C H A P T E R 1 Functions, Graphs, and Limits Section 1.1

The Cartesian Plane and the Distance Formula . . . . . . . . . . 32

Section 1.2

Graphs of Equations . . . . . . . . . . . . . . . . . . . . . . . . 35

Section 1.3

Lines in the Plane and Slope . . . . . . . . . . . . . . . . . . . . 40

Section 1.4

Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Section 1.5

Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

Section 1.6

Continuity

Review Exercises

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

C H A P T E R 1 Functions, Graphs, and Limits Section 1.1

The Cartesian Plane and the Distance Formula

Solutions to Even-Numbered Exercises 2. (a) a  
13  12

1  12  12

4. (a) a  
6  22

2
22  4

b  
13  132

6  12  5

b  
2  22

5
22  7

c  
13  12

6  12  13

c  
2  62

5
22  65

(b) a2
b2 
122

52  169  c2

(b) a2
b2 
42

72  65  c2 y

y 10

6

8

(13, 6)

6 4

c

(1, 1)

2

2

b (13, 1)

a

(2, 5)

4

c

b

x 4

x 2

4

6

8

10 12 14

−2

6. (a) a  3  1  2

(b)









22

(2, − 2)

(6, − 2)

(b) d  
3  32

2
22  52  213

c  
1 
42

1  32  52
22  29 b2

8

8. (a) See graph.

b  1 
4  5

a2

6

a

52

 29 

(c) Midpoint 

c2

32
3, 2
2
2 
0, 0

y

Subscribers (in millions)

y 3

(− 3, 2)

80

2

60

1

40

−3 −2 −1 −1

20

(0, 0)

x 2

−2

x 2

10. (a) See graph. (b) d 

(3, − 2)

−3

4 6 8 10 Year (2 ↔ 1992)

3

12. (a) See graph.

56  32
1
31  361
169  2

2


56

23 1 
13 3 1 (c) Midpoint  ,  , 2 2 4 3



  

65

(b) d  
3  12

7
12  16
64  45

6 (c) Midpoint 

32
1, 7 2 1 
1, 3

y

y

(−3, 7)

8 6

( 56 , 1( 1

4

(−1, 3)

( 34 , 13 (

−6 x

( 23 , − 13 (

32

1

−4

x

−2

2 −2

(1, −1)

4

Section 1.1

The Cartesian Plane and the Distance Formula 16. a  
2  32

4  22  29

14. (a) See graph. (b) d  
2  02

0  2  2  6

b  
3  12

2
32  29

22
0, 0
2 2   1, 22  

(c) Midpoint 

c  
2  12

4
32  58



Since a  b the figure is an isosceles triangle.

y

Note: It is also a right triangle since a2
b2  c2.

3 y

(− 2, 4)

2

(0,

(−1, ( 2 2

4

2(

d1

3

(3, 2)

2 −3

x

(− 2, 0) −1

d3

1 −1

x

−3 −2 −1

1

3

−2 −3

18. a  
3  02

7  12  35

4

d2 (1, − 3)

d1  
5  02

11  42  74

20.

b  
3  42

7  42  10

d2  
0  72

4
62  149

c  
4  12

4
22  35

d3  
5  72

11
62  433

d  
1  02

2  12  10

d1
d2 20.80888

Since a  c and b  d, the figure is a parallelogram.

d3 20.80865 Since d1
d2  d3, the points are not collinear.

y

(3, 7)

8 6 4

d1

y

(− 5, 11)

d2 (4, 4)

d1 8 (0, 1) −2

d4

d3

−2

6

d3

(0, 4)

x 4

2

8

(1, − 2)

x

−6 −4 −2

2

−4 −6

22.

d1  
1  32

1  32  25

(5, 5) 5 4

d3  
1  52

1  52  213

d2

d3

3 2

(3, 3) d1

(−1, 1)

d3 7.21110

(7, − 6)

y

d2  
3  52

3  52  22 d1
d2 7.30056

8 10 12

d2

−1

−1

x 1

2

3

4

5

Since d1
d2  3, the points are not collinear. 24. d  
x  22

2
12  5

26. d  
5  52

y  12  8

x2  4x
13  5


y  12  8

x2  4x
13  25 x2  4x  12  0


x
2
x  6  0 x  2, 6


y  12  64 y1 ± 8 y1 ± 8 y  7, 9

33

34

Chapter 1

28. To show that

Functions, Graphs, and Limits



2x1
x2 2y1
y2 is a point of trisection of the line segment joining
x1, y1 and
x2, y2, we must show that , 3 3



1 d1  d 2 and d1
d 2  d3. 2

    x x y y 1  
 
x  x 

y  y  3  3  3 2x
x 2y
y  x  
y  3  3 2x  2x 2y  2y   
 3   32
x  x 

y  y  3

d1 

2x1
x2  x1 3 2

d2

2

1

2

2y1
y2  y1 3




2

1

2

2

1



2

2

1

2

2

1

2

2

1

y

d2

2

2

2

d1 (x 1, y 1)

2

2

1

2

2

1

2



2

1

2

(x 2, y 2)

d3

2

1

( 2x 3+ x , 2y 3+ y ( 1

2

1

2

x

2

d3  
x2  x12

y2  y1 2 2x1
x2 2y1
y2 1 Therefore, d1  d 2 and d1
d2  d3. The midpoint of the line segment joining and
x2, y2 is , 2 3 3



Midpoint 

30. (a)





x1
2x2 y1
2y2 . , 3 3



32. c2  2002
1252 c2  55,625

1
2
4 2
2
1 , 
3, 0 3 3



c 235.8495 feet

2 2
2
0 3
2
0   , 1 , 3 3 3

 

125

 36. (a)

150

(b) 2

c

2
23
0, 2
33
0   34, 2 

34.



2
13
4, 2
23
1 
2, 1 

(b)



2x1
x2 2y1
y2
x2
y2 3 3 , 2 2



12 0

Let t  3 correspond to 1993. Answers will vary. The number of subscribers appears to be increasing rapidly (not linearly).

38. (a)

107  103 0.039 3.9% 103

(b)

159  148 0.074 7.4% 148

8550  10,400 0.178 17.8% 10,400 10,700  8,900 0.202 20.2% 8,900

200

Section 1.2

40. (a) Revenue midpoint 

Graphs of Equations

35

1999 2 2003, 256.6 2 508.6


2001, 382.6 Revenue estimate for 2001: $382.6 million Profit midpoint 

1999 2 2003, 34.3 2 74.8


2001, 54.55 Profit estimate for 2001: $54.55 million (b) Actual 2001 revenue: $379.8 million Actual 2001 profit: $48.2 million (c) The revenue increased in a linear pattern (382.6 is close to 379.8). The profit is somewhat linear (54.55 is close to 48.2). (d) 1999 Expenses: 256.6
34.3  $222.3 million 2001 2003 Expenses: 508.6
74.8  $433.8 million (e) Answers will vary. 42. (a)
0, 2 is translated to
0
3, 2
3 

3,
1

(b)

3


1, 3 is translated to
1
3, 3
3 

2, 0 −4


3, 1 is translated to
3
3, 1
3 
0,
2

4


2, 0 is translated to
2
3, 0
3 

1,
3 −3

Section 1.2

Graphs of Equations

2. (a) This is a solution point since 7
6  4

9
6  0 (b) This is not a solution point since 7

5  4
10
6 
1  0 1 5 (c) This is a solution point since 7
2   4
8 
6  0

4. (a) This is not a solution point since x 2y  x 2
5y  02
15   02
5
15  
1  0. (b) This is a solution point since x 2y  x 2
5y  22
4  22
5
4  0. (c) This is not a solution point since x 2y  x 2
5y 

22

4 

22
5

4  8  0. 6. (a) This is not a solution point since 3

5  2

7

5


72 
14  5 (b) This is a solution point since 3
6  2

1
6


12  5 6 6 (c) This is a solution point since 3
5   2
1
5 
12  5

8. The graph of y 
12 x  2 is a straight line with y-intercept at
0, 2. Thus, it matches (b). 12. The graph of y  x 3
x has intercepts at
0, 0,
1, 0, and

1, 0. Thus, it matches (d).

10. The graph of y  9
x2 is a semicircle with intercepts
0, 3,
3, 0, and

3, 0. Thus, it matches (f).

36

Chapter 1

Functions, Graphs, and Limits

14. Let y  0: 4x
5  0

16. Let x  0: y  3

5 4

y-intercept:
0, 3

x




x-intercept:

5 4,

0

Let y  0:


x
3
x
1  0

Let x  0:
2y
5  0 y y–intercept:
0,


52


52



x  3, 1 x-intercepts:
3, 0,
1, 0 20. The y-intercept is
0, 0. To find the x-intercepts, let y  0 to obtain

18. Let x  0: y2  0 y0

0

y-intercept:
0, 0 Let y  0:

x3
4x  0

0  x
x  3

x  0, 2,
2 x-intercepts:
0, 0,
2, 0,

2, 0 22. Let x  0: 8y  1

x2  3x
3x  12

0  x2  3x

x
x
2
x  2  0

y

x 2
4x  3  0

x  0,
3. Thus, the x-intercepts are
0, 0 and

3, 0. 24. The graph of y 
3x  2 is a straight line with slope
3 and y  intercept
0, 2.

1 8

y

y-intercept:
0, 18 

(0, 2)

2

Let y  0:
x2  1

1

( 23 , 0(

No x-intercepts

x

−1

1

2

3

−1 −2

26. The graph of y  x2  6 is a parabola with vertex at
0, 6, which is also the only intercept.

28. The graph of y 
5
x2 is a parabola with vertex at
5, 0. y

x

0

±1

±2

y

6

7

10

10 8 6

y

4 2

12

x

9

2

(0, 6) 3

−6

−3

(5, 0)

x 3

6

4

6

8

10

Section 1.2 30. Intercepts:
0, 1 and
1, 0 x

0

1


1

2

y

1

0

2


7

Graphs of Equations

32. The graph of y  x  1 is a translation of y  x one unit to the left. Intercepts:

1, 0,
0, 1 y

4

y

3 2

2

(−1, 0)

(0, 1)

(0, 1) x

−1

1

2

3

x

(1, 0)

−1

34. Intercepts:
2, 0 and
0,
2

36. Intercept:
0, 1

x

2

0

1

3

4

x

0

±1

±2

±3

y

0


2


1


1


2

y

1

1 2

1 5

1 10

y

y

1 2

(2, 0) x 1

2

3

4

(0, 1)

−1 −2

(0, − 2)

−3

x

−1

0

3

4

y

±2

±1

0

1

40.
x
02 
y
02  52

38. Intercepts:
0, 2,
0,
2,
4, 0 x

x2  y2  25 x2  y2
25  0

y

(0, 2) 1

(4, 0) x 1

2

3

−1

(0, − 2)

42.


x  42 
y
32  32

44. Radius  

1
32 
1  22  5

x2  8x  16  y2
6y  9  9


x
32 
y  22  52

x 2  y2  8x
6y  16  0

x2
6x  9  y2  4y  4  25 x2  y2
6x  4y
12  0

46. Center  midpoint 

37


42 4,
12 1 
0, 0

Radius  distance from the center to an endpoint 
4
02 
1
02  17


x
02 
y
02 
17

2

x2  y2
17  0

38

Chapter 1

Functions, Graphs, and Limits

48.
x2
2x  1 
y 2  6y  9  15  1  9

50.
x2
4x  4 
y2  2y  1 
3  4  1


x
22 
y  12  2


x
12 
y  32  25 1

3

−9

−1

9

5

(2, −1)

(1, − 3)

−3

−9

x2  y2
x  12 y
14  0

52.


x2
x  14  
y2  12 y  161   14  14  161
x
12 2 
y  14 2  169

x2  y2
2y
13  0

54.

x2 
y2
2y  1  13  1 x2 
y
12  43

2

3

(0, 1) −2

4

(

1, 1 2 4

−3

(

−2

3

−1

56. The first equation gives y  7
x. Hence, 3x
2
7
x  5x
14  11

58. Solving for y in the second equation yields y  2x
1 and substituting this value into the first equation gives us the following.

5x  25

x2 
2x
1  4

x  5, y  2

x2  2x
5  0 x 
1 ± 6 by the Quadratic Formula The corresponding y-values are y 
3 ± 26, so the points of intersection are

1  6,
3  26  and

1
6,
3
26 .

60. By equating the y-values for the two equations, we have

62. By equating the y-values for the two equations, we have

x  x

x3
2x2  x
1 
x2  3x
1

x  x2

x3
x2
2x  0

0  x
x
1

x
x  1
x
2  0

x  0, 1.

x  0,
1, 2.

The corresponding y-values are y  0, 1, so the points of intersection are
0, 0 and
1, 1. 64. (a) Cg 
initial price 
cost per mile  20,930 

1.759x 16

Ch  22,052 

1.759x 35

The corresponding y-values are y 
1,
5, and 1, so the points of intersection are
0,
1,

1,
5, and
2, 1. Cg  Ch

(b) 20,930 

1.759x 1.759x  22,052  16 35

1.759
x  1122 1.759 16 35  0.0596804x  1122 x  18,800 miles

Section 1.2 66.

Graphs of Equations

RC 35x  6x  500,000 29x  500,000 x  500,00029  17,242 units RC

68.

3.29x  5.5x  10,000


3.29x
10,0002 
5.5x 

2

10.8241x2
65,800x  100,000,000  30.25x 10.8241x2
65,830.25x  100,000,000  0 By using the Quadratic Formula, we have x 

65,830.25 ± 3,981,815.062 21.6482

x  3133 units.

Note: x  2949 units is an extraneous solution. You can also solve this problem with a graphing utility by determining the point of intersection of the two equations y1  R  3.29x and y2  C  5.5x  10,000. 70. p  190
15x  75  8x 115  23x x5

(Thousand)

Equilibrium point
x, p 
5, 115.

72. Model: y 


4.97  0.021t 1
0.025t


t  55 corresponds to 1955 (a)

t

55

60

65

70

75

80

85

90

95

100

Model

10.2

7.4

5.8

4.7

3.9

3.3

2.8

2.5

2.2

1.9

Exact

9.9

7.8

5.9

4.2

3.6

3.1

2.8

2.6

2.6

1.7

The model is a good fit. (b) For 2010, t  110 and y  1.5%. (c) Answers will vary. 74. Model: y  60.64t2
544.0t  12,624


t  8 corresponds to 1998 (a)

Year

1998

1999

2000

2001

2002

Transplants

12,153

12,640

13,248

13,977

14,824

(b) 1998: 12,244 transplants. 2002: 14,741 transplants. The model seems accurate. (c) For 2008, t  18 and y  22,479 transplants. Answers will vary.

39

40

Chapter 1

Functions, Graphs, and Limits

76. If C and R represent the cost and revenue for a business, the break-even point is that value of x for which C
R. For example, if C
100,000  10x and R
20x, then the break-even point is x
10,000 units. 78. Intercepts:
0, 6.25,
1.0539, 0,
10.5896, 0

80.

4

20

−4

−24

5

−2

12

Intercepts:
3.3256, 0,
1.3917, 0,
0, 2.3664

−4

82.

0.5 −5

15

−1.5

Intercepts:
0, 1,
13.25, 0

Section 1.3

Lines in the Plane and Slope

2. The slope is 2 since the line rises two units vertically for each unit of horizontal change from left to right. 4. The slope is 1 since the line falls one unit vertically for each unit of horizontal change from left to right. 6. The points are plotted in the accompanying graph and the slope is

8. The points are plotted in the accompanying graph and the slope is

22
0. 1 
2

m


10 
2 8
. 11 11 0  3 3

m


y

Undefined

4

The line is vertical.

3

(−2, 2)

(1, 2)

y

1 −3

−2 −1

2 x

x 1

−1

2

3

1

−2

2 3 11 , −2 3

(

−4

−2

4

5

)

−6 −8 −10

(113 , −10)

−12

10. The points are plotted in the accompanying graph and the slope is m


12. The points are plotted in the accompanying graph and the slope is

1 
5 4

1. 2 
2 4

m


y

−3

−2 −1

y x

−1 −2

1

2

3

−3

−2 −1

(2, −1)

−4 −5 −6

x −1

1

2

3

−2

−3

(−2, − 5)

5 
5
0. 2  3

−3

(−2, −5)

−4

−6

(3, −5)

Section 1.3 14. The points are plotted in the accompanying graph and the slope is

41

16. The points are plotted in the accompanying graph and the slope is

45 27
.
56 
32 7

m


Lines in the Plane and Slope

m



14 
34 8
 .
54 
78 3

y

y 6

2

4

( 56 , 4( x

−6 −4 −2

2

( 78 , 34 (

1

2 4

x

−1

6

1 −1

(− 32 , − 5(

( 45 , − 41 ( 3

−2

18. The equation of this horizontal line is y
1. Therefore, three additional points are
0, 1,
1, 1, and
2, 1. 20. The equation of this line is y2
y


5 2
x 5 2x

22. The equation of this line is y  6
1
x  10

 2

y
x  4.

3

Therefore, three additional points are
0, 3,
2, 8,
4, 13.

Therefore, three additional points are
11, 7,
9, 5, and
8, 4.

24. The equation of this vertical line is x
3. Therefore, three additional points are
3, 0,
3, 1, and
3, 2. 28. 6x  5y
15

26. 2x  y
40

y
65 x  3

y
2x  40

6

Therefore, the slope is m
2, and the y-intercept is
0, 40.

Therefore, the slope is m
5, and the y-intercept is
0, 3.

30. 2x  3y
24

32. x  5
0

y
13
2x  24
23 x  8 Slope is m


2 3,

x
5

y-intercept is
0, 8

The line is vertical. Slope is undefined and there is no y-intercept.

34. Since the line is horizontal, the slope is m
0, and the y-intercept is
0, 1. 36. The slope of the line is m


38. The slope of the line is

4 
4
2. 1 
3

m


Using the point-slope form, we have

Using the point-slope form, we have

y  4
2
x  1

y  2
1
x  1

y

y
2x  2

4

0
2x  y  2.

y
x  3

(1, 4)

y

(− 3, 6) 6 5

x  y  3
0.

2

−4

26
1. 1 
3

4

x

−2

2

3

4

(1, 2)

2

−2

1

(− 3, − 4)

−4 −3 −2 −1

x 1

2

42

Chapter 1

Functions, Graphs, and Limits

11
0. 10  6 The line is horizontal and its equation is

40. The slope of the line is m


42. Slope is undefined. Line is vertical: x
2 y 4

y
1

3

y  1
0.

2 1

y −3

−2 −1

6

x 1

−1

3

−2

4

(6, 1)

2

(10, 1) x

2

4

6

8

10

−2 −4

8
14 
34
 .
54 
78 3 Using the point-slope form, we have

44. The slope of the line is m




3 7 8 y
 x 4 3 8

46. The slope is m


y1




16
x  4 15

15y  15
16x  64

3 8 7 y
 x 4 3 3

15y  16x  79
0 y

12y  9
32x  28 32x  12y  37
0.

1 
5 4 16

. 4 
14
154 15

−1

x −1

1

2

3

5

(4, −1)

−2

y

4

−3 −4

2

−5

( 78 , 34 (

1

( 14 , −5)

x

−1

( 45 , − 41 ( 3

1 −1 −2

49. Using the slope-intercept form, we have y


2 3x

0

50. Since the slope is undefined, the line is vertical and its equation is x
0. 4

2x  3y
0. 4

−6

−2

5

(0, 0)

6

−4

−2

52. Since the slope is 0, the line is horizontal and its equation is y
4. 6

54. Using the point-slope form we have y  4
2
x  1 y
2x  6

4

−10

8

2x  y  6
0

(−2, 4) −6

6 −8 −2

Section 1.3

Lines in the Plane and Slope

56. Using the point-slope form, we have y  23
16
x  0

1

6y  4
x

−3

3

− 23

0
x  6y  4.

(0, ( −3

58. The slope of the line joining
5, 11 and
0, 4 is The slope of the line joining
0, 4 and
7, 6 is

11  4 7 7

 . 5  0 5 5

4 
6 10
 . 07 7

Since the slopes are different, the points are not collinear. d1

5  02 
11  42
25  49
74 8.60233 d2

7  02 
6  42
49  100
149 12.20656 d3

5  72  11 
62
144  289
433 20.80865 Since d1  d2  d3, the points are not collinear. 60. Since the line is horizontal, it has a slope of m
0, and its equation is

62. The line is vertical: x
5

y
0x 
5 y
5.

64. Given line: y
2x  32

5 66. Given line: y
 3 x 5 (a) Parallel: m1
 3

(a) Parallel: m1
2

y  34
 53
x  78 

y  1
2
x  2

y  34
 53 x  35 24

0
2x  y  3 1 (b) Perpendicular: m2
 2

y1


 12
x

24y  18
40x  35

 2

2y  2
x  2

40x  24y  53
0 (b) Perpendicular: m 2
35 y  34
35
x  78 

x  2y  4
0 4

y  34
35 x  21 40

4x − 2y = 3

40y  30
24x  21 0
24x  40y  9

(2, 1) −2

7 3

( 78 , 34 (

−2 −4

5

−3

5x + 3 y = 0

43

44

Chapter 1

Functions, Graphs, and Limits

68. Given line: y  4
0 is horizontal

70. Given line: x  4
0 is vertical

(a) Parallel: m
0, y
5

(a) Parallel: slope is undefined, x
12

(b) Perpendicular m is undefined, x
2

(b) Perpendicular: m
0, y
3

6

6

(2, 5) −8 −10

16

8

(12, − 3)

−6

−12

72. y
4 is a horizontal line with y-intercept
0, 4.

74. x  2y  6
0 has intercepts at
6, 0 and
0, 3.

76. 4x  5y  20
0 has intercepts at
5, 0 and
0, 4. y

y

y 2

5

2

4

−4

x

−2

2

−6

4

−2

x

−2

3

2 −2

2

−4

1 x 1

−6

−6

13 78. y
3x  13 has intercepts at
0, 13 and
 3 , 0.

80. (a) Slope: m


y

2

3

4

5

6

29,700  26,300 3400

1700 2004  2002 2

y  26,300
1700
t  2

18

y
1700t  22,900

15

(b) For 2008, t
8 and y
36,500

9 6 3 −4 −3 −2 −1

x 1

82. Use F
95C  32 and C
59
F  32. 5 (a) If F
102.5, C
9
102.5  32 39.2 5 (b) If F
74, the C
9
74  32 23.3

84. (a) W
0.80x  9.25 W
1.15x  6.85

(union plan) (corporation plan)

(b) 0.8x  9.25
1.15x  6.85

17

2.4
.35x x


240 6.857 35

W 14.736

(6.857, 14.736) 5

9 13

(c) The point of intersection indicates the number of units
6.857 a worker needs to produce for the two plans to be equivalent.

Section 1.3 86. Use the points
0, 825,000 and
25, 75,000. y  825,000


825,000  75,000
t  0 0  25

45

88. Let t
0 represent 2002. Slope
m


2702  2546
78 20

y
78t  2546

y  825,000
30,000t y
30,000t  825,000,

Lines in the Plane and Slope

0 ≤ t ≤ 25

For 2008, t
6 and y
3014 students.

90. (a) C

5.25  9.50t  26,500
14.75t  26,500 (b) R
25t (c) P
R  C
25t 
14.75t  26,500
10.25t  26,500 R
C

(d)

25t
14.75t  26,500 10.25t
26,500 t 2585.4 hours 92. (a) W
2000  .07S (c)

(b) W
2300  .05S (d) No. You will make more money
if sales are $20,000 at your current job
w
$3400) than in the offered job
w
$3300.

5000

0

30,000 0

The lines intersect at
15,000, 3050. If you sell $15,000, then both jobs would yield wages of $3050.

94. C
30,000  575x where C ≤ 100,000.

96. C
24,900  1785x ≤ 100,000

Thus, 30,000  575x ≤ 100,000

1785x ≤ 75,100 x ≤ 42.07

575x ≤ 70,000

x ≤ 42 units

x ≤ 121.739 122 units. Therefore, x ≤ 121 units.

200,000

200,000

0

100 0

0

200 0

98. C
83,620  67x ≤ 100,000

100. C
53,500  495x ≤ 100,000

67x ≤ 16,380 x ≤ 244.48

x ≤ 93.94

x ≤ 244 units

x ≤ 93 units

200,000

0

120,000

2000 0

495x ≤ 46,500

0

100 0

46

Chapter 1

Functions, Graphs, and Limits

102. C
75,500  1.50x ≤ 100,000

120,000

1.50x ≤ 24,500 x ≤ 16,333.3 x ≤ 16,333 units

0

20,000 0

Section 1.4

Functions 4. y


2. y
±
4  x y is not a function of x since there are two values of y for some x.

y is a function of x since there is only one value of y for each x.

6. x2  2x  1   y2  4y  4
1  1  4

8. yx2  4
x2

x  12   y  22
4

y


The graph is a circle; therefore, by the vertical line test, y is not a function of x.

10. Domain:  , 

3x  5 2

x2 x2  4

y is a function of x since there is only one value of y for each x. [Note: It is not a one-to-one function.]

12. Domain: 3, 3

Range:  , 

14. Domain: 1, 

Range: 0, 3

Range:  , 

6

2

−3

8

3

−2

6

6

2

−2

16. Domain:  , 1
1, 

18. Domain:

Range:  , 4
0, 

16

−4

 32, 

20. Domain:  , 

Range: 0, 

Range: 0, 

6

−12

12

−10



22. f x
x 2  2x  2 (a) f 

1 2


 

1 2 2

 2

24. f x
x  4 1 2

2


5 4

(b) f 1
12  21  2
5 (c) f c  2
c  22  2c  2  2
c2  4c  4  2c  4  2
c2  2c  2 (d) f x  x
x   x2  2x   x  2
x2  2x x   x2  2x  2 x  2

 (b) f 2
2  4
6 (c) f x  2
x  2  4 (d) f x   x  f x
x  x  4  x  4
x   x  x (a) f 2
2  4
6

Section 1.4

26.

Functions

47

h2  x  h2 2  x2  2  x  1  4  2  1
x x


4  4x  x2  2  x  1  3 x




x3  x x


3  x, x  0 1 1  f x  x  f x x  x  4 x  4 30.
x x

f x  f 2 1
x  1   1 28.
x2 x2


1 
x  1 1 
x  1  x  2
x  1 1 
x  1




x  4  x  x  4 x x  x  4x  4




1  x  1 x  2
x  11 
x  1




1 , x  0 x  x  4x  4





2x



x  2
x  1 1 
x  1



1


x  1 1 
x  1





, x2

32. y is a function of x.

34. y is not a function of x.

36. (a) f x  gx
2x  5  2  x
x  3

38. (a) f x  gx
x 2  5 
1  x,

(b) f x gx
2x  52  x
(c) f xgx


2x 2

 9x  10

2x  5 2x

(b) f x  gx
x  5
1  x, 2

(c)

(d) f gx
f 2  x
22  x  5
2x  1

f x x2  5
, gx
1  x

x ≤ 1

x ≤ 1

x < 1

(d) f gx
f 
1  x 

1  x   5

(e) g f x
g2x  5
2  2x  5
2x  7

2


6  x,

x ≤ 1

(e) g f x is not defined since the domain of g is  , 1 and the range of f is 5, . The range of f is not in the domain of g.

40. (a) f x  gx
(b) f x  gx
f x (c)
gx

x x 4  x3  x  x3
x1 x1

x x 1 x 
x x 1 4

3

x x 1 x3



12
12

2

1 3

1


3 4

 
12
f 12  1
f  21
2

(c) f g

x3 (d) f gx
f x 
3 x 1 3



(a) f g 2
f 22  1
f 3
(b) g f 2
g

1
2 x x  1

x x
(e) g f x
g x1 x1

1 42. f x
, gx
x2  1 x



3

x3
x  13

 
12
g
2 
2  1
1

(d) g f

(e) f g x
f x2  1
(f) g f x
g

1x
1x

1 x2  1 2

1

48

Chapter 1

Functions, Graphs, and Limits

44. The data fits the function (a) f x
14 x, with c
14.

1x
1x1
x, x  0

48. f gx
f



46. The data fits the function (c) hx
3
x , with c
3. 3 3 1  x 
1  
1  x
x 50. f  gx
f 
3

3 1  1  x 3
x g f x
g1  x 3





1 1 g f x
g

x, x  0 x 1x

See accompanying graph. y

See accompanying graph. f

y

2

2

f=g g

1 x

−1

1

x

−1

2

2

3

−1

−1

f x
6  3x
y

52.

3

f x
x3  1
y

54.

x
y3  1

x
6  3y y


6x 3

f 1x


6x 3

3 x  1
y
3 x  1 f 1x

y 4

y

f

3

f −1

2

6

f

5

x

−2

4 3

2

3

4

−2

f −1

1

x 1

3

4

5

6

f x

x2  4
y, x ≥ 2

56.

f x
x 35
y

58.

x

y 2  4

x
y 35

x2  4
y 2

x 53
y

y

x2  4, x ≥ 0

f 1x
x 53

f 1x

x2  4, x ≥ 0

y

2

y

f −1 f

1

5

f

4

−1

−2 3

x

−1

1 −1

f

2

−2

1 x 1

2

3

4

5

2

Section 1.4

Functions

62. f x
x 4 is not one-to-one since f 2
16
f 2.

60. f x

x  2 is one-to-one for x ≥ 2. f 1x
x2  2 where x ≥ 0.

650

12

−5

5 0

0

100

f x is not one-to-one.

0

f x is one-to-one. f 1x
x2  2,

x ≥ 0

64. f x
3 is not one-to-one since f 0
3
f 1. 5

−6

6

−3

f x is not one-to-one. y

66. (a)

y

(b) 3

3

5

2

2

4

1

3

−3 −2

2 1 x

−3 −2 −1

1

2

1 x 2

−1

3

−2

−3

−3

y

(e) 3

2 1

2 x

−4 −3 −2

1

2

(−1, −1)

−3 −2 −1 −1

−3

−2

−4

−3

x 1

2

3

68. Value
V
2500x  750,000 70. (a) C
0.95x  6000 C 0.95x  6000 6000

0.95  x x x

(c) 0.95 

−1 −1

−2 3

y

(d)

(b) C


y

(c)

6

6000 < 1.69 x 6000 < 0.74 x 6000 < x since x > 0. 0.74

8108.108 < x Must sell 8109 units before the average cost per unit falls below the selling price.

x 1

2

3

4

5

49

50

Chapter 1

Functions, Graphs, and Limits

72. Cost
Cost on land  Cost underwater

74. (a) Graphing utility graph


1052803  x  155280
x  2


5528023  x  3
x  2

1 4

(b) 25, 14 is equilibrium point.

1 4



(c) Demand exceeds supply for x < 25. (d) Supply exceeds demand for x > 25. 30

0

50 0

76. (a) Cost
C
98,000  12.30x (b) Revenue
R
17.98x (c) Profit
R  C
17.98x  12.30x  98,000
5.68x  98,000

78. Ft
98 

3 t1

80.

100

110

−5

5

−100 0

20



f x
2 3x2 

95

The function is valid for all t ≥ 0 because the patient’s temperature can only be affected by the drug from the time that it is administered. 82.

−2



Zero: x
1.2599 The function is not one-to-one. 84.

40

6 x

6

3 −6

−30

6

−2

hx
6x 3  12x2  4 Zeros: x 0.5419, 0.7224, 1.7925 The function is not one-to-one.

f x


  1 2 x 4 2

Zeros: x
± 2
2 The function is not one-to-one.

86.

0.4

−10

10

−0.2

f x



x2  16

x2



Domain: x ≥ 4 Zeros: x
± 4

Section 1.5

Section 1.5 2.

4.

6.

1.9

1.99

1.999

2

2.001

2.01

2.1

f
x


1.09


1.0099


1.000999


1


0.998999


0.9899


0.89

x

1.9

1.99

1.999

2

2.001

2.01

2.1

lim

x→2

f
x

72.39

79.20

x


0.1


0.01


0.001

0

0.001

0.01

0.1

f
x

0.3581

0.3540

0.3536

undefined

0.3535

0.3531

0.3492

lim

8.

Limits

x

x  2
2

x

x→0

Limits



79.92

1

undefined

80.08

80.80

x

0.5

0.1

0.01

0.001

0

f
x


0.1


0.1190


0.1244


0.1249

undefined

10. (a) lim f
x 
2

88.41

lim

x→0 

1
2  x

12 1 

0.125 2x 8

12. (a) lim h
x 
5

y

x→1

(b) lim h
x 
3

x

−1

1

y

x→
2

(3, 0)

x→3

x5
32  80 x
2

 0.354

22

(b) lim f
x  0

lim
x 2
3x  1 
1

x→2

2

2

x→0

−4

(1, − 2)

−2

−6

y = f ( x)

14. (a) lim  f
x  g
x  lim f
x  lim g
x  x→c



x→c

3 1  2 2 2

16. (a) lim f
x  9  3 x→c

(b) lim
3f
x  3
9  27

 lim g
x  2 2  4

(b) lim  f
x  g
x  lim f
x x→c

y = h ( x)

(− 2, − 5)

−4

x→c

4

(0, − 3)

−3

x→c

x

−2

3

1

3

x→c

x→c

(c) lim  f
x 2  92  81

lim f
x f
x 32 (c) lim  x→c  3 x→c g
x lim g
x 12

x→c

x→c

18. (a)

lim f
x 
2

20. (a)

x→
2

lim f
x  2

22. (a)

x→
2

lim f
x  0

x→
1

(b) lim
f
x 
2

(b) lim
f
x  2

(b) lim
f
x  2

(c) lim f
x 
2

(c) lim f
x  2

(c) lim f
x does not exist.

x→
2

x→
2

x→
2

x→
2

(− 2, 3) (− 2, 2)

(−π , − 2)

π

−2

y

4

2

−π

x→
1

y

y

− 2π

x→
1

2π

4

3

3

2

x

(−1, 2)

1 −5

x

−3 −2 −1

1 −1

(−1, 0) −3 −2 −1

x 1

2

3

4

51

52

Chapter 1

Functions, Graphs, and Limits

24. lim x 3 

23 
8

26. lim
2x
3  2
0
3 
3

28. lim

x2  x
2 
4  2
2 
4

3 x  4   3 4  4  2 30. lim 

x→0

x→
2

x→2

32. lim

x→
2

36. lim

3x  1 3

2  1
5   2
x 2


2 4

x  1

x→3

x→4

x
4



3  1

3
4



34. lim

x→
1

2 
2
1

38. lim

x  4
2

x

x→5

1 1 1 1

1 x2 2 4 2 40. lim  
x→2 2 2 8

4x
5 4

1
5
9 9   
3
x 3


1 4 4

42. lim

x→
1



3
2 1  5 5

2x2
x
3
x  1
2x
3  lim x→
1 x1 x1  lim
2x
3 
5 x→
1

44. lim

x→2

2
x

x
2  lim x2
4 x→2
x  2
x
2  lim

x→2

46. lim t→1


1 1 
x2 4

 lim t→1

x3
1
x
1
x2  x  1  lim x→1 x
1 x→1 x
1

50. lim


48. lim

 lim
x→1

t2  t
2
t
1
t  2  lim t→1
t  1
t
1 t2
1

x2

x→2

 x  1  3

lim

x→2 

x
2 
1 x
2

x
2  1 x
2

Therefore, lim

x→2

52. lim
f
s  lim
s  1 s→1

54. lim

x →0

s→1

lim f
s  lim
1
s  0

s→1 

s→1

Therefore, lim f
s does not exist. s→1

56. lim

x  x
x

x

x→0

 lim

x  x
x

x

x→0

 lim

x→0

 lim

x→0





x  x  x x  x  x


x  x
x x x  x  x  1 x  x  x

1 2x


t  t2
4
t  t  2

t 2
4t  2 t→0 t

58. lim

t 2  2t t 
t 2
4t
4t
t 2  4t t→0 t

 lim

 lim

t→0

2t t 
t 2
4t t

 lim
2t  t
4  2t
4 t→0

t2 3  t1 2

x
2 does not exist. x
2

4
x   x
5

4x
5 4 x  lim 4 x →0  x x

Section 1.6

60.

lim

20

x→1


−5

5   1x

62.

lim

5

x→0

Continuity x
1   x

5 −6

6

−3

−20

x

2

1.5

1.1

1.01

x

1

0.5

0.1

0.01

f
x

5

10

50

500

f
x

0

1

9

99

x

1.001

1.0001

1

x

0.001

0.0001

0

f
x

5000

50,000

Undefined

f
x

999

9999

Undefined

64. lim

x→1

x2
6x  7
x  1
x
7  lim x3  x2
2x  2 x→1
x  1
x2
2

66. lim

x→2


x
2
4x2  x
3 4x3
7x2
x
6  lim x→2 3x2  x  14
x
2
3x  7 1

4 −5

−1

2

3 −6

−1

21  1.615 13

8  2.667 3 68. Because 4  x2 ≤ f
x ≤ 4
x2, lim
4  x2 ≤ lim f
x ≤ lim
4
x2

x→0

x→0

x→0

4 ≤ lim f
x ≤ 4. x→0

Hence, lim f
x  4. x→0



70. lim A  lim 1000 1
r→0.06

r→0.06

Section 1.6

r 4



40

 1814.02 Yes, the limit exists.

Continuity

2. The polynomial f
x 
x 2  13 is continuous on the entire real line.

4. f
x 

1 1 is continuous on
 , 3,
3, 3 and
3, .  9  x2
3  x
3
x

6. f
x 

3x is continuous on
 , . x2
1

8. f is not continuous on the entire real line. f is not defined at x  1, 5. 10. g is not continuous, on the entire real line. g is not defined at x  ± 4.

53

54

Chapter 1

12. f
x 

Functions, Graphs, and Limits

1 is continuous on
 , 2,
2, 2 and
2, . x2  4

14. f
x is continuous on
 , 2 and
2, . 16. f
x is continuous on
 , . 18. f
x 

x3 is continuous on
 , 3,
3, 3 and
3, . x2  9

20. f
x 

1 is continuous on
 , . x2
1

22. f
x 

x1 x1  is continuous on
 , 2,
2, 1 and
1, . x2
x  2
x  1
x
2

24. f
x 

x
x is continuous on all intervals of the form
c, c
1 where c is an integer. 2

26. lim f
x  lim
3
x  5 x→2

x→2

lim f
x  lim

x 2
1  5

x→2


x→2

Since f
2  5, f is continuous on the entire real line.

30. lim x→4

lim


x→4

lim

x→4

4  x  1 4x

4  x  1 4x

28. lim f
x  4 x→0

lim f
x  1

x→0


f is not continuous at x  0. f is continuous on
 , 0 and
0, . 32. lim
x  x  c 
c  1  1, c is any integer x→c

lim
x  x  c  c  0, c is any integer

x→c


f is continuous on all intervals
c, c
1.

4  x does not exist. 4x

f is continuous on
 , 4 and
4, . 34. h
x  f
g
x  f
x2
5 

1 1 
x2
5  1 x2
4

36. f
x 

5 is continuous on 2, 2 . x2
1

Note: f is continuous on the entire real line.

Thus, h is continuous on the entire real line.

38. f
x 

x has nonremovable discontinuities at x  1 and x  3.
x  1
x  3

Section 1.6

40. f
x 

2x2
x x
2x
1  x x

42. f
x 

x3 x3 1   , 4x2  12x 4x
x  3 4x

Continuity

x3

f has a removable discontinuity at x  3, and a nonremovable discontinuity at x  0. f is continuous on
 , 0,
0, 3,
3, .

f has a removable discontinuity at x  0; continuous on
 , 0 and
0, . y

y 2 1

(3, 121 )

1

x

−3

1

2

3

x

−1

1

−1

44. f is continuous on
 , 0 and
0, .

46.

lim f
x  2

x→1

lim f
x  a
b

y

x→1
6

lim f
x  3a
b

x→3

4

lim f
x  2

2 −6

x→3


x

−4

2

4

6

Thus:

−4

a 3a 4a

−6

48. f
x 

x4 1 x4   , x 2  5x
4
x  4
x  1 x  1

x4


b 2
b  2  4 a  1 b 1

50. f is continuous on
 , . 7

is not continuous at x  1 and x  4. 3 −5 −3

10

6 −3

−3

1 There is a hole at
4, 3 .

52.

54. f
x  xx
3 is continuous on 3, .

3

−3

6

−3

f is not continuous at all 12c, where c is an integer.

56. f
x 

x
1 is continuous on
0, . x

55

56

Chapter 1

Functions, Graphs, and Limits

x3  8 x  2x2  2x  4
x2 x  2 appears to be continuous on
4, 4. But, it is not continuous at x
2 (removable discontinuity).

58. f x


8

60. C


2x 100  x

(a)
0, 100; Negative x values do not make sense in this context nor do values greater than 100. Also, C100 is undefined. (b) C is continuous on its domain because all rational functions are continuous on their domains. (c) For x
75,

−4

4 0

C


275 150

6 million dollars. 100  75 25

50

0

100 0



3, n
0 62. C
3  0.25n, n > 0, n is not an integer. 3  0.25n  1, n > 0, n is an integer.

64. (a) Nonremovable discontinuities at t
1, 2, 3, 4, 5 50,000

6

0

6 0

0

9 0

(b) For t
5, S
$43,850.78.

C is continuous at all intervals n, n  1, n is a nonnegative integer.


Note: C
3  0.251  n, n > 0 66. Yes, a linear model is a continuous function. No, actual revenue would probably not be continuous because the actual revenue would probably not follow the model exactly, which may introduce some discontinuities.

Review Exercises for Chapter 1 2. Matches (c)

4. Matches (d)

6. Distance
1  42  2  32
9  1
10. 8. Distance

6  3 2  8  7 2
81  1
82

10. Midpoint


0 2 4, 0 2 8
2, 4

12. Midpoint


7 2 3, 92 5
2, 2

Review Exercises for Chapter 1 14. 1999: R $120 thousand

2000: R $170 thousand

2001: R $70 thousand

C $70 thousand

C $92 thousand

C $33 thousand

P $50 thousand

P $78 thousand

P $37 thousand

2002: R $200 thousand

2003: R $260 thousand

C $110 thousand

C $135 thousand

P $90 thousand

P $125 thousand

16. 2, 1 → 2, 0

y

18.

1, 2 → 3, 1

y

20. 12

3 2

1, 0 → 5, 1

1

x

0, 1 → 4, 2

1

2

4

5

6

1

4

2

2

3

x 6





22. y
4  x

6

y

2

6

26. y
4x  1

y

24.

2

y

5 12

8

9

6

3

4

−2

2

4

6

8

3

1

x −2

6

2

2 10

−3 −3

x 3

2

1

1

2

3

x 3

6

9

12 15

−6

−4

28. y-intercept: x
0 ⇒ y
3 ⇒ 0, 3 x-intercept: y
0 ⇒ x
 4 ⇒  4, 0 3

30. x  02   y  02
r 2 x2  y2
r2

3

x 2  y 2  6x  8y
0

32.

x 2  6x  9   y 2  8y  16
9  16

22  5 
r 2 2

Equation:

x2

x  32   y  42
25

9
r2

Center: 3, 4

3
r

Radius: 5

y
3 2

2

y 2

x 2

8

2

4

)3,

4)

6

8

57

58 34.

Chapter 1

Functions, Graphs, and Limits

xy
2⇒y
2x

2  x
2x  1

2x  y
1 ⇒ y
2x  1

3
3x

36. y
x 3

x3
x

y
x

x3  x
0

1
x

xx  1x  1
0

Point of intersection: 1, 1

Points of intersection: 0, 0, 1, 1, 1, 1 40. p
91.4  0.009x
6.4  0.008x

38. (a) C
200  2x  8x
200  10x R
14x

85
0.017x C
R

(b)

x
5000 units

200  10x
14x

p
$46.40

200
4x x
50 shirts

x, R
x, C
50, 700. 1 5 42.  3 x  6 y
1 5 6y

44. x
3




1 3x

1

y


2 5x

6 5

Slope: m




46. 3.2x  0.8y  5.6
0 8y
32x  56

Slope: undefined (vertical line) No y-intercept

y
4x  7

y

2 5

y-intercept: 0,

6 5



Slope: 4 y-intercept: 0, 7

2 1

y

y −4

4

−2

x

−1

8 −1

3

6

2

−2

x

−3 −2 −1 −1

1

2

2

3

−2

−6

75 2 1

 5  1 4 2

48. Slope


52. y  3
12
x  3 y


1 2x



y
45 x  35

2 −6

−4

x

−2

4

−4 −6

54. (a) Slope
0 → y
3 (c) 4x  5y
3

4

x 2

3  3 0

0 (horizontal line) 1  11 10

(b) Slope undefined → x
1

3 2

y

(− 3, − 3)

50. Slope


−4

y  3
45 x  1 y
45 x  19 5 (d) 5x  2y
3 y
52 x  32 Slope of perpendicular is  25. y  3
 25 x  1 y
 25 x  13 5

Review Exercises for Chapter 1

59

56. 0, 117,000, 9, 0 m


117,000
13,000 9

(a) v
13,000t  9
13,000t  117,000

(c) v4
$65,000

(b) Graphing utility

(d) v
84,000 when t 2.54 years

200,000

0

10 0



58. x2  y 2
4



60. y
x  4

No

Yes

62. f x
x2  4x  3 (a) f 0
02  40  3
3 (b) f x  1
x  12  4x  1  3
x2  2x (c) f x  x  f x
x  x2  4x  x  3  x2  4x  3
2xx  x2  4x 66. f x


64. f x
2 Domain:  ,  Range: 2




3

x3 x  3
x2  x  12 x  3x  4 1 ,x3 x4

Domain:  , 4
4, 3
3, 

71 17, 

Range:  , 0
0, −3




3 4 −1

−8

4

−4

68. f x


7 12 x 13 8

70. (a) f x  gx
2x  3  x  1 (b) f x  gx
2x  3  x  1

Domain:  ,  Range:  , 

(c) f xg x
2x  3x  1

2

−4

(d) 2

f x 2x  3
gx x  1

(e) f g x
f x  1 
2x  1  3 (f) g f x
g2x  3
2x  3  1
2x  2

−2

60

Chapter 1



Functions, Graphs, and Limits



72. f x
x  1 does not have an inverse by the horizontal line test.

74. f x
x 3  1 has an inverse by the horizontal line test. y
x3  1 x
y3  1 x  1
y3 3 x  1 y
 3 x  1 f 1x


76. lim 2x  9
22  9
13

78. lim

x→2

x→2

7 5x  3 52  3

2x  9 22  9 13

80. lim t→0

lim

t→0

t2  1
 t t2  1
 t

t2  1 does not exist. t→0 t

lim

t1
 t2

82. lim t→2

t→2

x→3

x2  9
lim x  3
6 x→3 x3

86. lim

x→12

2x  1 1 1
lim
6x  3 x→12 3 3

t1
t2 

lim

t→2

lim

84. lim

t1 does not exist. t2

1 1  x4 4 4  x  4 8x 8x 8x
lim
lim ; lim
 ; lim
; 88. lim x→0 xx  44 x→0 x→0 4xx  4 x→0 4xx  4 x→0 4xx  4 x 

1 1  x4 4 lim does not exist. x→0 x

90. lim

s→0

11  s  1
lim 1  1  s  1  1  s s


lim

s→0

92. lim

x→0

s1  s

s→0

1  1  s 1 1
lim
 2 s1  s1  1  s s→0 1  s 1  1  s

1  x2  2xx  x2  1  x2
1  x  x2  1  x2
lim x→0 x x
lim

x→0

94.

1  1  s

x2x  x
lim 2x  x
2x x→0 x

x

1.1

1.01

1.001

1.0001

f x

0.3228

0.3322

0.3332

0.3333

lim

x→1

3 x 1  1
 x1 3

96. The statement lim x 3
0 is true. x→0

Review Exercises for Chapter 1 3 x
0 98. The statement lim  is true. x→0

100. The statement lim f x
1 is true since x→3

lim f x
lim x  2
1

x→3

x→3

lim f x
lim x2  8x  14
1.

x→3

x→3

102. f x


x2 is continuous on the intervals  , 0 and 0, . x

104. f x


x1 is continuous on the intervals  , 1 and 1, . 2x  2

106. f x
x  2 is continuous on all intervals of the form c, c  1, where c is an integer. 108. f x is continuous on  , . 110. lim f x
lim x  1
2 x→1

x→1

lim f x
lim 2x  a
2  a

x→1

x→1

Thus, 2
2  a and a
0.



t < 1 t > 1, t is not an integer. t ≥ 1, t is an integer.

2 112. C
2  0.1t, 2  0.1t  1, 4

0

6 0

114. Nonremovable discontinuities at t
1, 2, 3, . . . Yellow sweet maize:

White flint maize:

Intercepts: 0, 45, 5, 0

Intercepts: 0, 30, 5.5, 0

Line: y  45


45  0 x  0 05

y
9x  45

Line: y  30


30  0 x  0 0  5.5

y
5.45x  30

61