 3.01 [2(9) 3]

0.0118.1202 3.01 [18 3]

0.0118.1202 3.01 18 3

0.010.1302 13.02 ft/sec

2

+ − +=−

+ − +=

+ − +=

+ − −=

= =

e. It appears that the average velocity of the ball approaches 13 ft/sec.


Chapter Test 49

....................................................... Chapter Test 1. 2 3(1 ) 3 5

2 3 3 35 3 3 5

2 84

x x xx x x

x xxx

− − = +− + = +

− = +==

5 [1.1] 2. 2

23 5

3 5 0x x

x x− =

− − = [1.1]

2( 3) ( 3) 4(1)( 5)2(1)

3 9 202

3 292

x

x

x

− − ± − − −=

± +=

±=

3. 4 5 6 7

12 26

6

x xx

xx

− ≥ +− ≥

− ≥≤ −

[1.1] 4. 2 4 12 0( 6)( 2) 0

x xx x

+ − ≤+ − ≤

[1.1]

Critical values are 6 and 2.

6 2x− ≤ ≤ 5.

midpoint = )1 ,1(22 ,

22

2)1(3 ,

242

2 ,

22121 =⎟

⎠

⎞⎜⎝

⎛=⎟⎠

⎞⎜⎝

⎛ −++−=⎟

⎠

⎞⎜⎝

⎛ ++ yyxx [1.2]

length = 1325216364)6())1(3()42()()( 2222221

221 ==+=+−=−−+−−=−+−= yyxxd

6. 42 2 −= yx [1.2]

44)0(20 2 −=−=⇒= xy Thus the x-intercept is (−4, 0).

2

2

2

0 0 2 4

4 2

2 2

x y

y

yy

= ⇒ = −

=

=

± =

Thus the y-intercepts are (0, 2)−

and (0, 2).

7. 12 ++= xy [1.2]

8. 2 2

2 2

2 2

2 2

4 2 4 0

( 4 ) ( 2 ) 4

( 4 4) ( 2 1) 4 4

( 2) ( 1) 9

x x y y

x x y y

x x y y

x y

− + + − =

− + + =

1− + + + + = + +

− + + =

[1.2]

center (2, −1), radius 3

9. 2 16 0

( 4)( 4) 0x

x x− ≥

− + ≥

The product is positive or zero. The critical values are 4 and −4.

The domain is { }4or 4 −≤≥ xxx . [1.3]

10.

a. increasing on (−∞, 2] b. never constant c. decreasing on [2, ∞ ) [1.3]



11. a. R = 12x

b. P = revenue − cost

87525.11

)87575.6(12−=

+−=xP

xxP

c. break-even 0=⇒ P

x

xx

≈=

−=

7825.11875

87525.110

78 parcels must be sent to break even. [1.5]

12.

[1.6]

13. a. 4 2

4 2 4 2( )

( ) ( ) ( ) ( )( ) is an even function.

f x x x

f x x x x x f xf x

= −

− = − − − = − =

[1.6]

b. 3

3 3

3

( )

( ) ( ) ( )

( ) ( )( ) is an odd function.

f x x x

f x x x x x

x x f xf x

= −

− = − − − =− +

=− − =−

c. ( ) 1( ) 1 ( ) not an even function( ) 1 ( ) not an odd function

f x xf x x f xf x x f x

= −− =− − ≠− =− − ≠ −

14. 3 2 42 3

3 22

x yy x

y x

4− =− =− +

= −

[1.4]

Slope of perpendicular line is .32

−

32

32

36

38

32

38

322

)4(322

)( 11

+−=

−+−=

+−=+

−−=+

−=−

xy

xy

xy

xy

xxmyy

15.

2)1(2

42

=−

−=−a

b [1.5]

12884

8)2(42)2( 2

−=−−=

−−=f

The minimum value of the function is –12.

16. 2

2

2

( )( ) ( ) ( )

( 1) ( 2

3

( )( )( )

1, 22

f g x f x g x

x x

x x

f f xxg g x

x xx

)

+ = +

= − + −

= + −

⎛ ⎞=⎜ ⎟⎜ ⎟

⎝ ⎠

−= ≠−

[1.7]

17. 1)( 2 += xxf [1.7]

hxh

hxhh

hxh

hxhxhx

hxhx

hxfhxf

+=

+=

+=

−−+++=

+−++=

−+

2

)2(2

112

)1(1)()()(

2

222

22

18. 2( ) 2 ( ) 2 5f x x x g x x= − = + [1.7]

( )( ) ( )2

2

2

( )( ) [ ( )] 2 5

2 5 2 2 5

4 20 25 4 1

4 16 15

f g x f g x f x

x x

x x x

x x

= = +

= + − +

0= + + − −

= + +

o


Essentials of PreCalculus, 1e, AN SSM Chapter...

Documents

Transcript of Essentials of PreCalculus, 1e, AN SSM Chapter...