chapter 1 functions, graphs, and...

Copyright © Houghton Mifflin Company. All rights reserved.

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

Section 1.1 The Cartesian Plane and the Distance Formula ..................................14

Section 1.2 Graphs of Equations .............................................................................18

Section 1.3 Lines in the Plane and Slope ................................................................24

Mid-Chapter Quiz Solutions ......................................................................................30

Section 1.4 Functions...............................................................................................33

Section 1.5 Limits ....................................................................................................38

Section 1.6 Continuity .............................................................................................42

Review Exercises ..........................................................................................................45

Chapter Test Solutions ................................................................................................51

Practice Test ...............................................................................................................54

14 Copyright © Houghton Mifflin Company. All rights reserved.

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

Section 1.1 The Cartesian Plane and the Distance Formula

1.

Skills Review

1. ( ) ( ) ( )22 2 23 6 1 5 3 6

9 36

45

3 5

− + ⎡ − − ⎤ = − +⎣ ⎦

= +

=

=

2. ( ) ( ) ( ) ( )22 2 22 0 7 3 2 4

4 16

20

2 5

− − + ⎡− − − ⎤ = − + −⎣ ⎦

= +

=

=

3. ( )5 4 12 2

+ −=

4. ( )3 1 4 22 2

− + − −= = −

5. 27 12 3 3 2 3 5 3+ = + =

6. 8 18 2 2 3 2 2− = − = −

7.

( ) ( )

( ) ( ) ( )( ) ( )

( )

( )

( )

2 2

2 22 2

2 2

2 2

2

2

3 7 4 45

3 7 4 45

3 7 4 45

3 3 45

3 9 45

3 36

3 63 6

3 63, 9

x

x

x

x

x

x

xxxx

− + − =

⎛ ⎞− + − =⎜ ⎟⎝ ⎠

− + − =

− + =

− + =

− =

− = ±

− = − ±

=

= −

∓

8.

( ) ( )

( ) ( ) ( )( ) ( )

( )

( )

( )

2 2

2 22 2

2 2

22

2

2

6 2 2 52

6 2 2 52

6 2 2 52

4 2 52

16 2 52

2 36

2 66 26 28, 4

y

y

y

y

y

y

yyyy

− + − − =

⎛ ⎞− + − − =⎜ ⎟⎝ ⎠

− + − − =

+ − − =

+ − − =

− − =

− − = ±

− = ± +

= −

= −

∓

9. ( )

( )

57

25 14

19

x

x

x

+ −=

+ − =

=

10. 7 32

7 61

y

yy

− += −

− + = −

=

x

y

−1−2−3−4−5 2 3

−2

−3

−4

−5

−6

1

2

3

(1, −6)

(−2, −4)

(1, −1)

(2, 0)

(−5, 3)

Section 1.1 The Cartesian Plane and the Distance Formula 15


3. (a)

(b) ( ) ( )2 25 3 5 1 4 16 2 5d = − + − = + =

(c) Midpoint ( )3 5 1 5, 4, 32 2+ +⎛ ⎞= =⎜ ⎟

⎝ ⎠

5. (a)

(b) ( ) ( ) ( )2 23 2 1 2 5 1

4 36

2 10

d = ⎡ − ⎤ + −⎣ ⎦

= +

=

(c) Midpoint ( ) ( ) ( )1 2 3 2 1 5,

2 2

1, 22

⎛ + − + − ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞= − −⎜ ⎟⎝ ⎠

7. (a)

(b) ( ) ( )2 24 2 14 2

4 144

2 37

d = − + −

= +

=

(c) Midpoint ( )2 4 2 14, 3, 82 2+ +⎛ ⎞= =⎜ ⎟

⎝ ⎠

9. (a)

(b) ( ) ( )221 1 1 3

4 1 2 3 3

8 2 3

d = − − + −

= + − +

= −

(c) Midpoint ( )1 1 3 1 3 1, 0,2 2 2

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

11. (a)

(b) ( ) ( )( )220.5 0 6 4.8

0.25 116.64

116.89

d = − + − −

= +

=

(c) Midpoint ( )0 0.5 4.8 6, 0.25, 0.62 2+ − +⎛ ⎞= =⎜ ⎟

⎝ ⎠

13. (a) 4a =

3b =

( ) ( )2 24 0 3 0

16 95

c = − + −

= +

=

(b) 2 2 216 9 25a b c+ = + = =

15. (a) 10a =

3b =

( ) ( )2 27 3 4 1

100 9

109

c = + + −

= +

=

(b) 2 2 2100 9 109a b c+ = + = =

5

4

3

2

1

54321x

(5, 5)

(4, 3)

(3, 1)

y

2

−6

−4

42x

1 , 12

y

( )

−2−4

− , −2 12( )

− , −532( )

14

10

6

2

8642x

(4, 14)

(3, 8)

(2, 2)

y

x

y

6

6

4

2

−2

−4

−6

−2−4−6 42

(0, −4.8)

(0.25, 0.6)

(0.5, 6)

2

−1

2−2 1−1x

20,1 +

(1, )

(−1, 1)

y

3

3

16 Chapter 1 Functions, Graphs, and Limits


17. ( ) ( )2 21 3 0 7 1

9 36

45

3 5

d = − + −

= +

=

=

( ) ( )2 22 4 0 1 1

16 4

20

2 5

d = − + − −

= +

=

=

( ) ( ) 223 3 4 7 1

1 64

65

d = − + ⎡ − − ⎤⎣ ⎦

= +

=

Because 2 2 21 2 3 ,d d d+ = the figure is a right triangle.

19.

( ) ( )

( ) ( )

( ) ( )

( ) ( )

2 21

2 22

2 23

2 24

1 0 2 0

1 4

5

3 1 3 2

4 1

5

2 3 1 3

1 4

5

0 2 0 1

4 1

5

d

d

d

d

= − + −

= +

=

= − + −

= +

=

= − + −

= +

=

= − + −

= +

=

Because 1 2 3 4,d d d d= = = the figure is a parallelogram.

21. d = ( ) ( )

( )( )

2 2

2

2

2

1 4 0 5

2 17 5

2 17 25

2 8 0

4 2 0

4, 2

x

x x

x x

x x

x x

x

− + − − =

− + =

− + =

− − =

− + =

= −

23. d = ( ) ( )2 2

2

2

2

3 0 0 8

9 8

9 64

55

55

y

y

y

y

y

− + − =

+ =

+ =

=

= ±

25. (a) 2 2 2

2

16 5 , 0

281

281 16.76 feet

d d

d

d

= + >

=

= ≈

(b) ( )( )2 40 281

80 2811341.04 square feet

A =

=

≈

27.

Answers will vary. Let 6x = correspond to 1996. The number of subscribers steadily increased from 1996 to 2001 and then steadily decreased from 2001 to 2005.

29. From the graph you can estimate the values to be: (a) March 2005: 10,700 (b) December 2005: 10,900 (c) May 2006: 11,400 (d) January 2007: 12,500

31. (a) 1990: $92,000 (b) 1992: $100,000 (c) 1997: $122,000 (d) 2005: $208,000

6

4

2

62x

d2

d1

3d

(4, −1)

(3, 7)

(0, 1)

y

3

2

1

321x

d4

d2

d3

d1

(3, 3)

(2, 1)

(1, 2)

(0, 0)

y

560

16

70

Section 1.1 The Cartesian Plane and the Distance Formula 17


33. (a) Revenue midpoint

( )

2001 2005 25,269 31,944,2 2

2003, 28,606.5

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

Revenue estimate for 2003: $28,606.5 million

Profit midpoint

( )

2001 2005 2058.0 2729.0,2 2

2003, 2393.5

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

Profit estimate for 2003: $2393.5 million (b) Actual 2003 revenue: $27,061 million Actual 2003 profit: $1354 million (c) No, the increase in revenue from 2003 to 2005 is

greater than the increase in revenue from 2001 to 2003.

No, the profit decreased from 2001 to 2003 and then increased from 2003 to 2005.

(d) 2001 expenses: 25,269 2058.0 $23,211 million− =

2003 expenses: 27,061 1354 $25,707 million− =

2005 expenses: 31,944 2729.0 $29,215 million− =

(e) Answers will vary.

35. (a)

(b) The larger the clinic, the more patients a doctor can treat.

37. (a) (0, 0) is translated to ( ) ( )0 2, 0 3 2, 3 .+ + =

( )3, 1− − is translated to

( ) ( )3 2, 1 3 1, 2 .− + − + = −

( )1, 2− − is translated to ( ) ( )1 2, 2 3 1, 1 .− + − + =

(b)

39. Midpoint 1 2 1 2,2 2

x x y y+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

The point one-fourth of the way between ( )1 1,x y and

( )2 2,x y is the midpoint of the line segment from

( )1 1,x y to 1 2 1 2, ,2 2

x x y y+ +⎛ ⎞⎜ ⎟⎝ ⎠

which is

1 2 1 21 1

1 2 1 23 32 2, , .2 2 4 4

x x y yx y x x y y+ +⎛ ⎞+ +⎜ ⎟ + +⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

The point three-fourths of the way between ( )1 1,x y and

( )2 2,x y is the midpoint of the line segment from

1 2 1 2,2 2

x x y y+ +⎛ ⎞⎜ ⎟⎝ ⎠

to ( )2 2, ,x y which is

1 2 1 22 2

1 2 1 23 32 2, , .2 2 4 4

x x y yx y x x y y+ +⎛ ⎞+ +⎜ ⎟ + +⎛ ⎞=⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Thus,

1 2 1 2 1 2 1 23 3, , , ,4 4 2 2

x x y y x x y y+ + + +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

and

1 2 1 23 3,4 4

x x y y+ +⎛ ⎞⎜ ⎟⎝ ⎠

are the three points that divide the line segment joining ( )1 1,x y and ( )2 2,x y into four equal parts.

41. (a) ( ) ( )

( ) ( )

3 1 4 3 2 1 7 7, ,4 4 4 4

1 4 2 1 5 3, ,2 2 2 2

1 3 4 2 3 1 13 5, ,4 4 4 4

⎛ + − − ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

+ − −⎛ ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ + − + − ⎞ ⎛ ⎞= −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

(b) ( ) ( )

( ) ( )

3 2 0 3 3 0 3 9, ,4 4 2 4

2 0 3 0 3, 1,2 2 2

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

⎛ − + − + ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

− + − +⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ − + − + ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

Small clinic

Large clinic

Num

ber

of e

ar in

fect

ions

20

30

40

50

60

70

10

Number of doctors1 2 3 4

Medium clinic

4

66

4



Section 1.2 Graphs of Equations

1. (a) This is not a solution point because

( )2 1 2 3 3 0.− − = − ≠

(b) This is a solution point because

( ) ( )2 1 1 3 0.− − − =

(c) This is a solution point because

( )2 4 5 3 0.− − =

3. (a) This is a solution point because

( ) ( )221 3 4.+ − =

(b) This is not a solution point because

( ) ( )2 2 51

2 41 4.+ − = ≠

(c) This is not a solution point because

( ) ( )2 23 7 292 2 2 4.+ = ≠

Skills Review

1. 5 125 12

125

y xy x

xy

− =

= +

+=

2. 1515

y xy x

− = −

= −

3.

( )3

3

3

2 1

2 1

12

x y y

y x

yx

+ =

+ =

=+

4. 2 2

2 2

2 2

2

6 0

6

6

6

x x y

y x x

y x x

y x x

+ − − =

− = − −

= + −

= + −

5. ( ) ( )

( ) ( )

( )

( )

( )

2 2

2 2

2

2

2

2

2 1 9

1 9 2

1 9 2

9 2 1

9 4 4 1

5 4 1

x y

y x

y x

y x

x x

x x

− + + =

+ = − −

+ = − −

⎛ ⎞= − − −⎜ ⎟⎝ ⎠

= − − + −

= + − −

6. ( ) ( )

( ) ( )

( )

( )

( )

2 2

2 2

2

2

2

2

6 5 81

5 81 6

5 81 6

5 81 6

5 81 12 36

5 45 12

x y

y x

y x

y x

x x

x x

+ + − =

− = − +

− = − +

= + − +

= + − + +

= + − −

7.

( )

22

22

2

442

4 2

4 4

x x

x x

x x

−⎛ ⎞− + ⎜ ⎟⎝ ⎠

− + −

− +

8. ( )22

2 2

2

626

6 3

6 9

x x

x x

x x

+ +

+ +

+ +

9. 2

2

2

552

2554

x x

x x

−⎛ ⎞− + ⎜ ⎟⎝ ⎠

− +

10. ( )22

2

32

94

3

3

x x

x x

+ +

+ +

11.

( )( )

2 3 2

1 2

x x

x x

− +

− −

12.

( )( )

2 5 6

2 3

x x

x x

+ +

+ +

13.

( )

2

2

94

32

3y y

y

− +

−

14.

( )

2

2

494

72

7y y

y

− +

−

Section 1.2 Graphs of Equations 19


5. The graph of 2y x= − is a straight line with -intercepty at ( )0, 2 .− So, it matches (e).

7. The graph of 2 2y x x= + is a parabola opening up with vertex at ( )1, 1 .− − So, it matches (c).

9. The graph of 2y x= − has a y-intercept at

( )0, 2− and has x-intercepts at ( )2, 0− and ( )2, 0 .

So, it matches (a).

11. Let 0.y = Then,

( )32

2 0 3 0

.

x

x

− − =

=

Let 0.x = Then,

( )2 0 3 0

3.

y

y

− − =

= −

-intercept:x ( )32, 0

-intercept:y ( )0, 3−

13. Let 0.y = Then,

20 2x x= + −

( )( )0 2 1x x= + −

2, 1.x = −

Let 0.x = Then,

( ) ( )20 0 2

2

y

y

= + −

= −

x-intercepts: ( ) ( )2, 0 , 1, 0−

y-intercept: ( )0, 2−

15. Let 0.y = Then,

2

2

0 4

44.

x

xx

= −

=

= ±

Let 0.x = Then,

( )24 0y = −

2.y =

x-intercepts: ( ) ( )4, 0 , 4, 0−

y-intercept: ( )0, 2

17. Let 0.y = Then,

( )( )

2 402

0 2 2

2.

xxx x

x

−=

−= − +

= ±

Let 0.x = Then,

( )( )

20 40 2

2.

y

y

−=

−

=

x-intercept: Because the equation is undefined when 2,x = the only x-intercept is ( )2, 0 .−


19. Let 0.y = Then,

( ) ( )2 2

2

0 4 0 0

00.

x x

xx

− + =

=

=

Let 0.x = Then,

( ) ( )2 20 0 4 0

0.

y y

y

− + =

=

x-intercept: ( )0, 0


21. 2 3y x= +

−4 −3 −1 21 3 4

4

3

1

−2

−3

−4

y

x

(0, 3)

, 032( )−

23. 2 3y x= −

y

x

3, 0− )( 3, 0)(

(0, −3)

−4 −3−1

1

2

3

4

−4

1 3 4

x −2 32− −1 0 1 2

y −1 0 1 3 5 7

x −2 −1 0 1 2 3

y 1 −2 −3 −2 1 6



25. ( )21y x= −

(0, 1)(1, 0)

1 2 3 4 5−1−1

4

5

6

−2

−2−3

y

x

27. 3 2y x= +

321−3 −2

5

4

3

1

−1

x

(0, 2)),, 0( 2− 3

y

29. 1y x= − −

x

y

5

3

2

1

−1

−2

−3

−1 432

(1, 0)

31. 1y x= +

−4 −3 −2 −1

−1

1

2

3

4

5

6

−2

1 2 3 4

y

x

(0, 1)

(−1, 0)

33. 13

yx

=−

−3

−2

−1

3

2

1

654x

y

0, − 13

35. 2 4x y= −

1−1−2−3

−3

−1

3

1

x

(0, 2)

(−4, 0)

y

(0, −2)

37. ( ) ( )2 2 2

2 2

2 2

0 0 4

16

16 0

x y

x y

x y

− + − =

+ =

+ − =

39. ( ) ( )2 2 2

2 2

2 2

2 1 3

4 4 2 1 9

4 2 4 0

x y

x x y y

x y x y

− + + =

− + + + + =

+ − + − =

41. Since the point ( )0, 0 lies on the circle, the radius must

be the distance between ( )0, 0 and ( )1, 2 .−

r = ( ) ( )2 20 1 0 2 5+ + − =

( ) ( )2 2

2 2

1 2 5

2 4 0

x y

x y x y

+ + − =

+ + − =

x −2 −1 0 1 2

y 9 4 1 0 1

x −2 −1 0 1 2

y −6 1 2 3 10

x 1 2 3 4 5

y 0 −1 −1.41 −1.73 −2

x −3 −2 −1 0 1

y 2 1 0 1 2

x −1 0 1 2 2.5 3.5 4 5 6

y 14− 1

3− 12− −1 −2 2 1 1

2 13

x 5 0 −3 −4

y ±3 ±2 ±1 0



43. ( )0 6 0 8Center Midpoint , 3, 42 2+ +⎛ ⎞= = =⎜ ⎟

⎝ ⎠

( ) ( )2 2

distance from center to an endpoint

0 3 0 4 5

r =

= − + − =

( ) ( )2 2 2

2 2

2 2

3 4 5

6 9 8 16 25

6 8 0

x y

x x y y

x y x y

− + − =

− + + − + =

+ − − =

45. ( ) ( )( ) ( )

2 2

2 2

2 1 6 9 6 1 9

1 3 4

x x y y

x y

− + + + + = − + +

− + + =

−6

5−40

(1, −3)

47. ( ) ( )( ) ( )

2 2

2 2

4 4 2 1 1 4 1

2 1 4

x x y y

x y

− + + − + = − + +

− + − =

−2

7−2

4

(2, 1)

49.

( ) ( )( ) ( )

2 2

2 2

2 2

32

31 1 1 14 4 2 4 4

1 12 2 2

x y x y

x x y y

x y

+ − − =

− + + − + = + +

− + − =

, 12

12

3.5−2.5

2.5

−1.5

( (

51.

( ) ( )( ) ( )

2 2

2 2

2 2

414

9 9414 4 4

32

3 6 0

3 6 9 9

3 1

x y x y

x x y y

x y

+ + − + =

+ + + − + = − + +

+ + − =

53. Solving for y in the equation 2x y+ = yields 2 ,y x= − and solving for y in the equation

2 1x y− = yields 2 1.y x= − Then setting these two -valuesy equal to each other, we have

2 2 13 3

1.

x xx

x

− = −

=

=

The corresponding y-value is 2 1 1,y = − = so the point of intersection is ( )1, 1 .

55. Solving for y in the second equation yields 10 2y x= − and substituting this into the first

equation gives

( )

( )( )

22

2 2

2

2

10 2 25

100 4 40 25

5 40 75 0

8 15 0

3 5 0

3, 5.

x x

x x x

x x

x x

x x

x

+ − =

+ + − =

− + =

− + =

− − =

=

The corresponding y-values are 4y = and 0,y = so the points of intersection are ( )3, 4 and ( )5, 0 .

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

( )

3

3

2

2

2 0

2 0

0, 2.

x x

x x

x x

x

=

− =

− =

= ±

The corresponding y-values are 0,y = 2 2,y = −

and 2 2,y = so the points of intersection are

( )0, 0 , ( )2, 2 2 ,− − and ( )2, 2 2 .

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

( )( )

4 2 2

4 2

2

2 1 1

0

1 1 0

0, 1.

x x x

x x

x x x

x

− + = −

− =

+ − =

= ±

The corresponding y-values are 1, 0,y = and 0, so the

points of intersection are ( )1, 0 ,− ( )0, 1 , and ( )1, 0 .

3−6

−1

5

3− , 32( (



61. (a) 11.8 15,00019.30

C xR x

= +

=

(b) 11.8 15,000 19.30

15,000 7.52000 units

C Rx x

xx

=

+ =

=

=

(c)

( )1000 19.3 11.8 15,000

16,000 7.52133.3

P R C

x x

xx

= −

= − +

=

≈

So, 2134 units would yield a profit of $1000.

63. 1.55 0.85 35,0000.7 35,000

35,000 50,000 units0.7

R Cx xx

x

=

= +

=

= =

00 100,000

150,000

65. 9950 8650 250,0001300 250,000

250,000 193 units1300

R Cx xx

x

=

= +

=

= ≈

00 300

3,000,000

67. 180 4 75 3105 7

15

x xx

x

− = +

=

=

Equilibrium point ( ) ( ), 15, 120x p =

69. (a) Model: 3 20.796 8.65 312.9 4268y t t t= − + +

The model is a good fit. Answers will vary. (b) When 13:t = $8623y ≈ million

71. (a) Model: 20.77 27.3 587y t t= − + +

(b) Answers will vary. (c) When t = 9: $770.33y =

Yes, this prediction seems reasonable.

73.

The greater the value of ,c the steeper the line.

75.

Intercept: ( )0, 5.36

77.

Intercepts: ( )1.4780, 0 , ( )12.8553, 0 , ( )0, 2.3875

79.

Intercept: ( ) ( )5120, 0, 0.4167≈

t 2000 2001 2002 2003 2004 2005

Model 4268 4573 4866 5150 5432 5716

Actual 4265 4571 4899 5086 5479 5703

Year 2000 2001 2002 2003 2004 2007

Salary 587.00 613.53 638.52 661.97 683.88 740.37

−4.7

−3.1

4.7

3.1

8

−2

−12

18

−10

30−15

20

−6

−4

6

4



81. (a)

(b) 4.05 0.0131 0.024

tyt

− +=

−

(c) Yes, the model appears to be a good fit for the data. (d) Sample answer: A quadratic model may be a good fit for the data.

(e) A quadratic model that approximates the data is 20.00418 0.806 40.8.y t t= − +

(f )

(g) Rational model:

When t = 115: ( )( )

4.05 0.013 1151.5%

1 0.024 115y

− += ≈

−

Quadratic model:

When t = 115: ( ) ( )20.00418 115 0.806 115 40.8 3.4%y = − + ≈

The rational model should be used to predict future values, because its values keep decreasing as t increases. The quadratic model should not be used because its values increase for 96.t >

t 55 56 57 58 59 60 61 62 63 64 65 66 67 68

y 10.4 9.7 9.0 8.4 7.9 7.4 7.0 6.6 6.3 6.0 5.7 5.5 5.2 5.0

t 69 70 71 72 73 74 75 76 77 78 79 80 81 82

y 4.8 4.6 4.4 4.3 4.1 4.0 3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1

t 83 84 85 86 87 88 89 90 91 92 93 94 95 96

y 3.0 2.9 2.8 2.8 2.7 2.6 2.5 2.5 2.4 2.4 2.3 2.3 2.2 2.1

054 106

10

054 106

10

t 97 98 99 100 101 102 103 104 105

y 2.1 2.1 2.0 2.0 1.9 1.9 1.8 1.8 1.8



Section 1.3 Lines in the Plane and Slope

1. The slope is 1m = because the line rises one unit

vertically for each unit the line moves to the right.

3. The slope is 0m = because the line is horizontal.

5.

The slope is ( )0 3 1.9 0 3

m− −

= =−

7.

The slope is ( )2 43.

5 3m

− −= =

−

Skills Review

1. ( )5 2 7 13 4 7− −

= = −− − −

2. ( )7 0 74 1 3

− − −= −

−

3. 1 , 3

1 13 3

mm

− = −

− =−

4.

67

1 6,7

1 76

mm

− =

− = −

5. 4 74 7

x yy x

− + =

= +

6. 3 77 33 7

x yy xy x

− =

− = −

= −

7. ( )( )

2 3 4

3 4 2

3 12 23 10

y x

y x

y xy x

− = −

= − +

= − +

= −

8. ( ) ( )5 1 2

5 27

y x

y xy x

− − = − ⎡ − − ⎤⎣ ⎦+ = − −

= − −

9. ( ) ( )( )

( )

4 33 2

2 173 21

3 7 147 17

y x

y x

y xy x

− −− − = −

−

+ = −

+ = −

= −

10. ( ) ( )

( )

( )

3 11 17 1

41 16

21 132 213 32 53 3

y x

y x

y x

y x

y x

− −− = ⎡ − − ⎤⎣ ⎦− − −

−− = +

−

− = +

− = +

= +

x

y

10

6

4

2

−2

−4

−6

−2 842

(0, −3)

(9, 0)

62

2

−4

−2

x

(3, −4)

y

(5, 2)

Section 1.3 Lines in the Plane and Slope 25


9.

The slope is ( )

2 2 0.6 1 2

m −= =

−

So, the line is horizontal.

11.

The slope is undefined because ( )( )

5 38 8

m− − −

=− − −

and

division by zero is undefined. So, the line is vertical.

13.

The slope is ( )

3 1 4 2.4 2 6 3

m − − −= = = −

− −

15.

The slope is ( )3 518 4 8

1 2 3 24.5

m− −

= = = −− − −

17.

The slope is( )5 5 10

2 6 352 1

3 4 12

10 12 8.3 5

m− −

= = = ⋅ =−

19. The equation of this horizontal line is 1.y = So, three additional points are ( ) ( )0, 1 , 1, 1 , and ( )3, 1 .

21. The equation of this line is

( )24 632 8.3

y x

y x

+ = −

= −

So, three additional points are ( ) ( )3, 6 , 9, 2 ,− − and

( )12, 0 .

23. The equation of the line is

( )7 3 1

3 10.

y x

y x

− = − −

= − +

So, three additional points are ( ) ( )0, 10 , 2, 4 , and ( )3, 1 .

25. The equation of this vertical line is 8.x = − So, three additional points are ( ) ( )8, 0 , 8, 2 ,− − and ( )8, 3 .−

27. 15

5 20

4

x y

y x

+ =

= − +

So, the slope is 15,m = − and the y-intercept is ( )0, 4 .

29. 76

7 6 30

5

x y

y x

+ =

= − +

So, the slope is 76,m = − and the y-intercept is ( )0, 5 .

31. 3 153 15

x yy x

− =

= −

So, the slope is 3,m = and the y-intercept is ( )0, 15 .−

33. 4x =

Because the line is vertical, the slope is undefined. There is no y-intercept.

35. 4 04

yy

− =

=

So, the slope is 0,m = and the y-intercept is ( )0, 4 .

6

4

−2

642x

(6, 2)

y

( )12

, 2

−1−2 2 3 4

−2

1

2

3

4

−3

−4

−3−4

y

x(−2, 1)

(4, −3)

−1 1−2 2 3 4

−2

4

−3

−4

−3−4

y

x

14

, −2( )

38

, 1( )−

−1 1−2 2 3 4

−2

3

2

1

4

−3

−4

−3−4

y

x

14

56

,( )−

23

52

,( )

x

y

−2−4−6 2

−2

−4

−6

−8

2

(−8, −3)

(−8, −5)



37. The slope of the line is ( )3 52.

4 0m

− −= =

−

Using the point-slope form, we have

( )5 2 0

2 50 2 5.

y x

y xx y

+ = −

= −

= − −

39. The slope of the line is 3 0 3.1 0

m −= = −

− −

Using the point-slope form, we have 3

3 0.y x

x y= −

+ =

3

2

−1

1−1−2−3x

y

(−1, 3)

(0, 0)

41. The slope of the line is 2 32 2

m − −= =

−undefined.

So, the line is vertical, and its equation is 2

2 0.x

x=

− =

5431−1

4

3

2

1

−2

−3

−1

x

y

(2, 3)

(2, −2)

43. The slope of the line is ( )1 10.

2 3m

− − −= =

− −So, the line

is horizontal, and its equation is 1

1 0.y

y= −

+ =

−1 1−2 2 3 4

−2

3

2

1

4

−3

−4

−3−4

y

x

(−2, −1) (3, −1)

45. The slope of the line is ( )

1 5 6 1.1 3 2 3 2

m −= =

− +


1 112 31 72 6

3 6 7 0.

y x

y x

x y

⎛ ⎞− = +⎜ ⎟⎝ ⎠

= +

− + =

−1 1 2−2

−2

−1

2

y

x

23

56

,( )−

13

, 1( )−

47. The slope of the line is 8 4 4.1 2 1 2

m −= =

+


18 42

4 60 4 6.

y x

y xx y

⎛ ⎞− = −⎜ ⎟⎝ ⎠

= +

= − +

−4

−4

6

8

−6

−8

2 4 6 8−6−8

y

x

12

, 4( )−

12

, 8( )

−4

−2

−2

2

4 6x

y

(0, −5)

(4, 3)



49. Using the slope-intercept form, we have

34 3

4 3 123 4 12 0.

y x

y xx y

= +

= +

− + =

−6

−2

2

(0, 3)

6

51. Because the slope is undefined, the line is vertical and its equation is 1.x = −

1

−1

−3

3

(−1, 2)

53. Because the slope is 0, the line is horizontal and its equation is 7.y =

6

0−6

8

(−2, 7)

55. Using the point-slope form, we have

( )2 4 0

4 24 2 0.

y x

y xx y

+ = − −

= − −

+ + =

(0, −2)

4

−4

−5

2

57. Using the slope-intercept form, we have

3 24 3

12 9 89 12 8 0.

y x

y xx y

= +

= +

− + =

−4 2

−2

0,( )23

4

59. The slope of the line joining ( )2, 1− and ( )1, 0− is

( )

1 0 1 1.2 1 1−

= = −− − − −

The slope of the line joining ( )1, 0− and ( )2, 2− is

( )0 2 2 2.1 2 3 3− −

= = −− − −

Because the slopes are different, the points are not collinear.

61. The slope of the line joining ( )2, 7 and ( )2, 1− − is

1 7 2.2 2− −

=− −

The slope of the line joining ( )0, 3 and ( )2, 1− − is

1 3 2.2 0− −

=− −

Because the slopes are equal and both lines pass through ( )2, 1 ,− − the three points are collinear.

63. Because the line is vertical, it has an undefined slope, and its equation is

33 0.x

x=

− =

65. Because the line is parallel to a horizontal line, it has a slope of 0,m = and its equation is

10.y = −

67. Given line: 7, 1y x m= − + = −

(a) Parallel: 1 1m = −

( )2 1 3

1 0

y x

x y

− = − +

+ + =

(b) Perpendicular: 2 1m =

( )2 1 3

5 0

y x

x y

− = +

− + =

9

−2

−9

10

(−3, 2)

x + y = 7



69. Given line: 3 7 34 4 4,y x m= − + = −

(a) Parallel: 134m = −

( )7 3 32 18 4 3 4 2

8 7 6 46 8 3 0

y x x

y xx y

− = − + = − −

− = − −

+ − =

(b) Perpendicular: 243m =

( )7 84 2 48 3 3 3 9

72 63 96 6496 72 127 0

y x x

y xx y

− = + = +

− = +

− + =

−6

−4

6

4

23

78

,( )−3x + 4y = 7

71. Given line: 3y = − is horizontal, 0m =

(a) Parallel: 1 0m =

( )0 0 1

0

y x

y

− = +

=

(b) Perpendicular: 2m is undefined

1x = −

−4

−4

2(−1, 0)

y + 3 = 0

1

73. Given line: 2 0x − = is vertical, m is undefined

(a) Parallel: 1m is undefined, 1x =

(b) Perpendicular: ( )2 0, 1 0 1 , 1m y x y= − = − =

6

−2

−3

4

(1, 1)x − 2 = 0

75. 2y = −

−3

−1

1

2−1 1−2x

y

77. 2 3y x= −

−2

−1

1

−3

21−1−2 3x

y

79. 2 1y x= − +

3

2

1

−1

21−1−2x

y

81. 35 3y x= − −

x

y

(0, −3)

−1 1 2 3 4 5−1

−2

−4

−5

−6

x −2 −1 0 1

y −2 −2 −2 −2

x −1 0 1 2

y −5 −3 −1 1

x −1 0 1 2

y 3 1 −1 −3

x 0 2 4 5

y −3 215− 27

5− −6



83. 4 6y x= − −

−6

−2

−11−2−3−4 2

x

y

85. ( )0, 32 , ( )100, 212

( )212 3232 0100 0

91.8 32 325

F C

F C C

−− = −

−

= + = +

or

( )5 329

C F= −

87. (a) ( )0, 4024 , ( )5, 4255

4255 4024 231 46.25 0 5

m −= = =

−

( )4024 46.2 0

46.2 4024

y t

y t

− = −

= +

The slope 46.2m = tells you that the population is increasing by 46.2 thousand per year.

(b) When 2:t = 4116.4y =

In 2002, the population was about 4,116,400. (c) When 4:t = 4208.8y =

In 2004, the population was about 4,208,800. (d) 2002: 4,024,000 2004: 4,198,000 The estimates were close to the exact values. (e) The model could possibly be used to predict the

population in 2009 if the population continues to grow at the same linear rate.

89. (a) The equipment depreciates 10255 $205= per year,

so the value is 1025 205 ,y t= − where 0 5.t≤ ≤

(b)

(c) When 3,t = the value is $410.00.

(d) The value is $600 when 2.07t = years.

91. (a) Using the points ( )9, 7802 and ( )15, 10,239 ,

10,239 7802 2437.15 9 6

m −= =

−

So,

( )7802 406 9

2437 8293.6 2

y t

y t

− = −

= +

(b) When 11:t = 8614y =

In 2001, the personal income was about $8614 billion.

(c) When 17:t = 11,051y =

In 2007, the personal income was about $11,051 billion.

(d) 2001: $8724 billion 2007: $11,595 billion The estimates were relatively close to the actual

values.

93. (a) 50 350,000C x= +

(b) 120R x=

(c)

( )120 50 350,000

70 350,000

P R C

x x

x

= −

= − +

= −

(d) When 13,000:x = 560,000P =

If the company sells 13,000 units, the profit is $560,000.

(e) 120 50 350,00070 350,000

5000

R Cx xxx

=

= +

=

=

The company must sell 5000 units to break even.

95. (a) 2000 0.07W S= +

(b) 2300 0.05W S= +

(c)

The lines intersect at ( )15,000, 3050 . If you sell

$15,000, then both jobs would yield wages of $3050. (d) No. You will make more money (if sales are $20,000)

at your current job ( )$3400w = than in the offered

job ( )$3300 .w =

x 32− −1 0 1

y 0 −2 −6 −10

00

6

120000

30,000

5000



97. 23,500 3100 100,0003100 76,500

24.677

xxx

+ ≤

≤

≤

So, 24 units.x ≤

0300

120,000

99. C = 18,375 1150 100,0001150 81,625

70.978

xxx

+ ≤

≤

≤

So, 70 units.x ≤

0800

120,000

101. C = 75,500 89 100,00089 24,500

275.28

xxx

+ ≤

≤

≤

So, 275 units.x ≤

03000

120,000

103. C = 32,000 650 100,000650 68,000

104.62

xxx

+ ≤

≤

≤

So, 104 units.x ≤

0

0110

120,000

105. C = 50,000 0.25 100,0000.25 50,000

x+ ≤

≤

So, 200,000 units.x ≤

100,0000

0

100,000

Mid-Chapter Quiz Solutions 1. (a)

(b) ( ) ( )( )223 3 1 2

36 9

3 5

d = − − + − −

= +

=

(c) 3 3 1 2 1Midpoint , 0,2 2 2

− + −⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2. (a)

(b) 221 1 3 1 492

2 4 2 16 4

197 1 19716 4

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

= =

(c) 31 1

2 4 22 3 1Midpoint , ,2 2 8 4

⎛ ⎞+ − ⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

x

y

3

2

1

−1

−2

−3

−1−2−3 321

(−3, 1)

(3, −2)

, − 12( (0

x

y

2

1

−1

−2

−1−2 21

, 2 12( (

, − 14

32( (

,38

14( (

Mid-Chapter Quiz Solutions for Chapter 1 31


3. (a)

(b) ( ) ( )2 20 3 4 0

3 16

19

d = − + − −

= +

=

(c) 0 3 4 0 3Midpoint , , 22 2 2

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

4.

( ) ( )

( )( ) ( )( )

( ) ( )

2 2

2 2

2 2

2 4 1 0 5

2 1 1 5 3 5

1 4 5 0 5 2

a

b

c

= − + − =

= − − + − − =

= − − + − − =

( ) ( ) ( )2 2 22 2 25 3 5 5 2a b c+ = + = =

5. Use the points ( )2003, 5719 and ( )2005, 5800 .

( )

2003 2005 5719 5800Midpoint ,2 2

2004, 5759.5

+ +⎛ ⎞= ⎜ ⎟⎝ ⎠

=

The population in 2004 was about 5,759,500 people.

6. 5 2y x= +

x

y

(0, 2)

−1−2−3−4 1 2 3 4

1

2

3

4

5

6

7

, 0 25( (−

7. 2 6y x x= + −

x

y

−1−2−4 1 3−1

−2

−3

−4

−5

−7

8. 3y x= −

x

y

−1 1 2 3 4 5 6

1

2

3

4

5

6

9. ( ) ( )2 2

2 2

2 2

1 0 37

2 1 37

2 36 0

x y

x x y

x y x

+ + − =

+ + + =

+ + − =

10. Because the point ( )1, 2− lies on the circle, the radius

must be the distance between ( )1, 2− and ( )2, 2 .−

( ) ( )2 21 2 2 2 5r = − − + + =

( ) ( )2 2

2 2

2 2

2 2 25

4 4 4 4 25

4 4 17 0

x y

x x y y

x y x y

− + + =

− + + + + =

+ − + − =

11.

( ) ( )( ) ( )

2 2

2 2

2 2

8 6 16 0

8 16 6 9 16 16 9

4 3 9

x y x y

x x y y

x y

+ + − + =

+ + + − + = − + +

+ + − =

3−9

−1

7

(−4, 3)

x 25− 0 1

5 1

y 0 2 3 7

x −3 −2 −1 −0.5 0 1 2

y 0 −4 −6 −6.25 −6 −4 0

x 0 1 2 3 4 5 6

y 3 2 1 0 1 2 3

x

y

−1−2 1 2 3 4−1

−2

−3

−4

−5

1), 0

(0, −4)

( 3

, −2 23

x

y

−1−2 1 2 3 4−1

−2

−3

1

(−1, −5)

(4, 0)

(2, 1)a

c

b



x

y

(−4, 5)

−1−2−3−4 32−1

−2

2

3

4

5

(1, −1)

x

y

1

−3

−1

−2

−1−3 −2 321

, 2 52( ((0, 2)

x

y

(−2, 3)

(−2, 2)

−1−3−4−5 1−1

−2

−3

1

2

3

12.

( ) ( )( ) ( )

2 2

2 2

2 2

22

114

1 11 14 4 4

12

4 4 8 4 11 0

2 0

2 1 1

1 4

x y x y

x y x y

x x y y

x y

+ − + − =

+ − + − =

− + + + + = + +

− + + =

5−4

−4

2

12

1, − ( (

13. 4.55 12,5007.19

C xR x

= +

=

7.19 4.55 12,5002.64 12,500

4734.8

R Cx xxx

=

= +

=

≈

The company must sell 4735 units to break even.

14. ( ) ( )1, 1 , 4, 5

5 1 64 1 5

m

− −

+= = −

− −

( )61 156 15 5

y x

y x

+ = − −

= − +

15. ( ) ( )2, 3 , 2, 2

2 3 undefined2 2

m

− −

−= =

− +

Because the slope is undefined, the line is vertical and its equation is 2.x = −

16. ( )

52

5, 2 , 0, 22

2 2 00

m

⎛ ⎞⎜ ⎟⎝ ⎠

−= =

−

Because the slope is 0, the line is horizontal and its equation is 2.y =

17. Given line: 1 1 14 2 4,y x m= − − = −

(a) Parallel: 114m = −

( )14

1714 4

5 3y x

y x

+ = − −

= − −

(b) Perpendicular: 2 4m =

( )5 4 3

4 17

y x

y x

+ = −

= −

18. Let 5x = represent 2005, and y represent the sales, in thousands of dollars.

( ) ( )5, 1330 , 7, 1800

1800 1330 2357 5

m −= =

−

( )1330 235 5

235 155

y x

y x

− = −

= +

When 6: 1565t y= =

When 9: 2270t y= =

You can predict the sales to be $1,565,000 in 2006 and $2,270,000 in 2009.

19. An equation for the daily cost C in terms of x, the number of miles driven, is 0.42 175.C x= +

20. (a) ( ) ( )4, 28,300 , 6, 31,700

31,700 28,300 17006 4

m −= =

−

( )28,300 1700 4

1700 21,500

y t

y t

− = −

= +

(b) When 10: 38,500t y= =

In 2010, your salary will be about $38,500.

Section 1.4 Functions 33


Section 1.4 Functions

1. 24y x= ± − y is not a function of x since there are two values of y for

some x.

3. 12

1 112 2

6 3x y

y x

− = −

= +

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

5. 24y x= − y is a function of x since there is only one value of y for

each x.

7. 2y x= +

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

9. Domain: ( ),−∞ ∞

Range: [ 2.125, )− ∞

11. Domain: ( ) ( ), 0 0,−∞ ∪ ∞

Range: { 1, 1}−

Skills Review

1. ( ) ( ) ( )25 1 6 1 9 5 1 6 9 20− − − + = + + =

2. ( ) ( ) ( )3 22 7 2 10 8 7 4 10 18 28 10− + − − = − + − = − + =

3. ( )2 2 22 5 10 4 4 5 10 6x x x x x x x− + − = − + + − = + −

4. ( ) ( ) ( ) ( )( )( )

3 2

3 2 2

3 2

3 3 3 3 6 9

3 3 6 18 9 27

9 26 30

x x x x x x

x x x x x x

x x x

− + + = − + + + +

= − + + + + + +

= + + +

5. ( )1 1 1

1 1 1 1x x x= =

− − − +

6. 1 1 1 2 11 x x x x x xx x x x x− − + − −

+ = + = =

7. 2 6 112 17

x yy x

+ − =

= − +

8. 2

2

2

2

5 6 1 0

5 6 1

6 15

6 15 5

y x

y x

xy

x

− − =

= +

+=

= +

9. ( ) ( )

( )

2 2

2

2

2

3 5 1

3 5 1

3 5 2 1

2 6 3

y x

y x

y x x

y x x

− = + +

− = + +

− = + + +

= + + +

10. 2 2

2 2

2

4 2

2 4

2 4

y x

y x

y x

− =

= +

= +

11. 2 14

4 2 14 1 24 1

2122

yx

x yx yx y

x y

−=

= −

+ =

+=

+ =

12. 3

3

3

31 12 2

2 1

2 1

2 1

x y

x y

y x

y x

= −

= −

− = − −

= +

7

−3

−5

5

3

−2

−3

2



13. Domain: ( )4, ∞

Range: [4, )∞

15.

Domain: ( ) ( ), 4 4,−∞ − ∪ − ∞

Range: ( ) ( ), 1 1,−∞ ∪ ∞

17. Domain: ( ),−∞ ∞

Range: ( ),−∞ ∞

19. Domain: ( ),−∞ ∞

Range: ( ], 4−∞

21. (a) ( ) ( )0 3 0 2 0 2 2f = − = − = −

(b) ( ) ( )1 3 1 2

3 3 23 5

f x x

xx

− = − −

= − −

= −

(c) ( ) ( )3 2

3 3 2

f x x x x

x x

+ ∆ = + ∆ −

= + ∆ −

23. (a) 1 1 44 1 4

g⎛ ⎞ = =⎜ ⎟⎝ ⎠

(b) ( ) 144

g xx

+ =+

(c) ( ) ( )

( )( )

( )

1 1g x x g xx x xx x x

x x x

xx x x

+ ∆ − = −+ ∆− + ∆

=+ ∆

−∆=

+ ∆

25.

( ) ( )

( ) ( ) ( )

( )

( )

2 2

22 2

2

5 2 5 2

[ 2 5 5 2] [ 5 2]

2 52 5, 0

f x x f xx

x x x x x x

x

x x x x x x x xx

x x x xx x x

x

+ ∆ −∆

+ ∆ − + ∆ + − − +=

∆

+ ∆ + ∆ − + ∆ + − − +=

∆

∆ + ∆ + ∆= = + ∆ + ∆ ≠

∆

27. ( ) ( )

( ) ( )

1 1 1 11 1

1 11 1

1 , 01 1

g x x g xx

x x x x x xx x x x

x x xx x x x

xx x x

+ ∆ −∆

+ ∆ + − + + ∆ + + += ⋅

∆ + ∆ + + +

+ ∆ + − +=

⎡ ⎤∆ + ∆ + + +⎣ ⎦

= ∆ ≠+ ∆ + + +

29. ( ) ( )

( ) ( )( )( )

( )( )

1 12 2

2 22 2

1 , 02 2

f x x f x x x xx x

x x xx x x x

xx x x

−+ ∆ − + ∆ − −=∆ ∆

− − + ∆ −=

+ ∆ − − ∆

−= ∆ ≠

+ ∆ − −

31. y is not a function of x.

33. y is a function of x.

35. (a) ( ) ( ) ( )2 5 5 2f x g x x x+ = − + =

(b) ( ) ( ) ( )( )2 5 5 10 25f x g x x x⋅ = − = −

(c) ( )( )

2 55

f x xg x

−=

(d) ( )( ) ( ) ( )5 2 5 5 5f g x f= = − =

(e) ( )( ) ( )2 5 5g f x g x= − =

37. (a) ( ) ( ) ( ) ( )2 21 1f x g x x x x x+ = + + − = +

(b) ( ) ( ) ( )( )2 3 21 1 1f x g x x x x x x⋅ = + − = − + −

(c) ( )( )

2 1, 11

f x x xg x x

+= ≠

−

(d) ( )( ) ( ) ( )2

2

1 1 1

2 2

f g x f x x

x x

= − = − +

= − +

(e) ( )( ) ( ) ( )2 2 21 1 1g f x g x x x= + = + − =

39. ( ) ,f x x= ( ) 2 1g x x= −

(a) ( )( ) ( ) ( )21 1 1 0 0f g f f= − = =

(b) ( )( ) ( ) ( )1 1 1 0g f g g= = =

(c) ( )( ) ( )0 0 1g f g= = −

(d) ( )( ) ( )4 15 15f g f− = =

(e) ( )( ) ( )2 21 1f g x f x x= − = −

(f) ( )( ) ( ) 1,g f x g x x= = − 0x ≥

−20 10

−10

10

300

0

20



8−4

−2

6

f

f −1

3−3

−2

2

f

f −1

6−3

−2

4

f = f −1

00

6

4

f −1

f

41. ( )( ) 1 15 15 5

x xf g x f x− −⎛ ⎞ ⎛ ⎞= = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )( ) ( ) ( )5 1 15 1

5x

g f x g x x+ −

= + = =

2 31

2

3

1

x

f

g

y

43. ( )( ) ( ) ( )( )( ) ( ) ( )

2

2 2

2

9 9 9

9 9 9

, 0

f g x f x x x

g f x g x x

x x x

= − = − − =

= − = − −

= = ≥

x

6

6 9

9f

g

y

45. ( )

( )1

2 3

2 33

23

2

f x x y

y xxy

xf x−

= − =

− =

+=

+=

47. ( )

( )

5

5

5

1 5

f x x y

y x

y x

f x x−

= =

=

=

=

49. ( ) 29 ,f x x y= − = 0 3x≤ ≤

( )

2

2 2

2 2

2

1 2

9

9

9

9

9 , 0 3

y x

y x

y x

y x

f x x x−

− =

− =

= −

= −

= − ≤ ≤

51. ( )

( )

2 3

2 3

3 2

1 3 2

, 0f x x y x

y x

y x

f x x−

= = ≥

=

=

=

53.

( ) 3 7

3 73 73

7

f x x

y xx y

xy

= −

= −

= −

−=

is one-to-one.

55.

( ) 2f x x=

f is not one-to-one because ( ) ( )1 1 1 .f f= = −

57. ( ) 3f x x= +

2−7

−1

5

f is not one-to-one because ( ) ( )5 2 1 .f f− = = −

−3

−1

3

4

−3

−2

3

5



59. (a) 2y x= +

1 2 3 4

3

2

1

4

x

y

(b) y x= −

(c) 2y x= −

2 3 41

2

3

4

1

x

y

(d) 3y x= +

(e) 4y x= −

4 6 82

2

4

6

8

x

y

(f ) 2y x=

3 421

3

4

2

1

x

y

61. (a) Shifted three units to the left: ( )23y x= +

(b) Shifted six units to the left, three units downward, and reflected: ( )26 3y x= − + −

63. (a)

(b) ( ) ( ) ( )( ) ( )( ) ( )

27 0.68 7 0.3 7 45 76.22

10 16.7 10 45 122

14 16.7 14 45 188.8

d

d

d

= − + =

= − =

= − =

The amounts spent in 1997, 2000, and 2004 were $76.22 billion, $122 billion, and $188.8 billion, respectively.

65.

( ) ( )1 2

2

2

Total Sales

690 8 0.8 458 0.78

1148 7.22 0.8 , 1, 2, 3, 4, 5, 6, 7

R R R

t t t

t t t

= = +

= − − + +

= − − =

67. (a)

( )

( )

14.751 0.01

14.75 10.01

14.750.01

100 14.75

1475 100

xp

px

pp

pp

p

+ =

−=

−=

−=

= −

(b) ( )100 14.75 1047.5 48 units

10x

−= = ≈

1600

210

80800

1500

21 3

−2

−1

−3

x

y

3

2

1

−2 −1−3x

y



69. ( )( )( )( ) ( )

70 375

40

40 2800 375

C x x

x t t

C x t C t t

= +

=

= = +

C is the weekly cost per t hours of production.

71. (a) If 0 100,x≤ ≤ then 90.p = If 100 1600,x< ≤ then ( )90 0.01 100 91 0.01 .p x x= − − = −

If 1600,x > then 75.p =

Thus,

90, 0 10091 0.01 , 100 1600.75, 1600

xp x x

x

≤ ≤⎧⎪= − < ≤⎨⎪ >⎩

(b) 60P px x= −

( ) 2

90 60 , 0 100 30 , 0 10091 0.01 60 , 100 1600 31 0.01 , 100 1600

75 60 , 1600 15 , 1600

x x x x xP x x x x x x x

x x x x x

− ≤ ≤ ≤ ≤⎧ ⎧⎪ ⎪= − − < ≤ = − < ≤⎨ ⎨⎪ ⎪− > >⎩ ⎩

73. (a) Revenue R rn= = ( ) 215 0.05 80 19 0.05n n n n= ⎡ − − ⎤ = −⎣ ⎦

(b)

(c) The revenue increases and then decreases as n gets larger, so it is not a good formula for the bus company to use.

75. ( ) 29 4f x x x= −

6

−4

−4

6

Zeros: ( ) 9

49 4 0 0,x x x− = ⇒ =

The function is not one-to-one.

77. ( ) 31tg t

t+

=−

−6

9−9

6

Zero: 3t = −

The function is one-to-one.

79. ( ) 2 2 4g x x x= −

6

0−6

8

Domain: 2x ≥

Zeros: 2x = ±

The function is not one-to-one.

n 100 125 150 175 200 225 250

R 1400 1593.75 1725 1793.75 1800 1743.75 1625



x

y

−1−2−3−4−5 1 2 3 4

−2−3−4−5

1234

y

−2

12345678

x−1−2−3−4−5 1 2 3 4 5

x

y

−1 1 2 3 4 5 6

1

2

3

4

5

6

x

y(0, 1)

−1−3 −2 1 2 3

−2

−3

1

2

3

Section 1.5 Limits

Skills Review

1. ( ) 2 3 3f x x x= − +

(a) ( ) ( ) ( )21 1 3 1 3

1 3 37

f − = − − − +

= + +

=

(b) ( ) 2 3 3f c c c= − +

(c) ( ) ( ) ( )2

2 2

3 3

2 3 3 3

f x h x h x h

x xh h x h

+ = + − + +

= + + − − +

2. ( )2 2, 13 1, 1

x xf x

x x− <⎧

= ⎨+ ≥⎩

(a) ( ) ( )1 2 1 2

2 24

f − = − −

= − −

= −

(b) ( ) ( )3 3 3 1

9 110

f = +

= +

=

(c) ( ) ( )2 2

2

2

1 3 1 1

3 3 1

3 4

f t t

t

t

+ = + +

= + +

= +

3. ( ) 2 2 2f x x x= − +

( ) ( )

( ) ( ) ( )( )2 2

2

2

1 1

1 2 1 2 1 2 1 2

1 2 2 2 2 1 2 2

f h fh

h h

hh h h

hhh

h

+ −

+ − + + − − +=

+ + − − + − + −=

=

=

4. ( ) 4f x x=

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

8 4 8

4

4

f h f hh h

hh

hh

+ − + −=

+ −=

=

=

5. ( ) 5h x

x= −

Domain: ( ) ( ), 0 0,−∞ ∪ ∞

Range: ( ) ( ), 0 0,−∞ ∪ ∞

6. ( ) 225g x x= −

Domain: [ ]5, 5−

Range: [ ]0, 5

7. ( ) 3f x x= −

Domain: ( ),−∞ ∞

Range: [ )0, ∞

8. ( ) xf x

x=

Domain: ( ) ( ), 0 0,−∞ ∪ ∞

Range: 1, 1y y= − =

9. 2 2

2 2

22

2

9 4 49

4 49 9

49 94

49 92

x y

y x

xy

xy

+ =

= −

−=

± −=

Not a function of x (fails the vertical line test).

10.

( )

2

2

2

2

2 8 7

2 7 8

2 7 8

82 7

x y x y

x y y x

y x x

xyx

+ =

− = −

− = −

= −−

Yes, y is a function of x.

Section 1.5 Limits 39


1.

( )2

lim 2 5 9x

x→

+ =

3.

22

2 1lim 0.254 4x

xx→

−= =

−

5.

0

1 1lim 0.5x

xx→

+ −=

7.

0

1 114 4lim 0.0625

16x

xx−→

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠ = − = −

9. (a) ( )0

lim 1x

f x→

=

(b) ( )1

lim 3x

f x→−

=

11. (a) ( )0

lim 1x

g x→

=

(b) ( )1

lim 3x

g x→−

=

13. (a) ( ) ( ) ( ) ( )lim lim lim

3 912

x c x c x cf x g x f x g x

→ → →⎡ + ⎤ = +⎣ ⎦

= +

=

(b) ( ) ( ) ( ) ( )lim lim lim

3 927

x c x c x cf x g x f x g x

→ → →⎡ ⎤⎡ ⎤⎡ ⎤ =⎣ ⎦ ⎢ ⎥⎢ ⎥⎣ ⎦⎣ ⎦

= ⋅

=

(c) ( )( )

( )( )

lim 3 1limlim 9 3x c

x cx c

f xf xg x g x

→

→→

= = =

15. (a) ( )lim 16 4x c

f x→

= =

(b) ( ) ( )lim 3 3 16 48x c

f x→

⎡ ⎤ = =⎣ ⎦

(c) ( ) 2 2lim 16 256x c

f x→

⎡ ⎤ = =⎣ ⎦

17. (a) ( )3

lim 1x

f x+→

=

(b) ( )3

lim 1x

f x−→

=

(c) ( )3

lim 1x

f x→

=

19. (a) ( )3

lim 0x

f x+→

=

(b) ( )3

lim 0x

f x−→

=

(c) ( )3

lim 0x

f x→

=

21. (a) ( )3

lim 3x

f x+→

=

(b) ( )3

lim 3x

f x−→

= −

(c) ( )3

limx

f x→

does not exist.

23. 2 22

lim 2 4x

x→

= =

25. ( ) ( )3 3 3

lim 2 5 lim 2 lim 5 2 3 5 1x x x

x x→− →− →−

+ = + = − + = −

27. ( )2 2 21 1 1

lim 1 lim 1 lim 1 1 0x x x

x x→ → →

− = − = − =

29. 3

lim 6 3 6 3x

x→

+ = + =

x 1.9 1.99 1.999 2 2.001 2.01 2.1

f (x) 0.2564 0.2506 0.2501 ? 0.2499 0.2494 0.2439

x −0.1 −0.01 −0.001 0 0.001 0.01 0.1

f (x) 0.5132 0.5013 0.5001 ? 0.4999 0.4988 0.4881

x −0.5 −0.1 −0.01 −0.001 0

f (x) −0.0714 −0.0641 −0.0627 −0.0625 ?

x 1.9 1.99 1.999 2 2.001 2.01 2.1

f (x) 8.8 8.98 8.998 ? 9.002 9.02 9.2



31. 3

2 2lim 22 3 2x x→−= = −

+ − +

33. ( )( )

22

2

2 11 3 3lim2 2 2 4 4x

xx→−

− −−= = = −

− −

35. ( )7

5 75 35lim2 7 2 9x

xx→

= =+ +

37. 3

1 1 3 1 1 1lim3 3x

xx→

+ − + −= =

39. 1

1 1 1 11 14 4 5 4lim

1 20 20xx

x→

− − −+ = = = −

41. ( )( )

( )

2

1 1

1

1 11lim lim1 1

lim 1 2x x

x

x xxx x

x→− →−

→−

+ −−=

+ += − = −

43. ( )( )22 2

2

2 2lim lim4 4 2 2

1lim does not exist.2

x x

x

x xx x x x

x

→ →

→

− −=

− + − −

=−

45. ( )( )24 4

4

4 4lim lim16 4 4


t t

t

t tt t t

t

→ →

→

+ +=

− + −

=−

47. ( )( )

( )

23

2 2

22

2 2 48lim lim2 2

lim 2 4 12

x x

x

x x xxx x

x x

→− →−

→−

+ − ++=

+ +

= − + =

49. ( )2

2 2lim 1

2 2t

x xx x−→−

+ − += = −

+ +

( )2

2 2lim 1

2 2t

x xx x+→−

+ += =

+ +

So, 2

2lim

2t

xx→−

++

does not exist.

51. ( )2

lim 4 2x

x−→

− =

( )2

lim 4 2x

x+→

− =

So, ( )2

lim 2x

f x→

=

53. ( )3 3

1lim lim 2 13x x

f x x− −→ →

⎛ ⎞= − = −⎜ ⎟⎝ ⎠

( ) ( )3 3

lim lim 2 5 1x x

f x x+ +→ →

= − + = −

So, ( )3

lim 1.x

f x→

= −

55. ( )0 0

0

2 2 2 2 2lim lim

lim 2 2x x

x

x x x x x xx x∆ → ∆ →

∆ →

+ ∆ − + ∆ −=

∆ ∆= =

57.

( ) ( )

0 0

0

0

2 2 2 2 2 2lim lim2 2

2 2lim

2 2

1lim2 2

12 2

x x

x

x

x x x x x x x x xx x x x x

x x xx x x x

x x x

x

∆ → ∆ →

∆ →

∆ →

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

+ + ∆ − +=

⎡ ⎤∆ + + ∆ + +⎣ ⎦

=+ + ∆ + +

=+

59. ( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( )

( )

2 22 2 2

0 0

2

0

0

5 5 2 5 5 5lim lim

2 5lim

lim 2 5

2 5

t t

t

t

t t t t t t t t t t t t t tt t

t t t tt

t t

t

∆ → ∆ →

∆ →

∆ →

+ ∆ − + ∆ − − + ∆ + ∆ − − ∆ − +=

∆ ∆

∆ + ∆ − ∆=

∆= + ∆ −

= −

Section 1.5 Limits 41


61. 21

2lim1x x−→= −∞

−

Because ( ) 22

1f x

x=

−decreases without bound as x tends to 1 from the left, the limit is .−∞

63. 2

1lim2x x−→−= −∞

+

Because ( ) 12

f xx

=+

decreases without bound as x tends to 2− from the left, the limit is .−∞

65.

2

22

5 6lim4 4x

x xx x→

− +− +

does not exist.

67.

3 2

24

4 4lim 1.8892 7 4x

x x xx x→−

+ + +≈ −

+ −

69. (a) If 50,p =( )25,000 50

$25,000.100 50

C = =−

(b) If 25,000100,000 ,100

pCp

= =−

then

( )4 100

400 580 80%.

p p

pp

− =

=

= ⇒

(c) 100

limp

C−→

= ∞

The cost function increases without bound as p approaches 100%.

71. (a)

(b) When 0.25:x = 2685.06A =

When 1365:x = 2717.91A =

(c) Using the zoom and trace features, 0

lim $2718.28.x

A+→

≈ Because x, the length of the compounding period, is approaching

0, this limit represents the balance with continuous compounding.

x 0 0.5 0.9 0.99 0.999 0.9999 1

f (x) −2 −2.67 −10.53 −100.5 −1000.5 −10,000.5 undefined

x −3 −2.5 −2.1 −2.01 −2.001 −2.0001 −2

f (x) −1 −2 −10 −100 −1000 −10,000 undefined

−4

−10

1

10

−4

−10

0.5

10

−1

−10

4

10

−8

−10

2

10

102000

3000



73. (a)

( )10

lim 1 2.718x

xx

→+ ≈

(b) (c) Domain: ( ) ( )1, 0 0,− ∪ ∞

Range: ( ) ( )1, ,e e∪ ∞

( )10

lim 1 2.718x

xx

→+ ≈

Section 1.6 Continuity

Skills Review

1. ( )( )( )( )

2

2

4 26 86 16 8 2

48

x xx xx x x x

xx

+ ++ +=

− − − +

+=

−

2. ( )( )( )( )

2

2

6 15 69 18 6 3

13

x xx xx x x x

xx

− +− −=

− + − −

+=

−

3. ( )( )( )( )( )( )

( )

22

2 2

2 62 2 124 24 36 4 6 9

2 3 24 3 3

22 3

x xx xx x x x

x xx x

xx

− −− −=

− + − +

− +=

− −

+=

−

4. ( )

( )( )

( )( )( )( )( )( )

23

3 2 2

2

16162 8 2 8

16

4 2

4 44 2

42

x xx xx x x x x x

x x

x x x

x x xx x x

xx

−−=

+ − + −

−=

+ −

+ −=

+ −

−=

−

5.

( )

2 7 0

7 0

07 0 7

x x

x x

xx x

+ =

+ =

=

+ = ⇒ = −

6.

( )( )

2 4 5 0

5 1 0

5 0 51 0 1

x x

x x

x xx x

+ − =

+ − =

+ = ⇒ = −

− = ⇒ =

7.

( )( )

2

23

3 8 4 0

3 2 2 0

3 2 0

2 0 2

x x

x x

x x

x x

+ + =

+ + =

+ = ⇒ = −

+ = ⇒ = −

8.

( )( )( )

3 2

2

5 24 0

5 24 0

3 8 0

03 0 38 0 8

x x x

x x x

x x x

xx xx x

+ − =

+ − =

− + =

=

− = ⇒ =

+ = ⇒ = −

9. ( ) ( ) ( )

( )

2 23

lim 2 3 4 2 3 3 3 4

2 9 9 4

13

xx x

→− + = − +

= − +

=

10. ( ) ( ) ( )

( )

332

lim 3 8 7 3 2 8 2 7

3 8 16 7

24 231

xx x

→−− + = − − − +

= − + +

= − +

= −

x −0.01 −0.001 −0.0001 0 0.0001 0.001 0.01

f (x) 2.732 2.720 2.718 undefined 2.718 2.717 2.705

−2

−2

10

10

Section 1.6 Continuity 43


1. Continuous; The function is a polynomial.

3. Not continuous; The rational function is not defined at 2.x = ±

5. Continuous; The rational function's domain is the entire real line.

7. Not continuous; The rational function is not defined at 3x = or 5.x =

9. Not continuous; The rational function is not defined at 2.x = ±

11. ( )2 1xf x

x−

= is continuous on ( ), 0−∞ and

( )0, ∞ because the domain of f consists of all real

numbers except 0.x = There is a discontinuity at 0x = because ( )0f is not defined and ( )

0limx

f x→

does not exist.

13. ( )2 1

1xf xx

−=

+is continuous on ( ), 1−∞ − and

( )1,− ∞ because the domain of f consists of all real

numbers except 1.x = − There is a discontinuity at

1x = − because ( )1f − is not defined and

( ) ( )1

lim 1 .x

f x f→−

≠ −

15. ( ) 2 2 1f x x x= − + is continuous on ( ),−∞ ∞ because

the domain of f consists of all real numbers.

17. ( ) ( )( )2 1 1 1x xf x

x x x= =

− + −is continuous on

( ), 1 ,−∞ − ( )1, 1 ,− and ( )1, ∞ because the domain of f

consists of all real numbers except 1.x = ± There are discontinuities at 1x = ± because ( )1f and ( )1f − are

not defined and ( )1

limx

f x→

and ( )1

limx

f x→−

do not exist.

19. ( ) 21

1f x

x=

+is continuous on ( ),−∞ ∞ because the

domain of f consists of all real numbers.

21. ( ) ( )( )25 5

9 20 5 4x xf x

x x x x− −

= =− + − −

is continuous

on ( ), 4 ,−∞ ( )4, 5 , and ( )5, ∞ because the domain of

f consists of all real numbers except 4x = and 5.x = There is a discontinuity at 4x = and 5x = because ( )4f and ( )5f are not defined and

( )4

limx

f x→

does not exist and ( ) ( )5

lim 5 .x

f x f→

≠

23. ( ) 2 1f x x= + is continuous on all intervals of the

form ( )1 1 12 2 2, ,c c + where c is an integer. That is, f is

continuous on ( )12, , 0 ,−… ( )1

20, , ( )12, 1 , .… f is not

continuous at all points 12 ,c where c is an integer. There

are discontinuities at ,2cx = where c is an integer,

because ( )2

limx c

f x→

does not exist.

25. ( ) 2

2 3, 1, 1x x

f xx x− + <⎧

= ⎨≥⎩

is continuous on ( ),−∞ ∞ because the domain of f

consists of all real numbers, ( )1f is defined,

( )1

limx

f x→

exists, and ( ) ( )1

lim 1 .x

f x f→

=

27. ( )12 1, 23 , 2

x xf x

x x

⎧ + ≤⎪= ⎨− >⎪⎩

is continuous on ( ], 2−∞ and ( )2, .∞ There is a

discontinuity at 2x = because ( )2

limx

f x→

does not exist.

29. ( ) 11

xf x

x+

=+

is continuous on ( ), 1−∞ − and

( )1,− ∞ because the domain of f consists of all real

numbers except 1.x = − There is a discontinuity at 1x = − because ( )1f − is not defined, and

( )1

limx

f x→−

does not exist.

31. ( ) 1f x x= − is continuous on all intervals

( ), 1 .c c + There are discontinuities at ,x c= where c is

an integer, because ( )limx c

f x→

does not exist.

33. ( ) ( )( ) ( ) 11 ,1

h x f g x f xx

= = − =−

1x >

h is continuous on its entire domain ( )1, .∞

35. Continuous on [ ]1, 5− because ( ) 2 4 5f x x x= − − is a

polynomial.

37. Continuous on [ )1, 2 and ( ]2, 4 because

( ) 12

f xx

=−

has a nonremovable discontinuity at

2.x =



39. ( ) ( )( )2 4 416 4,4 4

x xxf x xx x

+ −−= = = +

− − 4x ≠

f has a removable discontinuity at 4;x = Continuous on ( ), 4−∞ and ( )4, .∞

4 62−2−6

8

6

2

10

x

y

41. ( ) ( )232

11,

x xx xf x xx x

++= = = + 0x ≠

f has a removable discontinuity at 0;x = Continuous on ( ), 0−∞ and ( )0, .∞

21−1−2

4

3

2

x

y

43. ( )2 1, 0

1, 0x x

f xx x

⎧ + <⎪= ⎨− ≥⎪⎩

f has a nonremovable discontinuity at 0;x = Continuous on ( ), 0−∞ and ( )0, .∞

−1

−3

−2

3

2

21−1 3−2−3x

y

45. ( ) 3

2 2lim lim 8

x xf x x

− −→ →= =

( ) 2

2 2lim lim 4

x xf x ax a

+ +→ →= =

So, 8 4a= and 2.a =

47.

From the graph, you can see that ( )2h and ( )1h − are not

defined, so h is not continuous at 2x = and 1.x = −

49.

From the graph, you can see that ( )3

limx

f x→

does not

exist, so f is not continuous at 3.x =

51.

From the graph, you can see that ( )lim 2 ,x c

x x→

− where

c is an integer, does not exist. So f is not continuous at all integers c.

53. ( ) 2 1xf x

x=

+is continuous on ( ), .−∞ ∞

55. ( ) 1 22

f x x= is continuous on all intervals of the form

1, ,2 2c c +⎛ ⎞

⎜ ⎟⎝ ⎠

where c is an integer.

57.

( ) ( )2 1x xx xf xx x

++= = appears to be continuous

on [ ]4, 4 .− But it is not continuous at 0x = (removable

discontinuity). Examining a function analytically can reveal removable discontinuities that are difficult to find just from analyzing its graph.

59. (a) ( ) 47500 1.015 ,tA = 0t ≥

t

A

Years

Am

ount

(in

dol

lars

)

1 2 3

7500

7750

8000

8250

8500

8750

9000

9250

The graph has nonremovable discontinuities at

3 51 14 2 4 4, , , 1, ,t = … (every 3 months).

(b) For 7,t = ( ) 4 77500 1.015 $11,379.17.A ⋅= =

After 7 years your balance is $11,379.17.

−2

3−3

2

−1

−4

7

12

3

−1

−3

3

−4

−3

4

3

Review Exercises for Chapter 1 45


61. 12.80 2.50 , 0, not an integer

12.80 2.50 112.80 2.50 1 , 0, an integer

x x xC x

x x x⎧ + >⎪= − − = ⎨

+ − >⎪⎩

0 5

0

25

C is not continuous at 1, 2, 3, .x = …

63. (a)

Nonremovable discontinuities at 1, 2, 3, 4, 5;t = s is not continuous at 1, 2, 3, 4,t = or 5.

(b) For 5,t = $43,850,78.S =

The salary during the fifth year is $43,850.78.

65. Yes, a linear model is a continuous function. No, actual revenue would probably not be continuous because revenue is usually recorded over larger units of time (hourly, daily, or monthly). In these cases, the revenue may jump between different units of time.

67. 2

30 , 0 10031 0.01 , 100 160015 , 1600

x xP x x x

x x

≤ ≤⎧⎪= − < ≤⎨⎪ >⎩

( )100 100

lim lim 30 3000x x

P x x− −→ →

= =

( ) ( )2

100 100lim lim 31 0.01 3000

x xP x x x

+ +→ →= − =

So, ( )100

lim 3000.x

P x→

=

Because ( )100P is defined, ( )100

limx

P x→

exists, and

( ) ( )100

lim 3000 100 ,x

P x P→

= = the function is

continuous at 100.x =

Review Exercises for Chapter 1

1.

3.

5. Matches (a)

7. Matches (b)

9. ( ) ( )2 2Distance 0 5 0 2

25 4

29

= − + −

= +

=

11. ( ) ( )2 2Distance 1 4 3 6

9 9

3 2

= ⎡− − − ⎤ + −⎣ ⎦

= +

=

13. ( )5 9 6 2Midpoint , 7, 42 2+ +⎛ ⎞= =⎜ ⎟

⎝ ⎠

15. ( )10 6 4 8Midpoint , 8, 62 2

− − +⎛ ⎞= = −⎜ ⎟⎝ ⎠

17. .P R C= − The tallest bars represent revenues. The middle bars represent costs. The bars on the left of each group represent profits because .P R C= −

19. The translated vertices are:

( ) ( )( ) ( )( ) ( )

1 3, 3 4 4, 7

2 3, 4 4 5, 8

5 3, 6 4 8, 10

+ + =

+ + =

+ + =

0

0 6

50,00

x

y

(0.5, −4)

(−1, −2)

−1−2−3 1 2 3−1

−3

−4

−5

1

x

y

(0, 6)

(2, 3)

−2−4 2 4 6−2

2

4

6

8



21. Bar graph for data: Mytilus 105 Gammarus 75 Littorina 66 Arbacia 7 Nassarius 113 Mya 19

23. 4 3y x= −

x

4

3

2

1

−14321−1

y

25. 21y x= −

x−2

−3

−2

2

−1

y

27. 2 3y x= −

x

3

2

1

−14321−1

y

29. 32 1y x= −

x

y

−1−2−3 1 2 3

−3

−4

1

2

31. 2y x=

x

3 421

2

3

4

1

y

33. Let 0.y = Then,

( ) ( )

( ) ( )

3 2

2

0 1 2 1

0 1 1

1.

x x

x x

x

= − + −

= − +

= ±

Let 0.x = Then,

( ) ( )3 20 1 2 0 1

1.

y

y

= − + −

=

x-intercepts: ( ) ( )1, 0 , 1, 0−


35. ( ) ( )

( ) ( )

( ) ( )

2 2 2

2 2 2

2

2

2 2

2 1

1 2 7 1

9 64

73

2 1 73

x y r

r

r

r

x y

− + + =

− − + + =

+ =

=

− + + =

37.

( ) ( )( ) ( )

2 2

2 2

2 2

10 4 7 0

10 25 4 4 7 25 4

5 2 36

x y x y

x x y y

x y

+ + + − =

+ + + + + = + +

+ + + =

Center: ( )5, 2− −

r: 6

−2−2

4

−4

−8

2−4−6−8

y

x

(−5, −2)

39. Solving for y in the second equation yields 1,y x= − and substituting this into the first

equation gives

( )

( )( )

22

2 2

2

1 5

2 1 5

2 2 4 0

2 2 1 0

1, 2.

x x

x x x

x x

x x

x

+ − =

+ − + =

− − =

− + =

= −

The corresponding y-values are 2y = and 1y = so the points of intersection are ( )1, 2− and ( )2, 1 .

0 60

120



41. Solving for y in the second equation yields 2 1,y x= − and substituting this into the first equation gives

2

2

2 1 4

2 5 0

x x

x x

+ − =

+ − =

( )2 4 4 5

22 2 6

21 6.

x− ± − −

=

− ±=

= − ±

The corresponding y-values are 3 2 6y = − +

and 3 2 6,− − so the points of intersection are

( )1 6, 3 2 6− + − + and ( )1 6, 3 2 6 .− − − −

43. (a) 6000 6.5013.90

C xR x

= +

=

(b) 6000 6.5 13.9

6000 7.4810.81, or 811 units

C Rx x

xx

=

+ =

=

≈

45. 3 23 2

x yy x

+ = −

= − −

Slope: 3m = −


−4

−3

−2

2

1

2 31−1−2−3x

y

47. 53y = −

Slope: 0m = (horizontal line)

y-intercept: ( )530, −

−2

−1

2

1

21−1−2x

y

49.

25

2 5 5 05 2 5

1

x yy x

y x

− − − =

= − −

= − −

Slope: 25m = −


−4

−2

4

2

2−4 4x

y

51. 6 0 6Slope7 0 7−

= =−

53. ( )( )

17 3 20Slope10 11 21

− −= =

− −

55. ( ) ( )1 2 3

2 5

y x

y x

− − = − −

= − +

−1

−10

5

10

57. ( ) ( )4 0 1.5

4

y x

y

− − = −

= −

6−6

−6

2



59. (a) ( )76 387 698 8

7 8 69 0

y x

y x

x y

− = ⎡ − − ⎤⎣ ⎦

= +

− + =

(b) 74 2 7 2 ;2

x y y x+ = ⇒ = − + slope 2= −

( )6 2 3

22 0

y x

y xx y

− = − ⎡ − − ⎤⎣ ⎦= −

+ =

(c) The line through ( )0, 0 and ( )3, 6− has slope

6 23

22 0

y xx y

= −−

= −

+ =

(d) 33 2 2 12

x y y x− = ⇒ = −

Slope of perpendicular is 2.3

−

( )26 332 43

2 3 12 0

y x

y x

x y

− = − ⎡ − − ⎤⎣ ⎦

= − +

+ − =

61. ( ) ( )32, 750 , 37, 700

750 700 50 1032 37 5

m −= = = −

− −

(a) ( )750 10 32

10 1070

x p

x p

− = − −

= − +

(b) If 34.50,p = ( )10 34.50 1070 725x = − + = units

(c) If 42.00,p = ( )10 42.00 1070 650x = − + = units

63. Yes, 2 2y x= − + is a function of x.

65. No, 2 214 4y x− = is not a function of x.

67. (a) ( ) ( )1 3 1 4 7f = + =

(b) ( ) ( )1 3 1 4 3 7f x x x+ = + + = +

(c) ( ) ( )2 3 2 4 10 3f x x x+ ∆ = + ∆ + = + ∆

69. Domain: ( ),−∞ ∞

Range: ( ),−∞ ∞

71.

Domain: [ )1,− ∞

Range: [ )0, ∞

73.

Domain: ( ),−∞ ∞

Range: ( ], 3−∞

75. (a) ( ) ( ) ( ) ( )2

2

1 2 1

2

f x g x x x

x x

+ = + + −

= +

(b) ( ) ( ) ( ) ( )2

2

1 2 1

2 2

f x g x x x

x x

− = + − −

= − +

(c) ( ) ( ) ( )( )2

3 2

1 2 1

2 2 1

f x g x x x

x x x

= + −

= − + −

(d) ( )( )

212 1

f x xg x x

+=

−

(e) ( )( ) ( )

( )2

2

2 1

1 2 1

4 4 2

f g x f x

x

x x

= −

= + −

= − +

(f ) ( )( ) ( )( )

2

2

2

1

2 1 1

2 1

g f x g x

x

x

= +

= + −

= +

77. ( ) 32f x x= has an inverse by the horizontal line test.

( )1

32322323

y x

x y

y x

f x x−

=

=

=

=

79. ( ) 2 12f x x= − + does not have an inverse by the

horizontal line test.

81. ( ) ( )2

lim 5 3 5 2 3 7x

x→

− = − = 6−6

−1

7

−1

4−2

3

−3

−6 6

5



83. ( )( ) ( ) ( )2

lim 5 3 2 3 5 2 3 2 2 3 49x

x x→

− + = ⎡ − ⎤⎡ + ⎤ =⎣ ⎦⎣ ⎦

85. ( )22

3

3 11 10lim3 3t

tt→

++= =

87. 1

1 1 1lim 22 1 2t

tt→

+ += = −

− −

89. ( )( )22 2

2

2 2lim lim4 2 2

1lim2

14

x x

x

x xx x x

x

→− →−

→−

+ +=

− + −

=−

= −

91. 2

0 0

1 1lim limx x

xxx x+ +→ →

−⎛ ⎞− = = −∞⎜ ⎟⎝ ⎠

93. ( ) ( )

( )

( )

0 0

0

1 2 1 1 2lim lim

2


x x

x

x xx x x

xx x

→ →

→

⎡ − ⎤ − − −⎣ ⎦ =−

−=

−

95.

( )

( )

[ ]

0 0

0

0

1 1 1 1 1 12 24 4 4 4lim lim1 1 1 1

24 244 4

lim1 1 4 4

241lim

1 1 4 1624

116

t t

t

t

t t tt t

t tt

t tt

tt

→ →

→

→

⎛ ⎞− + −⎜ ⎟+ + +⎜ ⎟⋅ =⎡ ⎤⎜ ⎟+ +⎜ ⎟ ⎢ ⎥+⎝ ⎠ +⎣ ⎦

− +=

⎡ ⎤+ ⎡ + ⎤⎣ ⎦⎢ ⎥+⎣ ⎦−

=⎡ ⎤

+ +⎢ ⎥+⎣ ⎦

= −

97. ( ) ( ) ( ) ( ) ( )

( ) ( )

( )

3 2 33 3 2 3

0 0

2 32

0

220

2

3 3lim lim

3 3lim

lim 3 3 1

3 1

x x

x

x

x x x x x x x x x x x x x x x xx x

x x x x x xx

x x x x

x

∆ → ∆ →

∆ →

∆ →

+ ∆ − + ∆ − − + ∆ + ∆ + ∆ − − ∆ − +=

∆ ∆

∆ + ∆ + ∆ − ∆=

∆

⎡ ⎤= + ∆ + ∆ −⎣ ⎦

= −

99.

1

2 1 3 1lim 0.57741 3x

xx+→

+ −= ≈

−

101. The statement is false since 0

lim 1.x

xx−→

= −

103. The statement is false since 0

limx

x−→

is undefined.

105. The statement is false since ( )2 2

lim lim 0 0.x x

f x+ +→ →

= =

107. ( )( )2

14

f xx

=+

is continuous on the intervals

( ), 4−∞ − and ( )4,− ∞ because the domain of f consists

of all real numbers except 4.x = − There is a

discontinuity at 4x = − because ( )4f is not defined.

109. ( ) 31

f xx

=+

is continuous on the intervals

( ), 1−∞ − and ( )1,− ∞ because the domain of f consists

of all real numbers except 1.x = − There is a discontinuity at 1x = − because ( )1f − is not defined.

x 1.1 1.01 1.001 1.0001

f(x) 0.5680 0.5764 0.5773 0.5773



111. ( ) 3f x x= + is continuous on all intervals of the

form ( ), 1 ,c c + where c is an integer. There are

discontinuities at all integer values c because ( )lim

x cf x

→does not exist.

113. ( ), 0

1, 0x x

f xx x

≤⎧= ⎨

+ >⎩ is continuous on the intervals

( ), 0−∞ and ( )0, .∞ There is a discontinuity at

0x = because ( )0

limx

f x→

does not exist.

115. ( ) ( )3 3

lim lim 1 2x x

f x x− −→ →

= − + = −

( ) ( )3 3

lim lim 8 3 8x x

f x ax a+ +→ →

= − = −

So, 2 3 8a− = − and 2.a =

117. (a)

The function is continuous for all 0x > except 25,x = 100,x = and 500.x = There are discontinuities at these values because ( )

25lim ,

xC x

→( )

100lim ,

xC x

→and ( )

500lim

xC x

→do not exist.

(b) ( ) ( )100 0.10 100 10C = =

The cost of making 100 copies is $10.

119. ( )1 0.1 , 0, not an integer

1 0.1 11 0.1 1 , 0, an integer

t t tC t t

t t t⎧ + >⎪= − − = ⎨

+ − >⎪⎩

0 10

0

2

The function is continuous for all noninteger values of 0.t >

121. (a)

(b)

(c) When 20,t = 15,007.9.D =

In 2010, the national debt will be about $15,007.9 billion.

2 150

9000

t 2 3 4 5 6 7 8

D (actual) 4001.8 4351.0 4643.3 4920.6 5181.5 5369.2 5478.2

D (model) 3937.0 4391.3 4727.4 4971.0 5147.8 5283.6 5404.0

t 9 10 11 12 13 14 15

D (actual) 5605.5 5628.7 5769.9 6198.4 6760.0 7354.7 7905.3

D (model) 5534.7 5701.5 5930.1 6246.1 6675.3 7243.3 7976.0

0 8000

40

Chapter Test Solutions for Chapter 1 51


Chapter Test Solutions

1. (a) ( ) ( )2 21 4 1 4

25 25

5 2

d = + + − −

= +

=

(b) 1 4 1 4 3 3Midpoint , ,2 2 2 2− − +⎛ ⎞ ⎛ ⎞= = −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(c) 4 1 14 1

m += = −

− −

2. (a) ( )2

25 0 2 22

25 04

52

d ⎛ ⎞= − + −⎜ ⎟⎝ ⎠

= +

=

(b) 52 0 2 2 5Midpoint , , 2

2 2 4

⎛ ⎞+ + ⎛ ⎞⎜ ⎟= = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

(c) 52

2 2 00

m −= =

−

3. (a) ( ) ( )2 23 2 2 2 1

8 13

d = − + −

= +

=

(b) 3 2 2 2 1 3Midpoint , 2 2,2 2 2

⎛ ⎞+ + ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

(c) 1 22 3 2

1 22 2 2

24

m −=

−

−= ⋅

−

=

4.

( ) ( )( ) ( )

2 2

2 2

2 2

4 2 4 0

4 4 2 1 4 4 1

2 1 9

x y x y

x x y y

x y

+ − − − =

− + + − + = + +

− + − =

5. 65 2.1 43 1.94 22

5.5

x xxx

− = +

− = −

=

Equilibrium point ( ) ( ), 5.5, 5500x p =

The equilibrium point occurs when the demand and supply are each 5500 units.

6. 15m =

When ( )150: 0 2 2x y= = − = −


x

y

−1−2−3 1 2 3

−3

−1

−4

1

2

(0, −2)

7. The line is vertical, so its slope is undefined, and it has no y-intercept.

x

y

3

2

1

−3

−2

−1−1−2−3 321

8. 2.5 6.252.5

y xm= − +

= −

When ( )0: 2.5 0 6.25 6.25x y= = − + =

y-intercept: ( )0, 6.25

x

y

−2−4−6−8 2 4 6 8

−4

−6

2

4

6(0, 6.25)

x

y

−2 2 4 6

−2

4

6

(2, 1)



9. (a)

(b) Domain: ( ),−∞ ∞

Range: ( ),−∞ ∞

(c)

(d) The function is one-to-one.

10. (a)

(b) Domain: ( ),−∞ ∞

Range: )94,⎡− ∞⎣

(c)

(d) The function is not one-to-one.

11. (a)

(b) Domain: ( ),−∞ ∞

Range: [ )4,− ∞

(c)

(d) The function is not one-to-one.

12.

( )1

314 2

314 2

4 64 6

y xx y

y x

f x x−

= +

= +

= −

= −

( )( ) ( )( )

1 314 2

314 24 6

6 6

f f x f x

x

xx

− = −

= − +

= − +

=

( )( ) ( )( )

1 1

314 2

3 32 2

4 6

4 6

f f x f x

x

x

x

− −= +

= + −

= + −

=

13.

( )

3

3

3

3

1 3

813 3

813 3

8 3

8 3

8 3

y x

x y

x y

y x

f x x−

= −

= −

= −

= − +

= − +

( )( ) ( )( )

1 3

33

3 3

813 3

813 38 3

8 8

f f x f x

x

xx

− = − +

= − − +

= + −

=

( )( ) ( )( )( )

31 1

33 813 3

813 38 83 3

8 3

8 3

8 3

f f x f x

x

x

x

x

− −= −

= − − +

= − − +

= − + +

=

14. 0 0

5 0 5lim lim 15 0 5x x

xx→ →

+ += = −

− −

15.

5

5

5lim55lim5

x

x

xxxx

−→

+→

+= −∞

−+

= ∞−

( )5

limx

f x→

does not exist.

x −3 −2 3

f (x) −1 1 11

x −3 −2 3

f (x) 10 4 4

x −3 −2 3

f (x) −1 −2 −1 x 4.9 4.99 4.999 5.001 5.01 5.1

f (x) −99 −999 −9999 10,001 1001 101

x

y

−4−6−8 2 4 6 8

−4

−2

−6

6

8

10

(0, 5)

(−2.5, 0)

x

y

4

3

1

2

−4

−3

−2−3−4 431

(−1, 0) (2, 0)

(0, −2)

x

y

1

2

−4

−5

−6

−2

−2 −1−4 421

(−4, 0) (4, 0)

(0, −4)

Chapter Test Solutions for Chapter 1 53


16. ( )( )( )( )

2

23 3

3

1 32 3lim lim4 3 1 3

1lim1

2

x x

x

x xx xx x x x

xx

→− →−

→−

− ++ −=

+ + + +

−=

+=

17.

0

9 3lim 0.16667x

xx→

+ −≈

18. ( )2 16

4xf xx−

=−

is continuous on the intervals ( ), 4−∞

and ( )4, ∞ because the domain of f consists of all real

numbers except 4.x = There is a discontinuity at 4x = because ( )4f is not defined.

19. ( ) 5f x x= − is continuous on the interval

( )5, ∞ because the domain of f consists of all 5.x >

20. ( ) ( )

( ) ( )1 1

2

1 1

lim lim 1 0

lim lim 0x x

x x

f x x

f x x x

− −→ →

+ +→ →

= − =

= − =

So, ( )1

lim 0.x

f x→

=

Because ( )1f is defined, ( )1

limx

f x→

exists, and

( ) ( )1

lim 1 ,x

f x f→

= the function is continuous on the

interval ( ), .−∞ ∞

21. (a)

The model fits the actual data very well. (b) When 9: 2071.14t y= =

In 2009, the number of farms will be about 2,071,140.

x −0.01 −0.001 −0.0001

f (x) 0.16671 0.16667 0.16667

x 0.0001 0.001 0.01

f (x) 0.16667 0.16666 0.16662 t 0 1 2

y (actual) 2167 2149 2135

y (model) 2166 2151 2137

t 3 4 5

y (actual) 2127 2113 2101

y (model) 2124 2113 2103



Practice Test for Chapter 1

1. Find the distance between ( )3, 7 and ( )4, 2 .−

2. Find the midpoint of the line segment joining ( )0, 5 and ( )2, 1 .

3. Determine whether the points ( )0, 3 ,− ( )2, 5 , and ( )3, 15− − are collinear.

4. Find x so that the distance between ( )0, 3 and ( ), 5x is 7.

5. Sketch the graph of 24 .y x= −

6. Sketch the graph of 2.y x= −

7. Sketch the graph of 3 .y x= −

8. Write the equation of the circle in standard form and sketch its graph.

2 2 8 2 8 0x y x y+ − + + =

9. Find the points of intersection of the graphs of 2 2 25x y+ = and 2 10.x y− =

10. Find the general equation of the line passing through the points ( )7, 4 and ( )6, 2 .−

11. Find the general equation of the line passing through the point ( )2, 1− − with a slope of 23.m =

12. Find the general equation of the line passing through the point ( )6, 8− with undefined slope.

13. Find the general equation of the line passing through the point ( )0, 3 and perpendicular to the line given by 2 5 7.x y− =

14. Given ( ) 2 5,f x x= − find the following.

(a) ( )3f

(b) ( )6f −

(c) ( )5f x −

(d) ( )f x x+ ∆

15. Find the domain and range of ( ) 3 .f x x= −

16. Given ( ) 2 3f x x= + and ( ) 2 1,g x x= − find the following.

(a) ( )( )f g x

(b) ( )( )g f x

17. Given ( ) 3 6,f x x= + find ( )1 .f x−

18. Find ( )4

lim 2 5 .x

x→−

−

19. Find 2

6

36lim .6x

xx→

−−

Practice Test for Chapter 1 55


20. Find 1

1lim .

1x

xx→−

++

21. Find 0

5 5lim .x

xx→

+ −

22. Find ( )1

lim ,x

f x→

where ( ) 2

2 3, 1.

4, 1x x

f xx x

+ ≤⎧= ⎨

+ >⎩

23. Find the discontinuities of ( ) 28 .64

xf xx

−=

−Which are removable?

24. Find the discontinuities of ( ) 3.

3x

f xx−

=−

Which are removable?

25. Sketch the graph of ( )2 5 6.

3x xf x

x− +

=−

Graphing Calculator Required

26. Solve the equation for y and graph the resulting two equations on the same set of coordinate axes.

2 2 6 5 0x y x+ + + =

27. Use a graphing calculator to graph ( )2 9

3xf xx−

=−

and find ( )3

lim .x

f x→

Is the graph displayed correctly at 3?x =

chapter 1 functions, graphs, and...

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