SOLU Precalculo, 7ma Edicion - Michael Sullivan

7/25/2019 SOLU Precalculo, 7ma Edicion - Michael Sullivan

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2/827

Chapter 1 Graphs

2

23. d(P1 ,P2 ) = (6 4)2 + 4 (3)( )2 = 22 + 72 = 4 + 49 = 53

25. d(P1,P2) = 2.3 0.2( )( )2

+ 1.10.3( )2 = 2.5( )2 + 0.8( )2 = 6.25 + 0.64 = 6.89

27. d(P1 ,P2 ) = (0 a)2 + (0 b)2 = a2 + b2

29. A = 2,5( ), B = 1, 3( ), C= 1,0( )d(A,B) = 1 (2)( )2 + (35)2 = 32 + (2)2

= 9+ 4 = 13

d(B,C) = 11( )2 + (0 3)2 = (2)2 + (3)2

= 4 + 9 = 13

d(A,C) = 1 (2)( )2

+ (0 5)2 = 12 + (5)2

= 1+ 25 = 26

Verifying that ? ABCis a right triangle by the Pythagorean Theorem:

d(A,B)[ ]2 + d(B,C)[ ]2 = d(A,C)[ ]2

13( )2 + 13( )2 = 26( )213+13 =26 26 = 26

The area of a triangle is A =1

2

bh. Here,

A = 12 d(A,B)[ ] d(B,C)[ ]= 1

2 13 13 = 1

213= 13

2square units

31. A = 5,3( ), B = 6,0( ), C= 5,5( )d(A,B) = 6 (5)( )2 + (0 3)2 = 112 + (3)2

= 121+ 9 = 130

d(B,C) = 5 6( )2 + (5 0)2 = (1)2 + 52

= 1+ 25 = 26

d(A,C) = 5 (5)( )2 + (5 3)2 = 102 + 22

= 100 + 4 = 104


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Section 1.1 Rectangular Coordinates

3

Verifying that ?ABCis a right triangle by the Pythagorean Theorem:

d(A,C)[ ]2 + d(B,C)[ ]2 = d(A,B)[ ]2

104( )2

+ 26( )2

= 130( )2

104 + 26 =130 130 =130


2 bh. Here,

A =1

2 d(A,C)[ ] d(B,C)[ ]= 1

2 104 26 =

1

2 2704 =

1

2 52 = 26 square units

33. A = 4,3( ), B = 0,3( ), C= 4,2( )d(A,B) = (0 4)2 + 3 (3)( )2 = (4)2 + 02

= 16 + 0 = 16 = 4

d(B,C) = 4 0( )2 + 2 (3)( )2 = 42 + 52

= 16 + 25 = 41

d(A,C) = (4 4)2 + 2 (3)( )2 = 02 + 52

= 0+ 25 = 25 = 5

Verifying that ? ABCis a right triangle by the Pythagorean Theorem:

d(A,B)[ ]2 + d(A,C)[ ]2 = d(B,C)[ ]2

4 2 + 52 = 41( )2 16 + 25 = 41 41= 41


2 bh. Here,

A =1

2 d(A,B)[ ] d(A,C)[ ]= 1

2 4 5 =10 square units

35. All points having an x-coordinate of 2 are of the form (2, y). Those which are 5 units from

(2, 1) are:

2 (2)( )2 + y (1)( )2 = 5 42 + (y +1)2 = 5Squaring both sides : 4

2 + (y +1)2 = 25

16+ y2 + 2y +1= 25

y2

+ 2y 8 = 0(y + 4)(y 2) = 0 y = 4 or y = 2

Therefore, the points are (2, 4) and (2, 2).

37. All points on thex-axis are of the form (x, 0). Those which are 5 units from (4, 3) are:

x 4( )2 + 0 (3)( )2 = 5 (x 4)2 + 32 = 5Squaring both sides : (x 4)2 + 9 = 25


4/827

Chapter 1 Graphs

4

x2 8x +16+ 9 = 25

x 2 8x = 0x(x 8) = 0 x = 0 or x = 8

Therefore, the points are (0, 0) and (8, 0).

39. The coordinates of the midpoint are:

(x,y)= x

1+x

2

2,y

1+y

2

2

=

5 + 32

, 4 + 2

2

=

8

2,22

= (4,1)


(x,y)= x

1+x

2

2,y

1+y

2

2

=

3+ 62

,2 + 0

2

=

3

2,2

2

=

3

2,1


(x,y)= x

1+x

2

2

,y

1+y

2

2

=4 + 6

2

,3+1

2

=10

2

,2

2

= (5,1)


(x,y)= x

1+x

2

2,y

1+y

2

2

=

0.2+ 2.32

,0.3+1.1

2

=

2.1

2,1.4

2

= (1.05, 0.7)


(x,y)= x

1+x

2

2,y

1+ y

2

2

=

a + 02

,b+ 0

2

=

a

2,

b

2

49. The midpoint of ABis: D=0+ 0

2 ,0 + 6

2

= 0, 3( )

The midpoint of ACis: E=0 + 4

2,0 + 4

2

= 2, 2( )

The midpoint of BCis: F=0+ 4

2,6+ 4

2

= 2, 5( )

d(C,D)= 0 4( )2 + (3 4) 2 = (4)2 + (1)2 = 16 +1 = 17

d(B,E)= 2 0( )2 + (26)2 = 22 + (4)2 = 4 +16 = 20 = 2 5

d(A,F)= (20)2 + (50)2 = 22 + 52 = 4 + 25 = 29

51. d(P1,P2) = (4 2)2 + (11)2 = (6)2 + 02 = 36 = 6

d(P2,P3) = 4 (4)( )2 + (31)2 = 02 + (4)2 = 16 = 4

d(P1,P3) = (4 2)2 + (31)2 = (6)2 + (4)2 = 36+16 = 52 = 2 13

Since d(P1,P2)[ ]2 + d(P2 ,P3)[ ]

2 = d(P1,P3)[ ]2, the triangle is a right triangle.


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Section 1.1 Rectangular Coordinates

5

53. d(P1,P2) = 0 (2)( )2 + 7 (1)( )2 = 22 + 82 = 4 + 64 = 68 = 2 17

d(P2,P3) = 30( )2 + (2 7)2 = 32 + (5)2 = 9+ 25 = 34

d(P1,P3) = 3 (2)( )2 + 2 (1)( )2 = 52 + 32 = 25 + 9 = 34

Since d(P2,P3) =d(P1,P3) , the triangle is isosceles.

Since d(P1,P3)[ ]2 + d(P2 ,P3)[ ]2 = d(P1,P2)[ ]2, the triangle is also a right triangle.Therefore, the triangle is an isosceles right triangle.

55. P1 = 1, 3( ); P2 = 5,15( )d P1,P2( )= 5 1( )

2+ 15 3( )2

= 4( )2 + 12( )2

= 16+144

= 160 = 4 10

57. P1 = 4,6( ); P2 = 4,8( )

d P1,P2( )= 4 4( )( )2 + 8 6( )2

= 8( )2 + 14( )2

= 64 +196

= 260 = 2 65

59. Plot the vertices of the square at(0, 0), (0, s), (s, s), and (s, 0). Find the midpoints of the diagonals.

M1 =0 +s

2,0+s

2

=

s

2,s

2

M2 =0 +s

2,s + 0

2

=

s

2,s

2

Since the coordinates of the midpoints

are the same, the diagonals of a square

intersect at their midpoints.

61. Using the Pythagorean Theorem:

902 + 902 =d2

8100 + 8100 =d2

16,200 =d2

d= 16,200 = 90 2 127.28 feet

90

9090

90

d

(0,s)

(s,0)

(0,0)

(s,s)

x

y


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Chapter 1 Graphs

6

63. (a) First: (90, 0), Second: (90, 90)

Third: (0, 90)

0,0

(0,90)

90,0

(90,90)

x

y

(b) Using the distance formula:

d= (310 90)2 + (15 90)2

= 2202 + (75)2

= 54,025 232.4 feet

(c) Using the distance formula:

d= (300 0)2 + (300 90)2

= 3002 + 2102

= 134,100 366.2 feet

65. The Intrepid heading east moves a distance 30tafter t

hours. The truck heading south moves a distance 40tafter thours. Their distance apart after thours is:

d= (30t)2 + (40 t)2

= 900t2

+1600t2

= 2500t2

= 50t

d40t

30t


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Chapter1

7

Graphs

1.2 Graphs of Equations; Circles

1. add, 4 3. intercepts

5. (3,4) 7. True

9. 11.

13. 15.

17.

19. (a) (1, 0), (1, 0) (b) symmetric with respect to thex-axis,y-axis, and origin


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Chapter 1 Graphs

8

21. (a) 2

, 0

,

2

, 0

, (0,1) (b) symmetric with respect to they-axis

23. (a) (0, 0) (b) symmetric with respect to thex-axis

25. (a) (1, 0) (b) not symmetric with respect tox-axis,

y-axis, or origin

27. (a) (1, 0), (1, 0), (0, 1) (b) symmetric with respect to they-axis

29. (a) none (b) symmetric with respect to the origin

31. y =x 4 x 0 = 04 00 = 0

1=14 11 0

0 = (1)4 1

0 1 1

(0, 0) is on the graph of the equation.

33. y 2 =x 2 + 932 = 02 + 9

9 = 9

02 = 32 + 9

0 18

02 = (3)2 + 9

0 18

(0, 3) is on the graph of the equation.

35. x 2 +y 2 = 4 02 + 22 = 4

4 = 4

(2)2 + 22 = 4

8 4 2( )

2

+ 2( )2 = 44 = 4

(0, 2) and 2 , 2( )are on the graph of the equation.37. x

2 = y y - intercept : Letx = 0, then 0 2 =y y = 0 (0,0)

x - intercept : Lety = 0, then x2 = 0 x = 0 (0,0)

Test for symmetry:

x - axis : Replace yby y : x2 = y, which is not equivalent to x 2 =y .

y - axis : Replace xby x : (x)2 =yorx2 =y, which is equivalent to x 2 =y .

Origin : Replacexby x and yby y : (x)2 = yorx2 = y,

which is not equivalent to x 2 =y.

Therefore, the graph is symmetric with respect to the y - axis .

39. y = 3 y - intercept : Let = 0, then y = 3 0 = 0 (0,0)x - intercept : Lety = 0, then 3x = 0 x = 0 (0,0)


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Section 1.2 Graphs of Equations; Circles

9

Test for symmetry:- axis : Replace yby y : y = 3 , which is not equivalent to y = 3 .

y - axis : Replace xby x : y = 3(x) ory = 3x,

which is not equivalent to y = 3x.Origin : Replacexby x andyby y : y = 3(x) ory = 3x,

which is equivalent to y = 3x.

Therefore, the graph is symmetric with respect to the origin.

41. x2 +y 9 = 0

y - intercept : Let = 0, then 0 +y 9 = 0 y = 9 (0,9)

x - intercept : Lety = 0, then x2 9 = 0x = 3 (3,0), 3,0( )Test for symmetry:

x - axis : Replace yby y : x2 + y( ) 9 = 0 orx2 y 9 = 0,which is not equivalent to x2 +y 9 = 0.

y - axis : Replace xby x : x( )2

+y 9 = 0 orx2

+y 9 = 0,which is equivalent to x

2 +y 9 = 0.

Origin : Replacexby x andyby y : x( )2 + y( )9 = 0 orx 2 y 9 = 0,which is not equivalent to x

2 +y 9 = 0.

Therefore, the graph is symmetric with respect to they-axis.

43. 9x2 + 4y2 = 36

y - intercept : Letx = 0, then 4y2 = 36 y 2 = 9 y = 3 (0,3), 0,3( )x - intercept : Lety = 0, then 9x 2 = 36x 2 = 4 x = 2 (2,0),(2,0)

Test for symmetry:

x - axis : Replace yby y : 9x2 + 4 y( )2 = 36 or 9x 2 + 4y 2 = 36,which is equivalent to 9x 2 + 4y 2 = 36.

y - axis : Replace xby x : 9 x( )2 + 4y2 = 36 or 9x 2 + 4y 2 = 36,which is equivalent to 9x 2 + 4y 2 = 36.

Origin : Replacexby x andyby y : 9 x( )2 + 4 y( )2 = 36 or 9x2 + 4y2 = 36,which is equivalent to 9x

2 + 4y 2 = 36.

Therefore, the graph is symmetric with respect to thex-axis, they-axis, and the origin.

45. y =x 3 27y - intercept : Letx = 0, then y = 03 27 y = 27 (0,27)

x - intercept : Lety = 0, then 0 =x 3 27 x 3 = 27 x = 3 (3,0)


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Chapter 1 Graphs

10

Test for symmetry:

x - axis : Replace yby y : y =x 3 27, which is not equivalent to y =x 3 27.

y - axis : Replace xby x : y = (x)3 27 ory = x3 27,

which is not equivalent to y =x 3 27.

Origin : Replacexby x andyby y : y = (x)3 27 or

y =x 3 + 27, which is not equivalent to y =x 3 27.

Therefore, the graph is not symmetric with respect to thex-axis, they-axis, or the origin.

47. y =x 2 3x 4y - intercept : Letx = 0, then y = 02 3(0) 4 y = 4 (0,4)

x - intercept : Lety = 0, then 0 =x 2 3x 4 (x 4)(x +1)= 0x = 4,x = 1 (4,0), (1,0)

Test for symmetry:

x - axis : Replace yby y : y =x 2 3x 4, which is not

equivalent to y =x 2 3x 4.

y - axis : Replace xby x : y = (x)2 3(x) 4 or y =x 2 + 3x 4,

which is not equivalent to y =x 2 3x 4.

Origin : Replacexby x andyby y : y = (x)2 3(x) 4 or

y = x 2 3x + 4, which is not equivalent to y =x 2 3x 4.


49. y =32 + 9

y - intercept : Letx = 0, then y =0

0 + 9= 0 (0,0)

x - intercept : Lety = 0, then 0 =3x2 + 9

3x = 0x = 0 (0,0)

Test for symmetry:

x - axis : Replace yby y : y =3

x2 + 9

, which is not

equivalent to y = 3xx2 + 9

.

y - axis : Replace xby x : y =3 x( )x( )2 + 9

ory =3x

x2 + 9

,

which is not equivalent to y =3x

x2 + 9.

Origin : Replacexby x andyby y : y =3x

x( )2 + 9or

y =3x2 + 9

, which is equivalent to y =3x2 + 9

.



11/827


11

51. y = x3

2 9

y - intercept : Letx = 0, then y =0

9= 0 0,0( )

x - intercept : Lety = 0, then 0 =x 32

9x 3 = 0 x = 0 0,0( )

Test for symmetry:

x - axis : Replace yby y : y = x3

2 9, which is not equivalent to y = x

3

2 9.

y - axis : Replace xby x : y = x( )3

x( )2 9ory =

x 3

x 2 9,

which is not equivalent to y =x 32 9

.

Origin : Replace xby x and yby y : y = x( )3

x( )2

9or

y = x 3

2 9, which is equivalent to y =

x 32 9

.


53. y =x3 55. y = x

57. y = 3 + 5 2 = 3a + 5 3a = 3 a = 1

59. 2 + 3y = 6

2a+ 3b = 6b = 22

3a


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Chapter 1 Graphs

12

61. Center = (2, 1)Radius = distance from (0, 1) to (2,1)

= (20)2 + (11)2

= 4 = 2

(x 2)2 + (y 1)2 = 4

63. Center = midpoint of (1,2) and (4,2)

=1+ 4

2,2 + 2

2

=

5

2, 2

Radius = distance from5

2,2

to (4,2)

= 4 52

2

+ (22)2

=9

4=

3

2

x 5

2

2

+ (y 2)2 =9

4

65. (x h)2 + (y k)2 =r2

(x 0)2

+ (y 0)2

= 22

x2 +y2 = 4

General form: x 2 +y 2 4 = 0

67. (x h)2 + (y k)2 =r2 (x 1)2 + (y 1( ))2 =12

x 1( )2 + (y +1)2 =1 General form:

x 2 2x +1+y 2 + 2y +1=1

x2 +y 2 2x + 2y +1= 0

69. (x h)2 + (y k)2 =r2 (x 0)2 + (y 2)2 = 22

x2 + (y 2)2 = 4

General form:x

2 +y2 4y + 4 = 4x

2 +y2 4y = 0


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13

71. (x h)2 + (y k)2 =r2 (x 4)2 + (y (3))2 = 52

(x 4)2 + (y + 3)2 = 25

General form:

x2 8x +16 +y2 + 6y + 9 = 25

x2

+y2

8x + 6y = 0

73. (x h)2 + (y k)2 =r2

x 3( )( )2 + y 6( )( )2 = 62

x + 3( )2 + y + 6( )2 = 36 General form:

x 2 + 6x + 9+y2 +12y + 36 = 36

x2 +y 2 + 6x +12y + 9 = 0

75. (x h)2

+ (y k)2

=r2

x 0( )2 + y 3( )( )2 = 32

x2 + y + 3( )2 = 9 General form:

x 2 +y 2 + 6y + 9 = 9

x2 +y2 + 6y = 0

77. x2 +y2 = 4

x 2 +y 2 = 22 (a) Center : (0,0); Radius = 2(b)

(c) x-intercepts: y = 0x 2 + 0 = 4

x = 2

2,0( ), 2,0( )y-intercepts: = 0

0 +y 2 = 4y = 2

0,2( ), 0,2( )


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15/827


15

83. x 2 +y 2 2x + 4y 4 = 0x

2 2x +y 2 + 4y = 4

(x2 2x +1)+ (y 2 + 4y + 4) = 4 +1+ 4

(x 1)2 + (y + 2)2 = 32

(a) Center: (1,2); Radius = 3

(b)

(c) x- intercepts: y = 0(x 1)2 + 4 = 9

x 1( )2 = 5x +1= 5

x = 5 1,x = 5 1 5 1,0( ), 5 1,0(

y-intercepts: = 01+ (y + 2)2 = 9

y + 2( )2 = 8y + 2 = 8

y = 8 2,y = 8 2

0, 8 2( ), 0, 8 2( )

85. x 2 +y 2 x + 2y +1= 0x 2 x +y 2 + 2y = 1

x 2 x +1

4

+ (y 2 + 2y +1)= 1+

1

4+1

x 1

2

2

+ (y +1)2 =1

2

2

(a) Center:1

2,1

; Radius =

1

2

(b)

(c) x- intercepts: y = 0

x 1

2

2

+1=1

4

x 1

2

2

= 3

4

no solution nox - interceptsy-intercepts: = 01

4

+ (y +1)2 =1

4y +1( )2 = 0y +1= 0

y = 1

0,1( )


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17/827


17

The equation of the orbit is

x +1( )2 + y + 2( )2 = 64.6( )2

x 2 +y 2 + 2x + 4y 4168.16 = 0

103105. Answers will vary


18/827

Chapter 1

18

Graphs

1.3 Lines

1. undefined, 0 3. y= b, y-intercept

5. False

7. m1=m2, y-intercepts, m1 m2= 1 9. 1

2

11. (a) Slope= 1 02 0=

12

(b) If increases by 2 units, y will increase by 1 unit.

13. (a) Slope=1 2

1 (2)=

1

3

(b) If increases by 3 units, y will decrease by 1 unit.

15. ( 1,y1 ) ( 2,y2 ) (2,3) (4,0)

Slope= y2y1x2x1

=0 34 2

= 3

2

17. ( 1,y1 ) ( 2,y2 ) (2,3) (2,1)

Slope= y2y1x2x1

= 1 32 (2)

=24

= 12

19. ( 1,y1 ) ( 2,y2 ) (3,1) (2,1)

Slope= y2 y1

x2 x1=1 (1)

2 (3) = 0

5

= 0


19/827

Section 1.3 Lines

19

21. ( 1,y1 ) ( 2,y2 ) (1,2) (1,2)

Slope= y2y1x2x1

= 2 21 (1)

= 4

0

Slope is undefined.

23.

25. 27.

29. 31. (0,0) and (2,1) are points on the line.

Slope = 1 02 0

= 12

y - intercept is 0; using y= mx + b:

y=1

2x + 02y= x

0= x 2y

x 2y= 0 or y= 12

x

33. (2,2) and (1,1) are points on the line.

Slope= 1 21 (2)

=13

= 13

Using y y1= m(x x1)

y1= 13 x1( )y1=

1

3x+

1

3

y= 1

3x+

4

3

x+ 3y=4 or y= 1

3x +

4

3

35. y y1=m( 1), m= 2 y 3= 2( 3)y 3= 2x 6

y= 2x 32x y= 3 or y= 2x 3


20/827

Chapter 1 Graphs

20

37. y y1=m(x x1), m = 1

2

y 2 = 1

2(x1)

y 2 = 1

2 x +1

2

y= 1

2x +

5

2

x+ 2y= 5 or y= 1

2x +

5

2

39. Slope = 3; containing (2,3)y y1= m( 1 )y 3= 3(x ( 2))y 3= 3x + 6

y= 3x + 93x y= 9 or y= 3x + 9

41. Slope = 2

3; containing (1,1)

yy1=m( 1)

y (1) = 2

3 (x 1)

y + 1= 2

3x+

2

3

y = 2

3x

1

3

2x + 3y = 1 or y= 2

3x

1

3

43. Containing (1,3) and (1,2)

m= 2 31 1

= 1 2

= 12

y y1=m( 1 )

y 3= 12

(x1)

y 3= 12

x 12

y= 12

x + 52

x 2y= 5 or y= 12

x + 52

45. Slope = 3;y-intercept =3y = mx + by = 3x + 3

3x + y = 3 or y = 3x + 3

47. x-intercept = 2; y-intercept = 1Points are (2,0) and (0,1)

m=1 00 2

=1 2

= 12

y= mx + b

y=12

x 1

x 2y= 2 or y= 12

x 1

49. Slope undefined; passing through (2,4)

This is a vertical line.

= 2No slope-intercept form.

51. Horizontal; containing the point (3,2)

slope = 0y= 2

53. Parallel to y= 2 ; Slope = 2Containing (1,2)y y1=m( 1)y 2= 2(x (1))y 2= 2x + 2

y= 2x + 42x y= 4 or y= 2x+ 4

55. Parallel to 2 y= 2 ; Slope = 2Containing (0,0)y y1=m( 1)y 0= 2(x 0)

y= 2x2x y= 0 or y= 2x


21/827

Section 1.3 Lines

21

57. Parallel to = 5Containing (4,2)This is a vertical line.

= 4

No slope - intercept form.

59. Perpendicular to y=1

2x + 4

Slope of perpendicular = 2

Containing (1,2)y y1= m( 1)

y (2)= 2(x 1)y + 2= 2x + 2

y= 2x2x +y= 0 or y = 2x

61. Perpendicular to 2 + y= 2 2 +y = 2y= 2 + 2

Slope of perpendicular =1

2

Containing (3,0)

y y1= m( 1)

y 0=1

2(x (3))y =

1

2x +

3

2

x 2y= 3 or y =1

2x +

3

2

63. Perpendicular to = 8Slope of perpendicular = 0horizontal lineContaining the point (3,4)y= 4

65. y= 2 + 3Slope = 2y-intercept = 3

67.1

2y= x 1

y= 2 2 Slope = 2

y-intercept = 2


22/827

Chapter 1 Graphs

22

69. y=1

2x+ 2

Slope =1

2

y-intercept = 2

71. + 2y= 4

2y= x + 4y = 1

2x + 2

Slope = 1

2

y-intercept = 2

73. 2 3y = 6

3y= 2x + 6 y = 23x 2

Slope =2

3

y-intercept = 2

75. + y= 1

y= + 1Slope = 1y-intercept = 1

77. = 4Slope is undefinedy-intercept: none

79. y= 5Slope = 0y-intercept = 5


23/827

Section 1.3 Lines

23

81. y = 0 y= Slope = 1y-intercept = 0

83. 2y 3 = 0

2y = 3x y =3

2x

Slope =3

2; y-intercept = 0

85. y= 2 3 slope= 2y= 2 + 4 slope= 2

The lines are parallel.

87. y= 4 + 5 slope= 4 y= 4 + 2 slope= 4

The lines are neither parallel norperpendicular.

89. Consider the points A2, 5( ), B 1,3( ) and C1,0( )

slope of AB =3 5

1 2( )=

2

3; slope of AC=

0 51 2( )

= 5; slope of BC=0 31 1

=3

2

Therefore, ABChas a right angle at vertex Bsince

slope AB = 2

3and slope BC=

3

2 ABis perpendicular to BC.

91. Consider the points A1,0( ), B 2,3( ), C1,2( ) and D 4,1( )

slope of AB = 3 02 1( )

=1; slope of CD =1 2( )

4 1= 1

slope of AC=2 01 1( )

= 1; slope of BD =1 342

= 1

Therefore, the quadrilateral ACDB is a parallelogram since

slope AB =1 and slope CD =1 AB is parallel to CD

slope AC= 1 and slope BD= 1 AC is parallel to BD

Furthermore,

slope AB=1 and slope BD = 1 AB is perpendicular to BD

slope AC= 1 and slope CD =1 AC is perpendicular to CDSo the quadrilateral ACDB is a rectangle.

93. Slope = 1; y-intercept = 2y= + 2 or y= 2

95. Slope = 1

3; y-intercept = 1

y= 1

3x +1 or x + 3y = 3


24/827

Chapter 1 Graphs

24

97. The equation of the x-axis is y= 0 . (The slope is 0 and they- intercept is 0.)

99. Let x= number of miles driven, and let C= cost in dollars.Total cost = (cost per mile)(number of miles) + fixed cost

C= 0.07 + 29Whenx= 110, C= 0.07( ) 110( ) + 29= $36.70.Whenx= 230, C= 0.07( ) 230( )+ 29= $45.10.

101. Let x= number newspapers delivered, and let C= cost in dollars.

Total cost = (delivery cost per paper)(number of papers delivered) + fixed costC= 0.53 + 1,070,000

103. (a) C= 0.08275 + 7.58 (b)

(c) For 100 kWh,

C= 0.08275(100) + 7.58= $15.86(d) For 300 kWh,

C= 0.08725(300) + 7.58= $32.41(e) For each usage increase of 1 kWh,

the monthly charge increasesby 8.27 cents.

105. (C,F)= (0, 32); (C,F)= (100, 212)

slope = 21232100 0

= 180100

= 95

F 32= 95

(C 0)

F 32=9

5 (C)C= 5

9(F 32)

If F= 70, the

C= 59

(70 32)= 59

(38)

C 21

107. (a) Letx= number of boxes sold, A= money, in dollars, spent on advertising.

We have the points ( 1, A1)= (100,000, 40,000);( 2, A2) = (200,000,60,000)

slope = 60,000 40,000200,000 100,000

= 20,000100,000

= 15

A 40,000 =1

5x 100,000( )

A 40,000 = 15

x 20,000

A =1

5x + 20,000

(b) If x= 300,000, then A=1

5300,000( )+ 20,000 = $80,000

(c) To increase the number of boxes sold by 5 units, the amount spent on advertising

needs to increase by 1 dollar.


25/827

Section 1.3 Lines

25

109. (b), (c), (e) and (g) 111. (c)

113. (a) Since the tangent line intersects the circle in a single point, there is exactly one point

on the line y= mx+ bsuch that x 2 +y 2 = r2. That is, x 2 + (mx + b)2 =r2 has exactlyone solution.

x2 + (mx + b)2 =r2

x2 + m 2x 2 + 2bmx + b2 =r2

(1+m2)x 2 + 2bmx + b2 r2 = 0

There is one solution if and only if the discriminant is zero.

(2bm)2 4(1+ m 2)(b2 r2)= 0

4b2m 2 4b2 + 4r2 4b2m2 + 4m 2r2 = 0

4b2 + 4r2 + 4m 2r2 = 0

b2 + r2 +m 2r2 = 0

r2(1 + m 2)= b2

(b) Use the quadratic formula, knowing that the discriminant is zero:

x= 2bm2 1+m2( )=

bmb2

r2

=bmr2

b2 =

mr2

b

y= mmr2

b

+ b=

m 2r2

b+ b =

m 2r2 + b2

b=

r2

b

(c) The slope of the tangent line is m .The slope of the line joining the point of tangency and the center is:

r2

b 0

mr2b

0 =

r2

b

b

mr2=

1

m

115. x2 +y 2 4x+ 6y + 4= 0

(x2 4x + 4)+ (y2 + 6y + 9)= 4+ 4+ 9

(x 2) 2 + (y + 3)2 = 9

Center: (2, 3)

Slope from center to 3,2 2 3( )is 2 2 3 (3)3 2

= 2 21

= 2 2

Slope of the tangent line is:

12 2 =

2

4


26/827

Chapter 1 Graphs

26

Equation of the tangent line:

y 2 2 3( )= 24

(x 3)

y 2 2+ 3= 2

4x +

3 2

4

4y 8 2+ 12 = 2x + 3 22x + 4y= 11 212

117. Find the centers of the two circles:

x2 + y2 4x+ 6y+ 4= 0

(x2 4x+ 4) + (y2 + 6y+ 9)= 4 + 4 + 9

(x 2) 2 + (y + 3)2 = 9 Center: (2,3)x2 + y2 + 6x+ 4y+ 9= 0

(x2 + 6x+ 9) + (y2 + 4y+ 4)= 9 + 9 + 4

(x+ 3)2

+ (y+ 2 )2

= 4 Center: (3, 2)

Find the slope of the line containing the centers:

m =2 (3)

3 2=

1

5

Find the equation of the line containing the centers:

y + 3 = 1

5(x 2)

5y+15= x + 2x + 5y= 13

119. 2 y= CGraph the lines:2 y= 4 y= 2 + 42xy= 0 y= 2x2xy= 2 y= 2x 2

All the lines have the same slope, 2.

The lines are parallel.

121. Consider the following diagram:

y= 2

The path of the center of the circle is the line y= 2.

(2,2)


27/827

Section 1.3 Lines

27

123. Answers will vary.

125. Not every line has two distinct intercepts since a horizontal line might not touch the x-axisand a vertical line might not touch the y-axis. Also, a non-vertical, non-horizontal line thatpasses through the origin will have only one intercept, (0, 0).

A line must have at least one intercept since a vertical line always crosses the x-axis, ahorizontal line always crosses the y-axis and a non-vertical, non-horizontal line always

crosses both axes.

127. Two lines with the same non-zerox- intercept and the same y-intercept must have the same

slope and therefore must be represented by equivalent equations.

129. Two lines that have the samey- intercept but different slopes can only have the samex-intercept if the y-intercept is zero.

Assume Line 1 has equation y= m1 + band Line 2 has equation y= m2 + b ,

Line 1 has x- intercept

b

m1 andy- intercept b .

Line 2 has x- intercept b

m2andy- intercept b .

Assume also that Line 1 and Line 2 have unequal slopes, that is m1 m2 .

If the lines have the same x- intercept, then b

m1=

b

m2.

b

m1=

b

m2

m2b = m1b

m2b + m1b = 0

But m2b + m1b = 0 b m1 m2( ) = 0 b = 0 or m1 m2= 0 m1=m2

Since we are assuming that m1 m2 , the only way that the two lines can have thesamex- intercept is if b = 0.



28/827

Chapter 1

28

Graphs

1.4 Scatter Diagrams; Linear Curve Fitting

1. scatter diagram 3. Linear, m > 0

5. Linear, m < 0 7. Nonlinear

9. (a), (c) (d) Using the LINear REGresssion

program, the line of best fit is:y = 2.0357 2.3571

(e)

10

20

00

(b) Answers will vary. We select (3,4) and (9,16). The slope of the line containing these

points is: m =16 4

9 3=

12

6= 2.

The equation of the line is:

y y1= m( 1)

y 4= 2(x 3)

y 4= 2x 6y= 2x 2

11. (a), (c) (d) Using the LINear REGresssionprogram, the line of best fit is:

y = 2.2 + 1.2(e)

3

6

3

6


29/827

Section 1.4 Scatter Diagrams; Linear Curve Fitting

29

(b) Answers will vary. We select (2,4) and (2,5). The slope of the line containing

these points is: m =5 ( 4)

2 (2)=

9

4.


y y1= m( 1)

y (4)= 94

(x (2))

y+ 4=9

4x+

9

2

y =9

4x+

1

2

13. (a), (c) (d) Using the LINear REGresssionprogram, the line of best fit is:

y = 3.8613x + 180 .2920

(e)

0

160

2590

(b) Answers will vary. We select (20,100) and (10,140). The slope of the line

containing these points is: m =140 100

10 (20)=

40

10= 4.


y y1 =m( 1)

y100 = 4 x (20)( )y100 = 4x+ 80

y= 4x+ 180


30/827

Chapter 1 Graphs

30

15. (a) (b) Using points 20,16( )and 50,39( ) ,

slope=39 16

50 20=

23

30.

The point-slope formula yields

C16 =23

30I 20( )

C=23

30I

460

30+16

C 0.77I+ 0.67 (c) As disposable income increases by

$1000, personal consumptionexpenditure increases by about $770.

(d) C= 0.77(42) + 0.67= $33.01

A family with disposable income of$42,000 has personal consumption

expenditure of about $33,010.

(e) C= 0.7549I+ 0.6266

17. (a) (Data used in graphs is inthousands.)

800

220

30

(b) L= 2.9814I 0.0761(c)

80

220

030

(d) As annual income increases by$1000, the loan amount increases

by about $2981.40.

(e) L = 2.9814(42) 0.0761 125.143

An individual with an annual income of

$42,000 would qualify for a loan ofabout $125,143.

19. (a)

110

70

1055

(b) T = 0.0782h + 59.0909(c)

11010

70

55 (d) As relative humidity increases

by 1%, the apparent temperatureincreases by 0.0782o.

(e) T = 0.0782(75) + 59.0909 64.96

A relative humidity of 75% wouldgive an apparent temperature of 65o.


31/827

Chapter 1

31

Graphs

1.R Chapter Review

1. 0,0( ), 4,2( )(a) distance = 4 0( )2 + 2 0( )2

= 16 + 4 = 20 = 2 5

(b) midpoint =0 + 4

2,0 + 2

2

=

4

2,2

2

= 2,1( )

(c) slope =y

=2 04 0

=2

4=

1

2

(d) Whenxincreases by 2 units,yincreasesby 1 unit.

3. 1,1( ), 2,3( )(a) distance = 21( )2 + 3 1( )( )2

= 9 +16 = 25 = 5

(b) midpoint =1+ 2( )

2,1+ 3

2

=12

,2

2

=

1

2,1

(c) slope = y

= 3 1( )2 1

= 43

= 43

(d) Whenxincreases by 3 units,y

decreasesby 4 units.

5. 4,4( ), 4,8( )(a) distance = 4 4( )2 + 8 4( )( )2

= 0 +144 = 144 =12

(b) midpoint =4 + 4

2 ,

4 + 82

=8

2,4

2

= 4,2( )

(c) slope = y

=8 4( )

4 4= 12

0, undefined

(d) Undefined slope means the points lie on avertical line.

7. y =x 2 + 4


32/827

Chapter 1 Graphs

32

9. 2x = 3y 2 x-intercept(s): 2x = 3 0( )2 2x = 0 x = 0 0,0( )y-intercept(s): 2 0( )= 3y 2 0 = 3y 2 y = 0 0,0( )Test for symmetry:

x - axis : Replace yby y : 2x = 3(y)2 or 2x = 3y 2, which is

equivalent to 2x = 3y 2 .

y - axis : Replace xby x : 2(x) = 3y 2or 2x = 3y 2,

which is not equivalent to 2x = 3y 2 .

Origin : Replacexby x andyby y : 2(x) = 3(y)2 or

2x = 3y 2, which is not equivalent to 2x = 3y 2.

Therefore, the graph is symmetric with respect to thex-axis.

11. x 2 + 4y 2 = 16

x-intercept(s): x 2 + 4 0

( )

2= 16 x 2 =16 x = 4 4,0

( ), 4,0

( )

y-intercept(s): 0( )2 + 4y2 = 16 4y 2 =16 y 2 = 4 y = 2 0,2( ), 0,2( )Test for symmetry:

x - axis : Replace yby y : x2 + 4 y( )2 =16 or x 2 + 4y 2 =16,which is equivalent to x 2 + 4y 2 =16.

y - axis : Replace xby x : x( )2 + 4y2 =16 orx 2 + 4y 2 =16,which is equivalent to x 2 +4y 2 =16.

Origin : Replacexby x andyby y : x( )2 + 4 y( )2 =16 orx 2 + 4y2 =16,

which is equivalent to x

2

+ 4y2

=16.

Therefore, the graph is symmetric with respect to thex-axis, they-axis, and theorigin.

13. y =x 4 +2x 2 +1x-intercept(s): 0 =x4 +2x 2 +1 0 = x 2 +1( )x 2 +1( )x 2 +1= 0 x 2 = 1

no solutio no - intercepts y-intercept(s): y = 0( )4 + 2 0( )2 +1=1 0,1( )Test for symmetry:

x - axis : Replace yby y : y =x4 + 2x 2 +1,

which is not equivalent to y =x 4 + 2x2 +1.

y - axis : Replace xby x : y = x( )4 + 2 x( )2 +1 ory =x 4 + 2x 2 +1,which is equivalent to y =x 4 + 2x 2 +1.

Origin : Replace xby x andyby y : y = x( )4 + 2 x( )2 +1 or y =x4 + 2x 2 +1,which is not equivalent to y =x4 + 2x2 +1.

Therefore, the graph is symmetric with respect to they-axis.


33/827

Section 1.R Chapter Review

33

15. x 2 +x +y 2 + 2y = 0x-intercept(s): x 2 +x + 0( )2 + 2 0( )= 0 x 2 +x = 0 x x +1( )= 0

= 0, = 1 1,0( ), 0, 0( )y-intercept(s): 0( )2 + 0+y 2 + 2y = 0 y 2 + 2y = 0 y y + 2( )= 0

y = 0,y = 2 0,2( ), 0, 0( )Test for symmetry:

x - axis : Replace yby y : x2 +x + (y)2 + 2(y) = 0 orx2 +x +y 2 2y = 0,

which is not equivalent to x2 +x +y 2 + 2y = 0.

y - axis : Replace xby x : (x)2 + (x)+y 2 + 2y = 0 orx 2 x +y 2 + 2y = 0,

which is not equivalent to x2 +x +y 2 + 2y = 0.

Origin : Replacexby x andyby y : (x)2 + (x)+ (y)2 + 2(y) = 0 or

x2 x +y 2 2y = 0, which is not equivalent to

x2 +x +y 2 + 2y = 0.


17. (x h)2 + (y k)2 =r2

x 2( )( )2 + y 3( )2 = 42

x + 2( )2 + y 3( )2 =16

19. (x h)2 + (y k)2 =r2

x 1( )( )2 + y 2( )( )2 =12

x +1( )2 + y + 2( )2 =1

21. x2 + y 1( )2 = 4

x 2 + y 1( )2 = 22Center: (0,1)Radius = 2

23. x2 +y2 2x + 4y 4 = 0

x2 2x +y2 + 4y = 4

(x

2

2x +1)+ (y2

+ 4y + 4) = 4 +1+ 4(x 1) 2 + (y + 2)2 = 32

Center: (1,2) Radius = 3


34/827

Chapter 1 Graphs

34

25. 3x2 + 3y2 6x + 12y = 0x

2 +y2 2x + 4y = 0x

2 2x +y2 + 4y = 0(x2 2x +1)+ (y2 + 4y + 4) = 1+ 4

(x 1)2 + (y + 2)2 = 5( )2

Center: (1,2) Radius = 5

27. Slope = 2; containing (3,1)y y1 =m( 1 )

y (1) = 2(x 3)y +1 = 2x + 6

y = 2x + 52x +y = 5 or y = 2x + 5

29. vertical; containing (3,4)= 3

No slope - intercept form.

31. y-intercept = 2; containing (5,3)Points are (5,3) and (0,2)

m=2 (3)

0 5=

1

5=

1

5

y = mx + b

y = 1

5x 2

x + 5y = 10 or y = 1

5x 2

33. Parallel to 2

3y

= 4

2x 3y = 4 y =2

3x +

4

3

Slope =2

3; containing (5, 3)

y y1 =m( 1)

y 3 =23

(x (5))

y 3 = 23

x + 103

y = 23

x + 193

2x 3y = 19 or y = 23x + 19

3

35. Perpendicular to +y = 2 +y = 2 y = + 2

Containing (4,3)Slope of perpendicular = 1

y y1 =m( 1)y (3) = 1(x 4)

y + 3 =x 4y =x 7

x y = 7 or y =x 7

37. 4 5y = 20

5y = 4x 20y =4

5x + 4

slope =4

5;y-intercept = 4


35/827


35

39.1

2x

1

3y =

1

6

1

3y =

1

2x

1

6y =

3

2x +

1

2

slope =3

2;y-intercept =

1

2

41. slope =2

3, containing the point (1,2)

43. Given the points A = (2, 0), B = (4, 4), C= (8,5) .(a) Find the distance between each pair of points.

dA ,B = (4 (2))2 + (4 0)2 = 4 +16 = 20 = 2 5

dB ,C = (8 (4))2 + (5 4)2 = 144 +1 = 145

dA ,C = (8 (2))2 + (50)2 = 100+ 25 = 125 = 5 5

20( )2

+ 125( )2

= 145( )2

20 +125 =145 145 =145The Pythagorean theorem is satisfied, so this is a right triangle.

(b) Find the slopes:

mAB =4 0

4 (2)=

4

2= 2; mBC =

5 48 4( )

=1

12

mAC =5 0

8 2( )= 5

10= 1

2

mAB mAC = 2 12= 1

Since the product of the slopes is 1, two sides of the triangle are perpendicular and

the triangle is a right triangle.

45. slope of AB=1 56 2

= 1; slope of AC=158 2

= 1; slope of BC=118 6

= 1

Since all the slopes are equal, the points lie on a line.


49. Set the axes so that the fields maximum dimension is along thex-axis.

Let 2h = widt , 2k= height , therefore the pointfarthest from the origin has coordinates P h, k( ).So the distance from the origin to pointPis

circletheofradiusthe22 =+= khr .

P(h,k)

hk

r


36/827

Chapter 1 Graphs

36

Using 1 sprinkler arm:

If we place the sprinkler at the origin, we get a circle with equation 222 ryx =+ , where22

khr += . So how much excess land is being watered ? The area of the field = hkAF 4= .

The area of the circular water pattern AC = r2 = h2 +k2( )2 = h2 + k2( ).

Therefore the amount of excess land being watered = AC AF = h2

+ k2

( ) 4hk.Using 2 sprinkler arms:We want to place the sprinklers so that they overlap

as little as possible while watering the entire field.The equation of the circle with center on the

positivex-axis that passes through the point P h, k( )

and Q 0,k( )is x h2

2

+y 2 =1

4h2 +k2 , since the

center ish

2,0

, and the radius is

1

4h2 +k2 .

The case for the other side is similar. Thus, the sprinklers should have their centers at

h

2,0

and

h

2,0

, with the arm lengths set at

1

4h2 +k2 .

Each sprinkler waters the same area, so the total area watered is

AC = 2r2 = 2

1

4h2 +k2

2

= 21

4h2 +k2

. The amount of excess land being watered is

AC AF = 21

4h

2 + k2

4hk.

Comparison:

In order to determine when to switch from 1 sprinkler to 2 sprinklers, we want todetermine when 2 sprinklers water less excess land than 1 sprinkler waters.

That is, we want to solve: 21

4h

2 +k2

4hk< h2 + k2( ) 4hk.

21

4h 2 +k2

4hk< h2 +k2( ) 4hk

2h

2 + 2k2 4hk< h2 + k2 4hk

2h 2 + 2k2 < h2 + k2

12

h 2 + 2 k2 2k

So 2 sprinklers is the better choice when the longer dimension of the rectangle exceeds

the shorter dimension by a factor of more than 2 1.414.

Q(0,k) P(h,k)


37/827

Chapter 2

37

Functions and Their Graphs

2.1 Functions

1. (1,3)

3. We must not allow the denominator to be 0.

+ 4 0 4 ; Domain: x x 4{ }.

5. independent, dependent

7. The intersection of the intervals [0,7] and [2,5], i.e., [0,5].

9. f( )( ) 11. True

13. True

15. FunctionDomain: {Dad, Colleen, Kaleigh,

Marissa}Range: {Jan. 8, Mar. 15, Sept. 17}

17. Not a function

19. Not a function 21. Function

Domain: {1, 2, 3, 4}Range: {3}

23. Not a function 25. Function

Domain: {2, 1, 0, 1}Range: {4, 1, 0}

27. f x( )= 3x2 + 2x 4 (a) f 0( )= 3 0( )2 + 2 0( ) 4 = 4(b) f 1( )= 3 1( )2 + 2 1( ) 4 = 3+ 2 4 =1

(c) f 1( )= 3 1( )2

+ 2 1( ) 4 = 32 4 = 3(d) f x( )= 3 x( )2 + 2 x( )4 = 3x2 2x 4 (e) f x( )= 3x 2 + 2x 4( )= 3x 2 2x + 4(f) f x +1( )= 3 x +1( )

2 + 2 x +1( ) 4 = 3 x 2 + 2x +1( )+ 2x + 2 4 = 3x 2 + 6x + 3+ 2x + 2 4

= 3 2 + 8 +1


38/827

Chapter 2 Functions and Their Graphs

38

(g) f 2x( )= 3 2x( )2 + 2 2x( ) 4 =12x2 + 4x 4 (h) f x +h( )= 3 x +h( )

2+ 2 x +h( ) 4 = 3 x 2 + 2xh+h 2( )+ 2x + 2h 4

= 3 2 + 6 h+ 3h2 + 2 + 2 h 4

29. f x

( )=

2 +1

(a) f 0( )= 00

2 +1=

0

1= 0

(b) f1( )= 11

2 +1=

1

2

(c) f 1( )= 11( )2 +1

=1

1+1=

1

2

(d) f x( )= x( )2 +1

=

x 2 +1

(e) f x( )= x

x2 +1

=

xx2 +1

(f) f x +1( )= +1x +1( )2 +1

=+1

x 2 + 2x +1+1=

+1x 2 + 2x + 2

(g) f 2x( )= 22x( )2 +1

=2

4x 2 +1

(h) f x +h( )= +hx +h( )2 +1

=+h

x2 + 2xh +h 2 +1

31. f( )= + 4

(a) f 0( )= 0 + 4 = 0 + 4 = 4 (b) f 1( )= 1 + 4 = 1+ 4 = 5(c) f 1( )= 1 + 4 = 1+ 4 = 5(d) f ( )= + 4 = + 4(e) f x( )= x + 4( )= x 4(f) f +1( )= + 1 + 4 (g) f 2( )= 2 + 4 = 2 + 4(h) f +h( )= +h + 4

33. f x( )=2 +13 5

(a) f 0( )= 2 0( )+13 0( )5

=0 +10 5

= 1

5

(b) f1( )= 2 1( )+13 1( ) 5

=2+13 5

=3

2=

3

2

(c) f 1( )= 2 1( )+13 1( ) 5

=2+13 5

=18

=1

8


39/827


40/827


40

51. g(x) = 2 16

x2 16 0

2 16 4

Domain: x x 4, x 4

{ }

53. F(x) = 2

3 +

x3 + x 0

x(x2 +1) 0

x 0, x2 1

Domain: x x 0{ }

55. h(x) = 3x 12 3 12 0

3x 12 4

Domain: x x 4{ }

57. f(x) = 4x 9

9 > 0> 9

Domain: x x > 9{ }

59. p(x) =2

x 1

=2

1

1> 0>1

Domain: x x > 1{ }

61. f( ) = 3 + 4 ( ) = 2 3(a) (f+ )( ) = 3 + 4 + 2 3 = 5 +1

The domain is x xis any real number{ }.(b) (f )( ) = (3 + 4) (2 3) = 3 + 4 2 + 3 = + 7

The domain is x xis any real number{ }.

(c) (fg)(x ) = (3x + 4)(2x 3) = 6x2

9x + 8x 12 = 6x2

x 12 The domain is x xis any real number{ }.

(d)f

g

(x) =

3x + 42x 3

2 3 0

2x 3 x 32

The domain is x x 3

2

.

63. f(x) = x 1 g(x) = 2x2 (a) (f+g)(x) = x 1 + 2x2 = 2x 2 +x 1

The domain is x xis any real number{ }.(b) (fg)(x) = (x 1) (2x2) = x 1 2x2 = 2x2 +x 1

The domain is x xis any real number{ }.


41/827

Section 2.1 Functions

41

(c) (fg)(x ) = (x 1)(2x2) = 2x3 2x2 The domain is x xis any real number{ }.

(d)f

g

(x) =

x 12x 2

The domain is x x 0{ }.

65. f(x) = x g(x) = 3x 5(a) (f+g)(x) = x + 3x 5

The domain is x x 0{ }.(b) (fg)(x) = x (3x 5) = x 3x + 5

The domain is x x 0{ }.(c) (fg)(x ) = x (3x 5) = 3x x 5 x

The domain is x x 0{ }.

(d)f

g

(x) =

x

3x 5

0 and 3 5 0

3x 5 x 5

3

The domain is x x 0 and x 53

.

67. f(x) = 1 + 1x

g(x) = 1x

(a) (f+g)(x) = 1+ 1x

+ 1x

= 1 + 2x

The domain is x x 0{ .(b) (fg)(x) = 1+ 1

x 1

x= 1

The domain is x x 0{ .(c) (fg)(x ) = 1+ 1

x

1x

= 1x+ 1

x2

The domain is x x 0{ .

(d)f

g

(x) =

1+1

x1

=

+1x1

=x +1

xx

1=x +1

The domain is x x 0{ .


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42

69. f(x) = 2 + 33x 2 g(x) =

43x 2

(a) (f+g)(x)=2 + 33 2

+4

3 2=

2 + 3+ 43 2

=6 + 33 2

3 2 0

3x 2 x

2

3

The domain is x x 23

.

(b) (fg)(x) =2 + 33x 2

43x 2 =

2 + 3 43x 2 =

2 + 33x 2

3 2 0

3x 2 x 23


.

(c) (fg)(x)= 2x + 33x 2

4x

3x 2

= 8x

2

+12x(3x 2)2

3 2 0

3x 2 x 23


.

(d)f

g

(x) =

2 + 33x 2

4x

3 2

=2x + 33x 2

3x 2

4x=

2x + 34x

3 2 0 and 03x 2

x 2

3

The domain is x x 2

3and x 0

.

71. f(x)= 3x +1 (f+g)(x)= 6 1

2x

6 12x = 3x +1+g(x)

5 7

2x =g(x) g(x) = 5

7

2x


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43

73. f( ) = 4 + 3f( +h) f( )

h=

4( +h)+ 3 4 3h

=4x + 4h + 3 4x 3

h=

4 h

h= 4

75. f(x) = x 2 x + 4 f(x +h) f(x)

h=

(x +h)2 (x +h )+ 4 (x2 x + 4)h

= x 2 + 2xh+h 2 x h + 4 x 2 + x 4

h

=2xh+h 2 h

h

= 2 +h 1

77.f(x) =x

3

2 f(x +h) f(x)

h=

x +h( )3 2 x3 2( )h

= x 3 + 3x2h + 3xh 2 +h 3 2 x 3 + 2

h

=3x 2h + 3xh2 +h 3

h

= 3 2 + 3 h +h 2

79. f(x) = 2x3 +Ax 2 + 4x 5 and f(2) = 5f(2) = 2(2)3 +A(2)2 + 4(2)5

5 =16+ 4A + 8 5 5 = 4A +19

14 = 4A A = 7

2

81. f(x)= 3 + 82 A

and f(0) = 2

f(0) =3(0)+ 82(0) A

2 =8

A 2A = 8 A = 4

83. f(x)=2 A

3 and f(4) = 0

f(4)=2(4) A

4 30 = 8A

1

0 = 8 AA = 8

fis undefined when = 3.


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44

85. Let xrepresent the length of the rectangle.

Then2

represents the width of the rectangle, since the length is twice the width.

The function for the area is: A(x) = x x2= x

2

2= 1

2x

2

87. Let xrepresent the number of hours worked.The function for the gross salary is: G( ) = 10

89. (a) H1( )= 20 4.9 1( )2 = 20 4.9 =15.1 meters H1.1( )= 20 4.9 1.1( )2 = 20 4.9 1.21( )= 20 5.929 =14.071 metersH1.2( )= 20 4.9 1.2( )2 = 20 4.9 1.44( )= 20 7.056 =12.944 metersH1.3( )= 20 4.9 1.3( )2 = 20 4.9 1.69( )= 20 8.281=11.719 meters

(b) H( )=15 H( )=10 H( )= 5 15 = 20 4.9x 2

5 = 4.9x 2

x2 1.02041.01 seconds

10 = 20 4.9x2

10 = 4.9x2

x2 2.0408x 1.43 seconds

5 = 20 4.9x2

15 = 4.9x2

x2 3.06121.75 seconds

(c) H( )= 0 0 = 20 4.9x 2

20 = 4.9x 2

x2 4.0816 2.02 seconds

91. C x( )=100+10

+36,000

(a) C 500( )=100+500

10+

36,000

500=100+ 50 + 72 = $222

(b) C 450( )=100+450

10+

36,000

450=100+ 45 + 80 = $225

(c) C 600( )=100+600

10+

36,000

600=100+ 60 + 60 = $220

(d) C 400( )=100+400

10+

36,000

400=100+ 40 + 90 = $230

93. R x( )= LP

x( )= L x( )

P x( ) 95. H( )= PI( )( )=P( )I( )

97. (a) h( )= 2 h a +b( )= 2 a +b( )= 2a + 2b =h a( )+h b( )h( )= 2 has the property.


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45

(b) g x( )=x 2 g a+b( )= a+b( )2 =a 2 + 2ab +b2 a2 +b2 =g a( )+g b( )g(x) = x2 does not have the property.

(c) F( )= 5 2F a +b( )= 5 a +b( ) 2 = 5a + 5b 2 5a 2 + 5b 2 =F a( )+F b( )F( )= 5 2does not have the property.

(d) G x( )=1

G a+ b( )= 1a + b

1

a+

1

b= G a( )+ G b( )

G x( )= 1 does not have the property.



46/827

Chapter 2

46


2.2 The Graph of a Function

1. x 2 + 4y 2 = 16x-intercepts: x 2 + 4 0( )2 =16 x 2 =16 x = 4 4,0( ), 4,0( )y-intercepts: 0( )2 + 4y 2 =16 4y 2 =16 y 2 = 4 y = 2 0,2( ), 0,2( )

3. vertical

5. f x( )=ax2 + 4 a

1

( )

2

+4

=2

a

= 2

7. False

9. (a) (0) = 3 since (0,3) is on the graph. (6) = 3 since (6,3) is on the graph.

(b) (6) = 0 since (6, 0) is on the graph. (11)=1 since (11,1) is on the graph.

(c) (3) is positive since f(3) 3.7.(d) (4) is negative since (4) 1.(e) ( )= 0 when = 3, = 6, and =10.(f) ( )> 0 when 3< < 6, and 10 < 11.(g) The domain of f is x 6 x 11{ } or 6,11[ ](h) The range of f is y 3 y 4{ } or 3,4[ ](i) Thex-intercepts are (3, 0), (6, 0), and (11, 0).

(j) They-intercept is (0, 3)

(k) The line y =1

2intersects the graph 3 times.

(l) The line = 5intersects the graph 1 time(m) ( ) = 3 when = 0 and = 4.(n) ( ) = 2 when = 5 and = 8.

11. Not a function since vertical lines will intersect the graph in more than one point.

13. Function (a) Domain: x x { }; Range: y 1 y 1{ }(b) intercepts:

2

,0

,

2

,0

, (0,1)

15. Not a function since vertical lines will intersect the graph in more than one point.


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Section 2.2 The Graph of a Function

47

17. Function (a) Domain: x x > 0{ }; Range: y yis any real number{ (b) intercepts: (1, 0)

19. Function (a) Domain: x xis any real number{ }; Range: y y 2{ }(b) intercepts: (3,0), (3,0), (0,2)

21. Function (a) Domain: x xis any real number{ }; Range: y y 3{ }(b) intercepts: (1,0), (3,0), (0,9)

23. f(x) = 2x 2 x 1(a) f(1) = 2 1( )2 1( )1= 2 (1,2) is on the graph of f.(b) f(2) = 2 2( )2 2( )1= 9 (2,9) is on the graph of f.(c) Solve for :

1= 2x 2 x 1 0 = 2x 2 x

0

=x 2x

1

( )x

=0,x

=

1

2

(0, 1) and1

2,1

are on the graph of .

(d) The domain of f is : x xis any real number{ }.(e) x-intercepts:

f x( )= 0 2x 2 x 1 = 0

2x +1( ) x 1( )= 0 x = 12

,x =1

1

2

,0

and 1,0( )

(f) y-intercept: f 0( ) = 2 0( )2 0 1= 1 0,1( )

25. f(x) =+ 2 6

(a) f(3) =3+ 23 6

= 5

314 (3,14) is not on the graph of f.

(b) f(4)=4 + 24 6

=6

2= 3 (4, 3) is on the graph of f.

(c) Solve for :

2 =+ 2

x 62x 12 = x + 2

= 14

(14, 2) is a point on the graph of f.

(d) The domain of f is: x x 6{ }.


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48

(e) x-intercepts:

f x( )= 0 + 2x 6

= 0

x + 2 = 0 x = 2 2,0( )

(f) y-intercept: f 0( )= 0 + 2

0 6=

1

3

0,1

3

27. f(x) =2x2

4 +1

(a) f(1) =2(1)2

(1)4 +1=

2

2=1 (1,1) is on the graph of f.

(b) f(2) =2(2)2

(2)4 +1=

8

17 2,

8

17

is on the graph of f.

(c) Solve for :

1= 2x2

x4

+1x 4 +1= 2x 2

x4 2x 2 +1= 0

(x2 1)2 = 0

2 1= 0 = 1(1,1) and (1,1) are on the graph of .

(d) The domain of f is : x xis any real number{ }.(e) x-intercept:

f x( )= 0 2x2

x4 +1= 0

2x 2 = 0 x = 0 0,0( )

(f) y-intercept: f 0( )=2 0( )2

04 +1=

0

0+1= 0 0,0( )

29. h(x) = 32x2

1302+ x

(a) h(100) =32(100)2

1302+100 =

320,00016,900

+100 18.93+100 81.07 feet

(b) h(300) =32(300)2

130

2+ 300=

2,880,000

16,900

+ 300 170.41+ 300 129.59 feet

(c) h(500) =32(500)2

1302+ 500 =

8,000,00016,900

+ 500 473.37+ 500 26.63 feet


49/827

Section 2.2 The Graph of a Function

49

(d) Solving h(x ) =32x 2

1302 + x = 0

32x 2

1302

+x = 0 x32x130

2+1

= 0 x = 0 or

32x130

2+1= 0

32x

1302 +1

=0

1

=

32x

1302 130

2

=32x

x

=

1302

32 =528.125 feet

Therefore, the golf ball travels 528.125 feet.

(e) y1 =32x 2

1302+x

(f) Use INTERSECT on the graphs of y1 = 32x2

1302 +x and y2 =90.

The ball reaches a height of 90 feet twice. The first time is when the ball has traveled

approximately 115 feet, and the second time is when the ball has traveledapproximately 413 feet.

(g) The ball travels approximately 275 feet before it reaches its maximum height ofapproximately 131.8 feet.

(h) The ball travels approximately 264 feet before it reaches its maximum height of

approximately 132.03 feet.

31. (a) III (b) IV (c) I (d) V (e) II


50/827


50

33.

35. (a) 2 hours elapsed; Kevin was between 0 and 3 miles from home.(b) 0.5 hours elapsed; Kevin was 3 miles from home.(c) 0.3 hours elapsed; Kevin was between 0 and 3 miles from home.

(d) 0.2 hours elapsed; Kevin was at home.(e) 0.9 hours elapsed; Kevin was between 0 and 2.8 miles from home.(f) 0.3 hours elapsed; Kevin was 2.8 miles from home.

(g) 1.1 hours elapsed; Kevin was between 0 and 2.8 miles from home.(h) The farthest distance Kevin is from home is 3 miles.

(i) Kevin returned home 2 times.

37. Points of the form (5, y) and of the form (x, 0) cannot be on the graph of the function.

39. A function may have any number of x-intercepts; it can have only one y-intercept.

41. The only such function is f( )= 0.


51/827

Chapter 2

51


2.3 Properties of Functions

1. 2 5

3. y= 2x 2 + 3Test for symmetry:

x - axis : Replace yby y : y= 2x 2 + 3 y= 2x2 3,

which is not equivalent to y= 2x 2 + 3.

y - axis : Replace xby x : y= 2x( )2 + 3 = 2x 2 + 3,

which equivalent to y= 2x2

+ 3.Origin : Replace xby x and yby y : y= 2x( )2 + 3 y= 2x 2 3,

which is not equivalent to y= 2x 2 + 3.

Therefore, the graph is symmetric with respect to the y-axis.

5. y= x 2 9y - intercept : Let x= 0, then y= 02 9 = 9 (0,9)

x - intercepts : Let y= 0, then 0=x 2 9 x2 = 9 x= 3 (3,0), 3,0( )

7. even, odd 9. True

11. Yes 13. No, it only increases on (5, 10).

15. fis increasing on the intervals 8,2( ), 0,2( ), 5,( ).

17. Yes. The local maximum at = 2 is 10.

19. fhas local maxima at = 2 and = 2. The local maxima are 6 and 10, respectively.

21. (a) Intercepts: (2,0), (2,0), and (0,3).

(b) Domain: x 4 x 4{ }; Range: y 0 y 3{ }.(c) Increasing: (2, 0) and (2, 4); Decreasing: (4, 2) and (0, 2).(d) Since the graph is symmetric with respect to the y-axis, the function is even.


52/827


52

23. (a) Intercepts: (0,1).

(b) Domain: x xis any real number{ }; Range: y y > 0{ }.(c) Increasing: (,) ; Decreasing: never.(d) Since the graph is not symmetric with respect to the y-axis or the origin, the function

is neither even nor odd.

25. (a) Intercepts: (,0), (,0), and (0,0).(b) Domain: x x { }; Range: y 1 y 1{ }.(c) Increasing:

2

,2

; Decreasing: ,

2

and

2

,

.

(d) Since the graph is symmetric with respect to the origin, the function is odd.

27. (a) Intercepts:1

2, 0

,

5

2, 0

, and 0,

1

2

.

(b) Domain: x 3 x 3{ }; Range: y 1 y 2{ }.(c) Increasing: 2, 3( ); Decreasing: 1, 1( ) ; Constant: 3,1( )and 1, 2( )(d) Since the graph is not symmetric with respect to the y-axis or the origin, the function

is neither even nor odd.

29. (a) has a local maximum of 3 at = 0.(b) has a local minimum of 0 at both = 2 and = 2.

31. (a) has a local maximum of 1 at x=

2.

(b) has a local minimum of 1 at x=

2.

33. f(x) = 2x2 + 4 (a) Average rate of change of ffrom x= 0 to x= 2:

f 2( ) f 0( )2 0

=2 2( )2 + 4( ) 2 0( )2 + 4( )

2=

4( ) 4( )2

=8

2= 4

(b) Average rate of change of ffrom x= 1 to x= 3:

f 3( ) f1( )31

=2 3( )2 + 4( ) 2 1( )2 + 4( )

2=

14( ) 2( )2

=16

2= 8

(c) Average rate of change of ffrom x= 1 to x= 4:

f 4( ) f1( )4 1

= 2 4( )2

+ 4( ) 2 1( )2

+ 4( )3 = 28( ) 2( )

3=30

3= 10


53/827

Section 2.3 Properties of Functions

53

35. ( )= 5 37. f( )= 1 3

(a)

f( ) (1)1

=5 5

1=

5( 1)1

= 5(a) f( ) f(1)

x 1=

1 3 (2)x1

=3x+ 3 1

=3(x1)1

= 3

(b)f(2) f(1)

2 1 = 5(b) f(2) f(1)

2 1 = 3(c) Slope = 5; Containing (1, 5):

y 5 = 5( 1)

y 5 = 5x5y= 5x

(c) Slope = 3; Containing (1, 2):y ( 2) = 3( 1)

y + 2 = 3x+ 3y= 3x + 1

39. f(x)= x2 2x 41. f(x)= x3 x (a) f(x) f(1)

x1=

x 2 2x (1)x1

= x2 2x + 1

x1

= (x1)2

x 1= 1

(a) f(x) f(1)x1

= x3 x 0x1

= x3 x

x1

= x(x 1)(x+1)

1= x2 +x

(b) f(2) f(1)2 1

= 2 1=1

(b) f(2) f(1)2 1

= 22 + 2 = 6

(c) Slope = 1; Containing (1, 1):y (1)=1( 1)

y+1=x1y= x2

(c) Slope = 6; Containing (1, 0):

y 0 = 6( 1)y= 6 6

43. f(x) =2

+1 45. f(x)= x

(a)

f(x) f(1)1

=

2

x+ 1

2

21

=

2

x+1 11

=

2 1x +1x 1

=1x

x 1( ) x +1( )

= 1

+ 1

(a) f(x) f(1)

1 =

x 1

1

(b) f(2) f(1)2 1

= 12 + 1

= 1

3

(b) f(2) f(1)2 1

= 212 1

= 21

(c)Slope =

1

3; Containing (1, 1):

y1= 1

3(x1)

y1= 1

3x+

1

3y=

1

3x+

4

3

(c) Slope = 2 1; Containing (1, 1):y 1 = 21( )(x 1)y 1 = 21( )x 21( )

y= 21( )x 2+ 2


54/827


54

47. f(x)= 4x3 f(x)= 4(x)3 = 4x3

fis odd.

49. g(x)= 3x2 5g(x)= 3(x)2 5= 3x2 5

is even.

51. F(x)= x3

F(x)= x3

= x3

Fis odd.

53. f( )= +

f( )= + = + fis neither even nor odd.

55. g(x)=1

2

g(x)= 1(x)2

= 1x 2

is even.

57. h(x)= x3

3 2 9

h(x)= ( )3

3( )2 9=

3

3 2 9

h is odd.

59. f x( ) =x 3 3x + 2 on the interval2,2( ) Use MAXIMUM and MINIMUM on the graph of y1=x

3 3x+ 2.

local maximum at: 1,4( ) ; local minimum at: 1,0( )fis increasing on: 2,1( )and 1,2( ); fis decreasing on: 1,1( )

61. f x( ) = x5 x 3 on the interval2,2( ) Use MAXIMUM and MINIMUM on the graph of y1=x

5 x 3 .

local maximum at: 0.77,0.19( ); local minimum at: 0.77,0.19( )fis increasing on: 2,0.77( )and 0.77,2( ); fis decreasing on: 0.77,0.77( )

63. f x( ) = 0.2x 3 0.6x2 + 4x6 on the interval6,4( ) Use MAXIMUM and MINIMUM on the graph of y1= 0.2x

3 0.6x 2 + 4x 6.

local maximum at: 1.77,1.91( ) ; local minimum at: 3.77,18.89( )fis increasing on: 3.77,1.77( ) ; fis decreasing on: 6,3.77( )and 1.77,4( )


55/827

Section 2.3 Properties of Functions

55

65. f x( ) = 0.25x4 + 0.3x3 0.9x 2 + 3on the interval3,2( )Use MAXIMUM and MINIMUM on the graph of y1= 0.25x

4 + 0.3x 3 0.9x2 + 3.

local maximum at: 0, 3( ); local minimum at: 1.87,0.95( ), 0.97,2.65( )fis increasing on: 1.87,0( )and 0.97,2( ); fis decreasing on: 3,1.87( )and 0,0.97( )

67. (a) length = 24 2width = 24 2height =

V(x)= x(24 2x)(24 2x)= x(24 2x)2 (b) V(3)= 3(24 2(3))2 = 3(18)2 = 3(324)= 972 cu. in. (c) V(10)= 3(24 2(10))2 = 3(4) 2 = 3(16)= 48 cu.in.(d) y1=x (24 2x)

2

120

1100

0

Use MAXIMUM.The volume is largest when = 4 inches.

69. (a) y1= 16x2 + 80x+ 6

(b) Use MAXIMUM. The maximum height occurs when t= 2.5 seconds. (c) From the graph, the maximum height is 106 feet.

71. f(x)=x 2 (a) Average rate of change of ffrom x= 0 to x= 1:

f 1( ) f 0( )1 0

= 12 02

1= 1

1= 1

(b) Average rate of change of ffrom x= 0 to x= 0.5:f 0.5( ) f 0( )

0.5 0=

0.5( )2 020.5

=0.25

0.5= 0.5

(c) Average rate of change of ffrom x= 0 to x= 0.1:

f 0.1( ) f 0( )0.1 0

=0.1( )2 02

0.1=

0.01

0.1= 0.1


56/827


56

(d) Average rate of change of ffrom x= 0 to x= 0.01:

f 0.01( ) f 0( )0.01 0

=0.01( )2 02

0.01=

0.0001

0.01= 0.01

(e) Average rate of change of ffrom x= 0 to x= 0.001:

f 0.001( ) f 0( )0.001 0

=0.001( )2 02

0.001

=0.000001

0.001

= 0.001

(f) Graphing the secant lines:

(g) The secant lines are beginning to look more and more like the tangent line to the

graph of fat the point where x= 0.

(h) The slopes of the secant lines are getting smaller and smaller. They seem tobe approaching the number zero.


75. One, at most, because if fis increasing it could only cross the x-axis at most one time.The graph of fcould not "turn" and cross it again or it would start to decrease.

77. The only such function is f( )= 0.


57/827

Chapter 2

57


2.4 Library of Functions; Piecewise-Defined Functions

1. y = x

3. y =x 3 8

y - intercept : Let x = 0, then y = 0( )3 8 = 8 (0,8)x - intercept : Let y = 0, then 0 = x3 8 x 3 = 8 x = 2 (2,0)

5. piecewise-defined 7. False

9. C 11. E

13. B 15. F

17. f( )= 19. f x( )= x3


58/827


58

21.f x( )=

1

23. f(x) = x3

25. (a) f(2) = (2)2 = 4 (b) (0) = 2

(c) (2) = 2(2) +1 = 5

27. (a) f(1.2) = int (2(1.2) ) = int (2.4 ) = 2 (b) f(1.6) = int (2(1.6)) = int (3.2) = 3

(c) f(1.8) = int (2(1.8)) = int (3.6) = 4

29. f(x) =2 if 01 if x = 0

(a) Domain: x xis any real number{ }(b) x-intercept: none

y-intercept: (0,1)(c)

(d) Range: y y 0{ }

31. f(x) =2 + 3 if < 13x 2 if x 1

(a) Domain: x xis any real number{ }(b) x-intercept: none

y-intercept: (0,3)(c)

(d) Range: y y 1{ }


59/827

Section 2.4 Library of Functions; Piecewise-Defined Functions

59

33. f(x) =+ 3 if 2 < 1

5 if x = 1x + 2 if x >1

(a) Domain: x x 2{ }(b) x-intercept: (2, 0)

y-intercept: (0, 3)(c)

(d) Range: y y < 4{ }and y = 5

35. f(x)=1+ x if x < 0

x 2 if x 0

(a) Domain: x xis any real number{ }(b) x-intercepts: (1,0), (0,0)

y-intercept: (0,0)

(c)

(d) Range: y yis any real number{ }

37. f(x)=

x if 2 x < 0

1 if x = 0

x3 if x > 0

(a) Domain: x x 2{ }(b) x-intercept: none


(d) Range: y y > 0{ }

39. f( )= 2int( ) (a) Domain: x xis any real number{ }(b) x-intercepts: all ordered pairs

( ,0) when 0 < 1.y-intercept: (0,0)

(c)

(d) Range: y yis an even integer{ }

41. f(x)=x if 1x 0

1

2x if 0


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Section 2.4 Library of Functions; Piecewise-Defined Functions

61

Distance (miles)

(100,50)

(400,170)

(800,270)

(b) For hauls between 100 and 400 miles the cost is: C( ) = 10+ 0.40 .(c) For hauls between 400 and 800 miles the cost is: C( ) = 70+ 0.25 .

53. Let x= the amount of the bill in dollars. The minimum payment due is given by

f x( )=

x if x


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62

59. Each graph is that of y = , but either compressed or stretched vertically.

If y =k and k >1, the graph is stretched; if y =k and 0 < k< 1, the graph is

compressed. The graph of y =1

4 x is the same as the graph of y = , but compressed.

The graph of y = 5 is the same as the graph of y = , but stretched.

61. The graph of y = x is the reflection about the y-axis of the graph of y = x .

The same type of reflection occurs when graphing y = 2 +1 andy = 2( )+1.

The graph of y = f( ) is the reflection about the y-axis of the graph of y = f( ) .

63. For the graph of y = xn , n a positive even integer, as n increases, the graph of the functionis narrower for > 1and flatter for < 1.

65. f x( )=1 if xis rational

0 if xis irrational

x xis any real number{ } Range = {0,1}y-intercept: = 0 is rational y =1So they- intercept is (0, 1).x-intercept: y = 0 is irrational So the graph has infinitely many x- intercepts, namely,there is an x-intercept at each irrational value of x.

f( )=1 = f( ) when xis rational; f( )= 0 = f( ) when xis irrational, Sofis even.The graph of fconsists of 2 infinite clusters of distinct points, extending

horizontally in both directions.

One cluster is located 1 unit above the x-axis, and the other is located along the x-axis.


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Chapter 2

63


2.5 Graphing Techniques; Transformations

1. horizontal, right 3. 5, 2, and 2

5. False 7. B

9. H 11. I 13. L 15. F

17. G 19. y = (x 4)3 21. y =x 3 + 4 23. y = x( )3 = x3

25. y = 4x3

27. (1) y = x + 2 29. (1) y = x (2) y = x + 2( )(3) y = x + 2( )= x 2

(2) y = x + 2

(3) y = x + 3 + 2

31. c 33. c

35. f(x) = x2 1Using the graph of y = x2 , verticallyshift downward 1 unit.

37. g(x) = x3 + 1Using the graph of y = x3 , verticallyshift upward 1 unit.


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64

39. h(x) = x 2 Using the graph of y = x , horizontallyshift to the right 2 units.

41. f(x) = (x 1)3 + 2Using the graph of y = x3 , horizontallyshift to the right 1 unit, then vertically

shift up 2 units.

43. g(x) = 4 x

Using the graph of y = x , verticallystretch by a factor of 4.

45. h(x) =1

2x

=1

2

1

x

Using the graph of y =1

, vertically

compress by a factor of1

2.

47. f(x) = x3 Reflect the graph of y = x3 , about thex-axis.

49. ( ) = Reflect the graph of y = about they-axis.


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Section 2.5 Graphing Techniques: Transformations

65

51. h(x) = x3 + 2Reflect the graph of y = x3 about thex-axis, then shift vertically upward 2units.

53. f(x) = 2(x + 1)2 3Using the graph of y = x2 , horizontallyshift to the left 1 unit, vertically stretchby a factor of 2, and vertically shift

downward 3 units.

55. g(x) = x 2 + 1

Using the graph of y = x , horizontallyshift to the right 2 units and verticallyshift upward 1 unit.

57. h(x) = x 2

Reflect the graph of y = x about they-axis and vertically shift downward 2units.

59. f(x) = (x +1)3 1Using the graph of y = x3 , horizontallyshift to the left 1 unit, reflect the graphabout thex-axis, and vertically shift

downward 1 unit.

61. g(x) = 2 1x = 2 1 +x( ) = 2 x 1 Using the graph of y = , horizontallyshift to the right 1 unit, and verticallystretch by a factor or 2.


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66

63. h( ) = 2 int( 1) Using the graph of y = int ( ) ,horizontally shift to the right 1 unit, andvertically stretch by a factor of 2.

65. (a) F( ) = f( ) + 3Shift up 3 units.

(b) G( ) = f( + 2) Shift left 2 units.

(c) P( ) = f( )

Reflect about the x-axis.

(d) ( ) = ( +1) 2

Shift left 1 unit and shift down 2units.


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67

(e) Q(x) =1

2 f(x)

Compress vertically by a factor of1

2.

(f) ( ) = ( ) Reflect about they-axis.

(g) h( ) = (2 )

Compress horizontally by a factor of1

2.




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68

(c) P( ) = f( ) Reflect about the x-axis.

(d) ( ) = ( +1) 2 Shift left 1 unit and shift down 2units.

(e) Q(x) =1

2 f(x)

Compress vertically by a factor of1

2.

(f) ( ) = f( ) Reflect about they-axis.

(g) h( ) = (2 )


2.


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69



(c) P( ) = f( ) Reflect about the x-axis.

(d) ( ) = ( +1) 2 Shift left 1 unit and shift down 2units.

(e) Q(x) =1

2 f(x)

Compress vertically by a factor of

1

2 .

(f) ( ) = f( ) Reflect about the y-axis.


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70

(g) h( ) = (2 )


2.

71. (a)y = +1

10

10

10

10y = + 1

(b) y = 4 x 2

6

10

6

10y = 4 x 2 (c) y = x3 +x

33

10

10y = x3 +x

(d) Any part of the graph of y = f( ) thatlies below the x-axis is reflected aboutthex-axis to obtain the graph of

y = f( ) .

73. (a) y = f( ) (b) y = f x( )


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71

75. f(x) = x2 + 2x f(x) = (x2 + 2x +1) 1f(x) = (x +1)2 1

Using f(x) = x2 , shift left 1 unit and shiftdown 1 unit.

77. f(x) = x2 8x + 1f(x) = x 2 8x + 16( )+ 116f(x) = x 4( )2 15

Using f(x) = x2 , shift right 4 units andshift down 15 units.

79. f(x)=

x2

+x

+1

f(x) = x 2 + x +1

4

+1

1

4

f(x) = x + 12

2

+ 34

Using f(x) =x 2 , shift left1

2unit and

shift up3

4unit.

81. y=

(x

c)

2

If c = 0,y = x 2.

If c = 3,y = (x 3)2 ; shift right 3 units .

If c = 2,y = (x + 2)2; shift left 2 units .

83. F=9

5C+ 32 F=

9

5(K 273) + 32

Shift the graph 273 units to the right.


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72

85. (a)

(b) Select the 10% tax since the profits are higher.

(c) The graph of Y1 is obtained by shifting the graph of p( ) vertically down 10,000

units. The graph of Y2 is obtained by multiplying the y-coordinate of the graph of

p( ) by 0.9. Thus, Y2 is the graph of p( ) vertically compressed by a factor of 0.9.

(d) Select the 10% tax since the graph of Y1 = 0.9p(x) Y2 = 0.05x2 +100x 6800for all in the domain.


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Chapter 2

73


2.6 Mathematical Models; Constructing Functions

1. V = r2h, h = 2r V r( )= r2 2r( )= 2r3

3. (a) R(x) =x 1

6x +100

=

1

6x

2 +100x

(b) R(200) = 1

6(200)

2 +100(200) =20, 000

3+ 20,000 =

40,000

3 $13, 333

(c)

600

16000

00

(d) = 300 maximizes revenue

R(300) = 1

6(300)2 +100(300) = 15,000+ 30,000 = $15,000 = maximum revenue

(e) p = 1

6(300) +100= 50 +100 = $50maximizes revenue

5. (a) If x = 5p +100, then p =100x

5

. R(x)=x100 x

5

= 1

5

x2 + 20x

(b) R(15) = 1

5(15)2 + 20(15) = 45+ 300= $255

(c)

100

600

00

(d) = 50 maximizes revenue

R(50)= 1

5(50)2

+ 20(50) = 500+1000 = $500 = maximum revenue

(e) p =10050

5=

50

5= $10maximizes revenue


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74

7. (a) Let = width and y = length of the rectangular area.

P= 2x + 2y = 400 y =4002

2= 200x

Then A(x) = (200 x)x = 200x x 2 = x2 + 200x (b) We need 0 and y 0 200 0 200

So the domain of A is x 0

x

200

{ }

(c)

200

10000

00

=100 yards maximizes area

9. (a) The distance dfrom Pto the origin is d = x 2 + y2 . Since Pis a point on the graphof y = x2 8, we have:d(x) = x2 + (x2 8)2 = x4 15x2 + 64

(b) d(0) = 04 15(0)2 + 64 = 64 = 8(c) d(1) = (1)4 15(1)2 + 64 = 1 15+ 64 = 50 = 5 2 7.07(d)

(e) dis smallest when 2.74 and when 2.74 .

11. (a) The distance dfrom Pto the point (1, 0) is d = (x 1)2 +y2 . Since Pis a point onthe graph of y = x , we have:

d(x) = (x 1)2 + x( )2 = x2 x +1 (b)

2

2

00

(c) dis smallest when is 0.50.

13. By definition, a triangle has area A =1

2b h, b =base, h = height. From the figure, we know

that b = and h = y.Expressing the area of the triangle as a function of , we have:

A(x) =1

2x y =

1

2x x

3( )= 12

x4.


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Section 2.6 Mathematical Models; Constructing Functions

75

15. (a) A(x) = xy =x 16 x 2( )= x3 +16x (b) Domain: x 0 < x < 4{ }(c)

4

30

00

The area is largest when is approximately 2.31.

17. (a) A(x) = (2x)(2y) = 4x 4 x 2( )1/ 2 (b) p(x) = 2(2x) + 2(2y) = 4x + 4 4 x 2( )1/ 2 (c) Graphing the area equation: (d) Graphing the perimeter equation:

2

10

00

2

12

00

The area is largest when is The perimeter is largest when isapproximately 1.41. approximately 1.41.

19. (a) C= circumference, TA = total area, r= radius, = side of square

C= 2r=10 4x r=5 2

Total Area = area of the square + area of the circle =x 2 + r2

TA(x) =x 2 + 5 2x

2

= x2 +25 20x + 4x2

(b) Since the lengths must be positive, we have:10 4 > 0 and > 0

4x > 10 and x > 0< 2.5 and > 0

Domain: x 0 < x < 2.5{ }(c)

2.5

8

00

The total area is smallest when is approximately 1.40 meters.

21. (a) Since the wire of length is bent into a circle, the circumference is .


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76

Therefore, C( ) = .

(b) Since C= x = 2r, r=2 .

A(x) = r2 = x2

2

= x2

4 .

23. (a) A = area, r= radius; diameter = 2r (b) p =perimeter A( r) = (2 r)(r) = 2r2 p(r) = 2(2r)+ 2r= 6r

25. Area of the equilateral triangle = 12

x 32

x = 34

x 2

Area of1

3of the equilateral triangle =

1

2x r2

x

2

2

=1

2x r2

x 2

4=

1

3

3

4x2

Solving for r2:

1

2

x r2 x2

4

=1

3

3

4

x 2 r2 x 2

4

=2

x

3

12

x 2

r2 x2

4=

3

6x r2

x 2

4=

3

36x2 r2 =

x 2

3

Area inside the circle, but outside the triangle:

A(x)= r2 3

4x 2 =

x2

3

3

4x 2 =

3

3

4

x 2

27. (a)

d

d =40t

d =30t1

2

d2 =d12 +d2

2

d2 = 2 30t( )2 + 3 40t( )2

d t( )= 2 30t( )2 + 3 40t( )2

= 4 120t+ 900t2 + 9 240t+1600t2

= 2500t2 360t+13

(b)

0.150

4

0 The distance is smallest at t 0.072hours.

29. r= radius of cylinder, h = height of cylinder, V= volume of cylinder


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Chapter 2

78


2.R Chapter Review

1. This relation represents a function. Domain = {1,2,4}; Range = {0,3}.

3. f(x)=32 1

(a) f(2) =3(2)

(2) 2 1=

6

4 1=

6

3= 2

(b) f(2) =3(2)

(2)2 1=

64 1

=63

= 2

(c) f(x) = 3( )(x)2 1 =

3x2 1

(d) f(x) = 3x

x 2 1

=

3xx2 1

(e) f(x + 2) =3( + 2)

(x + 2)2 1=

3 + 6x 2 + 4x + 4 1

=3 + 6

x 2 + 4x + 3

(f) f(2x) =3(2 )

(2x)2 1=

6

4x 2 1

5. f(x) = x 2 4

(a) f(2) = 22 4 = 4 4 = 0 = 0

(b) f(2) = 2( )2 4 = 4 4 = 0 = 0(c) f(x) = (x)2 4 = x 2 4

(d) f(x) = x 2 4

(e) f(x +2) = (x + 2)2 4 = x 2 + 4x + 4 4 = x 2 + 4x

(f) f(2x)= (2x)2 4 = 4x 2 4 = 4 x 2 1( )= 2 x 2 1

7.f(x) = x

2 42

(a) f(2) = 22 422

= 4 44

= 04= 0

(b) f(2) =2( )2 42( )2

=4 4

4=

0

4= 0

(c) f(x) = (x)2 4

(x) 2= x

2 4x2


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79

(d) f(x) = x 2 4

x 2

=

4 x 2

x 2

(e) f(x + 2) = (x + 2)2 4

(x + 2)2= x

2 + 4x + 4 4x2 + 4x + 4

= x2 + 4x

x2 + 4x + 4

(f) f(2x)=

(2x)2 4(2x)2 =

4x 2 44x 2 =

4 x 2 1( )4x2 =

x2 1x 2

9. f(x) = 2 9

The denominator cannot be zero:

x2 9 0

(x + 3)(x 3) 0 3 or 3

Domain: x x 3,x 3{

11. f(x) = 2 x The radicand must be non-negative:2 0

2

Domain: x x 2{ }or ,2( ]

13. f(x) = xx

The radicand must be non-negative andthe denominator cannot be zero: > 0Domain: x x > 0{ }or 0,( )

15. f(x) = 2 + 2 3 The denominator cannot be zero:

x 2 + 2x 3 0

x + 3( ) x 1( ) 0x 3 or 1

Domain : x x 3, x 1{ }

17. f( ) = 2 ( )= 3 +1(f+ )( ) = f( )+ ( ) = 2 + 3 +1= 2 + 3

Domain: x xis any real number{ }(fg)(x)= f x( )g(x) = 2 x 3x +1( )= 2 x 3x 1= 4x +1

Domain: x xis any real number{ }(fg)(x) = f(x) g x( ) = 2 x( ) 3x + 1( ) = 6x + 2 3x2 x = 3x2 + 5x + 2

Domain: x xis any real number{ }f

g

(x) =

f x( )g x( )

= 2x

3x +13 +1 0

3x 1x

1

3

Domain : x x 13


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80

19. f(x) = 3x 2 +x + 1 g(x) = 3x (f+g)(x) = f x( )+ g(x) = 3x 2 + x +1 + 3x = 3x2 + 4x + 1

Domain: x xis any real number{ }(fg)(x) = f x( )g(x) = 3x 2 +x + 1 3x = 3x 2 2x + 1

Domain: x xis any real number{ }(fg)(x) = f(x) g x( ) = 3x 2 +x +1( ) 3x( )= 9x

3 + 3x 2 + 3x Domain: x xis any real number{ }

f

g

(x) =

f x( )g x( )

=3x2 +x +1

3x

3 0 0Domain : x x 0{

21. f(x) =+11

g(x) =1

(f+g)(x)= f x( )+g(x) = x +1x 1 + 1x = x x +1( )+1 x 1( )x x 1( ) = x

2

+x + x 1x x 1( )

= x

2 + 2x 1x x 1( )

Domain : x x 0,x 1{ (fg)(x) = f x( )g(x)= x +1

x 1

1

x=

x x +1( )1 x 1( )x x 1( )

= x 2 + x x +1

x x 1( )

= x

2 +1x x 1( )

Domain : x x 0,x 1{ (fg)(x) = f(x) g x( )= x +1

x 1

1

x

=

x +1x x 1( )

Domain : x x 0,x 1{ f

g

(x) =

f x( )g x( )

=

+1x 1

1=

x +1x 1

x

1

=

x2 +x

x 1

Domain : x x 0,x 1{


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81

23. f(x) = 2x2 +x + 1

f x +h( ) f x( )h

=2 x +h( )2 + x + h( )+1 2x2 +x +1( )

h

=2 x2 + 2xh+h 2( )+x +h +1+ 2x 2 x 1

h

=2x2 4xh 2h 2 + x +h +1+ 2x2 x 1

h

=4xh 2h2 + h

h=

h 4x 2h +1( )h

= 4 2h +1

25. (a) Domain: x 4 x 3{ } Range: y 3 y 3{ }(b) x-intercept: 0,0( ); y-intercept: 0,0( )(c) f 2( )= 1

(d) f( )= 3when x= 4(e) ( )> 0when 0 < 3(f) To graph y = f 3( ), shift the graph of fhorizontally 3 units to the right.

(g) To graph y = f1

2x

, stretch the graph of horizontally by a factor of 2.

(h) To graph y = f( ), reflect the graph of fvertically about the y-axis.


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82

27. (a) Domain: x 4 x 4{ } Range: y 3 y 1{ }(b) Increasing: 4,1( )and 3, 4( ); Decreasing: (1, 3)(c) Local minimum 3,3( ); Local maximum 1,1( )(d) The graph is not symmetric with respect to the x-axis, the y-axis or the origin.

(e) The function is neither even nor odd.(f) x-intercepts: 2,0( ), 0,0( ), 4,0( ); y-intercept: 0,0( )

29. f(x) = x3 4x f(x) = (x)3 4(x) = x3 + 4x = x3 4x( )= f(x) is odd.

31. h(x) =1

4 +1

2 +1

h(x) = 1(x)4

+ 1(x)2

+1 = 1x4

+ 1x2

+ 1 =h(x) h is even.

33. G(x) = 1 x +x 3 G(x) =1 (x)+ (x)3 =1+x x 3 G(x) or G(x)

G is neither even nor odd.

35. f(x) =1+ 2

f(x) = 1+ (x)2

= 1+x 2

= f(x) fis odd.

37. f x( )= 2x3 5x +1on the interval 3,3( )Use MAXIMUM and MINIMUM on the graph of y1 = 2x

3 5x +1.

local maximum at: 0.91,4.04( ) ; local minimum at: 0.91,2.04( )fis increasing on: 3,0.91( )and 0.91,3( ); fis decreasing on: 0.91,0.91( )

39. f x( )= 2x4 5x3 + 2x +1on the interval 2,3( ) Use MAXIMUM and MINIMUM on the graph of y1 = 2x

4 5x 3 + 2x + 1.


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83

local maximum at: 0.41,1.53( ) ; local minima at: 0.34,0.54( ) , 1.80,3.56( ) fis increasing on: 0.34,0.41( )and 1.80,3( ); fis decreasing on: 2,0.34( )and 0.41,1.80( )

41. f(x) = 8x2 x

(a)f(2) f(1)

2 1=

8 2( )2 2 8 1( )2 1( )1

= 32 2 7( )= 23

(b)f(1) f(0)

10=

8 1( )2 1 8 0( )2 0( )1

= 8 1 0( )= 7

(c)f(4) f(2)

4 2=

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

=128 4 30( )

2

=94

2

= 47

43. f( ) = 2 5 f( ) f(2)

x 2 =2 5 (8)

x 2 =5 +10

x 2 =5( 2)

x 2 = 5

45. f(x) = 3x 4x2 f(x) f(2)

x 2=

3x 4x 2 (10)x 2

=4x 2 + 3x +10

x 2

= (4x2 3x 10)

2= (4x + 5)(x 2)

2= 4x 5

47. (b), (c), and (d) pass the Vertical Line Test and are therefore functions.

49. f(x) = x3 51. F( ) = 4 Using the graph of y = , vertically shiftthe graph downward 4 units.

Intercepts: (4,0), (4,0), (0,4)

Domain: x xis any real number{ }Range: y y 4{ }


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84

53. ( ) = 2 Reflect the graph of y = about thex-axis and vertically stretch the graph bya factor of 2.

Intercepts: (0,0)


55. h(x) = x 1 Using the graph of y = x ,horizontally shift the graph to the right1 unit.

Intercept: (1,0)

Domain: x x 1{ }Range: y y 0{ }

57. f(x) = 1 x = 1(x 1) Reflect the graph of y = x about they-axis and horizontally shift the graph tothe right 1 unit..

Intercepts: (1,0), (0,1)

Domain: x x 1{ }Range: y y 0{ }

59. h(x) = (x 1)2 + 2Using the graph of y = x2 , horizontallyshift the graph to the right 1 unit and

vertically shift the graph up 2 units.

Intercepts: (0,3)



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85

61. g(x) = 3(x 1)3 + 1Using the graph of y = x3 , horizontallyshift the graph to the right 1 unitvertically stretch the graph by a factor of

3, and vertically shift the graph up 1 unit.

Intercepts: (0,2), 1 33

3 , 0

Domain: x xis any real number{ }Range: y yis any real number{ }

63. f(x) =3x if 2 1

(a) Domain: x x > 2{ }(b) x-intercept: 0, 0( )

y-intercept: (0,0)

(c)

(d) Range: y> 6{ }

65. f(x) =x if 4 x < 01 if x = 03x if x > 0

(a) Domain: x x 4{ }(b) x-intercept: none


(d) Range: y y 4, y 0{ }

67. f(4)= 5 gives the ordered pair (4,5).(0) = 3 gives (0,3).

Finding the slope: m = 3 (5)0 4 =

84 = 2

Using slope-intercept form: ( ) = 2 + 3

69. f(x) = Ax + 56x 2 and f(1) = 4

A(1)+ 56(1) 2 =

4

A+ 54

= 4

A+ 5 = 16A = 11

71. We have the points h1 ,T1( )= 0,30( ) and h2,T2( )= 10000,5( ).

slope = Th

= 5 3010,000 0

= 2510,000

= 0.0025

Using the point-slope formula yields

TT1 =m h h1( )T 30 = 0.0025 h 0( )T 30 = 0.0025h T= 0.0025h + 30

T h( )= 0.0025h + 30


86/827


86

73. S= 4 r2 r= S

4 V(S)=

4

3 r3 =

43

S

4

3

=43

S

4S

4=

S

6

S

V(2S) = 2S6

2S

= 2 2 S

6

S

The volume is 2 2 times as large.

75. S=kxd3, x = width; d= depth In the diagram, depth = diameter of the log = 6

S x( )=k x 6( )3 = 216kx Domain: x 0 x 6{ }

77. (a) We are given that the volume is 500 centimeters, so we have

V = r2h = 500h =500

r2

Total Cost = cost of top + cost of botto + cost of bod

= 2 cost of top( )+ cost of body = 2 area of top( ) cost per area of top( )+ area of body( ) cost per area of body( )

= 2 r2( ) 0.06( )+ 2rh( ) 0.04( )= 0.12r2 + 0.08rh = 0.12r2 + 0.08r

500

r2

SOLU Precalculo, 7ma Edicion - Michael Sullivan

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