Chapter 9: Rates of Change - Weeblyachsprecalc.weebly.com/uploads/1/9/8/2/19823981/cpm_precalculus... · Lesson 9.1.1 9-1. b. A heart rate is relatively constant. c. Jumping up and

CPM Educational Program © 2012 Chapter 9: Page 1 Pre-Calculus with Trigonometry

Chapter 9: Rates of Change Lesson 9.1.1 9-1. b. A heart rate is relatively constant. c. Jumping up and down would increase heart rate. Taking a nap, or meditating might slow

down the rate. 9-2. Sample answers: Velocity in miles per hour, field goal percentage in goals per attempt,

batting average in hits per at bat. 9-3. a. 30 minutes = 1

2 hour!!!!!!!!r =dt =

281 2 = 56mph

b. No, his speed would have varied. He probably drove slower and faster than 56mph at times. It means that if he had driven at that speed for 30 minutes he would have covered the same 28 miles.

c. He must have driven the average speed at some time. 9-4. All answers should be in feet per second (ft/s). a. !d

!t =4533"7021"3 = 4463

18 = 247.944 b. !d!t =

3200"7018"3 = 2061

15 = 208.667

c. !d!t =

2131"7015"3 = 2061

12 = 171.75 d. !d!t =

1306"7012"3 = 2061

9 = 137.33

e. !d!t =

701"709"3 = 631

6 = 105.167 f. !d!t =

297"706"3 = 227

3 = 75.667 9-5. a. !A

!t =4900"2900650"360 = 2000

290 = 6.897 ft/s

b. !P!t =

500"8502750"650 =

"3502100 = "0.167 mb/s

c. Pressure is dropping. d.

!temp!t = "17.9"17.8

24900"0 = "35.724900 = "0.0014! / ft . Students should say that temperature is

dropping on average… 0.0014 degrees per foot as the balloon ascends.


Review and Preview 9.1.1 9-6. !y

!x =8"74"2 =

12 . The slope is always 12 because the function is linear.

9-7. a. See graph at right. b. First it increases, then it decreases. c. The slope represents the speed of the ball. First it decreases

(slope decreases). It stops momentarily (slope is at zero at the vertex) then it increases (slope gets steeper).

d. T seconds h in feet

0 !16(0 ! 0.5)2 + 8 = !16 14( ) + 8 = !4 + 8 = 4

0.25 !16(0.25 ! 0.5)2 + 8 = !16 116( ) + 8 = !1+ 8 = 7

0.5 !16(0.5 ! 0.5)2 + 8 = !16(0) + 8 = 8 0.75 !16(0.75 ! 0.5)2 + 8 = !16 1

16( ) + 8 = !1+ 8 = 7 1 !16(1! 0.5)2 + 8 = !16 1

4( ) + 8 = !4 + 8 = 4 1.25 !16(1.25 ! 0.5)2 + 8 = !16 9

16( ) + 8 = !9 + 8 = !1" 0 1.5 0

e. !d!t =

7"40.25"0 =

30.25 = 12

!d!t =

8"70.5"0.25 =

10.25 = 4

!d!t =

7"80.75"0.5 =

"10.25 = "4

!d!t =

4"71"0.75 =

"30.25 = "12

!d!t =

"1"41.25"1 =

"50.25 = "16

!d!t =

"8"10.25"0 =

"90.25 = 0

f. No. The speeds are the same, but velocities are different (except after the ball gets stuck in the mud).

9-8.

! dew point! height = "37"9

24900"0 = "0.0018° / ft 9-9. a. Dividing distance by time. b. Yes, if they drove below the speed limit on another part or if they made a stop. 9-10. a. See graph at right. b. Distance is the area under a velocity curve. In this case it is the area of a triangle:

A! = 1

2bh

d(t) = 1

2(t) v(t)( ) = 1

2(t)(2t) = t2

t

h(t)

v(t)

t 50

100


9-11. a.

b. The unit price decreases as the size of the bag decreases. c. The rate change is more drastic in the smaller sizes. Note: Sketching a graph of rate with respect to bag size may help here.

9-12. Graph (c) represents the graph of profit versus price because profits will increase as ticket

price increases, but at some point the cost will outweigh demand and tickets sales will begin to decrease.

9-13. 5+x

5+ 2= 3

3( 5 + 2) = 5 + x

3 5 + 3 2 ! 5 = x

x = 2 5 + 3 2

9-14. a. 25 ! 49x2 = (5 + 7x)(5 ! 7x) b. x4y2 + xy5 = xy2(x3 + y3) = xy2(x + y)(x2 ! xy + y2 ) c. (x !1)7/2 ! (x !1)3/2 = (x !1)3/2 (x !1)4/2 !1( ) = (x !1)3/2 (x !1)2 !1( )

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

9-15. Average rate is !d!t =

90miles2hours = 45mph , but they probably did not travel this speed the

entire time. 9-16. a. In the second year, or the fifth quarter. b. The company is currently experiencing an upswing. c. 13 million

12 lb bag $0.89 0.89

0.5 = 1.78 1 lb bag $1.29 1.29

1 = 1.29 2 lb bag $1.89 1.89

2 = 0.945 5 lb bag $4.60 4.60

5 = 0.92 10 lb bag $8.95 8.95

10 = 0.895 20 lb bag $17.80 17.80

20 = 0.89


Lesson 9.1.2 9-17. a. A linear model fits well. A reasonable answer will have a slope between 0.22 and 0.23.

A linear regression has an equation of y = 0.224x ! 428.6 . b. y = 0.224(2020) ! 428.6 = 23.88million

c. !p!t =

18.94"12.511999"1970 = 6.43

29 = 0.2217million = 221, 700people/year

d. !p!t =

23.6"12.512020"1970 =

11.0950 = 0.22million = 220, 000people/year

e. From 1986 to 1987 the population increased by 490,000 people. 9-18.

a. y = 1000 1+ 0.0612( )12 = 1000(1.0617) = 1, 061.68

b. 1, 061.68 !1000 = 61.6861.681000 = 0.06168" 6.168%

c. y = 1000e0.06!1

= 1000 !1.0618= $1, 061.84

1, 061.84 !1000 = 61.8461.841000 = 0.06184" 6.184%

9-19. a. Dropping by $368,288.8419 = $19, 383.62 / year . b. Students’ answers should have something to do with interest. c. The graph appears to be in the shape of half of an upside down parabola. e. The average rate of change is not equal to the actual rate of change each year. f. Students will likely say increasing since more money is being removed each year, but the

slope is becoming more negative. g. Concave down; the slope is decreasing. 9-20. a. V = Area of base/height

V = x2h

b. P = 4x + 4x + 2h24 = 8x + 2h

24 ! 8x = 2hh = 12 ! 4xV = x2(12 ! 4x)

c. 12 ! 4x > 012 > 4xx < 3

Domain: 0 < x < 3

d. Place of maximum value (2, 16).

V

x

4

8

1 2 3

12

16


Review and Preview 9.1.2 9-21. a. a(6) = 3(6) ! 5 = 18 ! 5 = 13

a(2) = 3(2) ! 5 = 1"a"x =

13!16!2 = 12

4 = 3

b. a(7) = 3(7) ! 5 = 21! 5 = 16a(5) = 3(5) ! 5 = 15 ! 5 = 10"a"x =

16!107!5 = 6

2 = 3

c. Yes, because it is a line. 9-22 a. This situation would have a fairly constant rate of change. b. This situation would have a fairly constant rate of change. c. This situation would not have a fairly constant rate of change. 9-23. a. log3 x ! log3 5 = 2

log3 x5( ) = 2

3log3 x

5( ) = 32x5 = 9x = 45

b. 2(1.5)x ! 3 = 33

2(1.5)x = 36

(1.5)x = 18

log1.5 (1.5)x = log1.5 18 =log 18log 1.5

x = 7.129

9-24. 2(!3) + c = 0!!!"!!!c = 6

ax2x = 5!!!"!!!a = 10

9-25. lim

h!0h2 + 3h " 4 = 02 + 3(0) " 4 = "4

9-26. Change the amplitude to 5. Change the period: Solution: y = 5 sin 4x 9-27. Change the amplitude to 3. Horizontal shift of !2 . Vertical shift of 5. Solution: y = 3 cos x ! "

2( ) + 5 9-28. Surface Area = x2 + 2xh + 2xh = x2 + 4xh Cost of Construction = ($3)x2 + ($1)4xh = 3x2 + 4xh

Period = !2

2!a = !

2

!a = 4!a = 4


9-29. f (3) = 32 = 9

f (2) = 22 = 4

m = 9!43!2 = 5

y ! 4 = 5(x ! 2)y ! 4 = 5x !10

y = 5x ! 6

9-30. a. 20 times. b. There is no max, continuing compounding will still round to 5.00%. c. Using guess and check, you get between 121 and 122 times per year (121.027). Lesson 9.1.3 9-31. a. d(3) = 10 ! 32 = 90

d(4) = 10 ! 42 = 160"d"t =

160#904#3 = 70ft/s

b. d(3) = 10 ! 32 = 90d(3.1) = 10 ! 3.12 = 96.1"d"t =

96.1#903.1#3 = 6.1

0.1 = 61 ft/s

c. d(3) = 10 ! 32 = 90d(3.01) = 10 ! 3.012 = 90.601"d"t =

90.601#903.01#3 = 0.601

0.01 = 60.1 ft/s

d. d(3) = 10 ! 32 = 90d(3.001) = 10 ! 3.0012 = 90.06001"d"t =

90.06001#903.001#3 = 0.06001

0.001 = 60.01 ft/s

e. d(3) = 10 ! 32 = 90d(2.99) = 10 !2.992 = 89.401"d"t =

90#89.4013#2.99 = 0.599

0.01 = 59.9 ft/s

f. d(3) = 10 ! 32 = 90d(2.999) = 10 !2.9992 = 89.94001"d"t =

90#89.940013#2.999 = 0.05999

0.001 = 59.99 ft/s

9-32. 60ft/s 9-33. a. It has a slope of 2 and a y-intercept at (0, –1). m = 3!(!1)

2!0 = 42 = 2

b. f (!1) = 2(!1) !1 = !3

f (1) = 2(1) !1 = 1"y"x =

!3!1!1!1 =

!4!2 = 2

c. f (1) = 2(1) !1 = 1f (5) = 2(5) !1 = 9"y"x =

9!15!1 =

84 = 2

d. f (2) = 2(2) !1 = 3f (a) = 2(a) !1 = 2a !1"y"x =

2a!1!3a!2 = 2a!4

a!2 = 2(a!2)a!2 = 2

e. It is constant.

y ! (!1) = 2(x ! 0)y +1 = 2x

y = 2x !1


f. The slope is the coefficient of x.


9-34. a. x2 ! x1 b. f (x2 ) ! f (x1) c. f (x2 )! f (x1)

x2 !x1

9-35. a. See graph at right. Average rate of change: [0, 0.1] !y

!x =f (0.1)" f (0)0.1"0 = 1.1052"1

0.1"0 = 0.10520.1 = 1.052

[1,1.1] !y!x =

f (1.1)" f (1)1.1"1 = 3.0042"2.7183

1.1"1 = 0.28590.1 = 2.859

[2,2.1] !y!x =

f (2.1)" f (2)2.1"2 = 8.1662"7.3891

2.1"2 = 0.77710.1 = 7.771

b. It increases as x increases. c. Negatives far from zero, positives far from zero. 9-36. a. !y

!x =32 "223"2 = 5

1 = 5 (Positive)

b. !y!x =

log 1"log 0.11"0.01 = 1

0.99 = 1.01 (Positive)

c. !y!x =

(7"3#52 )"(7"3#22 )5"2 = "68"("5)

3 = "633 = "21 (Negative)

d. !y!x =

0.50 "0.5"10"("1) = 1"2

1 = "11 = "1 (Negative)

e. !y!x =

1.50 "1.5"10"("1) =

1" 231 =

131 = 1

3 (Positive)

f. !y!x =

32 "("2)23"("2) = 5

5 = 1 (Positive) Review and Preview 9.1.3 9-37. a. [4, 5]!!!!!!!!!!!! !d!t =

d(5)"d(4)5"4 = 250"160

1 = 90 ft/s

b. [4, 4.1]!!!!!!!!! !d!t =d(4.1)"d(4)4.1"4 = 168.1"160

4.1"4 = 8.10.1 = 81 ft/s

c. [4, 4.01]!!!!!!! !d!t =d(4.01)"d(4)4.01"4 = 160.801"160

4.01"4 = 0.8010.01 = 80.1 ft/s

d. [3.99, 4]!!!!!!! !d!t =d(4)"d(3.99)4"3.99

160"159.2014"3.99 = 0.799

0.01 = 79.9 ft/s

e. [3.999, 4]!!!!! !d!t =d(4)"d(3.999)4"3.999 = 160"159.920

4"3.999 = 0.079990.001 = 79.99 ft/s

9-38.

x

y


a. 8 = 24x

23 = 24x3 = 4xx = 3

4

b. 21 = 8x

21 = 23x1 = 3xx = 1

3

c. 1 = ex

ln1 = ln ex

x = 0

d. log4 2 = x

log4 41 2 = x

x = 12


9-39. a. Area of base: x2 Cost of base: $3 ! x2 = 3x2 b. Area of one side: xh Area of 4 sides: 4xh Cost of the sides: $1 ! 4xh = 4xh c. Total Cost - 2( ) 3 4C x x xh= +

d. V = x2h

100 = x2h

h = 100x2

C(x) = 3x2 + 4x 100x2( )

C(x) = 3x2 + 400x

e. See graph at right. Minimum at: x = 4.055

C(x) = 3 ! (4.055)2 + 4004.055

C(x) = 49.3291+ 98.6436 = $147.97

9-40. a. y ! (!2) = ! 1

2 (x ! 5)

y + 2 = ! 12 x +

52

y = ! 12 x +

12

b. m = 0!43!1 = ! 4

2 = !2y ! 0 = !2(x ! 3)

y = !2x + 6

9-41.

ex (eh !1)h

9-42. a. g(6) = 6 ! 2(6) = !6

g(2) = 6 ! 2(2) = 2

m = !6!26!2 = !8

4 = !2

b. g(7) = 6 ! 2(7) = !8g(5) = 6 ! 2(5) = !4

m = !8!(!4)7!5 = !4

2 = !2

c. Yes, because it is linear. d. g(a) = 6 ! 2(a) = 6 ! 2ag(b) = 6 ! 2(b) = 6 ! 2b

m = 6!2b!(6!2a)b!a = !2b+2a

b!a = !2(b!a)b!a = !2

9-43.

a. x2 +y2!

"#$3

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

2(x2 + y2 )1 2= (x2 +y2 )

2

b. 2x5 !8x3x+2 = 2x3(x2 !4)

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

x

C(x)

1 2 3 4

100200

5

300


9-44. 180! ! 40! ! 60! = 80! x

sin 40º =10

sin 80º10 !0.6428 = 0.9848x

6.428 = 0.9848xx " 6.527

ysin 60º =

10sin 80º

10 !0.8660 = 0.9848y8.660 = 0.9848y

y " 8.794

9-45. a. f (3+ h) = 10(3+ h)2 = 10(9 + 6h + h2 )

f (3+ h) ! 90 = 90 + 60h +10h2 ! 90 = 60h +10h2

b. f (x + h) = 10(x + h)2 = 10(x2 + 2xh + h2 )f (x + h) !10x2 = 10x2 + 20xh +10h2 !10x2 = 20xh +10h2

c. g(20 + h) = 15 ! 3(20 + h) = 15 ! 60 ! 3h = !45 ! 3h d. f (g(x)) = 10(15 ! 3x)2 = 10(225 ! 90x + 9x2 ) = 2250 ! 900x + 90x2 9-46. m = 9+6h+h2 !9

3+h!3 = 6h+h2h = 6 + h

Lesson 9.1.4 9-47.

t = 5, t = 5.5!d!t =

d(5.5)"d(5)5.5"5

302.5"2505.5"5 = 52.5

0.5 = 105

t = 5, t = 5.1!d!t =

d(5.1)"d(5)5.1"5

260.1"2505.1"5 = 10.1

0.1 = 101

t = 5, t = 5.01!!!!!!! !d!t =d(5.01)"d(5)5.01"5

251.001"2505.01"5 = 1.001

0.01 = 100.1 9-48. a. 3+ h ! 3 = h b. 10(3+ h)2 ! 90 = 60h +10h2

c. !d!t =

60h+10h23+h"3 = 10h(6+h)

h = 60 +10h ft/s 9-49. a. d(4 + h) = 10(4 + h)2 = 10(16 + 8h + h2 )

!d!t = 160+80h+10h

2 "1604+h"4 = 80h+10h

24+h"4 =

10h(8+h)h = 80 +10h

b. t = 4, t = 4.1! h = 0.180 +10(0.1) = 80 +1 = 81

t = 3.99, t = 4! h = "0.0180 +10("0.01) = 80 + ("0.1) = 80 " 0.1 = 79.9

t = 4, t = 4.01! h = 0.0180 +10(0.01) = 80 + 0.1 = 80.1

t = 3.999, t = 4! h = "0.00180 +10("0.001) = 80 + ("0.01) = 80 " 0.01 = 79.99


9-50. V (x + h) = (x + h)2 (12 ! 4(x + h))

= (x2 + 2xh + h2 )(12 ! 4x ! 4h)= 12x2 ! 4x3 ! 4x2h + 24xh ! 8x2h ! 8xh2 +12h2 ! 4xh2 ! 4h3

= 12x2 ! 4x3 !12x2h !12xh2 + 24xh +12h2 ! 4h3

V (x + h) !V (x) = 12x2 ! 4x3 !12x2h !12xh2 + 24xh +12h2 ! 4h3 !12x2 + 4x3

= !12x2h !12xh2 + 24xh +12h2 ! 4h3

!V!t =

"12x2h"12xh2 +24xh+12h2 "4h3x+h"x = h("12x2 "12xh+24x+12h"4h2 )

h

= "12x2 "12xh + 24x +12h " 4h2

V (2) = !12 "22 !12 "2h + 24 "2 +12h ! 4h2 = !48 ! 24h + 48 +12h ! 4h2 = !4h2 !12h a. t = 2, t = 2.5! h = 0.5!!!!!!!!!!"4(0.5)2 "12(0.5) = "1" 6 = "7 b. t = 2, t = 2.1! h = 0.1!!!!!!!!!!!"4(0.1)2 "12(0.1) = "0.04 "1.2 = "1.24 c. t = 2, t = 2.01! h = 0.01!!!!!!!"4(0.01)2 "12(0.01) = "0.0004 " 0.12 = "0.1204 d. t = 2, t = 2.001! h = 0.001!!!!"4(0.001)2 "12(0.001) = "0.000004 " 0.012 = "0.012 e. t = 1.99, t = 2! h = "0.01!!!!!!"4("0.01)2 "12("0.01) = "0.0004 + 0.12 = 0.1196 f. t = 1.999, t = 2! h = "0.001!!!"4("0.001)2 "12("0.001) = "0.000004 + 0.012 = 0.01196 g. The slope is approaching zero. 9-51. a. R(20+h)!R(20)

h b. The average velocity gets closer and closer to exact velocity at t = 10 seconds. 9-52. a. A = 1

2 h(b1 + b2 ) =12 (x)(15 +15 ! 3x) =

12 x(30 ! 3x) = 15x !1.5x

2

b. A = 1

2 h(b1 + b2 ) =12 (x !1)(12 +15 ! 3x) =

12 (x !1)(27 ! 3x) =

32 (x !1)(9 ! x)

= !1.5x2 +15x !13.5


Review and Preview 9.1.4 9-53. a. The average rate of change of f on the interval [3,5] – formerly known as slope.

b. f (5)! f (3)5!3 = 52 +3(5)+2!32 !3(3)!2

5!3 = 25+15+2!9!9!22 = 22

2 = 11 9-54.

f (4+h)! f (4)4+h!4 = (4+h)2 +3(4+h)+2!42 !3(4)!2

h = 16+8h+h2 +12+3h+2!16!12!2h = 11h+h2

h = 11+ h 9-55. a. sin ! "sin! 2

! "! 2 = 0"1! 2 = " 2

! b. log 10!log 110!1 = 1!0

9 = 19

9-56. a. lim

x!3 x3"8x2 "4

= 33"832 "4

= 195

b. limx!2

x3"8x2 "4

= (x"2)(x2 +2x+4)(x"2)(x+2) = 4+4+4

4 = 3

c. limh!0

f (2+h)" f (2)h = lim

h!0

3(2+h)"4"(3(2)"4)2+h"2 = lim

h!06+3h"4"6+4

h = limh!0

3hh = lim

h!03 = 3

d. limh!0

g(3+h)" f (3)h = lim

h!0

6 3+h"6 33+h"3 = lim

h!0

(3)6 3(3+h)"6(3+h) 3(3+h)h

= limh!0

18"18"6h 3(3+h)h = lim

h!0

"2h (3+h)h = lim

h!0"2

(3+h) = " 23

9-57. a. 3ex ! 5 = 13

3ex = 18ex = 6

ln ex = ln 6x = ln 6

b. ln(x +1) = 3.2eln(x+1) = e3.2

x +1 = e3.2

x = e3.2 !1

9-58. 3x+h! 3x

h " 3x+h+ 3x3x+h+ 3x

= 3x+h!3xh 3x+h+ 3x( ) =

hh 3x+h+ 3x( ) =

13x+h+ 3x

9-59.

f (3+h)! f (3)h = 2(3+h)2 !2"32

3+h!3 = 2(9+6h+h2 )!18h = 18+12h+2h2 !18

h = 12h+2h2h = 12 + 2h

9-60.

limh!0

2(3+h)2 "18h = lim

h!0 (12 + 2h) = 12


Lesson 9.2.1 9-61. a. Average rate of change = slope b. We are working from patterns. 9-62. a. Average rate of change = slope b. f (d) = md + b!!!!!!!!!! f (c) = mc + b!!!!!!!!!! f (d )! f (c)d!c = md+b!(mc+b)

d!c = md!mcd!c = m(d!c)

d!c = m c. We used variables, so we proved it is always true. 9-63. a. m = 16!9

4!3 = 71 = 7 b. m = 3.12 !32

3.1!3 = 9.61!90.1 = 0.61

0.1 = 6.1

c. m = 32 !2.9923!2.99 = 9!8.9401

0.01 = 0.05990.01 = 5.99

9-64.

a. f (3+h)! f (3)h = (3+h)2 !32

3+h!3 = (9+6h+h2 )!9h = 9+6h+h2 !9

h = 6h+h2h = 6 + h

b. f (3)! f (3!h)h = 32 !(3!h)2

3+h!3 = 9!(9!6h+h2h = 9!9+6h!h2

h = 6h!h2h = 6 ! h

c. The variable h is arbitrary, so we can get as close as we need to. d. The slope gets closer to 6. 9-65.

a. f (2+h)! f (2)h = 2(2+h)3!2"23

3+h!3 = 2(8+12h+6h2 +h3)!16h = 24h+12h2 +2h3

h = 24h+12h2 2h3h

= 24 +12h + 2h2

b. Use y for r(t) and for x h . c. y = 24 +12x +12x2 d. When t = 2 , IROC lim

h!0(24 +12h + 2h2 ) = 24

9-66. a. 2(5+h)!25

5+h!5 = 2(5+h)!25h

b. It is not defined when 0h = c. See graph at right. d. The value as h! 0 is approximately 22. This is the instantaneous rate of change at x = 5 .

h

y

1 2

510


Review and Preview 9.2.1 9-67.

a. f (5+h)! f (5)(5+h)!5 = (5+h)2 +2(5+h)!1!(52 +2(5)!1)

h = 25+10h+h2 +10+2h!1!(34)h = 12h+h2

h = 12 + h b. Slope of the segment joining two points on the graph: (5, f (5)) and (5 + h, f (5 + h)). 9-68. a. log3 81 = x

log3 34 = xx = 4

b. 5x = 15

5x = 5!1

x = !1

c. 3x = 7

log3 3x = log3 7

x = log 7log 3

d. 92x = 27x!2

32(2x) = 33(x!2)

4x = 3x ! 6x = !6

e. log2 log9(log3 x)[ ] = !1

2log2 log9 (log3 x)[ ] = 2!1

log9(log3 x) = 12

9log9 (log3 x) = 912

log3 x = 3

3log3 x = 33

x = 27

f. (x ! 3)(ln x ! 2) = 0x = 3

ln x = 2eln x = e2

x = e2

9-69. a. g(x) = 10

x !!!!!!!!!!g(6) =106 = 5

3 !!!!!!!!!!g(5) =105 = 2

m = 5 3!26!5 = 5 3!6 3

1 = ! 13

b. g(5 + h) = 105+h !!!!!!!!!!g(5) =

105 = 2

m =105+h!25+h!5 =

105+h!

2(5+h)5+h

h =10!10!2h5+hh = !2h

5+h "1h = ! 2

5+h

c. See graph at right. x ! "0.4 9-70. a. s(x) = sin x!!!!!!!!!!s !

2( ) = sin !2 = 1!!!!!!!!!!s

!4( ) = sin !

4 =22

m =1" 2

2!2 "

!4=2" 2

2!4

= 2" 22 # 4! = 4"2 2

!

b. m =sin(!4 +h)"sin

!4

(!4 +h)"!4

=sin(!4 +h)"sin

!4

h

c. m =sin !4 cos(h)+cos

!4 sin(h)"sin

!4

h =12cos(h)+ 1

2sin(h)" 1

2h = 1

2cos(h)+sin(h)"1

h( )

h

y

1 21

-1


9-71. f (x+h)! f (x)

x+h!x = f (x+h)! f (x)h This is the average rate of change.

9-72.

12 f !3g6

3 f !9"#$

%&'2= 12 f 9g6

3 f 3"#$

%&'2= 4 f 6g6( )2 = 16 f 12g12

9-73. Slope of the line through center of circle and (3,4): m = 4!0

3!0 =43 !!!!!!!!!!" m = ! 3

4

Equation of the line: y ! 4 = ! 34 (x ! 3)

y = ! 34 (x ! 3) + 4

9-74. f (3) = 3+1 = 2

f (8) = 8 +1 = 3

m = 3!28!3 =

15

Equation of the line: y ! 2 = 15 (x ! 3)

y = 15 (x ! 3) + 2

Lesson 9.2.2 9-75. a. f (1) = !(1)2 + 4(1) +1 = !1+ 4 +1 = 4

f (3) = !(3)2 + 4(3) +1 = !9 +12 +1 = 4

m = 4!43!1 =

02 = 0

b. f (1) = !(1)2 + 4(1) +1 = !1+ 4 +1 = 4f (2) = !(2)2 + 4(2) +1 = !4 + 8 +1 = 5

m = 5!42!1 =

11 = 1

c. f (1) = !(1)2 + 4(1) +1 = !1+ 4 +1 = 4f (1.5) = !(1.5)2 + 4(1.5) +1 = !2.25 + 6 +1 = 4.75

m = 4.75!41.5!1 = 0.75

0.5 = 1.5

d. It approaches the slope of the tangent line. 9-76. a. They are the same. b. As the points that determine the secant line get closer together, they approach a point of

tangency, so the line tangent to the curve at that point will have the same slope as the curve.


9-77. a. See graph at right. b. g(1) = 3(1) +1 = 2!!!!!!!!!!g(5) = 3(5) +1 = 4

m = 4!25!1 =

24 =

12

c. g(1) = 3(1) +1 = 2!!!!!!!!!!g(5) = 3(5) +1 = 4

AROC = 4!25!1 =

24 =

12

d. g(1) = 3(1) +1 = 2!!!!!!!!!!g(3) = 3(3) +1 = 10

AROC = 10!23!1 = 10!2

2

e. g(1) = 3(1) +1 = 2!!!!!!!!!!g(2) = 3(2) +1 = 7

AROC = 7!22!1 = 7!2

1 = 7 ! 2

9-78. a. g(1) = 3(1) +1 = 2!!!!!!!!!!g(1+ h) = 3(1+ h) +1 = 4 + 3h

m = 4+3h!21+h!1 = 4+3h!2

h

b. This cannot be done. c. y = 4+3h!2

h As h! 0 , the slope of the secant line approaches 0.75. 9-79. a. f (1) = 2(1)2 + 5 = 7

g(1! h) = 2(1! h)2 + 5 = 2(1! 2h + h2 ) + 5 = 7 ! 4h + 2h2

m = 7!(7!4h+2h2 )1!(1!h) = 4h!2h2

h = 4 ! 2h

b. See graph at right. As h! 0 , the AROC approaches 4. c. Limit of slope of the secant line as h! 0 is the slope of the tangent line when x = 1 . 9-80. The secant line has a slope of 32 .

x

y

h

y

1 2

12


Review and Preview 9.2.2 9-81. a. f (x) = 2x2 !!!!!!!! f (2) = 2(2)2 = 8!!!!!!!! f (2 + h) = 2(2 + h)2 = 2(4 + 4h + h2 ) = 8 + 8h + 2h2

m = 8+8h+h2 !82+h!2 = 8h+2h2

h = 8 + 2h!!!!!!! limh"0

(8 + 2h) = 8

b. f (x) = 3x !!!!!!!! f (2) = 32 = 9!!!!!!!! f (2 + h) = 3(2+h)

m = 3(2+h)!322+h!2 = 3(2+h)!32

h !!!!!!!!!!! limh"0

3(2+h)!32h = 9.888

c. f (x) = log x!!!!!!!!! f (2) = log 2!!!!!!!!! f (2 + h) = log(2 + h)

m = log(2+h)!log 22+h!2 = log(2+h)!log 2

h !!!!!! limh"0

log(2+h)!log 2h = 0.217

d. f (x) = cos x!!!!!!!! f (2) = cos 2!!!!!!!! f (2 + h) = cos(2 + h)

m = cos(2+h)!cos 22+h!2 = cos(2+h)!cos 2

h !!!!! limh"0

cos(2+h)!cos 2h = !0.909

9-82. a. f (2) = 2 !22 = 8

Equation of line:y " 8 = 8(x " 2)y = 8(x " 2) + 8

b. f (2) = 32 = 9Equation of line:y ! 9 = 9.888(x ! 2)y = 9.888(x ! 2) + 9

c. f (2) = log 2 = 0.301Equation of line:y ! 0.301 = 0.217(x ! 2)y = 0.217(x ! 2) + 0.301

d. f (2) = cos 2 = !0.416Equation of line:y ! (!0.416) = !0.909(x ! 2)y = !0.909(x ! 2) ! 0.416

9-83. Use 5 and f (5) instead of 2 and f (2) . 9-84.

c2 = 1052 + 752 ! 2(105)(75) cos(35!) = 3748.3553

c2 = 3748.3553 = 61.224 nautical miles

9-85. y = 6

(x+1)5+ 2

Vertical asymptote at x = !1 . Horizontal asymptote at y = 2 . Vertical shift of 2 units.

x

y

1

1

2

2


9-86. a. f (x+h)! f (x)

h = tan(x+h)! tan(x)h = 1

h( ) sin(x+h)cos(x+h) !

sin(x)cos(x)( )

= 1h( ) cos(x) sin(x) cos(h)+cos2 (x) sin(h)!sin(x) cos(x) cos(h)+sin2 (x) sin(h)

cos2 (x) cos(h)!sin(x) sin(h) cos(x)

= 1h( ) cos2 (x) sin(h)+sin2 (x) sin(h)

cos2 (x) cos(h)!sin(x) sin(h) cos(x)

= 1h( ) sin(h) cos2 (x)+sin2 (x)( )

sin(h) cos2 (x) cot(h)!sin(x) cos(x)( ) =1h( ) 1

cos2 (x) cot(h)!sin(x) cos(x)

b. Not, because cot(h) is undefined. c. lim

h!0

tan(h)tan(h)( ) 1

h( ) 1cos2 (x) cot(h)"sin(x) cos(x)

= limh!0

tan(h)h( ) 1

cos2 x"sin x cos x tan(h)

d. limh!0

1cos2 x"sin x cos x tan(h)

= 1cos2 (x)

= sec2(x)

9-87. a. See diagram at right. b. P = 3x + 2y

100 = 3x + 2y2y = 100 ! 3xy = 50 !1.5x

c. Area = area of rectangle + area of equilateral triangle Area of rectangle: (50 !1.5x)(x) = 50x !1.5x2 Height of triangle: h2 = x2 + 1

2 x2

h2 = 32 x

2

h = 32 x

Area of triangle: 12 x32 x =

34 x2

Total area = 50x !1.5x2 + 0.25 3x2 = 50x + (0.25 3 !1.5)x2 9-88. Distance traveled: 30 ! 14 + 60 !

12 + 30 !

14 = 7.5 + 30 + 7.5 = 45 miles

Average speed: 45 miles1.25hours = 36mph 9-89. time (minutes)

spee

d (m

ph)

time (minutes)

dist

ance

(mile

s)

x

y y

x x


Lesson 9.3.1 9-90. a. b. c. d. 9-91. a. Steady rate. Stop. Couple of quick steps. Stop. b. See graph at right. 9-92.

9-93. a. See graph at right. b. Run for 1 second, then stop.

t (seconds)

v(t)

5 10 t (seconds)

p(t)

5 10

t (seconds)

v(t)

5 10 t (seconds)

p(t)

5 10

t (seconds)

v(t)

5 10 t (seconds)

p(t)

5 10

t (seconds)

v(t)

5 10 t (seconds)

p(t)

5 10


9-94. See graph at right. a. Distance traveled in feet. b. 5 !10 + 5 !0 + 5 !10 = 100 feet . Area represents distance. c. The end of the graph should be 100, since the distance is 100. 9-95. b. The slopes must be equal. 9-96. a. It becomes a tangent line. b. x = 2

f (2) = 2 !22 " 3(2) +1 = 3

c. d. f (2) = 3f (2 + h) = 2(2 + h)2 ! 3(2 + h) +1f (2+h)! f (2)

h = 2(2+h)2 !3(2+h)+1!3h

e. f (2) = 3

f (2 + h) = 2(2 + h)2 ! 3(2 + h) +1= 2(4 + 4h + h2 ) ! 6 ! 3h +1= 8 + 8h + 2h2 ! 6 ! 3h +1= 2h2 + 5h + 3

f. y ! 3 = 5(x ! 2)y ! 3 = 5x !10

y = 5x ! 7

f (2+h)! f (2)h = 2h2 +5h+3!3

h = 2h2 +5hh

limh"0

(2h + 5) = 5

t (seconds)

v(t)

5 10

2

4

t (seconds)

p(t)

5 10

255075

100

x

y

1 2

12


Review and Preview 9.3.1 9-97. a. b. f (2) = 3 !22 + 7 = 19

y "19 = 12(x " 2)y "19 = 12x " 24

y = 12x " 5

9-98. f (3) = 1

3

y ! 13 = ! 1

9 (x ! 3)

y ! 13 = ! 1

9 x +13

y = ! 19 x +

23

9-99. (3a ! 2)4 + (6a +1)3 = 81a4 +17

81a4 ! 216a3 + 216a2 ! 96a +16 +216a3 +108a2 +18a +1 = 81a4 +1781a4 + 324a2 ! 78a +17 = 81a4 +17

324a2 ! 78a = 06a(54a !13) = 0a = 0!!!!!54a !13 = 0

!!!!!!!!!!!!!!!!!!!!!!!!a = 1354

9-100.

x8/3!x2/3

x2/3+x!1/3" x1 3x1 3

= x9/3!x3/3

x3/3+x0= x3!x

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

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

9-101. Vertical shift of 4 units down. Period: 4! = b

2!4! = 1

2

y = sin x2( ) ! 4

9-102.

t (seconds)

v(t)

5 10 t (seconds)

p(t)

5 10

x

y

1 2

4812


9-103.

12+h !

12

h=1(2)!1(2+h)2(2+h)

h=

!h2(2+h)

h= !h2(2+h) "

1h = ! 1

4+2h

9-104. a. 110.11 = 1537K

K = 0.07This is the withholding rate.

b. 128.17 = 0.07 ! II = $1831

c. Kyleah’s mother earns more than Kyleah and has a higher rate withheld. Her rate is around 13.9%.

9-105. 4(2 cos x sin x) = 2

4(sin 2x) = 2

sin 2x = 12

2x = !6 + !n, 5!

6 + !n

x = !12 + !n, 5!

12 + !n

Lesson 9.3.2 9-106.

t (seconds)

v(t)

5 10

t (seconds)

p(t)

5 10 The major difference is that velocity can be negative. 9-107.

a.

t (seconds)

v(t)

5 10

t (seconds)

p(t)

5 10

b.

t (seconds)

v(t)

5 10

t (seconds)

p(t)

5 10


c.

t (seconds)

v(t)

5 10

t (seconds)

p(t)

5 10

9-108. a. f (3) = !16 " 32 + 80(3) + 5 = 101 feet b. f (4) = !16 " 42 + 80(4) + 5 = 69 feet c. The arrow goes up and then falls.

d. AROC = f (4) ! f (3)4 ! 3

= 69 !1011

= !32ft/sec

e. The arrow is falling. 9-109. a. f (3) = 101

f (3+ h) = !16(3+ h)2 + 80(3+ h) + 5 = !16(9 + 6h + h2 ) + 240 + 80h + 5= !144 ! 96h !16h2 + 240 + 80h + 5 = 101!16h !16h2

f (3+h)! f (3)3+h!3 = 101!16h!16h2 !101

h = !16h!16h2h

b. limh!0

"16h"16h2h = lim

h!0("16 "16h) = "16

c. !16 ft/sec d. 16 ft/sec 9-110. a. f (3) = 23 !!!!!!!! f (3+ h) = 2(3+h) !!!!!!!! f (3+h)! f (3)3+h!3 = 2(3+h)!23

h b. See graph at right. Approximately 5.545. c. Limit of slope of the secant line as h! 0 , or the slope of the tangent line. d. The instantaneous rate of change. Review and Preview 9.3.2 9-111. See graph at right. a. The area under the curve represents distance traveled. b. = 1

2 !10 !20 +10 !20 +12 !10 !20 = 400 ft

c. If p = 0 when t = 0 , then p = 400 when t = 30 .

h

y

1 2

123

t (seconds)

v(t)-(ft/sec)

5 10

510


9-112. a. f (2) = 1

2 !!!!! f (3) =13

f (3)! f (2)3!2 =

13!121 = 2

6 !36 = ! 1

6

b. 1 2+h!1 2h = 2 2(2+h)!2+h 2(2+h)

h = 2!2!h 2(2+h)h = !h

2(2+h) "1h = ! 1

2(2+h)

c. 1 x+h!1 xx+h!x = x x(x+h)!x+h x(x+h)

h = x!x!h x(x+h)h = !h

x(x+h) "1h = ! 1

x(x+h)

d. limh!0

" 1x(x+h) = " 1

x(x) = " 1x2

9-113.

t (seconds)

v(t)

5 10

t (seconds)

p(t)

5 10 9-114. a. See diagram at right. b. P = 2x + y

900 = 2x + yy = 900 ! 2x

Area of the corral: A(x) = x(900 ! 2x) c. See graph at right. d. At 225 ft the maximum area occurs.x = e. The slope at 225 ftx = is 0. 9-115. 1 5x+h!1 5x

h = 5x 5x(5x+h)!5x+h 5x(5x+h)h = 5x!5x!h 5x(5x+h)

h = !h5x(5x+h) "

1h =

!15x(5x+h)

9-116.

x ! 5 x3 + 4x2 !11x ! 30x3 ! 5x2

9x2 !11x9x2 ! 45x

34x ! 3034x !170

140

x2 + 9x + 34 + 140x!5

x ! 5 is not a factor because it leaves a remainder.

x x

y

x

y

50 100

14,500

29,000


9-117. See graph at right. a. The domain is all real numbers such that x ! "5 . b. y gets close to !" and ! depending on which side you come from. c. y gets close to 6. d. y gets close to 6. 9-118. a. 7

3! 2" 3+ 23+ 2

= 7(3+ 2 )32 !2

= 7(3+ 2 )7 = 3+ 2

b. 1x3

! xx= x

x3 !x= x

x2

9-119. See graph at right. Lesson 9.3.3 9-120. a.

f (3) = 13f (2) = 8

13! 8 = 5

b. f (2) = 8f (2 + h) = (2 + h)2 + 4 = 4 + 4h + h2 + 4

= h2 + 4h + 8f (2+h)! f (2)2+h!2 = h2 +4h+8!8

h = h2 +4hh = h + 4

c. 5.3! 5 = 0.3 d. f (5) = 29

f (5 + h) = (5 + h)2 + 4 = 25 +10h + h2 + 4= h2 +10h + 29

f (5+h)! f (5)5+h!5 = h2 +10h+29!29

h = h2 +10hh = h +10

e. f (x) = x2 + 4f (x + h) = (x + h)2 + 4 = x2 + 2xh + h2 + 4

f (x+h)! f (x)x+h!x = x2 +2xh+h2 +4!(x2 +4)

h = 2xh+h2h = 2x + h

9-121. a. lim

h!0h2 +4h

h = lim(h!0

h + 4) = 4

b. The car is moving at 4 ft/sec after 2 seconds. The IROC at 2 seconds represents the slope of the tangent line to the curve at x = 2 .

c. limh!0

(2x + h) = 2x

d. It is the slope of the tangent line at any point x, f (x)( ) .

x

y

48

13

23

v e l o c i t y

time


e. For: x = 2, slope = 2(2) = 4x = !1, slope = 2(!1) = !2x = 0, slope = 2(0) = 0

9-122. a. f (2) = 2,!! f (4) = 4 = 2

f (4)! f (2)4!2 = 2! 2

2

b. 2 2 2 2 4 2 2 12 2 2 2(2 2) 2(2 2) 2 2! + !

+ + + +" = = =

9-123. a. f (3) = 3

f (3+ h) = 3+ hf (3+h)! f (3)3+h!3 = 3+h! 3

h

b. 3+h! 3h " 3+h+ 3

3+h+ 3= 3+h!3h( 3+h+ 3)

= hh( 3+h+ 3)

= 13+h+ 3

c. limh!0

13+h+ 3

= 12 3

9-124. a. f (x) = x

f (x + h) = x + hf (x+h)! f (x)

x+h!x = x+h! xh

b. x+h! xh " x+h+ x

x+h+ x= x+h!xh( x+h+ x )

= hh( x+h+ x )

= 1x+h+ x

c. limh!0

1x+h+ x

= 12 x

d. For: x = 9, slope = 12 9

= 16

x = 6, slope = 12 6

x = 100, slope = 12 100

= 120

x = a, slope = 12 a

9-125. a. (x + h)3 ! 2(x + h) +1! (x3 ! 2x +1)

h= 3x2 + 3xh + h2 ! 2

b. limh!0

(3x2 + 3xh + h2 " 2) = 3x2 " 2

c. For: x = 3, slope = 3 ! 32 " 2 = 25x = "2, slope = 3 ! ("2)2 " 2 = 12 " 2 = 10

x = 0.5, slope = 3 ! (0.5)2 " 2 = "1.25


Review and Preview 9.3.3 9-126.

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

x2( ) " x2 ! 2 + 1x2( ) = x4 ! 4x2 + 6 ! 4

x2+ 1x4

9-127. a. See graph at right. b. 35 ! 130 + 0 !

1.560 + 25 !

112 +15 !

1120 =

140120 +

250120 +

15120 =

405120

= 8124 = 3.375 or 3

38 mi

9-128. See graph at right. 9-129. a. See graph at right below. b. The slope of the tangent line to g(x) when x = 2 is 0. 9-130. a. P = 4x + y

110 = 4x + yy = 100 ! 4x

b. A(x) = x(110 ! 4x) c. Maximum Area = 756.25m2

9-131.

3(x+h)! 3xh " 3(x+h)+ 3x

3(x+h)+ 3x= 3(x+h)!3xh 3(x+h)+ 3x( ) =

3h3(x+h)+ 3x

" 1h =3

3 x+h+ x( ) =3

x+h+ x

9-132. a. f (3.1) = 3.12 + 3 = 12.61

f (3) = 32 + 3 = 12 f (3.1)! f (3)

3.1!3 = 12.61!120.1 = 0.61

0.1 = 6.1

b. f (x) = x2 + 3

f (x + h) = (x + h)2 + 3 = x2 + 2xh + h2 + 3f (x+h)! f (x)

x+h!x = x2 +2xh+h2 +3!(x2 +3)h = 2xh+h2

h = 2x + hlimh"0

2x + h = 2x

When x = 3, slope = 2(3) = 6.

t (minutes)

v(t)

1 2

1020

3

v e l o c i t y

time

x

A(x)

150

300

450

600

750

3 6 9


9-133. g(2) = 32

g(2 + h) = 3(2+h) g(2+h)!g(2)

2+h!2 = 3(2+h)!32h

The IROC at x = 2 is approximately 9.888. 9-134. a. sin(50

! + 20!) = sin(70!) b. sin 2 ! 2"5( ) = sin 4"5( )

c. cos(110! + 50!) = cos(160!) d. cos 2 ! "7( ) = cos 2"

7( )

e. cos 20!

2( ) = cos(10!) f. sin 20!

2( ) = sin(10!)

Lesson 9.3.4 9-135. a. They both describe the difference between x and another point. b. !f (x) = lim

"x#0 f (x+"x)$ f (x)

"x

9-136. a. f (x) = x

!f (x) = 12 x

!f (9) = 12 9

= 16

b. g(x) = x2 + 4!g (x) = 2x!g (3) = 2 " 3 = 6

c. h(x) = 1x

!h (x) = " 1x2

!h (10) = " 1102

= " 1100

9-137. g(x) = x2 ! x

g(x + h) = (x + h)2 ! (x + h) = x2 + 2xh + h2 ! x ! hg(x+h)!g(x)

x+h!x = x2 +2xh+h2 !x!h!(x2 !x)h = 2xh+h2 !h

h = 2x + h !1limh"0

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

#g (3) = 2(3) !1 = 5


9-138. g !

4( ) = tan !4 = 1,!!g

!4 + h( ) = tan !

4 + h( ) !g "4( ) = g(" 4+h)#g(" 4)

" 4+h#" 4 = tan(" 4+h)#1h

Choose small values of h to see that !g "4( ) = 2 .

9-139. a. !f (9) = 1

2 9= 12"3 =

16 b.

c. Multiply by 10. 9-140. a. V = 4

3 ! r3 !"!V = 4

3 ! (2t)3 = 4

3 ! #8t 3 = 323 ! t 3 b. SA = 4! r2

SA = 4! (2t)2 = 4! " 4t2 = 16! t2

c. V (r) = 4!3 r3

V (r + h) = 4!3 (r + h)

3 = 4!3 (r

3 + 3r2h + 3rh2 + h3)

V (r+h)"V (r)r+h"r = 4! 3(r3+3r2h+3rh2 +h3)"4! 3r3

h = 4! 3(3r2h+3rh2 +h3)h = 4!

3 (3r2 + 3rh + h2 )

d. S(r) = 4! r2

S(r + h) = 4! (r + h)2 = 4! (r2 + 2rh + h2 )S(r+h)"S(r)

r+h"r = 4! (r2 +2rh+h2 )"4! r2h = 4! (2rh+h2 )

h = 4! (2r + h)

e. V (t) = 32!3 t 3

V (t + h) = 32!3 (t + h)3 = 32!

3 (t 3 + 3t2h + 3th2 + h3)

V (t+h)"V (t )t+h"t =

32!3 (t3+3t2h+3th2 +h3)" 4!3 t3

h =32!3 (3t2h+3th2 +h3)

h = 32!3 (3t2 + 3th + h2 )

9-141. a. time = distance

rate

12 + x2 = d2

d = 12 + x2

t = 12 +x2miles2mph = 12 +x2

2

b. time = distancerate

1 = (1! x) + x

t = (1!x)miles4mph = (1!x)

4

c. 12 1+ x2 + 1

4 1! x( ) This is the distance of the diagonal plus the part of the border that is walked.

d. See graph at right. e. Slope equals zero. If it were positive then there must be a point

before it that is less and therefore it would not be a minimum. If it were negative, then there must be a point after it that is a minimum.

g(x) = 10 x

!g (x) = 102 x

!g (9) = 102 9

= 106 = 5

3

x

y

1 2

1

2


Review and Preview 9.3.4 9-142. f (3) = 23

f (3+ h) = 2(3+h) !f (3) = lim

h"0

f (3+h)# f (3)3+h#3 = lim

h"02(3+h)#23

h

Testing small values of h yields !f (3) " 5.545 . The slope of the function is 5.545 at 3 or the slope of tangent line at x = 3 is 5.545. This

tells you the function is increasing at that point. 9-143. h(x) = x2

h(x + h) = x2 + 2h ! x + h2

h(x+h)"h(x)x+h"x = x2 +2xh+h2 "x2

h = 2xh+h2h = 2x + h

limh!0

(2x + h) = 2x

"h (x) = 2x

a. !h (3) = 2 " 3 = 6 b. !h ("2) = 2 # "2 = "4 c. !h (a) = 2 "a = 2a 9-144. d(t) = 10t2

d(t + h) = 10(t2 + 2h ! t + h2 )d(t+h)"d(t )

t+h"t = 10(t2 +2ht+h2 "t2 )h = 10(2ht+h2 )

h = 10(2t + h)

limh!0

(10(2t + h)) = 20t

"d (t) = 20t

a. !d (1) = 20 "1 = 20 ft/s b. !d (3) = 20 " 3 = 60 ft/s c. !d (s) = 20 " s = 20s ft/s 9-145. a. f (3) = 33 = 27

f (5) = 53 = 125 f (5)! f (3)

5!3 = 125!272 = 98

2 = 49

b. f (4) = 43 = 64f (4 + h) = (4 + h)3 = 64 + 48h +12h2 + h3

f (4+h)! f (4)4+h!4 = 64+48h+12h2 +h3!64

h = 48h+12h2 +h3h = 48 +12h + h2

c. lim(h!0

48 +12h + h2 ) = 48

d. f (x) = x3

f (x + h) = (x + h)3 = x3 + 3x2h + 3xh2 + h3

f (x+h)! f (x)x+h!x = x3+3x2h+3xh2 +h3!x3

h = 3x2h+3xh2 +h3h = 3x2 + 3xh + h2

e. limh!0

(3x2 + 3xh + h2 ) = 3x2


9-146. a. Using range = [10, 20] b. y = 15 ! 5 sin "

6 (x !1)( )

c. 14 = 15 ! 5 sin "6 (x !1)( )

!1 = !5 sin "6 (x !1)( )

15 = sin "

6 (x !1)( )sin!1 1

5( ) = sin!1 sin "6 (x !1)( )( )

15 = "

6 (x !1)0.382 = x !1

x = 1.382x = 5am +1hour 23minutes = 6:23 AM

Other values = 11:37 AM, 6:23 PM, 11:37 PM

9-147. sin4 x ! cos4 x = 1

2

(sin2 x + cos2 x)(sin2 x ! cos2 x) = 12

sin2 x ! cos2 x = 12

1! cos2 x ! cos2 x = 12

!2 cos2 x = ! 12

cos2 x = 14

cos x = ± 12

x = ± "3 + "n

9-148. See graph at right. Distance = 1

2 !10 ! 30 = 150 feet 9-149. d(t) = 4t2

d(t + h) = 4(t + h)2 = 4(t2 + 2ht + h2 )d(t )!d(t+h)

t+h!t = 4(t2 +2ht+h2 )!4t2h = 8ht+4h2

h = 8t + 4hlimh"0

(8t + 4h) = 8t, IROC = 8 #5 = 40 ft/sec

7 : 30!p.m.!14.5 hours after 5am

y = 15 " 5 sin #6 (14.5 "1)( )

y = 15 " 5 $ sin(7.0686)y = 15 " 5 $0.7071 = 15 " 3.5355 = 11.46ft

t (seconds )

v(t)

5

10

1 2

15


Lesson 9.3.5 9-150. a. d = 30t b. There is no acceleration. 9-151. a. d = 30t b. m = 60!30

2!1 = 30

c. v(t) = !d (t) or d(t) = v(t)dt! 9-152. See graph at right. a. v(1) = 2 !1 = 2

v(3) = 2 ! 3 = 6v(x) = 2 ! x = 2x

b. Area under the curve = 12 !6 !12 = 36

c. b2 d. The area under the curve represents distance or position. 9-153. a. d(t) = t2

d(t + h) = (t + h)2 = t2 + 2ht + h2

d(t )!d(t+h)t+h!t = t2 +2ht+h2 !t2

h = 2ht+h2h = 2t + h

limh"0

(2t + h) = 2t

#d (1) = 2 $1 = 2#d (3) = 2 $ 3 = 6#d (x) = 2 $ x = 2x

b. 2(6) 6 36d = =

c. The distance function is the area under the velocity curve and the velocity function is the derivative of the distance function.

9-154. a. Find a general expression for the area under the curve. b. Take the derivative of the distance function.

t

v(t)

2

4

1 2


Review and Preview 9.3.5 9-155. d(t) = 0.1t 3

d(t + h) = 0.1(t + h)3 = 0.1(t 3 + 3t2h + 3th2 + h3)d(t )!d(t+h)

t+h!t = 0.1(t3+3t2h+3th2 +h3)!0.1t3h = 0.1(3t2h+3th2 +h3)

h = 0.1(3t2 + 3th + h2 )

0.1 limh"0

(3t2 + 3th + h2 ) = 0.3t2

a. !d (5) = v(t) = 0.3 "25 = 7.5 ft/s b. !d (t) = v(t) = 0.3t2 ft/s 9-156. x = 4

y = 12 x = 12 4 = 24

m = limh!0

12 x+h"12 xh = 3

(Find this by graphing on calculator.)

y ! 24 = 3(x ! 4)y ! 24 = 3x !12

y = 3x +12

9-157. See graph at right. a. Area = 1

2 ! t !6t = 3t2

b. This area represents distance or position. 9-158. See graph at right. a. d(3) = 3 ! 32 = 27

d(3+ h) = 3 ! (3+ h)2 = 3(9 + 6h + h2 )d(3+h)"d(3)3+h"3 = 3(9+6h+h2 )"27

h = 18h+3h2h = 18 + 3h

limh#0

(18 + 3h) = 18

$d (3) = 18

b. d(t) = 3 ! t2 = 3t2

d(t + h) = 3 ! (t + h)2 = 3(t2 + 2th + h2 )d(t+h)"d(t )

t+h"t = 3(t2 +2th+h2 )"3t2h = 6th+3h2

h = 6t + 3hlimh#0

(6t + 3h) = 6t

$d (t) = 6t

c. They are the same. 9-159. y = 2 sin 2x ! 3,

y = 2 cos 2 x ! "4( )( ) ! 3

t

v(t)

6

12

18

24

1 2 3

30

t

d(t)

10

20

30

1 2 3


9-160. a. 5x2 = 15x

5x2 !15x = 05x(x ! 3) = 0

x = 0, 3

b. 2(x ! 6)2 + 5 = 232(x ! 6)2 = 18(x ! 6)2 = 9

x ! 6 = ±3x = 6 ± 3x = 3, 9

9-161. x!2/3(2 + 5x2 ) 9-162.

a = x + 1x

ax = x2 +1x2 ! ax +1 = 0

x = !(!a)± (!a)2 !4(1)(1)2(1)

x = a± a2 !42

9-163. Area = 1

2 ! t !10t = 5t2

9-164. d(t) = 20 ! t2 = 20t2

d(t + h) = 20 ! (t + h)2 = 20(t2 + 2th + h2 )d(t+h)"d(t )

t+h"t = 20(t2 +2th+h2 )"20t2h = 40th+20h2

h = 40t + 20hlimh#0

(40t + 20h) = 40t

$d (t) = v(t) = 40t


Closure Chapter 9 9-165. Stage I: a. The graph is parabolic. b. A(200) = 3 !2202 + 60(220) = 158, 400 m c. A(t) = 3t2 + 60t

A(t + h) = 3t2 + 6th + 3h2 + 60t + 60hA(t+h)!A(t )

t+h!t = 3t2 +6th+3h2 +60t+60h!3t2 !60th = 6th+3h2 +60h

h = 6t + 3h + 60limh"0

(6t + 3h + 60) = 6t + 60

#A (t) = v(t) = 6t + 60

We need to find the equation for the instantaneous rate of change with respect to time. d. The graph is linear. e. v(220) = 6(220) + 60 = 1320 + 60 = 1380 meters per second STAGE II: a. Graph is a horizontal line at y = 1380 . b. D = 1380 !130 = 179, 400 m c. Graph is linear since velocity is constant. d. d(t) = 1380(t ! 220) +158, 400

d(t) = 1380t !145, 200

e. v(t) =6t + 60 for 0 ! t ! 2201380 for 220 < t ! 350"#$

d(t) =3t2 + 60t for 0 ! t ! 2201380t "145, 200 for 220 < t ! 350#$%&

g. d(350) = 1380(350) !145, 200 = 337, 800 m CL 9-166. a. 1.8 to 1.9˚/min. Students should recognize that the temp drops 4.5 degrees in the first 2

minutes (2.75 degrees/min), but only 1.8 degrees/min. in the next 2 minutes and only 1.9 degrees from 4 to 5 minutes.

b. c(5) = 70 +120(0.983)5

c(5 + h) = 70 +120(0.983)5+h

c(5+h)!c(5)5+h!5 = 70+120(0.983)5+h !(70+120(0.983)5 )

h = 120(0.983)5+h !120(0.983)5h

Temperature is increasing at –1.888˚/min or cooling at 1.888˚/min. CL 9-167.

a. W (4) = 20(1.02)4

1+(1.02)4= 21.649

2.082 = 10.396

W (2) = 20(1.02)2

1+(1.02)2= 20.8082.0404 = 10.198

W (4)!W (2)4!2 = 10.396!10.198

2 = 0.1982 = 0.099

b. W (2.1) = 20(1.02)2.1

1+(1.02)2.1= 20.849

2.042 = 10.208

W (2) = 20(1.02)2

1+(1.02)2= 20.8082.0404 = 10.198

W (2.1)!W (2)2.1!2 = 10.208!10.198

0.1 = 0.010.1 = 0.1


CL 9-168. a. V (1.5) = 0.2(1.4)t

V (1.5 + h) = 0.2(1.4)t+h

V (1.5+h)!V (1.5)1.5+h!1.5 = 0.2(1.4)t+h !0.2(1.4)t

h

!V (1.5) = 0.111 cu. ft/hr

b. !V (1.5) CL 9-169. a. To get instantaneous velocity, we must take the limit as the time interval approaches 0.

We can not do this because we are only given distances for selected times. b. This is the slope of the distance function at a given time, which is the same as the

instantaneous velocity. CL 9-170.

! f (x) is the limit of the slope as two points get closer and closer together (the distance between the two points approach zero.) The slope that is found is the slope of the line that grazes the curve at a particular point.

CL 9-171. a. b. CL 9-172. a. f (x) = 3x2

f (x + h) = 3(x + h)2 = 3(x2 + 2xh + h2 )f (x+h)! f (x)

x+h!x = 3(x2 +2xh+h2 )!3x2h = 6xh+3h2

h = 6x + 3hlimh"0

(6x + 3h) = 6x,! #f (x) = 6x

b. g(x) = 1x

g(x + h) = 1x+h

g(x+h)!g(x)x+h!x =

1x+h!

1x

h =x!(x+h)x(x+h)h =

!hx(x+h)h = !h

x(x+h) "1h =

!1x(x+h)

limh#0

!1x(x+h) = ! 1

x2 , $g (x) = ! 1x2

c. h(x) = x2 + 3x = x2 + 3x

h(x + h) = (x + h)2 + 3(x + h) = x2 + 2xh + h2 + 3x + 3hh(x+h)!h(x)

x+h!x = x2 +2xh+h2 +3x+3h!(x2 +3x)h = 2xh+h2 +3h

h = 2x + h + 3limh"0

(2x + h + 3) = 2x + 3, #h (x) = 2x + 3

ft

sec

ft sec

sec


CL 9-173. !f (x) = 3 "2x = 6x

!f (3) = 6 " 3 = 18 y ! 27 = 18(x ! 3)

y ! 27 = 18x ! 54y = 18x ! 27

CL 9-174. a. tan! = SR

300 , ! = tan"1 SR300( ) b. ! = tan"1 4t2300 = tan

"1 t275

c. limh!0

tan"1 (10+h)2 75( )" tan"1(4 3)h d. y =

tan!1 (10+h)2 75( )! tan!1(4 3)h

The rate at which Tommy is turning his camera is 0.096 rad/sec. CL 9-175. a. f (x) = 12 x

!f (x) = 122 x

2 = 122 x

4 x = 12

x = 3x = 9

b. f (9) = 12 9 = 12 ! 3 = 36Point = (9, 36)

c. !f (9) = 2y " 36 = 2(x " 9)y " 36 = 2x "18

y = 2x +18

The y-intercept is at (0, 18). The equation of the tangent line is y = 2x +18 . CL 9-176.

x!1/3 – x!4/3

x!4/3 – x2/3" x4/3x4/3

= x3/3 – x0

x0 – x6/3= x –11– x2

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

1+x

CL 9-177. a. sec x = ! 7

5

sin x = ! 2 67

csc x = ! 7 612

tan x = ! 2 67 " ! 7

5 =2 65

cot x = ! 57 " !

7 612 = 5 6

12

b. sin 2x = 2 sin x cos x

sin 2x = 2 ! " 2 67( ) ! " 5

7( ) = 20 649

cos 2x = cos2 x " sin2 x

cos 2x = " 57( )2 " " 2 6

7( )2 = 2549 "

2449 =

149

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