CHAPTER P Preparation for Calculuscrunchymath.weebly.com/uploads/8/2/4/0/8240213/solutions.pdf8...
Transcript of CHAPTER P Preparation for Calculuscrunchymath.weebly.com/uploads/8/2/4/0/8240213/solutions.pdf8...
C H A P T E R PPreparation for Calculus
Section P.1 Graphs and Models . . . . . . . . . . . . . . . . . . . . . . 2
Section P.2 Linear Models and Rates of Change . . . . . . . . . . . . 11
Section P.3 Functions and Their Graphs . . . . . . . . . . . . . . . . . 22
Section P.4 Fitting Models to Data . . . . . . . . . . . . . . . . . . . . 31
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
7.
x
2
−4
−2
−6
6
−4−6 4 6
(−3, −5) (3, −5)
(−2, 0)
(0, 4)
(2, 0)
y
y � 4 � x2 8.
−6 −4 −2−2
2
2
4
4 6
6
8
10
y
x
y � �x � 3�2
x 0 2 3
y 0 4 0 �5�5
�2�3
C H A P T E R PPreparation for Calculus
Section P.1 Graphs and Models
2
1.
x-intercept:
y-intercept:
Matches graph (b).
�0, 2�
�4, 0�
y � �12 x � 2
3.
x-intercepts:
y-intercept:
Matches graph (a).
�0, 4�
�2, 0�, ��2, 0�
y � 4 � x2
5.
x2
2
−4
−6
−8
4
6
8
−4−6−8 4 6 8
(−4, −5)(−2, −2)
(0, 1)
(2, 4)
(4, 7)
y
y �32 x � 1
x 0 2 4
y 1 4 7�2�5
�2�4
2.
x-intercepts:
y-intercept:
Matches graph (d).
�0, 3�
��3, 0�, �3, 0�
y � �9 � x2
4.
x-intercepts:
y-intercept:
Matches graph (c).
�0, 0�
�0, 0�, ��1, 0�, �1, 0�
y � x3 � x
6.
−6 −4 −2 2
2
4
6
6
10
y
x
y � 6 � 2x
x 0 1 2 3 4
y 10 8 6 4 2 0 �2
�1�2
x 0 1 2 3 4 5 6
y 9 4 1 0 1 4 9
Section P.1 Graphs and Models 3
9.
x
2
−2
4
6
−4−6 2
(−3, 1) (−1, 1)
(−4, 2)
(−2, 0)
(0, 2)(1, 3)
(−5, 3)
y
y � �x � 2�
11.
x
4
6
8
2
−6
−8
−4
−10
10
−2 2 1614 1812
(0, −4)
(1, −3)
(4, −2) (16, 0)
y
(9, −1)
y � �x � 4
x 0 1
y 3 2 1 0 1 2 3
�1�2�3�4�5
x 0 1 4 9 16
y 0�1�2�3�4
10.
−3 3
3
4
−2
−2
2
2
−1−1
1
1
y
x
y � �x� � 1
12.
−5 5 10 15 20
2
3
4
5
y
x
y � �x � 2
x 0 1 2 3
y 2 1 0 0 1 2�1
�1�2�3
x 0 2 7 14
y 0 1 2 3 4�2
�1�2
16.
Note that when or x � 10.x � 0y � 10
Xmin = -30Xmax = 30Xscl = 5Ymin = -10Ymax = 40Yscl = 5
15.
Note that when x � 0.y � 4
Xmin = -3Xmax = 5Xscl = 1Ymin = -3Ymax = 5Yscl = 1
13.
−1−2 1 2 3−1
1
2
3
x
(1, 2)
(2, 1)
(−2, −1)
(−1, −2)
−3, − ) 23) ) 2
33, )
y
y �2
x14.
x
−3, − ) 14)
−1, − ) 12) ) 1
45, ) ) 1
23, )
y
2 3 4 5
−3
−4
2
3
4
(2, 0)
(0, −1)1
y �1
x � 1
x 0 1 2 3
y Undef. 2 1 23�2�1�
23
�1�2�3 x 0 1 2 3 5
y Undef. 1 14
12�1�
12�
14
�1�3
4 Chapter P Preparation for Calculus
18.
(a)
(b) and �x, �4� � �1, �4��x, �4� � ��1.65, �4�
��0.5, y� � ��0.5, 2.47�
− 9
−6
9
6y � x5 � 5x17.
(a)
(b) �x, 3� � ��4, 3� �3 � �5 � ��4� ��2, y� � �2, 1.73� �y � �5 � 2 � �3 � 1.73�
−6 6
−3
5
( 4.00, 3)−(2, 1.73)
y � �5 � x
19.
y-intercept:
x-intercepts:
x � �2, 1; ��2, 0�, �1, 0�
0 � �x � 2��x � 1�
0 � x2 � x � 2
y � �2; �0, �2�y � 02 � 0 � 2
y � x2 � x � 2
21.
y-intercept:
x-intercepts:
x � 0, ±5; �0, 0�; �±5, 0�
0 � x2��5 � x��5 � x�
0 � x2�25 � x2
y � 0; �0, 0�
y � 02�25 � 02
y � x2�25 � x2
20.
y-intercept:
x-intercepts:
x � 0, ±2; �0, 0�, �±2, 0�
0 � x�x � 2��x � 2�
0 � x3 � 4x
y � 0; �0, 0�
y2 � 03 � 4�0�
y 2 � x3 � 4x
22.
y-intercept:
x-intercept:
x � 1; �1, 0�
0 � �x � 1��x2 � 1
y � �1; �0, �1�
y � �0 � 1��02 � 1
y � �x � 1��x2 � 1
24.
y-intercept:
x-intercepts:
x � 0, �3; �0, 0�, ��3, 0�
0 �x�x � 3��3x � 1�2
0 �x2 � 3x
�3x � 1�2
y � 0; �0, 0�
y �02 � 3�0�
�3�0� � 1�2
y �x2 � 3x
�3x � 1�223.
y-intercept: None. x cannot equal 0.
x-intercept:
x � 4; �4, 0�
0 � 2 � �x
0 �3�2 � �x�
x
y �3�2 � �x�
x
25.
y-intercept:
x-intercept:
x � 0; �0, 0�
x2�0� � x2 � 4�0� � 0
y � 0; �0, 0�
02�y� � 02 � 4y � 0
x2y � x2 � 4y � 0
Section P.1 Graphs and Models 5
26.
y-intercept:
x-intercept:
Note: is an extraneous solution.x � ��3�3
x ��33
; �33
, 0
x � ±�33
x2 �13
3x2 � 1
4x2 � x2 � 1
2x � �x2 � 1
0 � 2x � �x2 � 1
y � �1; �0, �1�
y � 2�0� � �02 � 1
y � 2x � �x2 � 1 27. Symmetric with respect to the y-axis since
y � ��x�2 � 2 � x2 � 2.
29. Symmetric with respect to the x-axis since
��y�2 � y2 � x3 � 4x.
31. Symmetric with respect to the origin since
��x���y� � xy � 4.
33.
No symmetry with respect to either axis or the origin.
y � 4 � �x � 3
35. Symmetric with respect to the origin since
y �x
x2 � 1.
�y ��x
��x�2 � 1
37. is symmetric with respect to the y-axis
since y � ���x�3 � ��x�� � ���x3 � x�� � �x3 � x�.y � �x3 � x�
28.
No symmetry with respect to either axis or the origin.
y � x2 � x
30. Symmetric with respect to the origin since
y � x3 � x.
�y � �x3 � x
��y� � ��x�3 � ��x�
32. Symmetric with respect to the x-axis since
x��y�2 � xy2 � �10.
34. Symmetric with respect to the origin since
xy � �4 � x2 � 0.
��x���y� � �4 � ��x�2 � 0
36. is symmetric with respect to the y-axis
since y ���x�2
��x�2 � 1�
x2
x2 � 1.
y �x2
x2 � 1
38. is symmetric with respect to the x-axis
since
�y� � x � 3.
��y� � x � 3
�y� � x � 3 39.
Intercepts:
Symmetry: none
�23 , 0�, �0, 2�
0
y
)20( ,2
1
x321
1
,23
y � �3x � 2
6 Chapter P Preparation for Calculus
40.
Intercepts:
Symmetry: none
1 2 3 4
−1
−2
1
3
(4, 0)
(0, 2)
y
x
�4, 0�, �0, 2�
y � �x2
� 2 41.
Intercepts:
Symmetry: none
x108
)0,8(
y
42
)4,
2
(
2
0
2
8
6
10
�8, 0�, �0, �4�
y �12
x � 4 42.
Intercepts:
Symmetry: none
−1
−1
1 2
2
−2
y
x
32
, 0( )(0, 1)
�0, 1�, ��32, 0�
y �23
x � 1
43.
Intercepts:
Symmetry: y-axis
x)0(1,
0, 1)
y
2
),( 1 0
(
1
2
2 2−
�1, 0�, ��1, 0�, �0, 1�
y � 1 � x2 44.
Intercept:
Symmetry: y-axis
3 6−3−6
9
12
(0, 3)
y
x
�0, 3�
y � x2 � 3 45.
Intercepts:
Symmetry: none
x
2
−2
8
10
12
−8 −6−10 2 4(−3, 0)
(0, 9)
y
��3, 0�, �0, 9�
y � �x � 3�2
46.
Intercepts:
Symmetry: none
1 2 3−1−2−3
1
3
5
2
4
(0, 0)12
− , 0( )
y
x
�0, 0�, ��12 , 0�
y � 2x2 � x � x�2x � 1� 47.
Intercepts:
Symmetry: none
y
5
4
3
3
x32
,
1
1
1
,,
23
)02 )(0 2(
�� 3�2, 0�, �0, 2�
y � x3 � 2 48.
Intercepts:
Symmetry: origin
−3
−3
−1−1
−2
1 3
3
(−2, 0) (2, 0)(0, 0)
y
x
�0, 0�, �2, 0�, ��2, 0�
y � x3 � 4x
49.
Intercepts:
Symmetry: none
Domain: x ≥ �2
�0, 0�, ��2, 0�
x
1
2
3
4
5
−2
6
−4 −3 −1 41 2 3
(−2, 0) (0, 0)
yy � x�x � 2 50.
Intercepts:
Symmetry: y-axis
Domain: ��3, 3�
��3, 0�, �3, 0�, �0, 3�
−2−4 2 4
−2
2
4
6
(0, 3)
(3, 0)( 3,0)−
5
1
−1−1
1
y
x
y � �9 � x2
Section P.1 Graphs and Models 7
51.
Intercept:
Symmetry: origin
x1
−2
−3
−4
2
3
4
−2 −1−3−4 2 3 4
(0, 0)
y
�0, 0�
x � y3 52.
Intercepts:
Symmetry: x-axis
1
−3
3
−1−2−5
( 4, 0)−(0, 2)
(0, 2)−
y
x
�0, 2�, �0, �2�, ��4, 0�
x � y2 � 4 53.
Intercepts: none
Symmetry: originy
3
1
2
x321
y �1x
54.
Intercept:
Symmetry: y-axis
−6 −4 −2
2
2
12
10
4 6
(0, 10)
y
x
�0, 10�
y �10
x2 � 155.
Intercepts:
Symmetry: y-axis
x2
2
−4
−2
−6
−8
4
6
8
−4 −2−8 4 6 8
(−6, 0)
(0, 6)
(6, 0)
y
�0, 6�, ��6, 0�, �6, 0�
y � 6 � �x� 56.
Intercepts:
Symmetry: none
2
2
4
4
6 8
8
y
x
(0, 6)
(6, 0)
�0, 6�, �6, 0�
y � �6 � x�
57.
Intercepts:
Symmetry: x-axis
1
−4
−11
4
(−9, 0)
(0, −3)
(0, 3)
�0, 3�, �0, �3�, ��9, 0�
y � ±�x � 9
y2 � x � 9
y2 � x � 9 58.
Intercepts:
Symmetry: origin and both axes
Domain:
−3 3
−2
(0, 1)
(0, 1)−
(2, 0)( 2, 0)−
2
��2, 2�
��2, 0�, �2, 0�, �0, �1�, �0, 1�
x2 � 4y2 � 4 ⇒ y � ±�4 � x2
259.
Intercepts:
Symmetry: x-axis
8
−3
−1
3
(6, 0)
( )0, 2
( )0, − 2
�6, 0�, �0, �2�, �0, ��2�
y � ±�2 �x3
3y2 � 6 � x
x � 3y2 � 6
60.
Intercept:
Symmetry: x-axis
�83, 0�
y � ±�34 x � 2
4y2 � 3x � 8−6
−6
6
12
83
, 0( )
3x � 4y2 � 8 61.
The corresponding y-value is
Point of intersection: �1, 1�y � 1.
1 � x
3 � 3x
2 � x � 2x � 1
2x � y � 1 ⇒ y � 2x � 1
x � y � 2 ⇒ y � 2 � x
8 Chapter P Preparation for Calculus
62.
The corresponding y-value is
Point of intersection: �2, �3�
y � �3.
x � 2
7x � 14
2x � 13 � 1 � 5x
2x � 133
�1 � 5x
3
5x � 3y � 1 ⇒ y �1 � 5x
3
2x � 3y � 13 ⇒ y �2x � 13
363.
The corresponding y-values are (for )and (for ).
Points of intersection: �2, 2�, ��1, 5�
x � �1y � 5x � 2y � 2
x � 2, �1
0 � �x � 2��x � 1�
0 � x2 � x � 2
6 � x2 � 4 � x
x � y � 4 ⇒ y � 4 � x
x2 � y � 6 ⇒ y � 6 � x2
64.
The corresponding y-values are and
Points of intersection: ��1, �2�, �2, 1�
y � 1.y � �2
x � �1 or x � 2
0 � x2 � x � 2 � �x � 1��x � 2�
3 � x � x2 � 2x � 1
3 � x � �x � 1�2
y � x � 1
x � 3 � y2 ⇒ y2 � 3 � x 65.
The corresponding y-values are and
Points of intersection: ��1, �2�, �2, 1�
y � 1.y � �2
x � �1 or x � 2
0 � 2x2 � 2x � 4 � 2�x � 1��x � 2�
5 � x2 � x2 � 2x � 1
5 � x2 � �x � 1�2
x � y � 1 ⇒ y � x � 1
x2 � y 2 � 5 ⇒ y2 � 5 � x 2
66.
The corresponding y-values are and
Points of intersection: �3, 4�, �5, 0�
y � 0.y � 4
x � 3 or x � 5
0 � 5x2 � 40x � 75 � 5�x � 3��x � 5�
25 � x2 � 100 � 40x � 4x2
25 � x2 � �10 � 2x�2
2x � y � 10 ⇒ y � 10 � 2x
x2 � y2 � 25 ⇒ y2 � 25 � x2 67.
The corresponding y-values are and
Points of intersection: �0, 0�, ��1, �1�, �1, 1�
y � 1.y � 0, y � �1,
x � 0, x � �1, or x � 1
x�x � 1��x � 1� � 0
x3 � x � 0
x3 � x
y � x
y � x3
68.
The corresponding y-values are and
Points of intersection: �1, �3�, ��2, 0�
y � 0.y � �3
x � 1 or x � �2
�x � 1�2�x � 2� � 0
x3 � 3x � 2 � 0
x3 � 4x � ��x � 2�
y � ��x � 2� y � x3 � 4x 69.
��1, �5�, �0, �1�, �2, 1�
x � �1, 0, 2
x�x � 2��x � 1� � 0
x3 � x2 � 2x � 0
x3 � 2x2 � x � 1 � �x2 � 3x � 1
y � �x2 � 3x � 1
y � x3 � 2x2 � x � 1
−4 6
−8
4
y x x= + 3 1− −2
y x x x= 2 + 1− −3 2
(2, 1)
(0, 1)−
( 1, 5)− −
Section P.1 Graphs and Models 9
70.
−3 3
−2
2
(1, 0)( 1, 0)−
(0, 1)
y x= −1 2
y x x= − +2 124
��1, 0�, �0, 1�, �1, 0�
x � �1, 0, 1
0 � x2�x � 1��x � 1�
0 � x 4 � x2
1 � x2 � x 4 � 2x2 � 1
y � 1 � x2
y � x 4 � 2x2 � 1 71.
Points of intersection:
Analytically,
x � �2, y � 2 ⇒ ��2, 2�. x � �3, y � �3 ⇒ ��3, �3�
�x � 3��x � 2� � 0
x2 � 5x � 6 � 0
x � 6 � �x2 � 4x
�x � 6 � ��x2 � 4x
��3, �3� � ��3, 1.732���2, 2�,
−7 2
4
−2
x + 6y =
−x2 − 4x
(−2, 2)
y =
(3, )3
y � ��x2 � 4x
y � �x � 6
72. Points of intersection:
Analytically,
or
or
Hence, �1, 5�.�3, 3�,
x � 1. x � 3
2x � 3 � �x 2x � 3 � x
�2x � 3� � x
−4 8
7
−1
y = 6 − x
y = − 2x - 3 + 6
(1, 5)
(3, 3)
��2x � 3� � 6 � 6 � xy � 6 � x
�1, 5��3, 3�,y � ��2x � 3� � 6
73. (a) Using a graphing utility, you obtain
(b)
(c) For 2010, and y � 217.t � 40
35
−50
−5
250
y � �0.007t2 � 4.82t � 35.4.
74. (a)
(b)
(c) For 2010, andacres.� 405y � �0.13�602� � 11.1�60� � 207
t � 60
0 50
500
0
y � �0.13t2 � 11.1t � 207
75.
The other root, does not satisfy the equation
This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R.
R � C.x � 2949,
x � 3133 units
0 � 10.8241x2 � 65,830.25x � 100,000,000 Use the Quadratic Formula.
30.25x � 10.8241x2 � 65,800x � 100,000,000
�5.5�x �2� �3.29x � 10,000�2
5.5�x � 10,000 � 3.29x C � R
10 Chapter P Preparation for Calculus
76.
If the diameter is doubled, the resistance is changed by approximately a factor of For instance, andy�40� � 6.36125.
y�20� � 26.55514.
00
400
100
y �10,770
x2 � 0.37
77. (other answers possible)y � �x � 2��x � 4��x � 6� 78. (other answers possible)y � �x �52��x � 2��x �
32�
80. (a) If is on the graph, then so is by y-axissymmetry. Since is on the graph, then so is
by x-axis symmetry. Hence, the graph issymmetric with respect to the origin. The converse isnot true. For example, has origin symmetry but is not symmetric with respect to either the x-axisor the y-axis.
(b) Assume that the graph has x-axis and origin symmetry.If is on the graph, so is by x-axis symmetry. Since is on the graph, then so is
by origin symmetry.Therefore, the graph is symmetric with respect to the y-axis. The argument is similar for y-axis and origin symmetry.
��x, ���y�� � ��x, y��x, �y�
�x, �y��x, y�
y � x3
��x, �y���x, y�
��x, y��x, y�79. (i) matches (b).
Use
(ii) matches (d).
Use
(iii) matches (a).
Use
(iv) matches (c).
Use �1��36� � k ⇒ k � 36, thus, xy � 36.�1, 36�:
xy � k
3 � k�1�3�2 ⇒ k � 3, thus, y � 3x3�2.�1, 3�:
y � kx3�2
�9 � �1�2 � k ⇒ k � �10, thus, y � x2 � 10.
�1, �9�:
y � x2 � k
7 � k�1� � 5 ⇒ k � 2, thus, y � 2x � 5.�1, 7�:
y � kx � 5
85.
Circle of radius 2 and center �0, 4�
x2 � �y � 4�2 � 4
x2 � y2 � 8y � 12 � 0
3x2 � 3y2 � 24y � 36 � 0
4x2 � 4y2 � 24y � 36 � x2 � y2
4�x2 � �y � 3�2� � x2 � y2
x
y
−1−2 2 3
−1
1
2
4
(0, 3)
(x, y)
(0, 0)
2��x � 0�2 � �y � 3�2 � ��x � 0�2 � �y � 0�2
86. Distance from the origin Distance from
Note: This is the equation of a circle!
�1 � K 2�x 2 � �1 � K 2�y 2 � 4K 2x � 4K 2 � 0
x2 � y 2 � K 2�x2 � 4x � 4 � y2�
�x2 � y2 � K��x � 2�2 � y2, K � 1
�2, 0�� K �
82. True 83. True; the x-intercepts are
�b ± �b2 � 4ac2a
, 0.
84. True; the x-intercept is
�b
2a, 0.
81. False; x-axis symmetry means that if is on the graph, then is also on the graph.�1, 2��1, �2�
Section P.2 Linear Models and Rates of Change
Section P.2 Linear Models and Rates of Change 11
1. m � 1 2. m � 2 3. m � 0
4. m � �1 5. m � �12 6. m �403
7.
2
3
4
5
−1
1
x4 51 3
(2, 3)
m = −2
m = 1
32
m = −m isundefined.
y 8.
2
4
6
−2
−2−6
m = 0
m =
m = 3m = 3−
( 4, 1)−
13
3
1−1−5
y
x
9.
x
y
1 2 3 5 6 7−1
(3, 4)−
(5, 2)
−2
−3
−4
−5
1
2
3
�62
� 3
m �2 � ��4�
5 � 3
10.
4),( 2 4
), 2
2
(
1
12
1
12
y
x
� �23
m �4 � 2
�2 � 111.
undefined
x
1
2
−1
−2
3
4
5
6
−1−2 1 3 4 5 6
(2, 1)
(2, 5)
y
�40
m �5 � 12 � 2
12.
1
1
−1
−3
2
2
3 4 5
y
x
(3, −2) (4, −2)
m ��2 � ��2�
4 � 3� 0
13.
�1�21�4
� 2
x
−1
−2
−3
2
3
−2−3 21 3
y
( )12
23,− ( )3
416,−
m �2�3 � 1�6
�1�2 � ��3�4� 14.
�1
�3�8� �
83
1−1−2
1
2
3
−154
14
−( ),
78
34( ),
y
x
m ��3�4� � ��1�4��7�8� � �5�4�
15. Since the slope is 0, the line is horizontal and its equationis Therefore, three additional points are
and �3, 1�.�1, 1�,�0, 1�,y � 1.
16. Since the slope is undefined, the line is vertical and itsequation is Therefore, three additional points are , and ��3, 5�.��3, 3�,��3, 2�
x � �3.
17. The equation of this line is
.
Therefore, three additional points are and �3, 1�.
�2, 4�,�0, 10�,
y � �3x � 10
y � 7 � �3�x � 1�
18. The equation of this line is
.
Therefore, three additional points are and �0, 2�.
��1, 0�,��3, �4�,
y � 2x � 2
y � 2 � 2�x � 2�
12 Chapter P Preparation for Calculus
19. (a)
(b)
By the Pythagorean Theorem,
x � 10�10 � 31.623 feet.
x2 � 302 � 102 � 1000
30 ft
10 ftx
Slope ��y�x
�13
20. (a) indicates that the revenues increase by 400 in one day.
(b) indicates that the revenues increase by 100 in one day.
(c) indicates that the revenues do not change from one day to the next.m � 0
m � 100
m � 400
21. (a)
(b) The slopes of the line segments are:
The population increased least rapidly between 2000 and 2001.
285.0 � 282.311 � 10
� 2.7
282.3 � 279.310 � 9
� 3.0
279.3 � 276.19 � 8
� 3.2
276.1 � 272.98 � 7
� 3.2
272.9 � 269.77 � 6
� 3.2
tPopu
latio
n (i
n m
illio
ns)
Year (6 ↔ 1996)
6 7 8 9 10 11
260
270
280
290
y 22. (a)
(b) The slopes are:
The rate changed most rapidly between 20 and 25 seconds. The change is �4.6 mph�sec.
43 � 6130 � 25
� �3.6
61 � 8425 � 20
� �4.6
84 � 8520 � 15
� �0.2
85 � 7415 � 10
� 2.2
74 � 5710 � 5
� 3.4
t
r
5 10 15 20 25 30
20
40
60
80
100
23.
Therefore, the slope is and the y-intercept is�0, 4�.
m � �15
y � �15 x � 4
x � 5y � 20 24.
Therefore, the slope is and the y-intercept is�0, �3�.
m �65
y �65 x � 3
6x � 5y � 15
25.
The line is vertical. Therefore, the slope is undefined andthere is no y-intercept.
x � 4 26.
The line is horizontal. Therefore, the slope is andthe y-intercept is �0, �1�.
m � 0
y � �1
27.
0 � 3x � 4y � 12
4y � 3x � 12
x
y
1−1−2−3−4
1
2
4
5
(0, 3)
y �34 x � 3 28.
x � 1 � 0
1−2−3
−1
1
2
3
( 1, 2)−
y
x
x � �1
Section P.2 Linear Models and Rates of Change 13
29.
x
y
1 2 3 4
−1
2
3
4
(0, 0)
2x � 3y � 0
3y � 2x
y �23 x 30.
1 2−1−2−3
1
2
3
5
(0, 4)
y
x
y � 4 � 0
y � 4 31.
x
1
2
−3
−1
−2
−4
−5
3
−1−2 1 2 3 4 5 6
(3, −2)
y
y � 3x � 11 � 0
y � 3x � 11
y � 2 � 3x � 9
y � 2 � 3�x � 3�
32.
1 2−1−2−3
1
2
4
5
( 2, 4)−
y
x
3x � 5y � 14 � 0
5y � 20 � �3x � 6
y � 4 � �35
�x � 2� 33.
x2
2
−8
4
6
8
−4 −2−6−8 4 6 8
(2, 6)
(0, 0)
y
y � 3x
y � 0 � 3�x � 0�
m �6 � 02 � 0
� 3 34.
33)1,(
), 0
1
(
1
0
2 13
y
x
3x � y � 0
y � �3x
y � 0 � �3�x � 0�
m �3 � 0
�1 � 0� �3
35.
x
y
−2 −1 2 3 4 5−1
−2
−3
−5
1
2
(2, 1)
(0, 3)−
0 � 2x � y � 3
y � 1 � 2x � 4
y � 1 � 2�x � 2�
m �1 � ��3�
2 � 0� 2 36.
2− 4−6
−2
− 4
2
4 (1, 4)
( 3, 4)− −
1 3−3−5
−3
−5
3
5
−7
y
x
0 � 2x � y � 2
y � 4 � 2x � 2
y � 4 � 2�x � 1�
m �4 � ��4�1 � ��3� �
84
� 2 37.
12345
−2
6789
x−1 4 6 7 8 91 2 3
(5, 0)
(2, 8)
y
3y � 8x � 40 � 0
y � �83
x �403
y � 0 � �83
�x � 5�
m �8 � 02 � 5
� �83
38.
x � y � 3 � 0
y � 2 � �x � 1
y � 2 � �1�x � 1�
1 2 3−1−3 −2−4
1
3
2
5
7
6
(1, 2)
( 3, 6)−
y
x
m �6 � 2
�3 � 1�
4�4
� �1 39. Undefined
Vertical line x � 5
12345
−2
6789
x−1 4 6 7 8 91 2 3
(5, 1)
(5, 8)
y
m �8 � 15 � 5
14 Chapter P Preparation for Calculus
40.
1
2)( ,3),(1
1
4
32
1 1 2 3 4
y
x
y � 2 � 0
y � �2
m � 0 41.
x1
2
1
3
4
−2 −1−3−4 2 3 4
( )12
72,
( )340,
y
22x � 4y � 3 � 0
y �112
x �34
y �34
�112
�x � 0�
m �7�2 � 3�41�2 � 0
�11�41�2
�112
42.
1−1−2
1
2
3
−154
14
−( ),
78
34( ),
y
x
32x � 12y � 37 � 0
12y � 3 � �32x � 40
y �14
��83 �x �
54�
�1
�3�8� �
83
m ��3�4� � ��1�4��7�8� � �5�4�
45.
3x � 2y � 6 � 0
x2
�y3
� 1 46.
3x � y � 2 � 0
3x � y � �2
�3x
2�
y2
� 1
x�2�3
�y
�2� 1 47.
x � y � 3 � 0
a � 3 ⇒ x � y � 3
3a
� 1
1a
�2a
� 1
xa
�ya
� 1
43.
x � 3 � 0
1 2 4
−1
−2
1
2
(3, 0)
y
x
x � 3 44.
xa
�yb
� 1
ba
x � y � b
y ��ba
x � b
( , 0)a
(0, )b
y
x
m � �ba
48.
x � y � 1 � 0
a � 1 ⇒ x � y � 1
1a
� 1
�3a
�4a
� 1
xa
�ya
� 1 49.
x
y
−1−2−3 1 2 3 4 5
−2
−4
−5
−6
1
2
y � 3 � 0
y � �3 50.
1 2 3 5
−1
−2
1
2
3
y
x
x � 4 � 0
x � 4
Section P.2 Linear Models and Rates of Change 15
51.
x
y
−2 −1 1 2
−1
3
y � �2x � 1 52.
−3
−3
−4
3−2
−2
2
1
−1
y
x
(0, −1)
3y � x � 3 � 0
y �13 x � 1 53.
x1
2
1
3
4
−2
−2
−3
−4
−3−4 2 3 4
y
2y � 3x � 1 � 0
y �32 x �
12
y � 2 �32�x � 1�
54.
4 8−8−12−16−4
−8
12
16
y
x
y � 3x � 13
y � 1 � 3�x � 4� 55.
y
1
x32
1
2
3
12
y � 2x � 3
2x � y � 3 � 0 56.
−10 −8 −6 −2
−4
−6
2
4
y
x
y � �12 x � 3
x � 2y � 6 � 0
57. (a)
The lines do notappear perpendicular.10
−10
−10
10 (b)
The lines appearperpendicular.15
−10
−15
10
The lines are perpendicular because their slopes 1 and are negative reciprocals of each other.You must use a square setting in order for perpendicular lines to appear perpendicular. Answers depend on calculator used.
�1
58. (a)
The lines do notappear perpendicular.−5
−5
5
5
(b)
The lines appearperpendicular.−6
−4
4
6
The lines are perpendicular because their slopes 2 and are negative reciprocals of each other.You must use a square setting in order for perpendicular lines to appear perpendicular. Answers depend on calculator used.
�12
59.
(a)
(b)
x � 2y � 4 � 0
2y � 2 � �x � 2
y � 1 � �12 �x � 2�
2x � y � 3 � 0
y � 1 � 2x � 4
y � 1 � 2�x � 2�
m � 2
y � 2x �32
4x � 2y � 3 60.
(a)
(b)
x � y � 5 � 0
y � 2 � x � 3
y � 2 � 1�x � 3�
x � y � 1 � 0
y � 2 � �x � 3
y � 2 � �1�x � 3�
m � �1
y � �x � 7
x � y � 7
16 Chapter P Preparation for Calculus
61.
(a)
(b)
40y � 24x � 53 � 0
40y � 35 � �24x � 18
y �78 � �
35�x �
34�
24y � 40x � 9 � 0
24y � 21 � 40x � 30
y �78 �
53�x �
34�
m �53
y �53x
5x � 3y � 0 62.
(a)
(b)
4x � 3y � 36 � 0
3y � 12 � 4x � 24
y � 4 �43�x � 6�
3x � 4y � 2 � 0
4y � 16 � �3x � 18
y � 4 � �34 �x � 6�
m � �34
y � �34 x �
74
3x � 4y � 7
63. The given line is vertical.
(a)
(b) y � 5 ⇒ y � 5 � 0
x � 2 ⇒ x � 2 � 0
64. (a)
(b) x � �1 ⇒ x � 1 � 0
y � 0
65. The slope is 125. when
V � 125�t � 4� � 2540 � 125t � 2040
t � 4. V � 2540 66. The slope is 4.5. when
V � 4.5�t � 4� � 156 � 4.5t � 138
t � 4.V � 156
67. The slope is when
V � �2000�t � 4� � 20,400 � �2000t � 28,400
t � 4.V � 20,400�2000. 68. The slope is when
� �5600t � 267,400
V � �5600�t � 4� � 245,000
t � 4.V � 245,000�5600.
69.
You can use the graphing utility to determine that the points of intersection are and Analytically,
The slope of the line joining and is Hence, an equation
of the line is
y � 2x.
y � 0 � 2�x � 0�
m � �4 � 0���2 � 0� � 2.�2, 4��0, 0�
x � 2 ⇒ y � 4 ⇒ �2, 4�.
x � 0 ⇒ y � 0 ⇒ �0, 0�2x�x � 2� � 0
2x2 � 4x � 0
x2 � 4x � x2
�2, 4�.�0, 0�
−1
−3 6(0, 0)
(2, 4)
70.
You can use the graphing utility to determine that the points of intersection are and Analytically,
The slope of the line joining and is Hence, an equation
of the line is
y � �x � 3.
y � 3 � �1�x � 0�
m � �0 � 3���3 � 0� � �1.�3, 0��0, 3�
x � 3 ⇒ y � 0 ⇒ �3, 0�.
x � 0 ⇒ y � 3 ⇒ �0, 3� 2x�x � 3� � 0
2x2 � 6x � 0
x2 � 4x � 3 � �x2 � 2x � 3
�3, 0�.�0, 3�
−9 9
−6
6
(0, 3)(3, 0)
y � �x2 � 2x � 3y � x2 � 4x � 3,
Section P.2 Linear Models and Rates of Change 17
71.
The points are not collinear.
m1 � m2
m2 ��2 � 0
2 � ��1� � �23
m1 �1 � 0
�2 � ��1� � �1 72.
The points are not collinear.
m1 � m2
m2 �11 � 4�5 � 0
� �75
m1 ��6 � 47 � 0
� �107
73. Equations of perpendicular bisectors:
Setting the right-hand sides of the two equations equal and solving for x yields
Letting in either equation gives the point of intersection:
This point lies on the third perpendicular bisector, x � 0.
�0, �a2 � b2 � c2
2c �.
x � 0
x � 0.
y �c2
�a � b
�c �x �b � a
2 � y �
c2
�a � b
c �x �a � b
2 �( , )b c
( , 0)a( , 0)−a
a b+2
b a−2 c
2
c2 ( )( ) ,
,
y
x
74. Equations of medians:
Solving simultaneously, the point of intersection is
( , )b c
( , 0)a( , 0)−a
a b+2
c2( ),
b a−2
c2( ),
(0, 0)
y
x
�b3
, c3�.
y �c
�3a � b�x � a�
y �c
3a � b�x � a�
y �cb
x
75. Equations of altitudes:
Solving simultaneously, the point of intersection is
( , )b c
( , 0)a
( , 0)−a
y
x
�b, a2 � b2
c �.
y � �a � b
c�x � a�
x � b
y �a � b
c�x � a�
76. The slope of the line segment from to is:
The slope of the line segment from to is:
Therefore, the points are collinear.
m1 � m2
m 2 ����a2 � b2 � c2���2c� � �c�3�
0 � �b�3� ���3a2 � 3b2 � 3c2 � 2c2���6c�
�b�3�
3a2 � 3b2 � c2
2bc
�0, �a2 � b2 � c2
2c ��b3
, c3�
m1 ���a2 � b2��c � �c�3�
b � �b�3� ��3a2 � 3b2 � c2���3c�
�2b��3�
3a2 � 3b2 � c2
2bc
�b, a2 � b2
c ��b3
, c3�
18 Chapter P Preparation for Calculus
77. Find the equation of the line through the points and
or
For .F � 72�, C � 22.2�
5F � 9C � 160 � 0
C �19�5F � 160�
F �95 C � 32
F � 32 �95 �C � 0�
m �180100 �
95
�100, 212�.�0, 32� 78.
If x � 137, C � 0.34�137� � 150 � $196.58.
C � 0.34x � 150
80. (a) Depreciation per year:
where 0 ≤ x ≤ 5.
y � 875 � 175x
8755 � $175
00
6
1000 (b)
(c)
x � 3.86 years
175x � 675
200 � 875 � 175x
y � 875 � 175�2� � $525
81. (a) Two points are and The slope is
(b)
If
(c) If p � 595, x �1
15�1330 � 595� � 49 units.
p � 655, x �1
15�1330 � 655� � 45 units.
15000
0
50
or x �1
15�1330 � p�
p � �15x � 750 � 580 � �15x � 1330
p � 580 � �15�x � 50�
m �625 � 58047 � 50
� �15.
�47, 625�.�50, 580� 82. (a)
(b)
(c) If .
(d) The slope shows the average increase in exam scorefor each unit increase in quiz score.
(e) The points would shift vertically upward 4 units. Thenew regression line would have a y-intercept 4 greaterthan before: y � 22.91 � 3.97x.
x � 17, y � 18.91 � 3.97�17� � 86.4
00
20
100
�x � quiz score, y � test score�y � 18.91 � 3.97x
79. (a)
(b) Using a graphing utility, the point of intersection is Analytically,
(c) Both jobs pay $17 per hour if 6 units are produced. For someone who can produce more than 6 units per hour, the secondoffer would pay more. For a worker who produces less than 6 units per hour, the first offer pays more.
y � 0.75�6� � 12.50 � 17.
3.3 � 0.55x ⇒ x � 6
0.75x � 12.50 � 1.30x � 9.20
�6, 17�.
00
30
50
(6, 17)
W2 � 1.30x � 9.20
W1 � 0.75x � 12.50
Section P.2 Linear Models and Rates of Change 19
83. The tangent line is perpendicular to the line joining thepoint and the center
Slope of the line joining and is
The equation of the tangent line is
12y � 5x � 169 � 0.
y ��512
x �16912
y � 12 ��512
�x � 5�
125
.
�0, 0��5, 12�
x
y
(5, 12)
(0, 0)
4
−4 8 16−8
−8
−16
8
�0, 0�.�5, 12�84. The tangent line is perpendicular to the line joining the
point and the center of the circle,
Slope of the line joining and is
Tangent line:
4y � 3x � 24 � 0
y �34
x � 6
y � 3 �34
�x � 4�
1 � 31 � 4
��43
.
�4, �3��1, 1�
x
y
−2−6 2 4−2
−6
2
4
(1, 1)
(4, −3)
�1, 1�.�4, �3�
85. 4x � 3y � 10 � 0 ⇒ d � 4�0� � 3�0� � 10�42 � 32
�105
� 2 86. 4x � 3y � 10 � 0 ⇒ d � 4�2� � 3�3� � 10�42 � 32
�75
87. x � y � 2 � 0 ⇒ d � 1��2� � ��1��1� � 2�12 � 12
�5�2
�5�2
2
89. A point on the line is The distance from the point to is
d � 1�0� � 1�1� � 5�12 � 12
�1 � 5�2
�4�2
� 2�2.
x � y � 5 � 0�0, 1��0, 1�.x � y � 1
88. x � 1 � 0 ⇒ d � 1�6� � �0��2� � 1�12 � 02
� 7
90. A point on the line is The distance from the point to is
d � 3��1� � 4��1� � 10�32 � ��4�2
� �3 � 4 � 105
�95
.
3x � 4y � 10 � 0��1, �1���1, �1�.3x � 4y � 1
91. If then is the horizontal line The distance to is
If then is the vertical line The distance to is
(Note that A and B cannot both be zero.)
—CONTINUED—
d � x1 � ��CA � �
Ax1 � CA �
Ax1 � By1 � C�A2 � B2
.
�x1, y1�x � �C�A.Ax � C � 0B � 0,
d � y1 � ��CB � �
By1 � CB �
Ax1 � By1 � C�A2 � B2
.
�x1, y1�y � �C�B.By � C � 0A � 0,
20 Chapter P Preparation for Calculus
91. —CONTINUED—
The slope of the line is The equation of the line through perpendicularto is:
The point of intersection of these two lines is:
(1)
(2)
(By adding equations (1) and (2))
(3)
(4)
(By adding equations (3) and (4))
point of intersection
The distance between and this point gives us the distance between and the line
� Ax1 � By1 � C�A2 � B2
���A2 � B2��C � Ax1 � By1�2
�A2 � B2�2
����A�C � By1 � Ax1�A2 � B2 �2
� ��B�C � Ax1 � By1�A2 � B2 �2
����AC � ABy1 � A2x1
A2 � B2 �2� ��BC � ABx1 � B2y1
A2 � B2 �2
d ����AC � B2x1 � ABy1
A2 � B2 � x1�2� ��BC � ABx1 � A2y1
A2 � B2 � y1�2
Ax � By � C � 0.�x1, y1��x1, y1�
��AC � B2x1 � ABy1
A2 � B2 , �BC � ABx1 � A2y1
A2 � B2 �
y ��BC � ABx1 � A2y1
A2 � B2
�A2 � B2�y � �BC � ABx1 � A2y1
Bx � Ay � Bx1 � Ay1⇒ �ABx � A2 y � �ABx1 � A2 y1
Ax � By � �C ⇒ ABx � B2y � �BC
x ��AC � B2x1 � ABy1
A2 � B2
�A2 � B2�x � �AC � B2x1 � ABy1
Bx � Ay � Bx1 � Ay1 ⇒ B2x � ABy � B2x1 � ABy1
Ax � By � �C ⇒ A2x � ABy � �AC
Bx1 � Ay1 � Bx � Ay
Ay � Ay1 � Bx � Bx1
y � y1 �BA
�x � x1�
Ax � By � C � 0�x1, y1��A�B.Ax � By � C � 0
92.
The distance is 0 when In this case, the line contains the point �3, 1�.y � �x � 4m � �1.
� 3m � 3�m2 � 1
d � Ax1 � By1 � C�A2 � B2
� m3 � ��1��1� � 4�m2 � ��1�2 − 9 9
−4
( 1, 0)−
8y � mx � 4 ⇒ mx � ��1�y � 4 � 0
Section P.2 Linear Models and Rates of Change 21
93. For simplicity, let the vertices of the rhombus be and as shown in the figure. The
slopes of the diagonals are then
and
Since the sides of the rhombus are equal,and we have
Therefore, the diagonals are perpendicular.
m1m2 �c
a � b�
cb � a
�c2
b2 � a2 �c2
�c2 � �1.
a2 � b2 � c2,
m2 �c
b � a.m1 �
ca � b
�a � b, c�,�b, c�,�a, 0�,
( , 0)a
( , )a b c+( , )b c
(0, 0)
y
x
�0, 0�,
94. For simplicity, let the vertices of the quadrilateral be and as shown in the figure.
The midpoints of the sides are
and
The slope of the opposite sides are equal:
Therefore, the figure is a paralleogram.
0 �e2
a2
�d2
�
c2
�c � e
2a � b
2�
b � d2
� �e
a � d
c2
� 0
a � b2
�a2
�
c � e2
�e2
b � d2
�d2
�cb
�d2
, e2�.�a
2, 0�, �a � b
2,
c2�, �b � d
2,
c � e2 �,
�d, e�,�b, c�,�a, 0�,�0, 0�,
( , 0)a(0, 0)
( , )b c
( , )d e
a2
0,( )
e2
d2
,( )
b d+2
c e+2
,( )
a b+2
c2
,( )
y
x
95. Consider the figure below in which the four points arecollinear. Since the triangles are similar, the result imme-diately follows.
( , )x y11
( , )x y1*
1*
( , )x y22 ( , )x y2*
2*
y
x
y2� � y1
�
x2� � x1
��
y2 � y1
x2 � x1
97. True.
m2 � �1
m1
bx � ay � c2 ⇒ y �ba
x �c2
a ⇒ m2 �
ba
ax � by � c1 ⇒ y � �ab
x �c1
b ⇒ m1 � �
ab
96. If then Let be a line withslope that is perpendicular to Then Hence, and are parallel. Therefore,and are also perpendicular.L1
L2L3m2 � m3 ⇒ L2
m1m3 � �1.L1.m3
L3m1m2 � �1.m1 � �1�m2,
98. False; if is positive, then is negative.m2 � �1�m1m1
Section P.3 Functions and Their Graphs
22 Chapter P Preparation for Calculus
1. (a) Domain of
Range of
Domain of
Range of
(b)
(c) for
(d) for
(e) for 1 and 2x � �1,g�x� � 0
x � 1f �x� � 2
x � �1f �x� � g�x�
g�3� � �4
f ��2� � �1
�4 ≤ y ≤ 4g:
�3 ≤ x ≤ 3g:
�3 ≤ y ≤ 5f :
�4 ≤ x ≤ 4f : 2. (a) Domain of
Range of
Domain of
Range of
(b)
(c) for and
(d) for
(e) for x � �1g�x� � 0
x � �4, 4f �x� � 2
x � 4x � �2f �x� � g�x�
g�3� � 2
f ��2� � �2
�4 ≤ y ≤ 2g:
�4 ≤ x ≤ 5g:
�4 ≤ y ≤ 4f :
�5 ≤ x ≤ 5f :
3. (a)
(b)
(c)
(d) f �x � 1� � 2�x � 1� � 3 � 2x � 5
f �b� � 2b � 3
f ��3� � 2��3� � 3 � �9
f �0� � 2�0� � 3 � �3 4. (a)
(b)
(c) undefined
(d) f �x � �x� � �x � �x � 3
f ��5� � ��5 � 3 � ��2,
f �6� � �6 � 3 � �9 � 3
f ��2� � ��2 � 3 � �1 � 1
5. (a)
(b)
(c)
(d) g�t � 1� � 3 � �t � 1�2 � �t2 � 2t � 2
g��2� � 3 � ��2�2 � 3 � 4 � �1
g��3� � 3 � ��3�2� 3 � 3 � 0
g�0� � 3 � 02 � 3 6. (a)
(b)
(c)
(d)
� �t � 4�2t � t3 � 8t2 � 16t
g�t � 4� � �t � 4�2�t � 4 � 4�
g�c� � c2�c � 4� � c3 � 4c2
g�32� � �3
2�2�32 � 4� �
94��5
2� � �458
g�4� � 42�4 � 4� � 0
9.f �x � �x� � f �x�
�x�
�x � �x�3 � x3
�x�
x3 � 3x2�x � 3x��x�2 � ��x�3 � x3
�x� 3x2 � 3x�x � ��x�2, �x � 0
11.
�1 � �x � 1
�x � 2��x � 1�
1 � �x � 1
1 � �x � 1�
2 � x
�x � 2��x � 1�1 � �x � 1��
�1�x � 1�1 � �x � 1�
, x � 2
f �x� � f �2�x � 2
��1��x � 1 � 1�
x � 2
7. (a)
(b)
(c) f ��
3� � cos�2��
3�� � cos2�
3� �
12
f ���
4� � cos�2���
4�� � cos���
2� � 0
f �0� � cos�2�0�� � cos 0 � 1 8. (a)
(b)
(c) f �2�
3 � � sin�2�
3 � ��32
f �5�
4 � � sin�5�
4 � ���2
2
f ��� � sin � � 0
12.f �x� � f �1�
x � 1�
x3 � x � 0x � 1
�x�x � 1��x � 1�
x � 1� x�x � 1�, x � 1
10.f �x� � f �1�
x � 1�
3x � 1 � �3 � 1�x � 1
�3�x � 1�
x � 1� 3, x � 1
Section P.3 Functions and Their Graphs 23
13.
Domain:
Range: ���, 0�
x � 3 ≥ 0 ⇒ ��3, ��
h�x� � ��x � 3 14.
Domain:
Range: ��5, ��
���, ��
g�x� � x2 � 5 15.
Domain: all k an integer
Range: ���, �1� � �1, ��
t � 4k � 2,
�t4
��2k � 1��
2 ⇒ t � 4k � 2
f �t� � sec �t4
16.
Domain: all k an integer
Range: ���, ��
t � k�,
h�t� � cot t 17.
Domain:
Range: ���, 0� � �0, ��
���, 0� � �0, ��
f �x� �1x
18.
Domain:
Range: ���, 0� � �0, ��
���, 1� � �1, ��
g�x� �2
x � 1
19.
and
and
Domain: 0 ≤ x ≤ 1
x ≤ 1x ≥ 0
1 � x ≥ 0x ≥ 0
f �x� � �x � �1 � x 20.
Domain: or
Domain: ���, 1� � �2, ��
x ≤ 1x ≥ 2
�x � 2��x � 1� ≥ 0
x2 � 3x � 2 ≥ 0
f �x� � �x2 � 3x � 2 21.
Domain: all n an integerx � 2n�,
cos x � 1
1 � cos x � 0
g�x� �2
1 � cos x
22.
Domain: all
n integer5�
6� 2n�,
x ��
6� 2n�,
sin x �12
sin x �12
� 0
h�x� �1
sin x � �1�2� 23.
Domain: all x � �3
x � 3 � 0
x � 3 � 0
f �x� �1
x � 3 24.
Domain: all x � ±2
�x � 2��x � 2� � 0
x2 � 4 � 0
g�x� �1
x2 � 4
25.
(a)
(b)
(c)
(d)
(Note: for all t)
Domain:
Range: ���, 1� � �2, ��
���, ��
t2 � 1 ≥ 0
f �t2 � 1� � 2�t2 � 1� � 2 � 2t2 � 4
f �2� � 2�2� � 2 � 6
f �0� � 2�0� � 2 � 2
f ��1� � 2��1� � 1 � �1
f �x� � 2x � 1,
2x � 2,
x < 0
x ≥ 026.
(a)
(b)
(c)
(d)
(Note: for all s)
Domain:
Range: �2, ��
���, ��
s2 � 2 > 1
f �s2 � 2� � 2�s2 � 2�2 � 2 � 2s 4 � 8s2 � 10
f �1� � 12 � 2 � 3
f �0� � 02 � 2 � 2
f ��2� � ��2�2 � 2 � 6
f �x� � x2 � 2,
2x2 � 2,
x ≤ 1
x > 1
24 Chapter P Preparation for Calculus
27.
(a)
(b)
(c)
(d)
Domain:
Range: ���, 0� � �1, ��
���, ��
f �b2 � 1� � ��b2 � 1� � 1 � �b2
f �3� � �3 � 1 � �2
f �1� � �1 � 1 � 0
f ��3� � �3 � 1 � 4
f �x� � x � 1,
�x � 1,
x < 1
x ≥ 128.
(a)
(b)
(c)
(d)
Domain:
Range: �0, ��
��4, ��
f �10� � �10 � 5�2 � 25
f �5� � �5 � 4 � 3
f �0� � �0 � 4 � 2
f ��3� � ��3 � 4 � �1 � 1
f �x� � �x � 4,
�x � 5�2,
x ≤ 5
x > 5
29.
Domain:
Range:
−2−4 2 4
2
4
6
8
x
y
���, ��
���, ��
f �x� � 4 � x 30.
Domain:
Range:
2
6
4
2 4 6
y
x
���, 0� � �0, ��
���, 0� � �0, ��
g�x� �4x
31.
Domain:
Range:
1 2 3
1
2
x
y
�0, ��
�1, ��
h�x� � �x � 1
32.
Domain:
Range:
2 4 6− 4− 6
4
8
6
y
x
���, ��
���, ��
f �x� �12 x3 � 2 33.
Domain:
Range:
−1−2−3−4 1 2 3 4
−2
−3
1
2
4
5
x
y
�0, 3�
��3, 3�
f �x� � �9 � x2 34.
Domain:
Range:
y-intercept:
x-intercept:
(0, 2)− 2, 0( (
1 2 3 4−1
−2
−3
−4
4
3
−4 −3 −2
y
x
���2, 0��0, 2�
��2, 2�2� � ��2, 2.83�
��2, 2�
f �x� � x � �4 � x2
35.
Domain:
Range: ��2, 2�
���, ��
t
y
2 3
−1
1
2
g�t� � 2 sin � t 36.
Domain:
Range: ��5, 5�
���, ��
ππ
5
− 5
θ−2 2
4
321
yh�� � �5 cos
2
Section P.3 Functions and Their Graphs 25
37. The student travels during the first
4 minutes. The student is stationary for the following
2 minutes. Finally, the student travels
during the final 4 minutes.
6 � 210 � 6
� 1 mi�min
2 � 04 � 0
�12
mi�min 38.
t
27
t t t1 2 3
9
18
d
39.
y is not a function of x. Some vertical lines intersectthe graph twice.
x � y 2 � 0 ⇒ y � ±�x 40.
y is a function of x. Vertical lines intersect the graphat most once.
�x2 � 4 � y � 0 ⇒ y � �x2 � 4
41. y is a function of x. Vertical lines intersect the graphat most once.
42.
y is not a function of x. Some vertical lines intersectthe graph twice.
y � ±�4 � x2
x2 � y2 � 4
43.
y is not a function of x since there are two values of y forsome x.
x2 � y2 � 4 ⇒ y � ±�4 � x2 44.
y is a function of x since there is one value of y foreach x.
x2 � y � 4 ⇒ y � 4 � x2
45.
y is not a function of x since there are two values of y forsome x.
y2 � x2 � 1 ⇒ y � ±�x2 � 1 46.
y is a function of x since there is one value of y foreach x.
x2y � x2 � 4y � 0 ⇒ y �x2
x2 � 4
47. is a horizontal shift 5 units to the left. Matches d.y � f �x � 5� 48. is a vertical shift
5 units downward. Matches b.y � f �x� � 5 49. is a reflection in
the axis, a reflection in the axis,and a vertical shift downward 2 units. Matches c.
x-y-y � �f ��x� � 2
50. is a horizontalshift 4 units to the right, followedby a reflection in the axis.Matches a.
x-
y � �f �x � 4� 51. is a horizontalshift to the left 6 units, and a vertical shift upward 2 units.Matches e.
y � f �x � 6� � 2 52. is a horizontalshift to the right 1 unit, and a vertical shift upward 3 units.Matches g.
y � f �x � 1� � 3
53. (a) The graph is shifted 3 units to the left.
—CONTINUED—
x
−6
−2
−4
4
−4−6 −2 2 4
y
(b) The graph is shifted 1 unit to the right.
x
−6
−2
−4
4
2
−2 2 4 6 8
y
(c) The graph is shifted 2 unitsupward.
x
−2
4
6
2
−4 −2 2 4 6
y
26 Chapter P Preparation for Calculus
53. —CONTINUED—
(d) The graph is shifted 4 unitsdownward.
x
−2
−8
−6
−4
−4 −2 2 4 6
y
(e) The graph is stretched vertically by a factor of 3.
x
−2
−8
−10
−6
−4
−4 −2 4 6
y
(f ) The graph is stretchedvertically by a factor of
x
−6
4
2
−4 −2 2 4 6
y
14.
54. (a)
Shift f right 4 units
1 2 3 5 6 7−1
1
3
2
4
−2
(0, 3)−
(6, 1)
−4
y
x
g�0� � f ��4� � �3
g�6� � f �2� � 1
g�x� � f �x � 4� (b)
Shift f left 2 units
1−1
3
2
4
−2
−4
−4 −3
−3
−5−7 −6
(0, 1)
( 6, 3)− −
y
x
g�x� � f �x � 2� (c)
Vertical shift upwards4 units
1 2 3−1−3−5 −4 −2
−2
1
4
6
5
2
( 4, 1)−
(2, 5)
y
x
g�x� � f �x� � 4
(d)
Vertical shift down 1 unit
2 3−1−3
−3
−5
−5
−6
−4
−4
−2
1
2
(2, 0)
( 4, 4)− −
y
x
g�x� � f �x� � 1 (e)
1 2 3−1−3
−3
−5
−5
−6
−4
−4
−2
1
2(2, 2)
( 4, )− −6
y
x
g��4� � 2f ��4� � �6
g�2� � 2 f �2� � 2
g�x� � 2f �x� (f )
1 2 3−1
−3
−3
−5
−5
−6
−4
−4
−2
1
2
3
1
2
2
− −4,
2,
( )
( )
y
x
g��4� �12 f ��4� � �
32
g�2� �12 f �2� �
12
g�x� �12 f �x�
55. (a)
Vertical shift 2 units upward
4
3
2
1
4321
y
x
y � �x � 2 (b)
Reflection about the x-axis
4321
−1
−2
1
−3
x
y
y � ��x (c)
Horizontal shift 2 units to the right
643
3
2
1
1
−1
−2
2
4
5x
y
y � �x � 2
56. (a) is a horizontal shift units to the left, followed by a vertical shift 1 unit upwards.
(b) is a horizontal shift 1 unit to the right followed by a reflection about the x-axis.h�x� � �sin�x � 1�
��2h�x� � sin�x � ���2�� � 1
Section P.3 Functions and Their Graphs 27
58.
(a)
(b)
(c)
(d)
(e)
(f ) g� f �x�� � g�sin x� � � sin x
f �g�x�� � f ��x� � sin��x�
g� f ���4�� � g�sin���4�� � g��2�2� � � ��2�2� ���2
2
g� f �0�� � g�0� � 0
f �g�1�2�� � f ���2� � sin���2� � 1
f �g�2�� � f �2�� � sin�2�� � 0
g�x� � �xf �x� � sin x,
57. (a)
(b)
(c) g� f �0�� � g�0� � �1
g� f �1�� � g�1� � 0
f �g�1�� � f �0� � 0 (d)
(e)
(f ) g� f �x�� � g��x � � ��x �2� 1 � x � 1, �x ≥ 0�
f �g�x�� � f �x2 � 1� � �x2 � 1
f �g��4�� � f �15� � �15
59.
Domain:
Domain:
No. Their domains are different. � f g� � �g f � for x ≥ 0.
���, ��
�g f ��x� � g� f �x�� � g�x2� � �x2 � x�0, ��
� f g��x� � f �g�x�� � f ��x� � ��x�2� x, x ≥ 0
f �x� � x2, g�x� � �x 60.
Domain:
Domain:
No, f g � g f.
���, ��
�g f ��x� � g�x 2 � 1� � cos�x 2 � 1�
���, ��
� f g��x� � f �g�x�� � f �cos x� � cos2 x � 1
f �x� � x2 � 1, g�x� � cos x
61.
Domain: all
Domain: all
No, f g � g f.
x � 0
�g f ��x� � g� f �x�� � g �3x� � �3
x�2
� 1� 9x2 � 1 �
9 � x2
x2
x � ±1
� f g��x� � f �g�x�� � f �x2 � 1� �3
x2 � 1
f �x� �3x, g�x� � x2 � 1
62.
Domain:
You can find the domain of by determining the intervals where and are both positive, or both negative.
Domain: ���, �12�, �0, ��
0 1 2−1 12
−2 −x
+ + + ++++ − −
x�1 � 2x�g f
�g f ��x� � g�1x� ��1
x� 2 ��1 � 2x
x
��2, ��
� f g��x� � f ��x � 2� �1
�x � 2
63. (a)
(b)
(c) which is undefinedg� f �5�� � g��5�,
g� f �2�� � g�1� � �2
� f g��3� � f �g�3�� � f ��1� � 4 (d)
(e)
(f ) which is undefinedf �g��1� � f ��4�,
�g f ���1� � g� f ��1�� � g�4� � 2
� f g���3� � f �g��3�� � f ��2� � 3
28 Chapter P Preparation for Calculus
64.
represents the area of the circle at time t.�A r��t�
�A r��t� � A�r�t�� � A�0.6t� � ��0.6t�2 � 0.36�t 2
65.
Let and
Then,
[Other answers possible]
� f g h��x� � f �g�x � 1�� � f �2�x � 1�� � �2�x � 1� � �2x � 2 � F�x�.
f �x� � �x.g�x� � 2xh�x� � x � 1,
F�x� � �2x � 2
66.
Let and
Then,
[Other answers possible]
� f g h��x� � f �g�1 � x�� � f �sin�1 � x�� � �4 sin�1 � x� � F�x�.
h�x� � 1 � x.g�x� � sin xf �x� � �4x,
F�x� � �4 sin�1 � x�
67.
Even
f ��x� � ��x�2�4 � ��x�2� � x2�4 � x2� � f �x� 68.
Odd
f ��x� � 3��x � � 3�x � �f �x�
69.
Odd
f ��x� � ��x� cos��x� � �x cos x � �f �x�
70.
Even
f ��x� � sin2��x� � sin��x� sin��x� � ��sin x���sin x� � sin2 x
71. (a) If f is even, then is on the graph.
(b) If f is odd, then is on the graph.�32 , �4��3
2 , 4� 72. (a) If f is even, then is on the graph.
(b) If f is odd, then is on the graph.��4, �9�
��4, 9�
73. f is even because the graph is symmetric about the axis.
is neither even nor odd.
is odd because the graph is symmetric about the origin.h
g
y-
74. (a) If f is even, then the graph is symmetric about the axis.
x
y
f
2
2
−2
−4
−6
4
6
4 6−6 −4 −2
y- (b) If is odd, then the graph is symmetric about the origin.
x
y
f
2
2
−2
−4
−6
4
6
4 6−6 −4 −2
f
75.
�4 ≤ x ≤ 0 f �x� � �2x � 5,
y � �2x � 5
y � 5 � �2�x � 0�x
y
−4−6 2 4 6
−6
2
4
6
(−4, 3)
(0, −5)
Slope �3 � 5
�4 � 0� �2
Section P.3 Functions and Their Graphs 29
76.
1 ≤ x ≤ 5 f �x� �34
x �54
,
y �34
x �54
y � 2 �34
�x � 1�
x
y
(1, 2)
(5, 5)
−1−2 1 2 3 4 5−1
−2
1
2
3
4
5 Slope �
5 � 25 � 1
�34
77.
x ≤ 0 f �x� � ���x,
y � ���x
y2 � �x
x
y
−1−2−3−4−5 1
−2
−3
1
2
3 x � y2 � 0
78.
�2 ≤ x ≤ 2 f �x� � ��4 � x2,
y � ��4 � x2
y2 � 4 � x2
x
y
−1 1
−1
1
x2 � y2 � 4
79. Matches (ii). The function is Since satisfies the equation, Thus, g�x� � �2x2.c � �2.
�1, �2�g�x� � cx2. 80. Matches (i). The function is Since satisfies the equation, Thus, f �x� � �1�4�x.c � 1�4.
�1, 1�4�f �x� � cx.
81. Matches (iv). The function is since it must beundefined at Since satisfies the equation,
Thus, r�x� � 32�x.c � 32.�1, 32�x � 0.
r�x� � c�x, 82. Matches (iii). The function is Since satisfies the equation, Thus, h�x� � 3�x.c � 3.
�1, 3�h�x� � c�x.
83. (a)
(b) If then the program would turn on (and off) one hour later.
(c) If then the overall temperature would be reduced 1 degree.H�t� � T�t� � 1,
H�t� � T�t � 1�,
T�4� � 16, T�15� � 23
84. (a) For each time t, there corresponds a depth d.
(b) Domain:
Range:
(c)
t
d
5
10
15
20
25
30
1 2 3 4 5 6
0 ≤ d ≤ 30
0 ≤ t ≤ 5
85. (a)
(b) A�15� � 345 acres�farm
10
100
200
300
400
500
20 30 40 50t
A
86. (a)
(b)
� 0.00078125x 2 � 0.003125x � 0.029
H� x1.6� � 0.002� x
1.6�2
� 0.005� x1.6� � 0.029
0 1000
25
30 Chapter P Preparation for Calculus
87.
If then
If then
If then
Thus,
f �x� � 2�1 � x�,2, 2�x � 1�,
x < 00 ≤ x < 2 .x ≥ 2
f �x� � x � �x � 2� � 2x � 2 � 2�x � 1�.x ≥ 2,
f �x� � x � �x � 2� � 2.0 ≤ x < 2,
f �x� � �x � �x � 2� � �2x � 2 � 2�1 � x�.x < 0,
f �x� � x � x � 2
88. has one zero. has three zeros. Every cubic polynomial has at least one zero. Given we have
as and as if Furthermore, as and as if Since the graph has no breaks, the
graph must cross the x-axis at least one time.A < 0.x →�p → ��x → ��
p →�A > 0.x →�p →�x → ��p → ��p�x� � Ax3 � Bx2 � Cx � D,
− 3 3
−2
p
p
2
1
2p2�x� � x3 � xp1�x� � x3 � x � 1
89.
Odd
� �f �x�
� ��a2n�1x2n�1 � . . . � a3 x3 � a1x�
f ��x� � a2n�1��x�2n�1 � . . . � a3��x�3 � a1��x�
90.
Even
� f �x�
� a2n x2n � a2n�2 x2n�2 � . . . � a2 x2 � a0
f ��x� � a2n��x�2n � a2n�2��x�2n�2 � . . . � a2��x�2 � a0
91. Let where f and g are even. Then
.
Thus, is even. Let where f and g are odd. Then
Thus, is even.F�x�
F��x� � f ��x�g��x� � ��f �x����g�x�� � f �x�g�x� � F�x�.
F�x� � f �x�g�x�F�x�
F��x� � f ��x�g��x� � f �x�g�x� � F�x�
F�x� � f �x�g�x�
92. Let where f is even and g is odd. Then
Thus, is odd.F�x�
F��x� � f ��x�g��x� � f �x���g�x�� � �f �x�g�x� � �F�x�.
F�x� � f �x�g�x�
93. (a) (c)
Domain:
(b)
The dimensions for maximum volume are cm.
The dimensions for maximum volume appear to be cm.4 � 16 � 16
4 � 16 � 16
12
−100
−1
1100
0 < x < 12
V � x�24 � 2x�2 � 4x�12 � x2�x length and width volume
1
2
3
4
5
6 6�24 � 2�6��2 � 86424 � 2�6�
5�24 � 2�5��2 � 98024 � 2�5�
4�24 � 2�4��2 � 102424 � 2�4�
3�24 � 2�3��2 � 97224 � 2�3�
2�24 � 2�2��2 � 80024 � 2�2�
1�24 � 2�1��2 � 48424 � 2�1�
Section P.4 Fitting Models to Data 31
98. False; let Then and Thus, 3f �x� � f �3x�.3f �x� � 3x2.f �3x� � �3x�2 � 9x2f �x� � x2.
94.
,
.L � �x2 � y2 ��x2 � � 2xx � 3�
2
y �6
x � 3� 2 �
2xx � 3
y � 2 �6
x � 3
By equating slopes, y � 20 � 3
�0 � 2x � 3
95. False; let
Then but .�3 � 3f ��3� � f �3� � 9,
f �x� � x2.
96. True 97. True, the function is even.
99. First consider the portion of in the first quadrant:and shown below.
The area of this region is 1 �12 �
32.
x
y
−1 2
−1
1
2
(2, 1)(0, 1)
(0, 0) (1, 0)
x � y ≤ 1;0 ≤ y ≤ 1x ≥ 0,R By symmetry, you obtain the entire region
The area of is
[49th competition, Problem A1, 1988]
4�32� � 6.R
x
y
−2 1 2
−2
2
(2, 1)
(2, −1)
(−2, 1)
(−2, −1)
R:
100. Let be a constant polynomial.
Then and
Thus, Since this is true for all real numbers c, f is the identity function: f �x� � x.f �c� � c.
g� f �x�� � c.f �g�x�� � f �c�
g�x� � c
Section P.4 Fitting Models to Data
1. Quadratic function 2. Trigonometric function 3. Linear function 4. No relationship
5. (a), (b)
Yes. The cancer mortality increases linearly withincreased exposure to the carcinogenic substance.
(c) If x � 3, then y � 136.
x
y
3 6 9 12 15
50
100
150
200
250
6. (a)
No, the relationship does not appear to be linear.
(b) Quiz scores are dependent on several variables such asstudy time, class attendance, etc. These variables maychange from one quiz to the next.
0 200
20
32 Chapter P Preparation for Calculus
7. (a)
(b)
The model fits well.
(c) If then d � 0.066�55� � 3.63 cm.F � 55,
00
10
F d= 15.13 + 0.1
125
d � 0.066F or F � 15.1d � 0.1 8. (a)
(b)
The model fits well.
(c) If t � 2.5, s � 24.65 meters�second.
45
5
−5
−1
s � 9.7t � 0.4
9. (a) Using a graphing utility,
correlation coefficient
(b)
(c) The data indicates that greater per capita electricityconsumption tends to correspond to greater per capitagross national product.
The data for Hong Kong, Venezuela and South Koreadiffer most from the linear model.
(d) Removing the data andyou obtain the model
with r � 0.968.y � 0.134x � 0.28�167, 17.3�,
�113, 5.74��118, 25.59�,
30
00
180
y = 0.124x + 0.82
r � 0.838
y � 0.124x � 0.82. 10. (a) Linear model:
(b)
The fit is very good.
(c) When
H � �0.3323�500� � 612.9333 � 446.78.
t � 500,
00 1300
600
H � �0.3323t � 612.9333
11. (a)
(b)
For y1 � y2 � y3 � 31.06 cents�mile.t � 12,
80
0
y1
y2
y3
y1 + y2 + y3
15
y3 � 0.0917t � 0.7917
y2 � 0.1095t � 2.0667
y1 � 0.0343t3 � 0.3451t2 � 0.8837t � 5.6061 12. (a)
(b)
(c) When .x � 2, S � 583.98 pounds
00
14
25000
S � 180.89x 2 � 205.79x � 272
13. (a) Linear:
Cubic:
(b)
(c) The cubic model is better.
90
1325
0
y1
y2
y2 � �0.1289t3 � 2.235t2 � 4.86t � 35.2
y1 � 4.83t � 28.6 (d)
(e) For Linear model million
Cubic model million
(f ) Answers will vary.
y2 � 51.5
y1 � 96.2t � 14:
90
1325
0
y � �0.084t2 � 5.84t � 26.7
Section P.4 Fitting Models to Data 33
14. (a)
(b)
(c) The curve levels off for
(d)
(e) The model is better for low speeds.
0 1000
21
t � 0.002s2 � 0.0346s � 0.183
s < 20.
20 1000
21
t � 0.00271s2 � 0.0529s � 2.671 15. (a)
(b)
(c) If horsepower.x � 4.5, y � 214
00
7
300
y � �1.806x3 � 14.58x2 � 16.4x � 10
16. (a)
(b)
(c) For pounds per square inch.
(d) The model is based on data up to 100 pounds persquare inch.
T � 300�F, p � 68.29
0 110150
350
T � 2.9856 � 10�4 p3 � 0.0641 p2 � 5.2826p � 143.1 17. (a) Yes, y is a function of t. At each time t, there is oneand only one displacement y.
(b) The amplitude is approximately
The period is approximately
(c) One model is .
(d)
00
4
0.9
(0.125, 2.35)
(0.375, 1.65)
y � 0.35 sin�4� t� � 2
2�0.375 � 0.125� � 0.5.
�2.35 � 1.65��2 � 0.35.
18. (a)
One model is
(b)
00
13
100
C�t� � 58 � 27 sin�� t6
� 4.1�.
H�t� � 84.4 � 4.28 sin�� t6
� 3.86� (c)
(d) The average in Honolulu is 84.4.
The average in Chicago is 58.
(e) The period is 12 months (1 year).
(f ) Chicago has greater variability �27 > 4.28�.
00
13
100
19. Answers will vary. 20. Answers will vary.
34 Chapter P Preparation for Calculus
Review Exercises for Chapter P
1.
y-intercept
x-intercepty � 0 ⇒ 0 � 2x � 3 ⇒ x �32 ⇒ �3
2 , 0�x � 0 ⇒ y � 2�0� � 3 � �3 ⇒ �0, �3�
y � 2x � 3
2.
y-intercept
x-interceptsy � 0 ⇒ 0 � �x � 1��x � 3� ⇒ x � 1, 3 ⇒ �1, 0�, �3, 0�
x � 0 ⇒ y � �0 � 1��0 � 3� � 3 ⇒ �0, 3�
y � �x � 1��x � 3�
3.
y-intercept
x-intercepty � 0 ⇒ 0 �x � 1x � 2
⇒ x � 1 ⇒ �1, 0�
x � 0 ⇒ y �0 � 10 � 2
�12
⇒ �0, 12�
y �x � 1x � 2
4.
and are both impossible. No intercepts.y � 0x � 0
xy � 4
5. Symmetric with respect to y-axis since
x2y � x2 � 4y � 0.
��x�2y � ��x�2 � 4y � 0
6. Symmetric with respect to y-axis since
y � x 4 � x2 � 3.
y � ��x�4 � ��x�2 � 3
7.
−1 1 2 3
−1
−2
1
2
3
x
y
y � �12 x �
32 8.
Slope: 2
y-intercept:
2 3−1−2
−1
−2
−3
1
y
x
�0, �3�
y � 2x � 3
4x � 2y � 6 9.
Slope:
y-intercept:
−1−2−3 1
−1
2
3
x
y
65
25
y �25 x �
65
�25 x � y �
65
�13 x �
56 y � 1
10.
Slope:
y-intercept: �0, 53��
215
y � �2
15 x �53
2x � 15y � 25
− 4 4 8 12
−1
−2
1
3
y
x
0.02x � 0.15y � 0.25 11.
x
y
−5−10 5
5
10
y � 7 � 6x � x2
Review Exercises for Chapter P 35
12.
2 4 8
2
4
6
8
10
−2
y
x
y � x�6 � x� 13.
Domain:
1 2 3 4 5
1
2
3
4
5
x
y
���, 5�
y � �5 � x 14.
−1 2 3 4 5 6−1
−2
−3
−4
−5
−6
y
x
y � �x � 4� � 4
15.
Xmin = -5Xmax = 5Xscl = 1Ymin = -30Ymax = 10Yscl = 5
y � 4x2 � 25 16.
Xmin = -40Xmax = 40Xscl = 10Ymin = -40Ymax = 40Yscl = 10
y � 8 3�x � 6 17.
Point: �4, 1�
y � 1
x � 4
7x � 28
4x � 4y � 20
3x � 4y � 8
19. You need factors and Multiply by x to obtain origin symmetry.
� x3 � 4x
y � x�x � 2��x � 2�
�x � 2�.�x � 2� 20.
(a) and
(b) and
(c)
(d) �1 � k��1�3 ⇒ k � 1 ⇒ y � x3
0 � k�0�3 ⇒ any k will do!
y � �18 x31 � k��2�3 ⇒ k � �
18
y � 4x34 � k�1�3 ⇒ k � 4
y � kx3
18.
No real solutionNo points of intersectionThe graphs of and do not intersect.y � x2 � 7y � x � 1
0 � x2 � x � 6
�x � 1� � x2 � 7
y � x � 1
−12 12
14
−2
21.
Slope ��5�2� � 15 � �3�2� �
3�27�2
�37
x
y
1 2 3 4 5
1
2
3
4
5
, 132
52
5,( )
( )
22.
The line is vertical and hasno slope.
2
6
4
8
12
10
14
−21 2 3 4 5 6
(7, 12)
(7, 1)−
y
x
23.
t �73
1 � t � �43
1 � t1 � 0
�1 � 5
1 � ��2�
36 Chapter P Preparation for Calculus
24.
t � �533
�53 � 3t
�44 � 9 � 3t
4
�3 � t�
�3�11
3 � ��1��3 � t
�3 � 6
�3 � 825.
y
x
(0, −5)
−4
−4
−8
−2−2
2
2
4
4 6 8
2y � 3x � 10 � 0
y �32x � 5
y � ��5� �32�x � 0� 26.
Horizontal line
y
x−6 −4
−4
−2
2
4
8
−2 2 4 6
(−2, 6)
y � 6
y � 6 � 0�x � ��2��
27.
y
x−8
−8
−6
−4
−2
2
4
−6 −4 2 4
(−3, 0)
3y � 2x � 6 � 0
y � �23x � 2
y � 0 � �23�x � ��3�� 28. m is undefined. Line is vertical.
y
x−4
−4
−6
−2−2
2
4
6
2 4 6 8
(5, 4)
x � 5
29. (a)
(b) Slope of line is
(c)
(d)
x � 2 � 0
x � �2
2x � y � 0
y � �2x
m �4 � 0
�2 � 0� �2
0 � 5x � 3y � 22
3y � 12 � 5x � 10
y � 4 �53
�x � 2�
53
.
0 � 7x � 16y � 78
16y � 64 � 7x � 14
y � 4 �7
16�x � 2� 30. (a)
(b) Slope of perpendicular line is 1.
(c)
(d)
y � 3 � 0
y � 3
0 � x � y � 2
y � x � 2
y � 3 � 1�x � 1�
m �4 � 32 � 1
� 1
0 � x � y � 2
y � x � 2
y � 3 � 1�x � 1�
2x � 3y � 11 � 0
3y � 9 � �2x � 2
y � 3 � �23
�x � 1�
31. The slope is
V�3� � �850�3� � 12,500 � $9950
V � �850t � 12,500.�850.
Review Exercises for Chapter P 37
34.
Function of x since there is onevalue for y for each x.
1 2 3−1−2−3
1
3
2
4
6
5
y
x
x2 � y � 0 35.
Function of x since there is onevalue of y for each x.
x
y
−1−2 3 4
−2
3
4
y � x2 � 2x 36.
Not a function of x since there are two values of yfor some x.
3 6 12−3−9 −6−12−1
−2
−4
1
4
2
y
x
x � 9 � y2
37.
(a) does not exist.
(b)
��1
1 � �x, �x � �1, 0
f �1 � �x� � f �1��x
�
11 � �x
�11
�x�
1 � 1 � �x�1 � �x��x
f �0�
f �x� �1x
38. (a) (because )
(b)
(c) f �1� � �1 � 2� � 1
f �0� � �0 � 2� � 2
�4 < 0f ��4� � ��4�2 � 2 � 18
39. (a) Domain:
Range:
(b) Domain: all
Range: all
(c) Domain: all x or
Range: all y or ���, ��
���, ��
y � 0 or ���, 0� � �0, ��
x � 5 or ���, 5� � �5, ��
0, 6�
36 � x2 ≥ 0 ⇒ �6 ≤ x ≤ 6 or �6, 6� 40. and
(a)
(b)
(c) g� f �x�� � g�1 � x2� � 2�1 � x2� � 1 � 3 � 2x2
f �x�g�x� � �1 � x2��2x � 1� � �2x3 � x2 � 2x � 1
f �x� � g�x� � �1 � x2� � �2x � 1� � �x2 � 2x
g�x� � 2x � 1f �x� � 1 � x2
32. (a)
(b)
(c)
t 5034.48 hours to break even
7.25t � 36,500
30t � 22.75t � 36,500
R � 30t
� 22.75t � 36,500
C � 9.25t � 13.50t � 36,500 33.
Not a function of x since there are two values of y forsome x.
x
y
1 2 3 4 5 6−1
−2
−3
1
2
3
y � ±�x
x � y2 � 0
38 Chapter P Preparation for Calculus
42.
(a) The graph of g is obtained from f by a vertical shiftdown 1 unit, followed by a reflection in the x-axis:
� �x3 � 3x2 � 1
g�x� � � f �x� � 1�
−6 6
−6
(0, 0)
(2, 4)−
6
f �x� � x3 � 3x2
(b) The graph of g is obtained from f by a vertical shiftupwards of 1 and a horizontal shift of 2 to the right.
� �x � 2�3 � 3�x � 2�2 � 1
g�x� � f �x � 2� � 1
43. (a) Odd powers:
The graphs of f, g, and h all rise to the right and fall tothe left. As the degree increases, the graph rises andfalls more steeply. All three graphs pass through thepoints �0, 0�, �1, 1�, and ��1, �1�.
3
−2
−3
f
h
g2
f �x� � x, g�x� � x3, h�x� � x5 Even powers:
The graphs of f, g, and h all rise to the left and to theright. As the degree increases, the graph rises moresteeply. All three graphs pass through the points �1, 1�, and ��1, 1�.
�0, 0�,
30
−3
h
g
f
4
f �x� � x2, g�x� � x4, h�x� � x6
(b) will look like but rise and fall even more steeply.
will look like but rise even more steeply.h�x� � x6,y � x8
h�x� � x5,y � x7
41. (a)
(c)
42
1
1
2
2
x
y
2c
0c
c 2
f �x� � �x � 2�3 � c, c � �2, 0, 2
3
3
1
223x
y
2
c
c
0
c 2
f �x� � x3 � c, c � �2, 0, 2 (b)
(d)
3
3
2
1
21123x
y
0
2
c
c
2c
f �x� � cx3, c � �2, 0, 2
3
1
22
2
3
x
y0c
2c
2c
f �x� � �x � c�3, c � �2, 0, 2
44. (a)
−4 10
−25
100
f �x� � x2�x � 6�2 (b)
−2 10
−100
300
g�x� � x3�x � 6�2 (c)
−4 10
−800
200
h�x� � x3�x � 6�3
Problem Solving for Chapter P 39
45. (a)
A � xy � x�12 � x� � 12x � x2
y � 12 � x
2x � 2y � 24
x x
y
y (b) Domain:
(c) Maximum area is In general, the maximumarea is attained when the rectangle is a square. In thiscase, x � 6.
A � 36.
00
12
40
0 < x < 12
49. (a) Yes, y is a function of t. At each time t, there is oneand only one displacement y.
(b) The amplitude is approximately
.
The period is approximately 1.1.
(c) One model is
(d)
0 2.2
−0.5
0.5
(0.5, −0.25)
(1.1, 0.25)
y �14
cos�2�
1.1t� �
14
cos�5.7t�
�0.25 � ��0.25���2 � 0.25
46. For company (a) the profit rose rapidly for the first year, and then leveled off. For the second company (b), the profit dropped,and then rose again later.
48. (a)
(b)
(c) The data point is probably an error. Without this point, the new model is
y � �1.4344x � 66.4387.
�27, 44�
3300
70
y � �1.204x � 64.2667
47. (a) 3 (cubic), negative leading coefficient
(b) 4 (quartic), positive leading coefficient
(c) 2 (quadratic), negative leading coefficient
(d) 5, positive leading coefficient
Problem Solving for Chapter P
1. (a)
Center: Radius: 5
(c) Slope of line from to is
Slope of tangent line is Hence,
Tangent liney � 0 �34
�x � 6� ⇒ y �34
x �92
34
.
4 � 03 � 6
� �43
.�3, 4��6, 0�
�3, 4�
�x � 3�2 � � y � 4�2 � 25
�x2 � 6x � 9� � �y2 � 8y � 16� � 9 � 16
x2 � 6x � y2 � 8y � 0 (b) Slope of line from to is Slope of tangent line
is Hence,
Tangent line
(d)
Intersection: �3, �94�
x � 3
32
x �92
�34
x �34
x �92
y � 0 � �34
�x � 0� ⇒ y � �34
x
�34
.
43
.�3, 4��0, 0�
40 Chapter P Preparation for Calculus
2. Let be a tangent line to the circle from the point Then
Setting the discriminant equal to zero,
Tangent lines: and y � ��3x � 1.y � �3x � 1
m � ±�3
4m2 � 12
16m2 � 12m2 � 12
16m2 � 4�m2 � 1��3� � 0
b2 � 4ac
�m2 � 1�x2 � 4mx � 3 � 0
x2 � �mx � 1 � 1�2 � 1
x2 � �y � 1�2 � 1
�0, 1�.y � mx � 1
3.
(a)
(c)
x1
2
1
3
4
−2 −1−1
−2
−3
−4
−3−4 2 3 4
y�H�x�
x1
2
1
3
4
−2 −1−1
−3
−4
−3−4 2 3 4
yH�x� � 2
x1
2
1
3
4
−2 −1−1
−2
−3
−4
−3−4 2 3 4
yH�x� � �1,
0,
x ≥ 0
x < 0
(b)
(d)
x1
2
3
4
−2 −1−1
−2
−3
−4
−3−4 2 3 4
yH��x�
x1
2
1
3
4
−2 −1−1
−2
−3
−4
−3−4 2 3 4
yH�x � 2�
(e)
x1
2
1
3
4
−2 −1−1
−2
−3
−4
−3−4 2 3 4
y12H�x� (f )
x1
1
3
4
−2 −1−1
−2
−3
−4
−3−4 2 3 4
y�H�x � 2� � 2
Problem Solving for Chapter P 41
4. (a)
(c)
(e)
(g) y
x−4
−4
−2
−2
2
2
4
4
f �x�
y
x−4
−4
−2
−2
2
2
4
4
�f �x�
y
x−4 −2
−4
−2
42
4
2 f �x�
y
x
−4
−3 −1
−2
1 3
4
f �x � 1� (b)
(d)
(f ) y
x−4
−4
−2
−2
2
2
4
4
f �x�
y
x−4
−4
−2
−2
2
2
4
4
f ��x�
y
x−4
−4
−2
4
4
f �x� � 1
5. (a)
Domain:
(b)
Maximum of at x � 50 m, y � 25 m.1250 m2
1100
0
1600
0 < x < 100
A�x� � xy � x�100 � x2 � � �
x2
2� 50x
x � 2y � 100 ⇒ y �100 � x
2(c)
is the maximum.x � 50 m, y � 25 mA�50� � 1250 m2
� �12
�x � 50�2 � 1250
� �12
�x2 � 100x � 2500� � 1250
A�x� � �12
�x2 � 100x�
42 Chapter P Preparation for Calculus
7. The length of the trip in the water is and thelength of the trip over land is Hence,the total time is
T ��4 � x2
2�
�1 � �3 � x�2
4 hours.
�1 � �3 � x�2.�22 � x2,
6. (a)
Domain:
(b)
Maximum of at x � 50 ft, y � 37.5 ft.3750 ft2
25 50 75 100
500
1000
1500
2000
2500
3000
3500
4000
y
x
0 < x < 100
A�x� � x�2y� � x�300 � 3x2 � �
�3x2 � 300x2
4y � 3x � 300 ⇒ y �300 � 3x
4(c)
square feet is the maximum area,where x � 50 ft and y � 37.5 ft.A�50� � 3750
� �32
�x � 50�2 � 3750
� �32
�x2 � 100x � 2500� � 3750
A�x� � �32
�x2 � 100x�
8. Let d be the distance from the starting point to the beach.
� 80 km�hr
�2
1120
�1
60
�2d
d120
�d
60
Average speed �distance
time
9. (a) Slope of tangent line is less than 5.
(b) Slope of tangent line is greater than 3.
(c) Slope of tangent line is less than 4.1.
(d)
(e) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at is 4.�2, 4�
� 4 � h, h � 0
�4h � h2
h
��2 � h�2 � 4
h
Slope �f �2 � h� � f �2�
�2 � h� � 2
Slope �4.41 � 42.1 � 2
� 4.1.
Slope �4 � 12 � 1
� 3.
Slope �9 � 43 � 2
� 5.
Problem Solving for Chapter P 43
10.
(a) Slope of tangent line is greater than
(b) Slope of tangent line is less than
(c) Slope of tangent line is greater than
(d)
(e)
As h gets closer to 0, the slope gets closer to The slope is at the point �4, 2�.14
14
.
�1
�4 � h � 2, h � 0
��4 � h� � 4
h��4 � h � 2�
�4 � h � 2
h�
�4 � h � 2h
��4 � h � 2
�4 � h � 2
��4 � h � 2
h
Slope �f �4 � h� � f �4�
�4 � h� � 4
1041
.Slope �2.1 � 2
4.41 � 4�
1041
.
13
.Slope �2 � 14 � 1
�13
.
15
.Slope �3 � 29 � 4
�15
.
y
x1
1
−1
2
3
4
2 3 4 5
(4, 2)
11. (a)
(b)
Circle of radius and center ��3, 0�.�18
�x � 3�2 � y2 � 18
x2 � y2 � 6x � 9 � 0
x2 � 6x � 9 � y2 � 2x2 � 2y2
�x � 3�2 � y2 � 2�x2 � y2�
−2−4−8 2 4−2
−6
2
6
8
x
y I
x2 � y2 �2I
�x � 3�2 � y2
� �3 ± �18 � 1.2426, �7.2426 x ��6 ± �36 � 36
2
x2 � 6x � 9 � 0
x2 � 6x � 9 � 2x2
0 1 2
x
3
I
x2 �2I
�x � 3�2
44 Chapter P Preparation for Calculus
12. (a)
If then is a vertical line. Assume
Circle
(c) As becomes very large, and
The center of the circle gets closer to and its radius approaches 0.�0, 0�,
16k�k � 1�2 → 0.
4k � 1
→ 0k
�x �4
k � 1�2
� y2 �16k
�k � 1�2,
x2 �8x
k � 1�
16�k � 1�2 � y2 �
16k � 1
�16
�k � 1�2
x2 �8x
k � 1� y2 �
16k � 1
k � 1.x � 2k � 1,
�k � 1�x2 � 8x � �k � 1�y2 � 16
�x � 4�2 � y2�k�x2 � y2�
Ix2 � y2 �
kI�x � 4�2 � y2
13.
Let Then or
Thus, and are on the curve.���2, 0��0, 0�, ��2, 0�x2 � 2.x 4 � 2x2 ⇒ x � 0y � 0.
�x2 � y2�2 � 2�x2 � y2�
�x4 � 2x2y2 � y4� � 2x2 � 2y2 � 0
x 4 � 2x2 � 1 � 2x2y2 � 2y2 � y4 � 1
�x2 � 1�2 � y22x2 � 2� � y4 � 1
�x � 1�2�x � 1�2 � y2�x � 1�2 � �x � 1�2� � y4 � 1
�x � 1�2 � y2��x � 1�2 � y2� � 1
(− 2 , 0) ( 2 , 0)
(0, 0)
x
y
−2
−2
−1
1
2
2
d1d2 � 1
14.
(a) Domain: all
Range: all
(b)
Domain: all
(c)
Domain: all
(d) The graph is not a line. It has holes at and y
x21−2
−2
1
2
�1, 1�.�0, 0�
x � 0, 1
f � f � f �x��� � f �x � 1x � �
1
1 � �x � 1x �
�11x
� x
x � 0, 1
f � f �x�� � f � 11 � x� �
1
1 � � 11 � x�
�1
1 � x � 11 � x
�1 � x
�x�
x � 1x
y � 0
x � 1
f �x� � y �1
1 � x
(b) If
x
y
−2−4−6 2 4
−2
−4
2
4
6
�x � 2�2 � y2 � 12k � 3,