Chapter 2: Rocket Launch Lesson 2.1.1. 2-1.

2-2. a.

b.

Domain: !" # x # " Range: 2 ! y ! " y-intercept ! y = 2 no x-intercepts

Time Hours sitting 8PM 1 9PM 2 10PM 3 11:30PM 4.5 12:00 5 12:30AM 5.5 1AM 6 2AM 7

f (x) =

{

4t 6(t " 5) + 20

Amount Earned \$4 \$4*2hrs = \$8 \$4*3hrs = \$12 \$4*4.5hrs = \$18 \$4*5 = \$20 \$4*5+\$6*½ = \$23 \$4*5+\$6*1 = \$26 \$4*5+\$6*2 = \$32

0!t !5 t>5

2-3. a. x –5 –4 –3 –2 –1 0 1

For x ≤ 1 y = x2 + 2 y = (!5)2 + 2 = 25 + 2 = 27 y = (!4)2 + 2 = 16 + 2 = 18 y = (!3)2 + 2 = 9 + 2 = 11 y = (!2)2 + 2 = 4 + 2 = 6 y = (!1)2 + 2 = 1 + 2 = 3 y = (0)2 + 2 = 0 + 2 = 2 y = (1)2 + 2 = 1 + 2 = 3

b.

x 1 2 3 4 5 6 7 c.

Closed at (1, 3), open at (1, 9).

For x > 1 y = 2x + 7 y = 2 !1 + 7 = 9 y = 2 ! 2 + 7 = 11 y = 2 ! 3 + 7 = 13 y = 2 ! 4 + 7 = 15 y = 2 ! 5 + 7 = 17 y = 2 ! 6 + 7 = 19 y = 2 ! 7 + 7 = 21

D = (!", "); R = [2, ") ;

2-5.

x =1 2-6. a. b.

D = (!", "); R = (!", 5) No, jump at x = 2 .

Review and Preview 2.1.1 2-7. a. b. c. d.

\$8.00 per hour * 5 hours = \$40.00 Rate of pay = \$8.00 per hour Sample graph at right. 5 Each rectangle represents eight dollars of Kristofâ&#x20AC;&#x2122;s earnings. (\$8/hour) ! (hours) = \$8

2-8.

f (x + 2) = 3(x + 2)2 + 5(x + 2) ! 2 = 3(x 2 + 4x + 4) + 5x + 10 ! 2 = 3x 2 + 12x + 12 + 5x + 10 ! 2 = 3x 2 + 17x + 20 2-9. f (x + h) = 3(x + h)2 + 5(x + h) ! 2 = 3(x 2 + 2xh + h 2 ) + 5x + 5h ! 2 = 3x 2 + 6xh + 3h 2 + 5x + 5h ! 2 = 3x 2 + 6xh + 5x + 3h 2 + 5h ! 2 = 3x 2 + x(6h + 5) + (3h 2 + 5h ! 2)

2-10. a.

b.

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

c.

(x + 2)(x + 2) + 6(x + 2) = (x + 2)(2(x + 2) + 6) = (x + 2)(2x + 4 + 6) = (x + 2)(2x + 10) = 2(x + 2)(x + 5)

2-11. a. 5x 2 ! 15x = 0 5x(x ! 3) = 0 Either 5x = 0 or x ! 3 = 0 x = 0 or x = 3

2-12. a. (32 )! 3 2 = 3!3 = c.

( ) 53 43

2 3

=! "

1 = 1 27 33 2 3 5 3# 5 2 = 4 \$ 4

( )

( )

=

25 16

b.

2(x ! 6)2 = 18 (x ! 6)2 = 9 x ! 6 = Âą3 Either x ! 6 = 3 or x ! 6 = !3 x = 9 or x = 3

b.

(33 )2 3 = 32 = 9

d.

( 3!2 )!1 2 = 31 = 3

2-13. a.

d=

( 5 ! ( !2 ) )2 + ( !2 ! 5 )2

= 7 2 + ( !7 ) = 49 + 49 = 98 = 7 2 2

b.

2-14. a.

!2 ! 5 !7 = = !1 5 ! ( !2 ) 7 point-slope!form:!!y ! 5 = !(x + 2) y + 2 = !(x ! 5)

slope =

! 6

slope ! intercept form:!!y ! 5 = !x ! 2 y = !x ! 2 + 5 y = !x + 3 3" 4

b.

!

b.

q(x) = (x + 2)2 + (x + 2) ! 2

Lesson 2.1.2 2-15. a. Shift to the left two units.

= x 2 + 4x + 4 + x + 2 ! 2 = x 2 + 5x + 4 2-16. g(x) should be shifted to the right one unit and up three units.

2-17. a and b.

c. d.

2-18. a. Shifts left 2 and down 3. # (x + 2)2 ! 3 !!for x < 0 b. k(x) = \$ %& x ! 1 !!!!!!!!!!!!!for x " 0

At x = 3 # (x ! 1)2 !!!for!x < 3 h(x) = \$ %& (x + 1)!!!!!for!x " 3

c.

2-19. a. b. c. d.

Any multiple of 5 will work; 5n where n is any integer. D = (!", "); R = [2, 4] (0, 2) No asymptotes – translational symmetry is present. The function is made up of line segments but it does not belong to any previously studied family.

2-20. a. Yes, because the difference in the x-values, 17 – 2 = 15 which is a multiple of the period, 5. b. Yes, shifting the function horizontally by its period will result in a graph that is identical to the original. !2, 12 , 3, 5 12 , 8, 10 12 , 13, 15 12 c.

2-21. a. p=7 b. The values visible on the graph are ≈ –2.5, .5, 4.5, and 7.5. We want students to find two more solutions using the idea that –2.5 +7n and –0.5 + 7n in which n is an integer are also solutions. c. g(53) = g(4 + 49) = g(4 + 7p) = g(4) = 2

Review and Preview 2.1.2 2-22. a. b.

2-23. a.

See graph at right. f (x) = x 3 ! 3x g(x) = y = (x + 2)3 ! 3(x + 2)

If f (x) = x 3 ! 3x f (x + 2) = (x + 2)3 ! 3(x + 2) g(x) = f (x + 2).

f (x) = 2 x ! 3 + 1 f (x) = x f (x) = x ! 3 (right 3 units) f (x) = 2 x ! 3 (change slope to 2) f (x) = 2 x ! 3 + 1 (up 1 unit)

b.

f (x) = 2 f (x + 5) = 2 f (x + 5) ! 3 = 2 g(x) = 2

x ! 3 +1 x ! 3+ 5 +1 x + 2 +1! 3 x+2 !2

c.

2-24. y = x 5 ! 2x y = (x ! 2)5 ! 2(x ! 2) (right 2 units) y = (x ! 2)5 ! 2(x ! 2) + 3 (3 units up)

2-25. a. b. c.

The graphs are both cubic functions, but are in different places on the axes. The second graph is the first graph shifted to the left 2 units. f (x) = x 3

f (x + 2) = (x + 2)3 (shifted left 2 units) 2-26. a. (2x + 3)(2x + 3)

b.

(x ! a)(x ! b)

= 4x 2 + 6x + 6x + 9

= x 2 ! ax ! bx + ab

= 4x 2 + 12x + 9

= x 2 ! x(a + b) + ab

2-27. a.

c 2 = a 2 + b 2 ! 2 " a " b " cos(x)

b.

6 2 = 8 2 + 10 2 ! 2 " 8 "10 " cos(x) 36 = 164 ! 160 " cos(x) 128 = 160 " cos(x) 0.8 = cos(x)

a sin A x sin 60Â° x 0.866 x 0.866

= = =

b sin B 28 sin 70Â° 28 0.940

= 29.787

x = 25.8!

cos!1 (0.8) = x 36.9! = x 2-28. 2x 2 + 14x ! 16 a.

b.

= 2(x 2 + 7x ! 8) = 2(x + 8)(x ! 1) 2-29. a.

2x 3 ! 128x = 2x(x 2 ! 64) = 2x(x + 8)(x ! 8)

a 2 + a 2 = 10 2

b.

cos(60! ) =

x 8 x 8

2a 2 = 100

0.5 =

a 2 = 50

4cm = x

tan(30! ) =

c.

1 3

=

3 x 3 x

x = 3 3cm

a = 50 = 5 2cm 2-30. 4, 9, 16, â&#x20AC;Ś, 512

Lesson 2.2.1 2-31.

a. b. c.

2 3 + 2 4 + 2 5 + 2 6 = 8 + 16 + 32 + 64 = 120 (4 ! 32 + 2) + (4 ! 4 2 + 2) + (4 ! 5 2 + 2) = 38 + 66 + 102 = 206

( 12 ) + 16( 12 ) + 16( 12 ) + 16( 12 ) + 16( 12 ) = 1 + 16 1 = 8 + 4 + 2 + 1 + 1 = 15.5 16 ( 12 ) + 16 ( 14 ) + 16 ( 18 ) + 16 ( 16 ) ( 32 ) 2 16

1

2

3

4

5

d.

(3 ! 2 + 1) + (3 ! 3 + 1) + (3 ! 4 + 1) + (3 ! 5 + 1) = 7 + 10 + 13 + 16 = 46

e.

1 1(1+1)

f.

Argument is

+

1 2(2+1)

+

1 1 1 + 4(4+1) = 12 + 16 + 12 3(3+1) 4n 2 + 2 . The index is n.

+

1 20

=

30 60

+ 10 + 60

2-32. Did not use integer values for p. Need to have p = 5, 6, 7.

5 60

+

3 60

=

30+10+5+ 3 60

=

48 60

=

4 5

2-34.

two coordinates = (1, 2.2) and (5, 3) slope =

2-35.

3!2.2 5!1

=

0.8 4

8 " 1 = 0.2 = 10 4

point-slope y ! 2.2 = 0.2(x ! 1) or y ! 3 = 0.2(x ! 5)

2.2 + 2.4 + 2.6 + 2.8 + 3.0 = (1! 0.2 + 2) + (2 ! 0.2 + 2) + (3 ! 0.2 + 2) + (4 ! 0.2 + 2) + (5 ! 0.2 + 2) =

slope-intercept y = 0.2x ! 1 + 3 y = 0.2x + 2

5

" 0.2k + 2

k =1

2-36. 3.6 + 4.0 + 4.4 + 4.8 + 5.2 = a.

3.6 + (1! 0.4 + 3.6) + (2 ! 0.4 + 3.6) + (3 ! 0.4 + 3.6) + (4 ! 0.4 + 3.6) =

4

"

0.4k + 3.6

k =0

b.

Start the index at 1 and end it at 5.

2-37. a. The difference between each term is 0.4 and the first term is 3.6. b. 0.4 ! 50 + 3.6 = 20 + 3.6 = 23.6, so k = 50 . 50

c.

!

0.4k + 3.6

k =0

d.

2.5 + (0.2 + 2.5) + (2 ! 0.2 + 2.5) + (3 ! 0.2 + 2.5) + (4 ! 0.2 + 2.5) + â&#x20AC;Ś + (100 ! 0.2 + 2.5) = 100

"

0.2k + 2.5

k =0

Review and Preview 2.2.1 2-38. y = ! 2(x + 1)3 ! 5 y = !2((x + 4) + 1)3 ! 5 = !2(x + 5)3 ! 5 (left 4 units) y = !2(x + 5)3 ! 5 ! 7 = ! 2(x + 5)3 ! 12 (7 units down)

2-39.

4x + 2y = 7

6x ! 2y = 10 2(3x ! y = 5) " 4x + 2y = 7 10x = 17 x = 17 = 1.7 10

3(1.7) ! y = 5 5.1 ! y = 5 !y = !0.1 y = 0.1

2-40. a. x 4 ! 81y 4

b.

8x 3 + 2x 7 = 2x 3 (4 + x 4 )

= (x 2 + 9y 2 )(x 2 ! 9y 2 ) = (x 2 + 9y 2 )(x + 3y)(x ! 3y) 2-41.

f (x) = 5x ! 1 " y = 5x ! 1 x = 5y ! 1 x + 1 = 5y x+1 5

= y " f !1 (x) =

x+1 5

2-42. 12 + 2 2 + 32 + 4 2 = 1 + 4 + 9 + 16 = 30 a. b. (6(2) ! 2 2 ) + (6(3) ! 32 ) + (6(4) ! 4 2 ) + (6(5) ! 5 2 ) + (6(6) ! 6 2 ) = 8 + 9 + 8 + 5 + 0 = 30 33 + 34 + 35 + 36 = 27 + 81 + 243 + 729 = 1080 c. 2-43. 12 + 2 2 + 32 + ... + 9 2 + 10 2 =

10

! k2

k =1

2-44.

( )1/3 = 52/3 1/2 7 3 = ( 7 3 ) = 7 3/2

a.

3

c.

c.

e.

7 ! 7 = 71 ! 71/2 = 7 3/2

52 = 52

d.

( 4 2 ) = ( 21/4 )5 = 25/4 ( 7 )3 = ( 71/2 )3 = 7 3/2

f.

9 ! 3 3 = 32 ! 31/3 = 37/3

b.

5

2-45. a.

sin(30! ) =

x 2 + x 2 = 52 2x 2 = 25 x 2 = 25 2 x=

25 2

=

5 2

0.5 = 4=x =

5 2 2

x 8

x 8

cos(30! ) = 3 2

=

y 8

8 3 = 2y 4 3=y

y 8

Lesson 2.2.2 2.46. 20

! ( 0.01j + 1.7 ) = 1.72 + 1.73 + 1.74 + 1.74 + 1.76 + 1.78 + 1.79 + 1.80 + 1.81 + 1.82 j =2

!!!!!!!!!!!!!!!!!!!!!!!!!!!!+1.83 + 1.84 + 1.85 + 1.86 + 1.87 + 1.88 + 1.89 + 1.90 = 34.39 2-47. 12 + 2 2 + 32 + 4 2 + 5 2 a. b. Index Expression Cumulative 2 (k) (k ) Sum 1 1 1 2 4 5 3 9 5+9=14 4 16 14+16=30 5 25 25+30=55 c. k = 1 and k = 5 d. At the start the sum is zero and at the end the sum is 55. 2-48. a. Initializes the sum to 0. d. The loop would go on forever because k would always be ≤ 5. 2-49. In step 2, change 1 to 7; in step 5, change 5 to 50. 2-50. In step 3, change k 2 to

10 k

.

2-51. b. Program keeps looping through these commands. c. Since we are using Y1, we need the independent variable X. d. Change the test value of k from 5 to 10. 2-52.

4354 she was changing the 1 ! x to !4 " x and If x ! 5 to If x ! 23 .

2-53. They need to change the line 1→X to B→X and the line if X≤5 to X≤E. 2-54. Change Y1 to

10 x

.

2-55. 34.39: Sum is same as in problem 2-46, but should take less time to calculate.

Review and Preview 2.2.2 2-56. 10 12 2 2 32 i2 + + + ... = ! (i + 1)2 2 2 32 4 2 i=1

2-57. 6

" (k ! 2)2 =(2 ! 2)2 + (3 ! 2)2 + (4 ! 2)2 + (5 ! 2)2 + (6 ! 2)2

k =2

!!!!!!!!!!!!!!!!!!!= 0 2 + 12 + 2 2 + 32 + 4 2 = 0 + 1 + 4 + 9 + 16 = 30

2-58. 3 4 ! 1 = 3 (length of interval)!!!!!!!! 5 pieces = 0.6 (length of each piece)

x0 = 1 x1 = 1 + .6 = 1.6!!!!!!!!!!!!!!x2 = 1.6 + 0.6 = 2.2 x3 = 2.2 + 0.6 = 2.8!!!!!!!!x4 = 2.8 + 0.6 = 3.4

2-59. a. These summations are the same, just written in different ways. 7

!

j2

=

33

+

42

+ 52

+ 62

+

7

7 2 ;!!

j=3

b.

! k 2 = 33 + 4 2 + 52 + 62 + 72

k=3

These summations are the same, just written in different ways. 7

! j 2 = 33 + 4 2 + 52 + 62 + 72 j=3 8

! ( j " 1)2 = (4 " 1)2 + (5 " 1)2 + (6 " 1)2 + (7 " 1)2 + (8 " 1)2 = 33 + 4 2 + 52 + 62 + 72 j=4

2-60. Change Y1 to (k + 1)2 , the starting index to B = 2 (or 2â&#x2020;&#x2019;X), and the end index to E = 7 (or Xâ&#x2030;¤7). 2-61. a

x+2 4

+

x! 3 5

b.

4 2! x

+

x 5

c.

6 x+2

!

=

5(x+2) 5(4)

+

4(x! 3) 4(5)

=

(5 x+10)+(4 x!12) 20

=

9 x!2 20

2 +2 x+20 5(4) (2! x)(x) 20+(2 x! x 2 ) + = = ! x 10!5 5(2! x) (2! x)(5) 10!5 x x 6(x!2) 4(x+2) (6 x!12)!(4 x+8) 4 = ! = = 2 x!20 x!2 (x+2)(x!2) (x!2)(x+2) x2 ! 4 x2 ! 4

=

2-62. a. b.

2 x!5 6

+

x+1 8

x x 2 +2 x!8

+

=

4(2 x!5) 4(6)

2 x+ 4

=

+

3(x+1) 3(8)

x (x+ 4)(x!2)

=

(8 x!20)+(3x+ 3) 24

+

2 x+ 4

= 11x!17 24

CD = (x + 4)(x ! 2) x (x+ 4)(x!2)

2(x!2)

+ (x+ 4)(x!2) =

x+(2 x! 4) (x+ 4)(x!2)

=

3x! 4 (x+ 4)(x!2)

Lesson 2.3.1 2-63. a. g(x)

c.

b.

10+3+32 = 45

2-64. a. and b. See graph at right. d. Answer will likely be between 7.75 and 10.75. 2-65.

Answers should range from 4.4 to 4.9. 2-66. Students should add the results of problems 2 and 3.

It looks like steps.

2-67. a. See graph at right. miles b. 路 hr = miles hr c.

25(0.5) + 25(0.25) + 12 ! 50 ! (0.25) + 75(1.25) = 12.5 + 6.25 + 6.25 + 93.75

d.

= 118.75 miles The distance Erin traveled from 30 minutes into her trip until 15 minutes later.

2-68. slope = a.

75!25 0.75!0.5

=

50 0.25

= 200

y ! 25 = 200(x ! 0.5) y ! 25 = 200x ! 100 y = 200x ! 75 b.

It is linear because Erin has a constant rate of change between 30 minutes and 45 minutes.

for 0 < x ! 0.5 # 25 % E(x) = \$ 200x " 75 for 0.5 < x ! 0.75 % 75 for 0.75 < x ! 2 &

Review and Preview 2.3.1 2-69.

y = 2.7 3 = 19.683 2-70. slope =

3!1 = 2 = 1 4!0 4 2 1 (y ! 1) = 2 (x ! 0) y ! 1 = 12 x y = 12 x + 1

2-71. The equation needs to be entered and the starting and ending values need to be adjusted. SUM = 1520 2-72. a.

4(1 + 2 + 3 + 4 + 5) =

5

! 4k = 60

b.

k =1

c.

2(1 + 2 + 3 + 4 + ... + 100) =

100

! 2k = 10, 100

k =1

2 0 + 21 + 2 2 + 2 3 + 2 4 + 2 5 =

5

! 2 k = 63

k =0

2-73. a. See graph at right. b. The domain is time elapsed, and Lew walks for one hour. The range is 0 to 4 miles. ihr = miles . c. The units are in miles since mi hr d. Length of rectangle (1 hr â&#x20AC;&#x201C; 0 hrs) = 1 hr Height of rectangle (4 mph â&#x20AC;&#x201C; 0 mph) = 4 mph Area of rectangle = 1hr * 4mph = 4 miles 2-74. a. (2x ! 1)(2x ! 1 + 3) = (2x ! 1)(2x + 2) = 2(2x ! 1)(x + 1) b. (2x ! 1)2 (2x ! 1 ! 4) = (2x ! 1)2 (2x ! 5) c.

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

2-75.

x(x 2 ! 3x ! 10) = 0 x(x + 2)(x ! 5) = 0 Either x = 0, x + 2 = 0, or x ! 5 = 0 x = 0, !2, 5 2-76. a.

" x 2 + 2 for x < 2 g(x) = # \$% 4 + 2 for x ! 2 " (x & 3)2 for x < 5 k(x) = # for x ! 5 %\$ 4

b.

2-77. x+2 x!1

+

x x+1

=

(x+2)(x+1) (x!1)(x+1)

+ (x+1)(x!1) =

b.

c a!b

+

c a+b

=

c(a+b) (a!b)(a+b)

+ (a+b)(a!b) =

c.

x+2 3x

!

x+1 6 x2

=

x(x!1)

(x 2 +1x+2 x+2)+(x 2 ! x) x 2 + x! x!1

a.

(x+2)(2 x) (3x)(2 x)

c(a!b)

!

x+1 6 x2

=

(ca+cb)+(ca!cb) a 2 +ab!ab!b 2

(2 x 2 + 4 x)!(x+1) 6 x2

=

=

=

2 x 2 +2 x+2 x 2 !1

2ca a 2 !b 2

2 x 2 + 3x!1 6 x2

Lesson 2.3.2 2-78. a. (1, 1), (1.5, 2.25), (2, 4), (2.5, 6.25) b. The height . c. Width of each rectangle is 0.5. x0 = 1 x1 = 1.5 x2 = 2 x3 = 2.5 x4 = 3 d. The height of the rectangle is the function value at the left endpoint of each interval. 0.5 !1 + 0.5 ! 2.25 + 0.5 ! 4 + 0.5 ! 6.25 = 0.5 + 1.125 + 2 + 3.125 = 6.75 e. This area is smaller than the true area. 2-79. a. The height. b. They start from the right endpoint of the interval. 0.5 ! 2.25 + 0.5 ! 4 + 0.5 ! 6.25 + 0.5 ! 9 = 1.125 + 2 + 3.125 + 4.5 = 10.75 c. This area is larger then the true area. 2-80. Some possibilities are averaging the two values or using more rectangles. 2-81. a. 0.25 b. height rectangle 1 = 12 = 1 height rectangle 2 =1.25 2 = 1.5625 height rectangle 3 =1.5 2 = 2.25 height rectangle 4 =1.75 2 = 3.0625 height rectangle 5 =2 2 = 4 height rectangle 6 =2.25 2 = 5.0625 height rectangle 7 =2.5 2 = 6.25 height rectangle 8 =2.75 2 = 7.5625 c. They are the y-values of the left endpoints. d. The width of each rectangle. e. The 3rd and the 6th. 2-82. xk = 1 + 0.5 k

2-83. 4

a.

! 0.5 (1 + 0.5k )2

b.

Right-hand endpoints: x1 = 1, x8 = 8

k =1

8

Left-hand endpoints: x0 = 0, x7 = 7

! 0.25 ( 0.25k + 1)

7

! 0.25 ( 0.25k + 1)2

2

k =1

k =0

2-84. a. See graph at right. height rectangle 1 = 0.25 b.

c. d.

height rectangle 2 = 1 height rectangle 3 = 2.25 height rectangle 4 = 4 0.5(0.5 2 + 12 + 1.5 2 + 2 2 ) 0.5(0.5 2 + 12 + 1.5 2 + 2 2 ) = 0.5(7.5) = 3.75

Review and Preview 2.3.2 2-85.

2 + 2.5 + 3 + 3.5 + ... + 9.5 =

4 2

19

+ 52 + 62 + â&#x20AC;Ś + 17 + 18 + 19 = ! 2 2 2

k=4

Other answers are acceptable as well. 2-86.

6 2 + h 2 = 12 2 36 + h 2 = 144 h 2 = 108 h = 108 = 3 ! 36 = 6 3 A = 12 ! b ! h = 12 !12 ! 6 3 = 36 3

12cm

k 2

2-87. 5 ! 2 = 3 (length of interval) a. 3 4 peices

f (x) = x 2

b.

f (x2 ) = f (3.5) = (3.5)2 = 12.25

= 0.75 (length of each piece)

f (x4 ) = f (5) = 5 2 = 25

x0 = 2 x1 = 2 + 0.75 = 2.75 x2 = 2.75 + 0.75 = 3.5 x3 = 3.5 + 0.75 = 4.25 x4 = 4.25 + 0.75 = 5 2-88.

y = !x 2 + 6 x 2 ! 2(!x 2 + 6) = !3 x 2 + 2x 2 ! 12 = !3 3x 2 = 9 " x 2 = 3 " x = Âą 3 y = !(Âą 3)2 + 6 " y = !3 + 6 = 3 2-89. x + 2 ! 0 " x ! #2 a. b. 3 ! x " 0 # !x " !3 # x \$ 3 c. d.

x 2 ! 1 " 0 # (x + 1)(x ! 1) " 0 # x + 1 " 0, x ! 1 " 0 # x " 1, !1 All reals.

2-90. a.

# 4 ! x 2 if x " 1 f (x) + 2 = g(x) = \$ %& x + 2 if x > 1

b.

2-91.

(

a.

100 3/2 = 1001/2

c.

( )

125 2/3 27

=

((

)3 = (10)3 = 1000

)

2 125 1/3 27

)

=

( )

5 2 3

=

25 9

(

b.

27 !2/3 = 271/3

d.

( ) a6 27

!2/3

" =\$ #

)!2 = (3)!2 = 312 = 19

( ) a6 27

1/3 % !2

'&

=

( ) =( ) a2 3

!2

3 a2

2

=

9 a4

2-92. a.

x x!1

+

x x+1

=

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

=

(x 2 + x)+(x 2 ! x) x 2 + x! x!1

x(x!1)

+ (x+1)(x!1) =

b.

2x !

4 x+2

=

2 x(x+2) (x+2)

!

4 x+2

=

2 x 2 + 4 x! 4 x+2

2 x2 x 2 !1

Lesson 2.3.3 2-93. a. b.

See graph at right. 4!2 = 25 = 0.4 5

x0 = 2.0

x1 = 2.0 + 0.4 = 2.4 x2 = 2.0 + 2 ! 0.4 = 2.8 x3 = 2.0 + 3 ! 0.4 = 3.2 x4 = 2.0 + 4 ! 0.4 = 3.6 x5 = 2.0 + 5 ! 0.4 = 4.0 4

d. e. f.

0.4 ! 2 2+0.4 k

k =0 0.4(2 2

+ 2 2.4 + 2 2.8 + 2 3.2 + 2 3.6 ) = 0.4(37.558) = 15.023 grams Lower because the rectangles lie below the curve.

2-94. See graph at right. 5

a. b. c.

0.4 ! 2 0.4 k +2 Change the indices. k =1 0.4(2 2.4

+ 2 2.8 + 2 3.2 + 2 3.6 + 2 4.0 ) = 0.4(49.558) = 19.823 grams Students should realize that the indices shift so that we use the same number of rectangles, but of different heights.

2-95. a. 60 miles ! 2 hours = 120 miles hour 40 miles !1hours = 40 miles hour 120 + 40 = 160 miles b. See graph at right. = mi c. Area under the curve = the distance traveled; i.e. hr ! mi hr

2-96. a. Gain of about 40 million. b. 20 + 25 + (!20) + 15 + (!5) + 5 = 65 ! 25 = 40 c. They should be counted as negative. 2-97. a. b. c. d.

See graph at right. 2 + (–0.5) = 1.5 f(x) is positive or negative based on the equation. In this case f(x) is negative when x > 3 . 1. 2.5 + 1.5 + 0.5 = 4.5 2. 4.5 + (–2) = 2.5 3. 2.5 + (–6) = –3.5 There is more area below the x-axis than above.

Review and Preview 2.3.3 2-98. 5 ! 3 = 2 (length of interval) a. 2 4 rectangles

= 0.5 (length of each sub-interval)

x0 = 3 x1 = 3 + 0.5 = 3.5 x2 = 3.5 + 0.5 = 4 x3 = 4 + 0.5 = 4.5 x4 = 4.5 + 0.5 = 5 L = 0.5

(

= 0.5 R = 0.5

(

= 0.5

b.

( (

32 + 2 + … + 4.5 2 + 2

11 + 14.25 + 18 + 22.25 3.5 2 + 2 + … + 5 2 + 2

)

)

11 + 14.25 + 18 + 22.25 + 27 3

left endpoints =

!

( 3 + 0.5k )2 + 2 !!!!!right endpoints = ! 0.5 ( 3 + 0.5k )2 + 2 k =1

The left endpoint indices go from k = 0 to k = 1 whereas the right endpoint indices go from k = 1!to k = 4.

2-99. a.

) 4

0.5

k =0

c.

)

1 x

!

1 y

=

1y xy

! 1x = yx

y! x xy

b.

(

1 x

!

1 y

) (x + y) = ( ) y! x xy

=

(x+ y) 1

xy+ y2 ! x 2 ! xy xy

=

y2 ! x 2 xy

2-100. slope =

5! 3.5 4!1

= 1.5 = 0.5 3

( g(x) ! 5 ) = 0.5(x ! 4) g(x) = 0.5x ! 2 + 5 g(x) = 0.5x + 3 a. b.

4

4

i=1

i=1

! 3 + 0.5i or ! 0.5i + 3 3.5 + 4 + 4.5 + 5 = 17

2-101. a.

6 ! 4 = 2 (length of interval)

2 5 rectangles

b.

= 0.4 (length of each sub ! interval)

x0 = 4

!x1 = 4 + 0.4 = 4.4

x2 = 4.4 + 0.4 = 4.8 x3 = 4.8 + 0.4 = 5.2 x4 = 5.2 + 0.4 = 5.6 x5 = 5.6 + 0.4 = 6

( 24 + 4.42 + 4.82 + 5.22 + 5.62 ) = 0.0845 2 + 2 + 2 + 2 + 2 = 0.779 upper = 0.4 ( 4.4 4.8 5.2 5.6 6)

c.

lower = 0.4

d.

Use more rectangles .

2-102. f (x) =

2-103. a.

f (x + 2) =

1 x! 4

â&#x20AC;&#x201C;f(x)

b.

=

1 x!2

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

1 x!2

f(x+2)â&#x20AC;&#x201C;1

2-104. Third angle = 180! ! (102! + 26! ) = 52! Area =

6 2 sin(102! ) sin(52! ) = 31.65cm 2 sin(26! )

1 (x+2)! 4

c.

!1

1 2

f (x)

Lesson 2.3.4 2-105. Students should recall, lower 15.023 and upper 19.823. 17.312 + 2 !1 = 17.312 + 2 = 19.312 a. b. The shift added a 2 by 1 rectangle with area 2. c. A(h, 2 ! x ! 4) = 10 + 17.312 = 37.312 The shift added a 2 by 10 rectangle with area 20. d. 2a + 17.312, the width is always 2 and the height added is a. 2-106. a. f(x)

b.

g(x) Shifts the graph up two units.

d.

h(x) (See graph below.) A(h, 0 ! x ! 3) = 0.5(2)(4) " 0.5(1)(2) = 4 "1= 3

A( f , 0 ! x ! 3) = 0.5(3)(6) = 9 c.

A(g, 0 ! x ! 3) = 9 + 6 = 15 Area added is a 2 by 3 rectangle of area 6.

2-107. a. No change b. Adding a 2 by 2 rectangle, which adds an area of 4 units. c. Nothing as long as the interval shifts also. d. Add a 2 by k rectangle. 2-108. Adds a 5 by 3 rectangle with an area of 15 units. A( j, 0 ! x ! 5) = W + 15 2-109. Subtracts a 2 by 2 rectangle with an area of 4 units. A(n, 1 ! x ! 3) = 11.190 " 4 = 7.190 2-110. See graph at right. A(g, 0 ! x ! 3) = 23.66622 " 10.09887 = 13.567

Review and Preview 2.3.4 2-111. 3 (length of interval) 6 (rectangles)

= 0.5

a.

Width of the rectangles

b. c.

3 ! 2 = 6, 6 + 7.579 = 13.579 3 !1000 = 3000, 3000 + 7.579 = 3007.579

2-112. a.

f (x) = x 2 + 5

b.

6 ! 1 = 5 (length of interval) = 1 (length of each sub ! interval)

5 5 rectangles

x0 = 1 x3 = 4

x1 = 2 x2 = 3 x 4 = 5 x5 = 6

L =1

2

( 1 + 5 +â&#x20AC;Ś+ R = 1( 2 + 5 + â&#x20AC;Ś + 2

2-113. CJ

60m

)

9 + 14 + 21 + 30 = 19.25

62

14 + 21 + 30 + 41 = 23.2

)

Amy

d

John !60 + 40t a. d.

20m

) ( 6+ + 5 ) = 1( 9 +

52 + 5 = 1

b.

!20 + 30t

c.

d 2 = (!60 + 40t)2 + (!20 + 30t)2

d = (40t ! 60)2 + (30t ! 20)2 t = 1.2 hours or at 1:12p.m; The distance between them is 20 miles.

2-114. a.

4 x 2 !8 x x 2 ! x!12 " 3 2 x 2 !9 2 x !8 x

b.

(x 2 ! y 2 )

4 x 2 + 4 xy 2 x +2 xy+ y2

2-115. a.

(1 + ) "\$#

4 y2 2 x ! y2

b.

x 2 !6 x+8 x 2 !16

รท

x y

4 x(x!2) (x! 4)(x+ 3) " (x+ 3)(x! 3) 2 x 2 (x! 4)

=

% '& =

2(x!2) x(x! 3)

4 x(x+ y)

= (x + y)(x ! y) (x+ y)(x+ y) = 4x(x ! y)

( ) "\$#

x 2 ! 4 x+ 4 x+ 4

=

y+ x y

=

4 y2 % (x+ y)(x! y) ' &

=

4y (x! y)

(x! 4)(x!2) (x+ 4) " (x+ 4)(x! 4) (x!2)(x!2)

=

1 x!2

2-116. a.

2-117. a.

x sin(50! ) x 0.766

=

6 sin(35! )

= 10.46

x = 8.01 2-118. 4 ! 180 " ! = 4 " 60 = 240! a. 3 b. c.

18! 180 " ! = 18 " 36 = 648! 5 4 ! 180 = 720 = 229.2! 1 " "

2-119. a. BC =

b. c.

5 =5 3 tan 30! AC = 5 ! = 10 sin 30 5 = 1 cos(A) = 10 2 sin(A) = 5103 = 23

b.

# x 2 ! 3 for x " !2 j(x) = \$ &% 3x + 7 for x < !2

b.

x 2 = 5 2 + 7 2 ! 2 " 7 " 5 " cos(120! ) = 109 x = 109 = 10.4

Lesson 2.3.5 2-120. a. 1! (2 + 5 + 10) = 17 b. 1! (5 + 10 + 17) = 32 c. 0.5 !1! (2 + 5) + 0.5 !1! (5 + 10) + 0.5 !1! (10 + 17) = 3.5 + 7.5 + 13.5 = 24.5 d. c, the sum in c is the average of the sums in a and b. e. It is too high. 2-121. b. Students answers should be between 126-130. 2-122. a. See graph at right. b. Increasing. c. (20)(2) + 0.5(40)(2) = 40 + 40 = 80 d. The units are miles. 2-123. a. See graph at right. b. Increasing. c. Not as far, at there is less area underneath the curve. d. Example of area using trapezoidal rule with four rectangles with length 0.5. 0.5 ! 0.5 ! (20 + 2(22.5) + 2(30) + 2(42.5) + 60) =67.5 True area â&#x2030;&#x2C6; 66.7 miles 2-124. See graph at right. 0.5 ! 0.5(5 + 2(4.75) + 2(4) + 2(2.75) + 1) a. b.

= 0.25(29) = 7.25 It is too low.

2-125. a. 0.5(5 + 4.75 + 4 + 2.75) = 8.25 square units b. 0.5(4.75 + 4 + 2.75 + 1) = 6.25 square units c.

8.25+6.25 2

= 7.25 They are the same.

2-126. The area underneath the curve is 80. a. Half of 80 miles, which is 40 miles. b. Vertical distances are halved. c. Draw the picture again or use the fact that the first car always travels twice as fast, so goes twice as far.

Review and Preview 2.3.5 2-127. 4 ! 1 = 3 (length of interval) 3 6 rectangles

= 0.5 (length of each sub ! interval)

x0 = 1 x4 = 3

x1 = 1.5 x5 = 3.5

x2 = 2 x6 = 4

x3 = 2.5

(

)

a.

left-hand approximation = 0.5 21 + 21.5 + 2 2 + 2 2.5 + 2 3 + 2 3.5

b.

right-hand approximation = 0.5 21.5 + 2 2 + 2 2.5 + 2 3 + 2 3.5 + 2 4

c.

trapezoids 1 2

( 0.5 ( 2

1

(

(

) ( 0.5 ( 2

+ 21.5 + 2 2 + 2 2.5 + 2 3 + 2 3.5 ) +

1 1.5 = 0.5 2 +2 2

+

21.5 +2 2 2

+

2 2 +2 2.5 2

+

2 2.5 +2 3 2

1 2

+

1.5

2 3 +2 3.5 2

+

)

+ 2 2 + 2 2.5 + 2 3 + 2 3.5 + 2 4 ) 2 3.5 +2 4 2

)

)

2-128. a. See graph at right. 3 30+60 = 3 ( 45 ) = 135 miles b. 2

(

c.

)

It is the units.

2-129. 5 ! 0 = 5 (length of interval) a. 5 20 rectangles

= 0.25 (length of each sub-interval)

b.

10 ! 1 = 9 (length of interval)!!!!!!!!!! 25

c.

5 ! (!3) = 8 (length of interval)!!!!!!!!!! 100

d.

E ! B = E ! B (length of interval)!!!!!!!!!! N

9 rectangles

= 0.36 (length of each sub-interval)

8 rectangles

= 0.08 (length of each sub-interval)

E!B rectangles

=

E!B N

(length of each sub-interval)

2-130. 9 10 ! 1 = 9 (length of interval)!!!!!!!!!! 25 rectangles = 0.36 (length of each sub-interval) a. b. c. d. e.

x0 = 1 ! height of rectangle = f (1) x1 = 1 + 0.36 = 1.36 ! height of rectangle = f (1.36) x25 = 10 ! height of rectangle = f (10) x24 = 10 ! 0.36 = 9.64 " height of rectangle = f (9.64) x0 = B ! height of rectangle = f (B) x1 = B + W ! height of rectangle = f (B + W ) xlast = E ! height of rectangle = f (E) xone before last = E " W ! height of rectangle = f (E " W )

2-131. 1 = a. cos ! c.

1 tan !

1 c/b 1 a/c

=

= 1" bc = = 1" ac =

b c c a

1 sin !

b.

=

1 a/b

= 1" ba =

b a

2-132. 2 ! 1 = 1 (length of interval) 1 4 rectangles

x0 = 1 x3 = 1.75

a.

= 0.25 (length of each sub-interval) x1 = 1.25 x4 = 2.0

x2 = 1.50

( ) Lower = 0.25 ( (!1.25 2 + 6) + (!1.5 2 + 6) + (!1.75 2 + 6) + (!2 2 + 6) ) = 3.281

Upper = 0.25 ( ! 12 + 6) + (!1.25 2 + 6) + (!1.5 2 + 6) + (!1.75 2 + 6) = 4.031

4

b.

0.25 " !(0.25k + 1)2 + 6 k =1

2-133. The Hypotenuse Leg Postulate a.

DC = BC (def square)

b.

DF = BE (corr. Parts of ! " ' s are ! ) FC = CE (subtraction)

AD ! AB, AF ! AE c.

side of equilateral triangle = Area =

( 3.106 )2

3

3 cos 15!

= 3.106

= 4.177cm 2

4

2-134. a.

3!5 15 3 5 ! = = x+ 3 x"5 2 2 x "5 x+ 3x"15 x "2 x"15

x+ y x! y y(x+ y) x(x! y) ! = ! x y xy xy

b.

=

c.

(

x y

!

y x

) "\$#

x 2 y2 2 x y! xy2

% " '& = \$#

2-135. xy = 12 ! y = 12 x 3x " 2 3x "

( 12x ) = "6

24 x

= "6

3x 2 " 24 = "6x 3x 2 + 6x " 24 = 0

x 2 ! y2 xy

%" '& \$#

x 2 y2 % xy(x! y) ' &

=

x 2 ! y2 x! y

=

(x+ y)(x! y) x! y

x 2 + 2x ! 8 = 0 (x + 4)(x ! 2) = 0 Either x = !4 or x = 2 ! 4y = 12 2y = 12 y = !3 y=6 Solution : (!4, !3), (2, 6)

xy+ y2 ! x 2 + xy y2 +2 xy! x 2 = xy xy

= x+y

Lesson 2.3.6 2-136. a. 20 rectangles 19

c.

Left = ! 0.25(0.25k + 1)2 k =0

b.

6!1 = 5 20 20 20

= 0.25

Right = ! 0.25(0.25k + 1)2 k =1

2-137. a. Except for the first left-hand rectangle and the last right-hand rectangle, the other rectangles are in both areas. 19

b.

! 0.25( 0.25k + 1)2

k =1

2-138. W=

E!B N

2-139. Left-hand area = 67.34375

Right-hand area = 76.09375

2-140. Add: (R + L)/2 → Τ and Disp ″TRAP=″ T 2-141. The trapezoidal area should be best. Students may mention that there are fewer gaps between the trapezoids and the curve, or that the trapezoidal area is the average of areas that are too high and too low. To get a better estimate, increase the number of partitions. Note: Remind students of calculator limitations for rounding. 2-142. If students use n = 20, L = 3.22846, R = 3.02846, T = 3.128. a. Using more partitions will improve your estimate. b.

c.

y = 4 ! x2 y2 = 4 ! x 2 x 2 + y 2 = 4, y " 0 Students should see this as one fourth of a circle with radius 2, or

4! 4

=!.

Review and Preview 2.3.6 2-143. a. See graph at right. b. Too low, as rectangles will be under the curve. c. Too high, as rectangles will be above the curve. d. Left = 8.360, Right = 8.960 2-144. a. Gallons of water that passed through the pipe in the interval from 20 minutes to 50 minutes. b. Left-hand approx. ! 1.5 "10 + 1"10 + 1.5 "10 = 15 + 10 + 15 = 40 gallons Other answers are acceptable; Between 40 and 50 gallons. 2-145. 10+12+18 2

Area =

= 20

( 20 ! (20 " 10)(20 " 12)(20 " 18) ) = ( 20 !10 ! 8 ! 2 ) =

2-146. x 2 + 1 = !2ax + 7 for x = 1

12 + a = !2a "1 + 7 1 + a = !2a + 7 3a = 6 a=2 2-147. 1 x

!

3 y

=5"

y ! 3x = 5xy y ! 5xy = 3x y(1 ! 5xy) = 3x

(for x, y # 0)

3x y = 1!5 x

2-148.

(

1 y

!

x y

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

1! x y

1+ x x

1+ x! x! x 2 xy

= 1!xyx

2

3200 = 56.57cm 2

2-149. a. Flipped over x-axis then up 2.

b.

Left 2 and compressed vertically.

1 f(x+2) 2

â&#x20AC;&#x201C;f(x)+2

2-150. 20

! 0.5

0.5k + 2

k =0

2-151.

Lesson 2.3.7 2-152. The area under a velocity curve represents distance traveled. a. Students should understand that the area is found by multiplying miles/hr times hours to get miles. b. (30)(2) + (20)(1) = 60 + 20 = 80 miles c. The second car had half the velocity of the first and the height of the rectangles has changed. 2-153. Area represents 80 miles. a. The height of the trapezoid is t. Area of the trapezoid: c. D(t) = 0.5(20 + (20t + 20))(t)

0.5t(20t + 40) = 10t 2 + 20t e.

D(3) = 10 ! 32 + 20 ! 3 = 10 ! 9 + 60 = 150 V (3) = 20 ! 3 + 20 = 60 + 20 = 80 mph

b. d.

20 and 20t + 20 D(2) = 10 ! 2 2 + 20 ! 2

= 10 ! 4 + 40 = 80 miles

2-154. a. She used the midpoints of the sections. One might refer to them as midpoint rectangles. c. Each rectangle would have width 1. d. Rectangles are still one unit wide, change the 1 to 1.5. e. Left = 1(21 + 2 2 + 2 3 + 2 4 ) = (2 + 4 + 8 + 16) = 30

Right = 1(2 2 + 2 3 + 2 4 + 2 5 ) = (4 + 8 + 16 + 32) = 60 Trap = 0.5 !1(21 + 2 ! 2 2 + 2 ! 2 3 + 2 ! 2 4 + 2 5 ) = 0.5(90) = 45 Connie = 1(21.5 + 2 2.5 + 2 3.5 + 2 4.5 ) = 42.426 2-155. a. 42.426

Review and Preview 2.3.7 2-156. For the first six hours, Jack makes \$5/hr, meaning at the end of six hours, he has earned \$30. For anytime after six hours (where he has already earned \$30) he adds to the \$30 by making \$8 times the number of hours worked over six hours. 2-157. a. b. c.

d.

for 0 ! t ! 6 # 5t J(t) = \$ % 30 + 10(t " 6) for t > 6 for 0 ! t ! 8 # 6t C(t) = \$ % 48 + 12(t " 8) for t > 8 30 + 10(t ! 6) = 46 30 + 10t ! 60 = 46 !30 + 10t = 46 10t = 76 76 = 7.6 hours t = 10

When is 30 + 10(t ! 6) > 48 + 12(t ! 8) 30 + 10t ! 60 > 48 + 12t ! 96 !30 + 10t > !48 + 12t 18 + 10t > 12t 18 > 2t t<9

AND 6t < 30 + 10(t ! 6) 6t < 30 + 10t ! 60 6t < 10t ! 30 !4t < !30 t > 7.5 " 7.5 < t < 9

2-158.

3 ! 1 = 2 (length of interval) 2 10 rectangles

(

=

(

)

(

)

(

)

9

% 0.2 !" 3(0.2k + 1)3 + 1#\$

k =0

(

)

(

)

(

)

(

R = 0.2 3(1.2)3 + 1 + 0.2 3(1.4)3 + 1 + â&#x20AC;Ś + 0.2 3(2.8)3 + 1 + 0.2 3(3.0)3 + 1 =

c.

)

L = 0.2 3(1)3 + 1 + 0.2 3(1.2)3 + 1 + â&#x20AC;Ś + 0.2 3(2.6)3 + 1 + 0.2 3(2.8)3 + 1

a.

b.

= 0.2 (length of each sub-interval)

M=

)

10

% 0.2 !" 3(0.2k + 1)3 + 1#\$

k =1 9

% 0.2 !" 3(0.2k + 1.1)3 + 1#\$

k =0

2-159. 16 3/2 = 161.5 = 64 They are the same because 1.5 =

(

16 3/2 = 16 1/2

2-160. a.

)3 = 4 3 = 64

( )1/2

2 2(x+1) = 2 x

b.

2 2 x+2 = 2 x /2 ! 2x + 2 =

x 2

4x + 4 = x 4 = "3x x="

31 (32 x ) = 33(x+1) 31+2 x = 33x+ 3 ! 1 + 2x = 3x + 3 "2 = x

4 3

2-161.

( q )2 = 512 + 34 2 ! 2 " 51" 34 cos 25! ( q )2 = 2601 + 1156 + 3468 " (0.906) ( q )2 = 614.99 q = 24.8 24.8 34 34 = \$ 58.68 = sin 25 sin #P sin #P 34 \$ sin #P = = 0.579 58.68 #P = 35.4 ! \$ #R = 180! ! 25! ! 35.4 ! = 119.6!

3 2

c.

( 5 x )1/3 = ( 5!2 )x!2 5 x /3 = 5 !2 x+ 4 "

x 3

= !2x + 4

x = !6x + 12 7x = 12 x = 12 7

2-162. 2! a. 3

b.

!

5" 6

2-163. a. It shows that the number of customers increases approaching lunchtime and decreases after lunchtime. b. The total number of customers served from 11 a.m. to 1 p.m. c. From the graph, it can be seen that during each hour, more than eight people are served, so the total must be more than 16. 2-164. a.

If k(x) is continuous, ! 2x 2 " 4 = "3x + 10

for x = 2

b.

# 2(x + 1)2 ! 4 for x < 1 j(x) = \$ %& !3(x + 1) + 10 !for x " 1

2 # 2 2 " 4 = "3 # 2 + 10 8 " 4 = "6 + 10 4=4 So, k(x) is continuous at x = 2.

Lesson 2.3.8 2-165. a. 17162 ft (left endpoint) 19941 ft (right endpoint) 18551.5 ft (trapezoidal) b. See graph at right. c. Altitude = 0.5 ! 23.5x ! (x) = 11.75x 2 d. e.

11.75(40.32)2 " 19100 (rounded) h = 11.75t 2 Students will draw the line in part b. and find the area of the triangle which has a base of t and a height of 23.5t. h(t) = 11.75t 2 h(t) = 11.75t 2

30, 000 = 11.75t 2

150, 000 = 11.75t 2

2553.19 = t 2

12765.96 = t 2

50.53 = t v(50.53) = 23.5 ! 50.53 = 1188 ft/sec when it enters the ozone layer.

112.99 = t v(112.99) = 23.5 !112.99 = 2655 ft/sec when it leaves the ozone layer.

2.3.8 Review and Preview 2-167. â&#x2030;&#x2C6; 3.12 to 3.16 depending on left or right rectangles. 2-169. a. g(x) = 2(x 3 + 2) = 2x 3 + 4 b. See graph at right. c. The area under g(x) is twice the area under f(x). 2-170. The interval is 8 units long and you add a height of 4 units. Therefore the new area will be 9 + (8)(4) = 41. 2-171.

!M = 180! " 63! " 50! = 67! 7 o n = = ! ! sin 67 sin 50 sin 63! o n 7.605 = and 7.605 = 0.766 0.891 !!!!!!!!!!!!o = 5.8 and n = 6.8 2-172. a. g(3) = 11 ! 32 = 11 ! 9 = 2 c. g(!1) = 2(!1) = !2

b. d.

g(!2) = 2(!2) = !4 g(4) = 11 ! 4 2 = 11 ! 16 = !5

2-173. a. At around 50 minutes the number of fish entering the hatchery decreases greatly. b. The total number of salmon entering the hatchery during the 60-minute interval. 2-174. log 4 16 = 2 a.

b.

log 4 4 2 = 2

)

c.

6 !2 =

b.

( 361 )

log 2 16 = log 2 2 4 = 4

log 1000 = 3 log10 10 3 = 3

log 6 6 !2 = !2

4 2 = 16 2-175. a. log 7 49 = log 7 7 2 = 2

(

log 6 36 !1 = !2

10 3 = 1000

c.

log 3 38 = 8

Closure Problems 2-176. a. b. c.

d. e.

x!3 # 1 R(x) = \$ x" 3 x>3 %2 She started taking the potion at (3,1). The area represents the total length of Rapunzel’s hair after a certain period. We are multiplying inches per month by months. Trapezoidal approximation with height 1 unit. 0.5(1! 2 0 + 2 ! 21 + 2 ! 2 2 + 2 3 ) = 0.5(1 + 4 + 8 + 8) = 10.5 inches True value ≈ 10.1 inches. From 3 to 9 months, hair growth is 90.9 inches so total is about 93.9 inches (don’t forget the original 3 inches). About 8.34 months.

2-177. a. A(3x ! 7, 2 " x " 7) 8

b.

! 2x 2 dx

c.

Answers will vary but, 1.6875 < A < 2.6875 .

8

CL 2-178. a.

b.

Time

0.5 1

1.5 2

2.5 3

Distance 30 60 90 120 140 160

c.

d.

for!!!!!!!!t ! 0 # 60t d(t) = \$ %120 + 40(t " 2) !for !2 < t ! 3

CL 2-179. a.

10

b.

liters minute ! minute = liters

d.

Yes, only about 41 liters have leaked.

c.

! 0.2 (1.2 ) dt t

0

CL 2-180. 3!1 2 a. = = 5 5 sub-intervals of width 0.4 0.4 0.4

b.

1 ! x ! 31

d.

left-hand

4

c.

S = 0.4 ! 2 0.4 k +1 k =0

CL 2-181. for 0 ! t ! 3 #150 C(t) = \$ %150 + 40(t " 3) for t > t 3

CL 2-182. a. (3 !12 + 5) + (3 ! 2 2 + 5) + (3 ! 32 + 5) + (3 ! 4 2 + 5) = 8 + 17 + 32 + 53 4

b.

0.2( 4 3 + 4 3+0.2!1 + 4 3+0.2!2 + 4 3+0.2!3 + 4 3+0.2!4 ) = 0.2 " 4 0.2k + 3

c.

A(4 x , 3 !

k =0

x ! 4)

CL 2-183. a. See graph at right. b. Yes, it is continuous c. See graph in part (a). # (x ! 1)2 ! 1 if x " 3 d. h(x) = \$ if x > 3 &% 9 ! 2x CL 2-184. f (x + 2) = 2 x!1+2 + 4(x + 2)2 = 2 x+1 + 4(x + 2)2

CL 2-185. a. b. Left =

(

) ( )

) (

)

! 4 + 2 + 4 + 2.5 + 4 + 3 \$ # & 2 # + 4 + 3.5 + 6 + 4 + 4.5 & " % 1

' 17.360

Right =

(

(

(

) (

)

) (

)

! 4 + 2.5 + 4 + 3 + 4 + 3.5 \$ # & 2 # +6 + 4 + 4.5 + ( 4+ 5 ) & " % 1

(

)

' 17.771

CL 2-186. a. f (x ! 1) shifts right 1.

b.

CL 2-187. a.

x+1 x

!

x x!1

=

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

=

x 2 + x! x!1! x 2 x(x!1)

CL-2-188. a.

!

x"x x(x!1)

=!

b. 1 x(x!1)

! f (x) is a reflection over the x-axis.

(1 ! ) "\$# =( ) "\$# x y

y! x y

x 2 y2 % ' 3 y x! x 3 y &

=

( ) "\$# y! x y

x 2 y2 xy(y! x)(y+ x)

15 ft 10 ft

b. c.

! = 15ft = 1.5 radians 10ft !

1.5 " 180 = #

270! #

= 85.94 !

15 radians is about 2.4 times around the circle. The central angle is then (15 ! 4" ) # 2.34 radians # 139.44ยบ .

% '& =

x 2 y2 % ' 2 2 xy(y ! x ) &

x y+ x

Chapter 2 Solutions

Chapter 2 Solutions