essentials of precalculus, 1e, an ssm chapter...
TRANSCRIPT
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
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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)−∞ −
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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
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4 Chapter 1: Functions and Graphs
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.
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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
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6 Chapter 1: Functions and Graphs
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= − < = −
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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
Copyright © Houghton Mifflin Company. All rights reserved.
8 Chapter 1: Functions and Graphs
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
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.2 9
....................................................... Connecting Concepts 77.
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
= − − −=
Copyright © Houghton Mifflin Company. All rights reserved.
10 Chapter 1: Functions and Graphs
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
Copyright © Houghton Mifflin Company. All rights reserved.
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 ) ,( ∞−∞
Copyright © Houghton Mifflin Company. All rights reserved.
12 Chapter 1: Functions and Graphs
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).
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
14 Chapter 1: Functions and Graphs
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)
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
16 Chapter 1: Functions and Graphs
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.
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
18 Chapter 1: Functions and Graphs
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
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.4 19
....................................................... Connecting Concepts 83.
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
Copyright © Houghton Mifflin Company. All rights reserved.
20 Chapter 1: Functions and Graphs
....................................................... 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
( 4) 11 standard form,
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
( 2) 6 standard form,
f x x x
x x
x x
x
=− + +
=− − +
=− − + + +
=− − +
vertex (2, 6), axis of symmetry x = 2
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
22 Chapter 1: Functions and Graphs
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
Copyright © Houghton Mifflin Company. All rights reserved.
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 −= = .
Copyright © Houghton Mifflin Company. All rights reserved.
24 Chapter 1: Functions and Graphs
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 $24.25.
Copyright © Houghton Mifflin Company. All rights reserved.
Section 1.5 25
65.
2
( ) ( ) ( )(102.50 0.1 ) (52.50 1840)
0.1 50 1840
P x R x C xx x x
x x
= −= − − +
=− + −
The break-even points occur when or P(x) = 0. )()( xCxR =
Thus, 1840501.00 2 −+−= xx250 50 4( 0.1)( 1840)2( 0.1)
50 17640.2
50 420.2
x− ± − − −
=−
− ±=
−− ±
=−
40 or 460x x= = The break-even points occur when x = 40 or x = 460.
67. Let x = the number of people that take the tour. a.
2
( ) (15.00 0.25(60 ))(15.00 15 0.25 )
0.25 30.00
R x x xx x
x x
= + −= + −
=− +
b. 2
2
( ) ( ) ( )
( 0.25 30.00 ) (180 2.50 )
0.25 27.50 180
P x R x C x
x x x
x x
= −
= − + − +
=− + −
c. 55)25.0(2
50.272
=−
−=−a
b
25.576$
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
Copyright © Houghton Mifflin Company. All rights reserved.
26 Chapter 1: Functions and Graphs
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+
Copyright © Houghton Mifflin Company. All rights reserved.
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 ++=
++=
++=
−+++=
−+−+
−−−
=
Copyright © Houghton Mifflin Company. All rights reserved.
28 Chapter 1: Functions and Graphs
....................................................... 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.
Copyright © Houghton Mifflin Company. All rights reserved.
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 +
Copyright © Houghton Mifflin Company. All rights reserved.
30 Chapter 1: Functions and Graphs
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.
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
32 Chapter 1: Functions and Graphs
....................................................... 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)
21 45 Domain all real numbers
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 )
2 8 Domain all real numbers
f x g x x x x x
x x x
+ = − + +
= − +
3 2
3 2( ) ( ) ( 2 7 )
2 6 Domain all real numbers
f x g x x x x x
x x x
− = − + −
= − +
3 2
4 3 2( ) ( ) ( 2 7 )
2 7 Domain all real numbers
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
= + − + −
⎧ ⎫−= + ≠ ≠ −⎨ ⎬⎩ ⎭+ −
Copyright © Houghton Mifflin Company. All rights reserved.
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
+ + − ++ − =
+ + − −=
=
=
Copyright © Houghton Mifflin Company. All rights reserved.
34 Chapter 1: Functions and Graphs
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
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
36 Chapter 1: Functions and Graphs
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− − −= = = = −
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
38 Chapter 1: Functions and Graphs
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
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
40 Chapter 1: Functions and Graphs
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
= −
= −
= −
= −
Copyright © Houghton Mifflin Company. All rights reserved.
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]
Copyright © Houghton Mifflin Company. All rights reserved.
42 Chapter 1: Functions and Graphs
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
− − = − −
+ = +
+ = +
= + −
= + −
= +
Copyright © Houghton Mifflin Company. All rights reserved.
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).
Copyright © Houghton Mifflin Company. All rights reserved.
44 Chapter 1: Functions and Graphs
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]
Copyright © Houghton Mifflin Company. All rights reserved.
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
Copyright © Houghton Mifflin Company. All rights reserved.
46 Chapter 1: Functions and Graphs
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]
Copyright © Houghton Mifflin Company. All rights reserved.
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
The domain is all real numbers.
2
3 2( )( ) ( 9)( 3)
3 9 2
fg x x x
x x x
= − +
= + − −
The domain is all real numbers.
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]
The domain is all real numbers.
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
)− = + − − +
= + − + −
= − + +
The domain is all real numbers.
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= − + + − +
The domain is all real numbers.
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]
Copyright © Houghton Mifflin Company. All rights reserved.
48 Chapter 1: Functions and Graphs
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.
Copyright © Houghton Mifflin Company. All rights reserved.
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]
Copyright © Houghton Mifflin Company. All rights reserved.
50 Chapter 1: Functions and Graphs
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
Copyright © Houghton Mifflin Company. All rights reserved.