Chapter 7: Algebra for College Mathematics Courses · CPM Educational Program © 2012 Chapter 7: Page 1 Pre-Calculus with Trigonometry Chapter 7: Algebra for College Mathematics Courses

CPM Educational Program © 2012 Chapter 7: Page 1 Pre-Calculus with Trigonometry

Chapter 7: Algebra for College Mathematics Courses Lesson 7.1.1 7-1. a. See graph at right.

Increasing: x > 2 ; Decreasing: x < 2 b. As the x-values get larger, the y-values get larger. Or, the slope

of the tangent line is positive. 7-2. a. First x-value is less than the 2nd and both are in the interval [a, b]. b. First y-value is less than the 2nd. c. See graph at right. d. Yes, f (x1) < f (x2 ) . 7-3. a. x1 = 2!!!!!x2 = 3

22 < 23

4 < 8

x1 = 1.7!!!!!x2 = 1.821.7 < 21.8

3.25 < 3.48

b. g(x) is an increasing function on the interval [a, b] if, for every two points x and x + h with a ! x < x + h ! b, h > 0 , g(x) < g(x + h) .

c. 2x+h = 2h !2x . Since h > 0, 2h > 1 ; therefore 2h !2x > 2x . 7-4. a. Something like, “as x gets larger, y gets smaller.” b. If x2 > x1 , then f (x2 ) < f (x1) . c. f (x) is a decreasing function on the interval [a, b] if, for every two points 1x and 2x with

a ! x1 < x2 ! b , then f (x2 ) < f (x1) . d. Given h > 0 , 5 ! (x + h)2 = 5 ! h(2x + h) .

If !" < x < x + h < 0 , then 2x + h < 0 ; therefore !h(2x + h) > 0 and 5 ! (x + h)2 > 5 ! x2 . If 0 < x < x + h < ! , then 2x + h > 0 ; therefore !h(2x + h) < 0 and 5 ! (x + h)2 < 5 ! x2 .

7-6. A graph is concave down over an interval [a, b] if a line segment joining any two points on

the graph over that interval lies completely below the graph. 7-7. See graph at right. Increasing on (!", !1) and (1,!) , decreasing on (!1, 0) and

(0,1) ; concave up on (0,!) , concave down on (!", 0) .

x

y

a bx1 x2

f(x1)f(x2)

x

y f(x)

x

y


7-8. Any line or odd function that passes through the origin, for example:

y = 5x,!y = x5,!y = sin x . Review and Preview 7.1.1 7-9. a. Increasing: (–∞, –3) ∪ (1, 5), Decreasing: (–3, 1) ∪ (5, ∞), Concave Up: (–1, 3),

Concave Down: (–∞, –1) ∪ (3, ∞); b. Decreasing: (–∞, ∞), Concave Up (–∞, 0), Concave down (0, ∞) 7-10. See sample graph at right. 7-11. Brittany; segments connecting any two points on the graph

are above the graph. 7-12. a. b(x) = a(x ! 2) + 5 = (x ! 2)3 ! 3(x ! 2) + 5

b(x) = (x ! 2)(x2 ! 4x + 4) ! 3x + 6 + 5b(x) = x3 ! 4x2 + 4x ! 2x2 + 8x ! 8 ! 3x + 6 + 5b(x) = x3 ! 6x2 + 9x + 3

b. c. W + 25 7-13. a. Amplitude : !7!(!1)

2 = 3

Period : 4"b = 2" !!#!!b = 12

y = 3sin 12 x( ) ! 4 !!or !!

y = 3 cos 12 (x ! " )( ) ! 4

b. Amplitude: 3!(!1)2 = 2

Period: "b = 2" !!#!!b = 2

y = 2 sin 2 x ! "4( )( ) +1!!or !!

y = !2 cos(2x) +1

x

y

x

y

x

y


7-14. 2(x ! 3) + 3

x+1 = 72(x ! 3)(x +1) + 3 = 7(x +1)2(x2 ! 2x ! 3) + 3 = 7x + 7

2x2 ! 4x ! 3 = 7x + 72x2 !11x !10 = 0

7-15. tan!1 2≈ 1.107 radians, (63.435˚) or 2.034 radians (180º – 63.435º = 116.565˚). 7-16.

slope = m = s!ur!t

distance = (r ! t)2 + (s ! u)2

distance = r ! t (r!t )2 +(s!u)2

(r!t )2

d = r ! t (r!t )2

(r!t )2+ (s!u)2

(r!t )2

d = r ! t m2 +1 Lesson 7.1.2 7-17. a. b. Changing of sign of x does not affect f(x). c. y = (a)2 = a2

y = (!a)2 = a2 They are equal.

d. f (a) = f (!a) e. Symmetric about the y-axis. 7-18. a. These functions have even power exponents. b. f (!x) = f (x) c. They are symmetric about the y-axis. d. y = cos x or y = x are good choices.

x

y

(t, u)

(r, s) y = mx + b

y = (2)2 = 4y = (!2)2 = 4y = (3)2 = 9y = (!3)2 = 9y = (1.721)2 = 2.962y = (!1.721)2 = 2.962

x = !(!11)± (!11)2 !4(2)(!10)2(2)

x = 11± 2014


7-19. a. b. Changing the sign of x changes only the sign of f (x) .

c. y = (a)3 = a3

y = (!a)3 = !a3

d. f (!a) = ! f (a) e. Symmetric around the origin. 7-20. a. These functions have odd exponents. b. f (!x) = ! f (x) c. They are symmetric about the origin. d. y = sin x is a good choice. 7-21. a. (–2, 5) b. (3, –5) c. unknown 7-23. a. Even g(!x) = (!x)2 + cos2(!x)

g(!x) = x2 + cos2 x

b. Neither f (!x) = (!x)2 + 3(!x)3

f (!x) = x2 ! 3x3

c. Odd h(!x) = (!x)!1 + 2 sin(!x)h(!x) = !x!1 ! 2 sin x

Review and Preview 7.1.2 7-24. a. cos(!x) = cos(x) and sin(!x) = ! sin x b. Sine is odd, cosine is even. 7-25. a. Any parabola with a vertex on the y-axis. Example: f (x) = x2 ! 3 b. Impossible c. Any parabola with a vertex not on the y-axis. Example: f (x) = 2x2 ! 8x + 5 7-26. Tangent is odd. tan(!x) = sin(!x)

cos(!x) =! sin xcos x = ! sin x

cos x = ! tan x 7-27. Increasing: (–2, 4); Decreasing: (–∞, –2) ∪ (4, ∞);

Concave Up: (–∞, 1); Concave Down: (1, ∞)

y = (2)3 = 8y = (!2)3 = !8y = (3)3 = 27y = (!3)3 = !27y = (1.721)3 = 5.097y = (!1.721)3 = !5.097


7-28. a. 2.9 – 1.7 = 1.2 seconds b. 46 – 20 = 26 c. y = 13 cos 2!

1.2 (x "1.7)( ) + 33 (other answers are possible) d. Because the period of the graph is less than the horizontal shift, one solution is x = 2.0445 !1.2 = 0.8445 . The graph will have another maximum at x = 1.7 !1.2 = 0.5 . Due to the symmetry of the graph, the second solution is x = 0.5 ! (0.8445 ! 0.5) = 0.1555 . 7-29.

logbNPM 2 = 1

2 logb N + 12 logb P ! 2 logb M

= 12 "0.6 +

12 " !1.8 ! 2 "2.1

= 0.3! 0.9 ! 4.2 = !4.8

7-30. a. b. c. 7-31. Interval length = 4!1

10 = 310 = 0.3 x0 = 1, x1 = 1.3, x2 = 1.6, x3 = 1.9, x4 = 2.2, x5 = 2.5,

x6 = 2.8, x7 = 3.1, x8 = 3.4, x9 = 3.7, x10 = 4.0

30 = 13 cos 2!1.2 (x "1.7)( ) + 33

"3 = 13 cos 2!1.2 (x "1.7)( )

" 313 = cos

2!1.2 (x "1.7)( )

cos"1 " 313( ) = 2!

1.2 (x "1.7)

1.804 = 2!1.2 (x "1.7)

0.3445 = x "1.72.0445 = x

!2 f (x)

f (!x) +1

1f (x)


0.3 40.3k+1

k=0

9

! = 6.023 . This is an upper bound because the rectangles are above the curve.


Lesson 7.2.1 7-32. a. See sketch at right. b. h = height, x = width c. V = 4500 = 2x ! x !h = 2x2h

S = 2x ! x + 2x !h + 2x !h + x !h + x !hS = 2x2 + 6xh

d. We want to know the smallest surface area represented by the variable, S. e. 4500 = 2x2h!!!!!h = 4500

2x2= 2250

x2

S = 2x2 + 6x " 2250x2

S = 2x2 + 13500x

f. x = 15, S = 1350

h = 2250x2

= 2250225 = 10

7-33. a. s

6 =(s+20)10

10s = 6s +1204s = 120s = 30

b. s6 =

(s+4t )10

10s = 6s + 24t4s = 24ts = 6t feet

7-34. V = ! r2d = 16!d

16!d = 3t

d(t) = 3t16!

7-35.

rd = 3

88r = 3d

r = 3d8

V = 2t = ! r2d3 = ! (3d 8)2 d

3

2t = ! (3d 8)2 d3

2t = ! 9d2 "d3"64

384t = 9!d3384t9! = d3

d(t) = 4 2t3!

3

10

20 s

6

h

2x x

r

d


Review and Preview 7.2.1 7-36. Graph or average the x-intercepts (x = 0 and x = 60) to find that the product will be a maximum when x = 30. 30 + y = 60!!!!!y = 30

30 + 30 = 60, product = 900

7-37. S = 2x2 + 2x2 + 6xh = 4x2 + 6xh

4500 = 2x2h

7-38. x2 + x2 = (x + 2)2

2x2 = x2 + 4x + 40 = x2 ! 4x ! 4

x = !(!4)± (!4)2 !4(1)(!4)2(1)

x = 4± 16+162 = 4±4 2

2 = 2 ± 2 2

Area = 12 ! (2 + 2 2) ! (2 + 2 2)

Area = 1+ 2( ) 2 + 2 2( )Area = 2 + 2 2 + 2 2 + 4

Area = 6 + 4 2

7-39. a. (7 ! 5 cos")2 = (7 ! 5 cos")(7 ! 5 cos")

= 49 ! 35 cos" ! 35 cos" + 25 cos2 "= 49 ! 70 cos" + 25 cos2 "

b. (sin! + cos!)2 = (sin! + cos!)(sin! + cos!)= sin2 ! + sin! cos! + sin! cos! + cos2 != 1+ 2 sin! cos!= 1+ sin 2!

7-40. y + y + x + x + x = 300

2y + 3x = 3002y = 300 ! 3x

y = 300!3x2

A = xy

A = x 300!3x2( )

A = 150x ! 32 x

2 This is a maximum when x = 50.

x = 502y + 3(50) = 300

2y = 150y = 75 They should be 37.5 ft wide and 50 ft long.

y

y

x x x

x + y = 60!!!!!P = xy

y = 60 ! x!!!!!P = x(60 ! x) = !x2 + 60x

x

x x+2

Diagram for 7-38.


7-41. a. 5(x!2) = 5x "5!2 = 1

52"5x = 1

25 "5x

b. 91 2 x+1 = 32(1 2 x+1) = 3x+2 = 3x ! 32 = 9 ! 3x

c. 60 23( )2x!2 = 60 2

3( )2x 23( )!2 = 60 " 3222 " 2

3( )2#$

%&x= 60 " 94 "

49( )x = 135 4

9( )x

7-42. a. b. 7-43. sin(2A) cos(2A) = 1

4

2 sin(2A) cos(2A) = 12

sin(2 !2A) = 12

sin(4A) = 12

4A = "6 + 2"n,

5"6 + 2"n

A = "24 +

"n2 ,

5"24 +

"n2


Lesson 7.2.2 7-44. a. u = x2 + 2 , 3u ! u = 5 b. u = 4! " 2 or u = sin(4! " 2) , sin2 u ! sin u +1 = 0 or u2 ! u +1 = 0 c. u = 3x

x2 +3, log u + 2u = 7

7-45. a. u = y!5 2

2x + u = 63x ! 2u = !5

u = 6 ! 2x

3x ! 2(6 ! 2x) = !53x !12 + 4x = !5

7x = 7x = 1

u = 6 ! 2(1) = 44 = y!5 2

y = 4!2 5

b. u = x2 + 3xu2y = 5!!!!!u = 10y

3u " 3y = 27

3(10y) ! 3y = 2727y = 27y = 1u = 10(1) = 10

(10)2 = x2 + 3x( )2100 = x2 + 3x0 = x2 + 3x !100

x = !3± 32 !4(1)(!100)2(1)

x = !3± 4092

7-46. a. u + u ! 6 = 0 b. v2 + v ! 6 = 0 c. (v + 3)(v ! 2) = 0 v ! 2 = 0 or v + 3 = 0

v = 2 v = !3

d. !3 " (M 2 + 3M !1)12 No value associated with v = –3.

(2)2 = (M 2 + 3M !1)1 2( )24 = M 2 + 3M !10 = M 2 + 3M ! 5

M = !3± 32 !4(1)(!5)2(1)

M = !3± 292

7-47. Joey needs 16 unit squares. 7-48.


a. y = x2 + 8xy = x2 + 8x + (16 !16)y = (x2 + 8x +16) !16y = (x + 4)2 !16

b. y = x2 + 6x !1y = x2 + 6x + (9 ! 9) !1y = (x2 + 6x + 9) ! (9 +1)y = (x + 3)2 !10


7-49. a. 9 b. c. V = (!3, !6) 7-50.

x2 + 6x + 9 + y2 ! 4y + 4 ! (9 + 4) = 51(x + 3)2 + (y ! 2)2 !13 = 51

(x + 3)2 + (y ! 2)2 = 64 center: ( 3, 2)! ; radius: 8 Review and Preview 7.2.2 7-51. a. 27 b. y = 3x2 !18x + 27 ! 27 +1

y = 3(x2 ! 6x + 9) ! 26y = 3(x ! 3)2 ! 26V = (3, !26)

7-52. y = x2 ! 8x +1

y = x2 ! 8x +16 !16 +1y = (x ! 4)2 !15V = (4, !15)

7-53. a. Let u = x2 + x !1

u2 ! 2u ! 8 = 0u ! 4( ) u + 2( ) = 0u = !2 or 4

!2 = x2 + x ! 2

x2 + x = 0x x +1( ) = 0x = 0 or !1

4 = x2 + x ! 2

x2 + x ! 6 = 0x + 3( ) x ! 2( ) = 0x = !3 or 2

! x = "3, "1, 0, 2

b. Let u = x2 + 5

12u + u = 7! u2 " 7u +12 = 0

u " 3( ) u " 4( ) = 0u = 3 or 4

x2 + 5 = 3 or x2 + 5 = 4

x2 + 5 = 9! x = ±2

x2 + 5 = 16! x = ± 11

" x = ±2 or ± 11

y ! 3 = x2 + 6x + 9 ! 9y ! 3 = (x + 3)2 ! 9

y = (x + 3)2 ! 6


7-54. 2x + y = 120 ! y = 120 " 2x

A = xyA = x(120 " 2x)A = "2x2 +120x

x = 30y = 120 ! 2(30)= 120 ! 60 = 6060ft x 30ft

This is a max when x = 30. 7-55. a. (1! cos")2 + (sin")2 =

1! 2 cos" + cos2 " + sin2 " =1! 2 cos" +1 =2 ! 2 cos"

b. (2 sin!)2 + (2 cos!)2 =4 sin2 ! + 4 cos2 ! =

4(sin2 ! + cos2 !) =4(1) = 4

7-56. a. 4x + 3 ! 12

4x ! 9

x ! 94

b. !6 < 12 x +1 < 8

!7 < 12 x < 7

!14 < x < 14

7-57. y = k

x+6

1 = k1+6

k = 7

y = 7x+6

f (!3) = 7!3+6 =

73

f (0) = 70+6 =

76

f 13( ) = 7

1 3+6 =7

1 3+18 3 =719 3 = 7 "

319 =

2119

f 1a( ) = 7

1 a+6 =7

1 a+6a a =7

(1+6a) a = 7 "a

1+6a =7a1+6a

Lesson 7.2.3 7-58. a. b. 2x4 ! 6x3 + x3 ! 3x2 ! 2x2 + 6x + x ! 3

2x4 ! 5x3 ! 5x2 + 7x ! 3 2x4 + x3 + 4x3 + 2x2 ! 6x ! 3

2x4 + 5x3 + 2x2 ! 6x ! 3

Solution continues on next page. →

B A R N

x

y

x

2x3 +x2 !2x + 1

x 2x4 x3 –2x2 x

–3 –6x3 –3x2 6x –3

x3 +2x2 + 0x –3

2x 2x4 4x3 0x2 –6x

+ 1 x3 2x2 0x –3


7-58. Solution continued from previous page. c. d. x4 + 2x3 ! 3x3 ! 6x2 + 2x2 + 4x + 4x + 8

x4 ! x3 ! 4x2 + 8x + 8 4x3 + 6x2 ! 2x2 ! 3x ! 4x ! 6

4x3 + 4x2 ! 7x ! 6

7-59.

2x !1 6x3 ! 5x2 + 5x ! 26x3 ! 3x2

! 2x2 + 5x! 2x2 + x

4x ! 24x ! 2

0

3x2 ! x + 2

7-60.

x +1 x5 + 0x4 ! 3x3 + 0x2 + 2x ! 5x5 + x4

! x4 ! 3x3

! x4 ! x3

! 2x3 + 0x2

!2x3 ! 2x2

2x2 + 2x !!!!!!!2x2 + 2x !!!!!!! ! 5

x4 ! x3 ! 2x2 + 2x ! 5x+1

7-61. a. b. c. The graph in Y1 follows the graph in Y2 except that it has an asymptote at x = 2. d. The quotient tells you about the general or global shape of the graph.

x3 –3x2 2x 4

x x4 –3x3 2x2 4x

+2 2x3 –6x2 4x 8

2x2 –x –2

2x 4x3 –2x2 –4x

+3 6x2 –3x –6

x ! 2 x3 ! 3x2 + 0x + 5x3 ! 2x2

! x2 + 0x! x2 + 2x

! 2x + 5!2x + 4

1

x2 ! x ! 2 + 1x!2


e. The remainder shows where the asymptotes occur.


Review and Preview 7.2.3 7-62. 7-63.

x !1 x5 + 0x4 + 0x3 + 0x2 + 0x !1x5 ! x4

x4 + 0x3

x4 ! x3

x3 + 0x2

x3 ! x2

x2 + 0x x2 ! x x !1 !x !1 0

x4 + x3 + x2 + x +1 x ! 3 x3 + x2 !14x + 2

x3 ! 3x2

4x2 !14x4x2 !12x

! 2x + 2!2x + 6

! 4

x2 + 4x ! 2 ! 4 x ! 3

7-64. Error is in the following line: y + 5 = !2(x2 + 2x) The line should be: y + 5 = !2(x2 ! 2x) 7-65. a. y = 2x2 ! 8x + 7

y = 2(x2 ! 4x + 4) + 7 ! 8y = 2(x ! 2)2 !1

b. Vertex is at(2, !1).

7-66. Using a graphing calculator, calculate the maximum value (ytr). The maximum value of 108 occurs when x = 3. 7-67. a. (1.02)x = 2

x log1.02 (1.02) = log1.02 2

x = log 2log 1.02 = 35

35 years

b. (1.05)x = 2x log1.05 (1.05) = log1.05 2

x = log 2log 1.05 = 14.207

14 years

c. (1.07)x = 2x log1.07 (1.07) = log1.07 2

x = log 2log 1.07 = 10.245

10 years

d. (1.1)x = 2x log1.1(1.1) = log1.1 2

x = log 2log 1.1 = 7.273

7 years

x + y = 9! y = 9 " xP = xy2

P(x) = x(9 " x)2


e. “The Rule of 70” is called as such because the number of years to double is close to 70 divided by the annual percent growth rate.

7-68. a. x2 ! 7x " !6

x2 ! 7x + 6 " 0(x ! 6)(x !1) " 0[1, 6]!!or !1 " x " 6

b. (x ! 2)(x !1)(x + 3) < 0(!", !3)# (1, 2),! " < x < !3 !or !1 < x < 2

7-69.

a. ( x2 + y2 )3

2 x2 + y2= (x2 + y2 )3 2

2(x2 + y2 )1 2

= (x2 + y2 )3 2!1 2

2= x2 + y2

2

b. 2x5 ! 8x3

x + 2= 2x

3(x2 ! 4)x + 2

= 2x3(x + 2)(x ! 2)(x + 2)

= 2x3(x ! 2)

Lesson 7.2.4 7-70. (x + y)1 = x + y

(x + y)2 = x2 + xy + xy + y2 = x2 + 2xy + y2

(x + y)3 = (x + y)(x2 + 2xy + y2 )= x3 + 2x2y + xy2 + x2y + 2xy2 + y3

= x3 + 3x2y + 3xy2 + y3

(x + y)4 = (x + y)(x3 + 3x2y + 3xy2 + y3)= x4 + 3x3y + 3x2y2 + xy3 + x3y + 3x2y2 + 3xy3 + y4

= x4 + 4x3y + 6x2y2 + 4xy3 + y4

7-71. a. Decrease by 1 each time. b. Increase by 1 each time. c. Each time the sum is the same as the exponent of expansion. 7-73. (x + y)0 = 1 It goes in “Row 0.” 7-74. a. Row 9 b. x9 + 9x8y c. x6 + 6x5y +15x4y2 + 20x3y3 +15x2y4 + 6xy5 + y6


d. 1, 8, 28, 56, 70, 56, 28, 8, 1


7-75.

(x + y)15 = 150

!"

#$ x

15 + 151

!"

#$ x

14y + 152

!"

#$ x

13y2 + 153

!"

#$ x

12y3

= x15 +15x14y +105x13y2 + 455x12y3

7-76. a. x3 + 3x2y + 3xy2 + y3 b. y = 2z

x3 + 3x2(2z) + 3x(2z)2 + (2z)3

x3 + 6x2z +12xz2 + 8z3

c. (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

(x + 3w)4 = x4 + 4x3(3w) + 6x2(3w)2 + 4x(3w)3 + (3w)4

(x + 3w)4 = x4 +12x3w + 54x2w2 +108xw3 + 81w4

7-77. a. (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 b. y = !3w c. (x ! 3w)4 = x4 + 4x3(!3w) + 6x2(!3w)2 + 4x(!3w)3 + (!3w)4 d. (x ! 3w)4 = x4 !12x3w + 54x2w2 !108xw3 + 81w4 e. The signs alternate in the expansion. Review and Preview 7.2.4 7-78. x6 + 6x6!1y0+1 +15x6!2y1+1 + 20x6!3y2+1 +15x6!4y3+1 + 6x6!5y4+1 + y5+1 =

x6 + 6x5y +15x4y2 + 20x3y3 +15x2y4 + 6xy5 + y6

7-79. a. a3 + 3a2b + 3ab2 + b3 b. (2x)3 + 3(2x)2 (!3y) + 3(2x)(!3y)2 + (!3y)3 c. 8x3 ! 36x2y + 54xy2 ! 27y3 7-80. a. x2 + y2 = r2

(!6)2 + (!8)2 = r2

36 + 64 = r2

100 = r2

r = 10

b. (x ! 7)2 + (y ! 5)2 = r2

(3! 7)2 + (!2 ! 5)2 = r2

16 + 49 = r2

65 = r2

(x ! 7)2 + (y ! 5)2 = 65


7-81. a. x2 !10x + y2 + 8y + 5 = 0

x2 !10x + 25 + y2 + 8y +16 = !5 + 25 +16(x ! 5)2 + (y + 4)2 = 36C = (5, !4), r = 6

b. x2 ! 8x + y2 + 6y ! 56 " 0x2 ! 8x +16 + y2 + 6y + 9 " 56 +16 + 9(x ! 4)2 + (y + 3)2 " 81C = (4, !3), r = 9

7-82. 7-83.

x !1 2x4 + 0x3 ! x2 + 3x + 52x4 ! 2x3

2x3 ! x2

2x3 ! 2x2

x2 + 3xx2 ! x

4x + 5 4x ! 4 9

2x3 + 2x2 + x + 4 + 9x!1

2x +1 2x5 + x4 ! 2x3 + 7x2 + 5x ! 22x5 + x4

!! 2x3 + 7x2

!!!!!!!2x3 ! x2

8x2 + 5x 8x2 + 4x

x ! 2

!!!!!! x + 12

!!! ! 52

x4 ! x2 + 4x + 12 ! 5/2

2x+1

7-84. Amplitude: 85!372 = 24 Period: 365b = 2! !!" b = 2!

365 Possible equations: y = !24 cos 2"

365 (x !17)( ) + 61

y = 24 sin 2"365 (x !107)( ) + 61

y = !24 cos 2"365 (44 !17)( ) + 61

y = !24 cos(0.4648) + 61

y = !24 "0.8939 + 61y = !21.45 + 61 = 39.5°

7-85.

a. x2 = x2( )2 + h2

h = x2 ! x24 = 3x2

4 = 32 x

b. V (x) = base ! length

V (x) = 12 xh2 = 6x 3

2 x( ) = 3 3x2

c. 200 = 3 3x2

x2 = 200 39

h = 5.373 ft


Lesson 7.3.1 7-86. a. Each pair equals 101. b. 50 pairs c. 101 x 50 = 5050 7-87. a. 12 x 4 = 48 b. It is twice as large as A. c. 1000 by 999 7-88. a. 1100 by 899 b. 1100!899

2 = 494, 450 7-89. n = number of terms, a

1= first term of the sequence, and an = nth term of the sequence.

7-90. a. 10.2 !10 = 0.2 b. 49 times c. 49 !0.2 = 9.8

n50 = 19.825 pairs, each pair = 19.8 +10 = 29.825 !29.8 = 745

7-91.

57!298 = 3.5!!!!!S = n a1+an( )

2 = 965!!!!!a1 = 29 ! 4(3.5) = 15!!!!!an = 15 + 3.5(n !1)

965 =n 15+ 15+3.5 n!1( )( )( )

2 = n 30+3.5n!3.5( )2 = n 26.5+3.5n( )

2 = 26.5n+3.5n22

1930 = 26.5n + 3.5n2

3.5n2 + 26.5n !1930 = 0

Review and Preview 7.3.1 7-92. a. (–∞, –1) ∪ (1, ∞) b. (0, ∞) c. (–∞, 0)

d. It is odd. f (!x) = (!x)2 +1!x = ! x2 +1

x = ! f (x) 7-93. SA = 2! rh + ! r2 = 200!!"!!2! rh = 200 # ! r2 !!or !!h = 200#! r2

2! r

V (r) = ! r2h = ! r2 200"! r22! r( ) = r

2 (200 " ! r2 ) = 100r " ! r3

2

Using a graphing calculator yields a maximum value of 307.106 cm3 when r = 4.607.

If r = 4.607 then h = 200!" (4.607)22" (4.607) = 4.607 .

n = !26.5± 26.52 !4(3.5)(!1930)2(3.5) = !26.5±166.5

7 = 20


7-94.

x + 4 2x3 + x2 !19x + 362x3 + 8x2

! 7x2 !19x!7x2 ! 28x

9x + 36 9x + 36

0

2x2 ! 7x + 9

7-95. a. 30(3+90)

2 = 1395 b. 41(20+100)2 = 2460

c. 41(20+100)2 = 2460 d. 46(37+262)

2 = 6877 7-96. a. x3 ! xy2 = x(x2 ! y2 ) = x(x + y)(x ! y) b. x3 + xy2 = x(x2 + y2 ) c. 4x2(x2 + y2 )1 2 ! 4(x2 + y2 )3 2

(x2 + y2 )1 2(4x2 ! 4(x2 + y2 ))!4y2(x2 + y2 )1 2

7-97. (1+ 2 sin!)2 + (2 cos!)2 =

1+ 4 sin! + 4 sin2 ! + 4 cos2 ! =

1+ 4 sin! + 4(sin2 ! + cos2 !) =1+ 4 sin! + 4 =5 + 4 sin!

7-98.

a. limx!4

f (x) = 4 b. limx!0"

f (x) = "2 c. limx!"#

f (x) = "3

d. limx!0

f (x) " does not exist e. limx!"3

f (x) = #

7-99. a = 200,!r = 1.01,!n = 12

200(1.0112 !1)0.01 = $2536.50


Lesson 7.3.2 7-100. a. 1+ 3+ 9 + 27 + 81+ 243+ 729 = 1093 b. 3 ! 729 = 2187 , 3 is the multiplier. c. 3S = 3 + 9 + 27 + 81 + 243 + 729 + 2187 d. 2S = 2186 S = 1 + 3 + 9 + 27 + 81 + 243 + 729 S = 1093 3S – S = 2S = 3279 – 1093 = 2186 This is twice as much as what was found in part (a). 7-101. a. S = 1+ 5 + 25 +125 + 625 + 3125 +15625

5S = 5 + 25 +125 + 625 + 3125 +15625 + 781255S ! S = 78124 !1 = 781234S = 78123S = 19531

b.

S = 1+ 6 + 36 +!+ 77766S = 6 + 36 + 216 +!+ 466566S ! S = 46656 !1 = 466555S = 46655S = 9331

c.

S = 5 +15 + 45 +…+ 885735S = 5 ! 30 + 5 ! 31 + 5 ! 32 +…+ 5 ! 311

3S = 5 ! 31 + 5 ! 32 + 5 ! 33 +…+ 5 ! 312

3S " S = 5 ! 312 " 5 ! 302S = 2657205 " 5 = 2657200S = 1, 328, 600

d.

S = 10000 +1000 +…+ 0.001S = 104 +103 +…+10!3

10S = 105 +104 +…+10!2

10S ! S = 105 !10!39S = 100000 ! 0.0019S = 99, 999.999S = 11,111.111

7-103. The first term has no power of r, so we need to stop at (n !1) . 7-104. a. 200(1+ 0.01)11 = 200(1.01)11 b. 200(1.01)

11 + 200(1.01)10 +!+ 200 c. a = 200,!r = 1.01,!n = 12

200(1.0112 !1)0.01 = $2536.50

7-105. a. a = 200,!r = 1.01,!n = 24

200(1.0124 !1)0.01 = $5394.69

b. a = 200,!r = 1.01,!n = 60200(1.0160 !1)

0.01 = $16, 333.93

c. a = 200,!r = 1.01,!n = 120200(1.01120 !1)

0.01 = $46, 007.74

7-106. a. a = 200,!r = 1.01,!n = 240

200(1.01240 !1)0.01 = $197, 851

b. a = 200,!r = 1.01,!n = 360200(1.01360 !1)

0.01 = $698, 992.83


c. a = 200,!r = 1.01,!n = 480200(1.01480 !1)

0.01 = $2, 352, 954.50


Review and Preview 7.3.2 7-107. S = 1! 3+ 9 ! 27 + 81! 243+ 729 ! 2187 = !1640 . The method still works when r < 0. 7-108. a. 4(5+x)

2 = 2004(5 + x) = 4005 + x = 100

x = 9595!53 = 30

Series = 5 + 35 + 65 + 95

b. 200 = 5 + 5r + 5r2 + 5r3

195 = 5r3 + 5r2 + 5r39 = r3 + r2 + rr = 3

Series = 5 +15 + 45 +135

7-109.

a. S = 3(211!1)2!1 = 3(2047) = 6141

b. 20(800+1560)2 = 23, 600

c. S = 0.02(311!34 )3!1 = 0.02(177,066)

2 = 3541.322 = 1770.66

d. 12 +

22 +

32 +

42…+ 20

2 =20 1

2+10!

"#$%&

2 =20 21

2( )2

= 10(21)2

= 105

7-110.

S = 100(1.00512 !1)1.005!1 = 100(0.0617)

0.005 = 1233.56 7-111.

2x ! 3 4x4 ! 2x3 + 0x2 ! 7x ! 54x4 ! 6x3

!!4x3 ! 0x2

!!!! 4x3 ! 6x2

!!!6x2 ! 7x!!!6x2 ! 9x

!!!!!2x ! 5 !!!!!!!2x ! 3 !!!! ! 2

2x3 + 2x2 + 3x +1! 22x!3

7-112. 500 = 4x + 2y!!!!!y = 250 " 2x A = xy = x(250 ! 2x) This is a maximum when x = 62.5 ft. Therefore y = 125 ft and the maximum area is 7812.5 ft2.

x

y

Diagram for 7-112.

di


7-113. a. sin A = 4

5 b. cos B = 1213

c. sin(A + B) = sin A cos B + sin B cos A= 4

5( ) 1213( ) + 5

13( ) 35( ) = 48

65 +1565 =

6365

d. cos(A + B) = cos A cos B ! sin A sin B= 3

5( ) 1213( ) ! 4

5( ) 513( ) = 36

65 !2065 =

1665

e. tan(A + B) = sin(A+B)cos(A+B) =

63/6516/65 =

6365( ) 65

16( ) = 6316

7-114. a. d = (2 ! 0)2 + (5 ! (!3))2 = 4 + 64 = 68 = 2 17

b. midpoint = 0+22 , !3+52( ) = (1,1)

slope = 5!(!3)2!0 = 8

2 = 4

slope "= ! 14

(y !1) = ! 14 (x !1)

y = ! 14 (x !1) +1

7-115. a. ar ! a = 24 !!!!!!!!!!a(r !1) = 24

ar4 ! ar3 = 648!!!!!!ar3(r !1) = 648 Dividing the equations yields:

r = 3!!!!!a(3"1) = 24 !!!!!a = 12 b. Since we are looking at the difference the equations in part (a) can be written as: ar ! a = 24

ar4 ! ar3 = 648 Other solutions will come from: ar ! a = !24

ar4 ! ar3 = !648

In this case a = –12. If r = –3, then a(!3!1) = 24 !!"!!a = !6!!!or!!!!a(!3!1) = !24 !!"!!a = 6 .

Thus all four solutions in the form (a, r) are (12, 3), (–12, 3), (–6, 3), (6, 3).

A 3

4 5

B

5

12

13

1r3

= 24648 =

127 !!!!!r = 3


Lesson 7.3.3 PROBLEM SET A

1. 103

!"

#$

12( )3 1

2( )10%3 = 120 12( )10 & 0.117

2. 54

!"

#$ (0.6)

4 (0.4)5%4 = 5(0.6)4 (0.4)1 & 0.259

3. 44

!"

#$ (0.8)

4 (0.2)4%4 = (0.8)4 & 0.410

4. 43

!"

#$

34( )4%3 1

4( )3 = 4 34( )1 1

4( )3 & 0.0469

5. 42

!"

#$

15( )2 4

5( )2 = 6 15( )2 4

5( )2 % 0.154

6. 63

!"

#$

16( )3 5

6( )6%3 = 20 16( )3 5

6( )3 & 0.054

7. 1816

!"

#$

910( )16 1

10( )18%16 = 153 910( )16 1

10( )2 & 0.284

8. 52

!"

#$ 0.3( )2 0.7( )5%2 = 10 0.3( )2 0.7( )3 & 0.309

7-117.

a. P(R, R) = 23( ) 2

3( ) = 23( )2

b. Tracing along the branches of the tree: R, then B = 23 !13 .

c. Tracing along the branches of the tree: B, then R = 13 !23 .

d. Using the results from parts (b) and (c) indicates that the probability of getting one red and one blue = 23 !

13 +

13 !23 = 2

13 !23( ) .

7-118. a. P(B, B) = 14 !

14 =

116

b. You can get red then blue or blue then red. 7-119. a. See diagram at right. b. Using the diagram, P(2 reds) = p2 .

c. Using the diagram, P(2 blues) = q2 . d. Using the diagram, P(one red and one blue) = pq + pq = 2pq . e. p2 + 2pq + q2 = (p + q)2 = 12 = 1

p

p

q

q p

q

= p2

= pq

= pq

= q2


7-120. a. sin2 u ! sin u + 0.24 = 0 b. Let v = sin u . c. v2 ! v + 0.24 = 0 d. v = sin(3x ! 5) 7-121. (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 Let x = a and y = bc.

a4 + 4a3(bc) + 6a2(bc)2 + 4a(bc)3 + (bc)4 =a4 + 4a3bc + 6a2b2c2 + 4ab3c3 + b4c4

7-122. a. See diagram at right. b. 0.73 = 0.343 c. 0.33 = 0.027 d. 3(0.7)2(0.3) = 0.441 e. 3(0.7)(0.3)2 = 0.189 7-123. a. y = k x

11 = k ! 411 = 2k

k = 112

y = 112 x

b. f (x) = 112 x

f (0) = 112 0 = 0

f (4) = 112 4 = 11

f (8) = 112 8 = 11

2 !2 2 = 11 2

f (a2 ) = 112 a2 = 11 a

2

7-124.

sin!1+ cos!

"1# cos!1# cos!

= sin!(1# cos!)1# cos2 !

= sin!(1# cos!)sin2 !

= 1# cos!sin!

1! cos"sin"

= 1sin"

! cos"sin"

= csc" ! cot" or tan "2( )

7-125.

0.7 0.7

0.3 0.3

0.3

0.3

0.3 0.3

0.3

0.7

0.7 0.7

0.7

0.7


a. x2 + x ! 6 = 0(!3)2 + (!3) ! 6 = 0

9 ! 3! 6 = 0

x2 + x ! 6 = 0(2)2 + (2) ! 6 = 0

4 + 2 ! 6 = 0

b. x2 + 2x ! 6 = 0

(!1+ 7)2 + 2(!1+ 7) ! 6 = 0

1! 2 7 + 7 ! 2 + 2 7 ! 6 = 0

!2 7 + 2 7 +1+ 7 ! 2 ! 6 = 00 = 0


7-126. d1 = d2 !!!!!t1 + t2 = 10!!!!!t2 = 10 " t1

r1t1 = r2t210t1 = 15(10 " t1)10t1 = 150 "15t125t1 = 150t1 = 6d = r1t1 = 10 #6 = 60 miles

7-127. If the y-axis is a line of symmetry then there is not a horizontal shift. The line y = 15

touches either the top or the bottom of the graph. Since the point (20, 50) is on the graph, the line y = 15 must touch the bottom. If (20, 50) is the next point of symmetry and in the middle, then the period is 80. Therefore the amplitude is 50 !15 = 35 and 80b = 2! !or !b = !

40 . Hence a possible equation is y = 35 cos ! x40( ) + 50 .

7-128. a. The zeros are at x = 1, 3, and 5. x < 1 or 3 < x < 5 The intervals to check are (!",1),!(1, 3), (3, 5),!and!(5,") . Choose a point in each interval and check to see if it makes the inequality true. (!",1)!!choose!x = 0!!#!!(1! 0)(0 ! 3)(0 ! 5) = 15 > 0!!true

(1, 3)!!choose!x = 2!!#!!(1! 2)(2 ! 3)(2 ! 5) = !3 /> 0!!false(3, 5)!!choose!x = 4 !!#!!(1! 4)(4 ! 3)(4 ! 5) = 3 > 0!!true(5,")!!choose!x = 6!!#!!(1! 6)(6 ! 3)(6 ! 5) = !15 /> 0!!false

Therefore the solution set is x < 1!!or !!3 < x < 5 . b. x2 ! 2x !15 < 0!!"!!(x ! 5)(x + 3) < 0 The zeros are at x = –3 and 5. The intervals to check are (!", !3), (!3, 5),!and!(5,") . Choose a point in each interval and check to see if it makes the inequality true. (!", !3)!!choose!x = !4 !!#!!(!4 ! 5)(!4 + 3) = 9 /< 0!!false

(!3, 5)!!choose!x = 0!!#!!(0 ! 5)(0 + 3) = !15 < 0!!true(5,")!!choose!x = 6!!#!!(6 ! 5)(6 + 3) = 9 /< 0!!false

Therefore the solution set is 3 < x < 5 . 7-129. a. d = kf

2 = k !10k = 0.2d = 0.2 f

b. d = 0.2 f

3 = 15 f

f = 15 pounds


Chapter 7 Closure 7-130. a. The function must be cosine because it is even. If the increasing regions repeat every 4

units, then the period is 8 units. Since amplitude = 10 and 8b = 2! !or !b = !4 , a possible

equation is y = 10 cos !4 x( ) .

b. y = 12 x

odd!# c. The given information indicates that there is a vertical asymptote at x = 2 and a horizontal

asymptote at y = –1. Therefore this is a rational function. A possible equation is y = 1

x!2 !1 . d. A function that has only one horizontal asymptote is an exponential function. Since it is

concave down it is reflected over the x-axis. Since the horizontal asymptote is y = –4, the function has been shifted down 4 units. Thus a possible equation is y = !(2)x ! 4 .

e. Since the asymptotes are at x = –2 and x = 2, a possible equation is y = 1(x+2)(x!2) =

1x2 !4

.

f. This will be an odd power function centered at x = 4. A possible equation is y = !(x ! 4)3 . CL 7-131. See graph below right. a. !2 < x < 2 b. x < !2 and x > 2 c. x < 0 d. x > 0 CL 7-132. SA = 2! r2 + 2! rh

V = ! r2h

h = V! r2

SA = 2! r2 + 2! r V! r2( )

SA = 2! r2 + 2Vr

CL 7-133. 15, 050 = 100(2+a2 )

230,100 = 100(2 + a2 )301 = 2 + a2299 = a2

CL 7-134. 100 + 90 + 80 + ...! 20 !18 !16 ! ...

10(100+10)2 ! 10(20+2)2 = 1100

2 ! 2202 = 550 !110 = 440

Separate the positive terms from the negative. Each forms an arithmetic sequence.


CL 7-135.

x ! 2 3x2 ! 2x +13x2 ! 6x

4x +1 4x ! 8

9

3x + 4 + 9x!2

The graph looks like the line 3 4x + globally, but has an asymptote at 2x = . CL 7-136. (3+ 2x !1)2 + 24 = 10(3+ 2x !1) Let u = 3+ 2x !1 . u2 + 24 = 10u

u2 !10u + 24 = 0(u ! 6)(u ! 4) = 0

u = 63+ 2x !1 = 6

2x !1 = 32x !1 = 92x = 10x = 5

u = 43+ 2x !1 = 4

2x !1 = 12x !1 = 12x = 2x = 1

u = 6!or !u = 4 Both answers check. CL 7-137. The signs are alternating so this must be subtraction. By looking at the pattern on the

exponents, the powers of x are decreasing by 1 and the power of y are increasing by 2. Therefore start with (ax ! by2 )? . The missing power must be 7 because x5(y2 )2 !!!!!5 + 2 = 7 . Now use the binomial formula to find a and b. 75

!"

#$ (ax)

5 (by2 )2 = 84x5y4

21a5x5b2y4 = 84x5y4a5b2 = 4

74

!"

#$ (ax)

4 (by2 )3 = %280x4y6

%35a4x4b3y6 = %280x4y6a4b3 = 8

Dividing the 2 new equations yields: ab =12 !!or !!2a = b .

Substitute: a5b2 = 4a5(2a)2 = 44a7 = 4a = 1!!!!!b = 2

Therefore the binomial is (x ! 2y2 )7 .


CL 7-138. a. The series is arithmetic with a difference of 0.5. There are 161 terms in the series. S161 =

161(10+90)2 = 8050

b. The series is geometric with a ratio of 1.05. There are 21 terms in the series.

an = 20(1.05)n!120(1.05)20 = 20(1.05)n!1

20 = n !121 = n

S21 =20(1!1.0521)1!1.05 = 714.385

c. The series is geometric with a ratio of ! 12 .

There are 7 terms in the series.

S7 =400 1! ! 12( )7"

#$%

&'

1! ! 12( ) = 268.75

7-139. After 1 hour the area of the base = 81!

59 = 12

base radius (x)

5x = 108x = 21.6

After 3 hours the area of the base = 1465.74 ft2

CL 7-140. x2 ! 4x + y2 + 6y = 12

x2 ! 4x + 4 + y2 + 6y + 9 = 12 + 4 + 9(x ! 2)2 + (y + 3)3 = 25

Center (2, –3) CL 7-141. a. 5 ! 3(x+2) = k ! 3x

5!3x32

3x= k

5 !9 = kk = 45

b. 6 !2(x+k) = 24 !2x

6!2x2k

2x= 24

6 !2k = 242k = 4k = 2

c. 20 !23x"1 = 10 ! kx

2 !23x"1 = kx

2(3x"1)+1 = kx

23x = kx

23( )x = kx8x = kx

k = 8

an = 10 + 0.5(n !1)90 = 10 + 0.5(n !1)80 = 0.5(n !1)160 = n !1161 = n

an = 400 ! 12( )n!1

6.25 = 400 ! 12( )n!1

164 = ! 1

2( )n!112( )6 = ! 1

2( )n!16 = n !17 = n

slope = 1!(!3)5!2 = 4

3

" slope = ! 34

y !1 = ! 34 (x ! 5)!!or !!y + 3 = ! 3

4 (x ! 2)


Note: x ! 0


CL 7-142. a. 1+cos(2x)

sin(2x) =

1+1!2 sin2 x2 sin x cos x =

2!2 sin2 x2 sin x cos x =

1!sin2 xsin x cos x =

cos2 xsin x cos x =

cos xsin x = cot x

b. cos2 x4( ) ! sin2 x

4( ) =cos 2 x

4( )( ) = cos x2( )

c. (1+ cot2 y)(cos 2y +1) =

1+ cos2 ysin2 y

!"#

$%& cos2 y ' sin2 y +1( ) =

cos2 y ' sin2 y +1+ cos4 ysin2 y

' cos2 y + cos2 ysin2 y

=

1' sin2 y + cos4 y+cos2 ysin2 y

=

cos2 y + cos4 y+cos2 ysin2 y

=

cos2 y sin2 y+cos4 y+cos2 ysin2 y

=

cos2 y(sin2 y+cos2 y+1)sin2 y

=

cos2 y(2)sin2 y

= 2 cot2 y

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Chapter 7: Algebra for College Mathematics Courses · CPM Educational Program © 2012 Chapter 7: Page 1 Pre-Calculus with Trigonometry Chapter 7: Algebra for College Mathematics Courses