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3.1

1 4 1 4

3.2

6 6

3.3

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

3.4

−32−32 ft/s

3.5

P(3.25)=20>0;P(3.25)=20>0; raise prices

3.6

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

3.7

( 0 , + ) ( 0 , + )

3.8

a=6a=6 and b=−9b=−9

3.9

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

3.10

a ( t ) = 6 t a ( t ) = 6 t

3.11

0

3.12

4 x 3 4 x 3

3.13

f ( x ) = 7 x 6 f ( x ) = 7 x 6

3.14

f ( x ) = 6 x 2 12 x . f ( x ) = 6 x 2 12 x .

3.15

y = 12 x 23 y = 12 x 23

3.16

j ( x ) = 10 x 4 ( 4 x 2 + x ) + ( 8 x + 1 ) ( 2 x 5 ) = 56 x 6 + 12 x 5 . j ( x ) = 10 x 4 ( 4 x 2 + x ) + ( 8 x + 1 ) ( 2 x 5 ) = 56 x 6 + 12 x 5 .

3.17

k ( x ) = 13 ( 4 x 3 ) 2 . k ( x ) = 13 ( 4 x 3 ) 2 .

3.18

g ( x ) = −7 x −8 . g ( x ) = −7 x −8 .

3.19

3 f ( x ) 2 g ( x ) . 3 f ( x ) 2 g ( x ) .

3.20

5 8 5 8

3.21

−4.4 −4.4

3.22

left to right

3.23

3,300

3.24

$2

3.25

f ( x ) = cos 2 x sin 2 x f ( x ) = cos 2 x sin 2 x

3.26

cos x + x sin x cos 2 x cos x + x sin x cos 2 x

3.27

t = π 3 , t = 2 π 3 t = π 3 , t = 2 π 3

3.28

f ( x ) = csc 2 x f ( x ) = csc 2 x

3.29

f ( x ) = 2 sec 2 x + 3 csc 2 x f ( x ) = 2 sec 2 x + 3 csc 2 x

3.30

4 3 4 3

3.31

cos x cos x

3.32

cos x cos x

3.33

v(5π6)=3<0v(5π6)=3<0 and a(5π6)=−1<0.a(5π6)=−1<0. The block is speeding up.

3.34

h ( x ) = 4 ( 2 x 3 + 2 x 1 ) 3 ( 6 x 2 + 2 ) = 8 ( 3 x 2 + 1 ) ( 2 x 3 + 2 x 1 ) 3 h ( x ) = 4 ( 2 x 3 + 2 x 1 ) 3 ( 6 x 2 + 2 ) = 8 ( 3 x 2 + 1 ) ( 2 x 3 + 2 x 1 ) 3

3.35

y = −48 x 88 y = −48 x 88

3.36

h ( x ) = 7 cos ( 7 x + 2 ) h ( x ) = 7 cos ( 7 x + 2 )

3.37

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

3.38

h ( x ) = 18 x 2 sin 5 ( x 3 ) cos ( x 3 ) h ( x ) = 18 x 2 sin 5 ( x 3 ) cos ( x 3 )

3.39

a ( t ) = −16 sin ( 4 t ) a ( t ) = −16 sin ( 4 t )

3.40

28 28

3.41

d y d x = −3 x 2 sin ( x 3 ) d y d x = −3 x 2 sin ( x 3 )

3.42

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

3.43

g ( x ) = 1 5 x 4 / 5 g ( x ) = 1 5 x 4 / 5

3.44

s ( t ) = ( 2 t + 1 ) 1 / 2 s ( t ) = ( 2 t + 1 ) 1 / 2

3.45

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

3.46

h ( x ) = −3 6 x 9 x 2 h ( x ) = −3 6 x 9 x 2

3.47

y = x y = x

3.48

d y d x = 5 20 x 4 sec 2 y 2 y d y d x = 5 20 x 4 sec 2 y 2 y

3.49

y = 5 3 x 16 3 y = 5 3 x 16 3

3.50

h ( x ) = e 2 x + 2 x e 2 x h ( x ) = e 2 x + 2 x e 2 x

3.51

996

3.52

f ( x ) = 15 3 x + 2 f ( x ) = 15 3 x + 2

3.53

9 ln ( 3 ) 9 ln ( 3 )

3.54

d y d x = x x ( 1 + ln x ) d y d x = x x ( 1 + ln x )

3.55

y = π ( tan x ) π 1 sec 2 x y = π ( tan x ) π 1 sec 2 x

Section 3.1 Exercises

1.

4 4

3.

8.5 8.5

5.

3 4 3 4

7.

0.2 0.2

9.

0.25 0.25

11.

a. −4−4 b. y=34xy=34x

13.

a. 33 b. y=3x1y=3x1

15.

a. −79−79 b. y=−79x+143y=−79x+143

17.

a. 1212 b. y=12x+14y=12x+14

19.

a. −2−2 b. y=−2x10y=−2x10

21.

5 5

23.

13 13

25.

1 4 1 4

27.

1 4 1 4

29.

−3 −3

31.

a. (i)5.100000,(i)5.100000, (ii)5.010000,(ii)5.010000, (iii)5.001000,(iii)5.001000, (iv)5.000100,(iv)5.000100, (v)5.000010,(v)5.000010, (vi)5.000001,(vi)5.000001, (vii)4.900000,(vii)4.900000, (viii)4.990000,(viii)4.990000, (ix)4.999000,(ix)4.999000, (x)4.999900,(x)4.999900, (xi)4.999990,(xi)4.999990, (x)4.999999(x)4.999999 b. mtan=5mtan=5 c. y=5x+3y=5x+3

33.

a. (i)4.8771,(i)4.8771, (ii)4.9875(iii)4.9988,(ii)4.9875(iii)4.9988, (iv)4.9999,(iv)4.9999, (v)4.9999,(v)4.9999, (vi)4.9999(vi)4.9999 b. mtan=5mtan=5 c. y=5x+10y=5x+10

35.

a. 13;13; b. (i)0.3(i)0.3 m/s, (ii)0.3(ii)0.3 m/s, (iii)0.3(iii)0.3 m/s, (iv)0.3(iv)0.3 m/s; c. 0.3=130.3=13 m/s

37.

a. 2(h2+6h+12);2(h2+6h+12); b. (i)25.22(i)25.22 m/s, (ii)24.12(ii)24.12 m/s, (iii)24.01(iii)24.01 m/s, (iv)24(iv)24 m/s; c. 2424 m/s

39.

a. 1.25;1.25; b. 0.50.5

41.

lim x 0 x 1 / 3 0 x 0 = lim x 0 1 x 2 / 3 = lim x 0 x 1 / 3 0 x 0 = lim x 0 1 x 2 / 3 =

43.

lim x 1 1 1 x 1 = 0 1 = lim x 1 + x 1 x 1 lim x 1 1 1 x 1 = 0 1 = lim x 1 + x 1 x 1

45.

a. (i)61.7244(i)61.7244 ft/s, (ii)61.0725(ii)61.0725 ft/s (iii)61.0072(iii)61.0072 ft/s (iv)61.0007(iv)61.0007 ft/s b. At 44 seconds the race car is traveling at a rate/velocity of 6161 ft/s.

47.

a. The vehicle represented by f(t),f(t), because it has traveled 22 feet, whereas g(t)g(t) has traveled 11 foot. b. The velocity of f(t)f(t) is constant at 11 ft/s, while the velocity of g(t)g(t) is approximately 22 ft/s. c. The vehicle represented by g(t),g(t), with a velocity of approximately 44 ft/s. d. Both have traveled 44 feet in 44 seconds.

49.

a.

The function starts in the third quadrant, passes through the x axis at x = −3, increases to a maximum around y = 20, decreases and passes through the x axis at x = 1, continues decreasing to a minimum around y = −13, and then increases through the x axis at x = 4, after which it continues increasing.


b. a1.361,2.694a1.361,2.694

51.

a. N(x)=x30N(x)=x30 b. 3.33.3 gallons. When the vehicle travels 100100 miles, it has used 3.33.3 gallons of gas. c. 130.130. The rate of gas consumption in gallons per mile that the vehicle is achieving after having traveled 100100 miles.

53.

a.

The function starts in the second quadrant and gently decreases, touches the origin, and then it increases gently.


b. −0.028,−0.16,0.16,0.028−0.028,−0.16,0.16,0.028

Section 3.2 Exercises

55.

−3 −3

57.

8 x 8 x

59.

1 2 x 1 2 x

61.

−9 x 2 −9 x 2

63.

−1 2 x 3 / 2 −1 2 x 3 / 2

65.


The function starts in the third quadrant and increases to touch the origin, then decreases to a minimum at (2, −16), before increasing through the x axis at x = 3, after which it continues increasing.
67.


The function starts at (−3, 0), increases to a maximum at (−1.5, 1), decreases through the origin and to a minimum at (1.5, −1), and then increases to the x axis at x = 3.
69.

f ( x ) = 3 x 2 + 2 , a = 2 f ( x ) = 3 x 2 + 2 , a = 2

71.

f ( x ) = x 4 , a = 2 f ( x ) = x 4 , a = 2

73.

f ( x ) = e x , a = 0 f ( x ) = e x , a = 0

75.

a.

The function is linear at y = 3 until it reaches (1, 3), at which point it increases as a line with slope 3.


b. limh133hlimh1+3hhlimh133hlimh1+3hh

77.

a.

The function starts in the third quadrant as a straight line and passes through the origin with slope 2; then at (1, 2) it decreases convexly as 2/x.


b. limh12hhlimh1+2x+h2xh.limh12hhlimh1+2x+h2xh.

79.

a. x=1,x=1, b. x=2x=2

81.

0 0

83.

2 x 3 2 x 3

85.

f(x)=6x+2f(x)=6x+2

The function f(x) is graphed as an upward facing parabola with y intercept 4. The function f’(x) is graphed as a straight line with y intercept 2 and slope 6.
87.

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

The function f(x) is in the first quadrant and has asymptotes at x = 0 and y = 0. The function f’(x) is in the fourth quadrant and has asymptotes at x = 0 and y = 0.
89.

f(x)=3x2f(x)=3x2

The function f(x) starts is the graph of the cubic function shifted up by 1. The function f’(x) is the graph of a parabola that is slightly steeper than the normal squared function.
91.

a. Average rate at which customers spent on concessions in thousands per customer. b. Rate (in thousands per customer) at which xx customers spent money on concessions in thousands per customer.

93.

a. Average grade received on the test with an average study time between two values. b. Rate (in percentage points per hour) at which the grade on the test increased or decreased for a given average study time of xx hours.

95.

a. Average change of atmospheric pressure between two different altitudes. b. Rate (torr per foot) at which atmospheric pressure is increasing or decreasing at xx feet.

97.

a. The rate (in degrees per foot) at which temperature is increasing or decreasing for a given height x.x. b. The rate of change of temperature as altitude changes at 10001000 feet is −0.1−0.1 degrees per foot.

99.

a. The rate at which the number of people who have come down with the flu is changing tt weeks after the initial outbreak. b. The rate is increasing sharply up to the third week, at which point it slows down and then becomes constant.

101.
Time (seconds) h(t)(m/s)h(t)(m/s)
00 22
11 22
22 5.55.5
33 10.510.5
44 9.59.5
55 77
103.

G(t)=2.858t+0.0857G(t)=2.858t+0.0857

This graph has the points (0, 0), (1, 2), (2, 4), (3, 13), (4, 25), and (5, 32). There is a quadratic line fit to the points with y intercept near 0.


This graph has a straight line with y intercept near 0 and slope slightly less than 3.
105.

H(t)=0,G(t)=2.858andf(t)=1.222t+5.912H(t)=0,G(t)=2.858andf(t)=1.222t+5.912 represent the acceleration of the rocket, with units of meters per second squared (m/s2).(m/s2).

Section 3.3 Exercises

107.

f ( x ) = 15 x 2 1 f ( x ) = 15 x 2 1

109.

f ( x ) = 32 x 3 + 18 x f ( x ) = 32 x 3 + 18 x

111.

f ( x ) = 270 x 4 + 39 ( x + 1 ) 2 f ( x ) = 270 x 4 + 39 ( x + 1 ) 2

113.

f ( x ) = −5 x 2 f ( x ) = −5 x 2

115.

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

117.

f ( x ) = x 2 18 x + 64 ( x 2 7 x + 1 ) 2 f ( x ) = x 2 18 x + 64 ( x 2 7 x + 1 ) 2

119.


The graph y is a slightly curving line with y intercept at 1. The line T(x) is straight with y intercept 3 and slope 1/2.


T(x)=4x+7T(x)=4x+7

121.


The graph y is a two crescents with the crescent in the third quadrant sloping gently from (−3, −1) to (−1, −5) and the other crescent sloping more sharply from (0.8, −5) to (3, 0.2). The straight line T(x) is drawn through (0, −5) with slope 4.


T(x)=4x5T(x)=4x5

123.

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

125.

h ( x ) = 3 f ( x ) ( g ( x ) + 2 ) 3 f ( x ) g ( x ) ( g ( x ) + 2 ) 2 h ( x ) = 3 f ( x ) ( g ( x ) + 2 ) 3 f ( x ) g ( x ) ( g ( x ) + 2 ) 2

127.

16 9 16 9

129.

Undefined

131.

a. 2,2, b. does not exist, c. 2.52.5

133.

a. 23, b. y=23x28y=23x28

The graph is a slightly deformed cubic function passing through the origin. The tangent line is drawn through (0, −28) with slope 23.
135.

a. 3, b. y=3x+2y=3x+2

The graph starts in the third quadrant, increases quickly and passes through the x axis near −0.9, then increases at a lower rate, passes through (0, 2), increases to (1, 5), and then decreases quickly and passes through the x axis near 1.2.
137.

y = −7 x 3 y = −7 x 3

139.

y = −5 x + 7 y = −5 x + 7

141.

y = 3 2 x + 15 2 y = 3 2 x + 15 2

143.

y = −3 x 2 + 9 x 1 y = −3 x 2 + 9 x 1

145.

1212112121 or 0.0992 ft/s

147.

a. −2t42t3+200t+50(t3+50)2−2t42t3+200t+50(t3+50)2 b. −0.02395−0.02395 mg/L-hr, −0.01344 mg/L-hr, −0.003566 mg/L-hr, −0.001579 mg/L-hr c. The rate at which the concentration of drug in the bloodstream decreases is slowing to 0 as time increases.

149.

a. F(d)=−2Gm1m2d3F(d)=−2Gm1m2d3 b. −1.33×10−7−1.33×10−7 N/m

Section 3.4 Exercises

151.

a. v(t)=6t230t+36,a(t)=12t30;v(t)=6t230t+36,a(t)=12t30; b. speeds up (2,2.5)(3,),(2,2.5)(3,), slows down (0,2)(2.5,3)(0,2)(2.5,3)

153.

a. 464ft/s2464ft/s2 b. −32ft/s2−32ft/s2

155.

a. 5 ft/s b. 9 ft/s

157.

a. 84 ft/s, −84 ft/s b. 84 ft/s c. 258s258s d. −32ft/s2−32ft/s2 in both cases e. 18(25+965)s18(25+965)s f. −4965ft/s−4965ft/s

159.

a. Velocity is positive on (0,1.5)(6,7),(0,1.5)(6,7), negative on (1.5,2)(5,6),(1.5,2)(5,6), and zero on (2,5).(2,5). b.

The graph is a straight line from (0, 2) to (2, −1), then is discontinuous with a straight line from (2, 0) to (5, 0), and then is discontinuous with a straight line from (5, −4) to (7, 4).


c. Acceleration is positive on (5,7),(5,7), negative on (0,2),(0,2), and zero on (2,5).(2,5). d. The object is speeding up on (6,7)(1.5,2)(6,7)(1.5,2) and slowing down on (0,1.5)(5,6).(0,1.5)(5,6).

161.

a. R(x)=10x0.001x2R(x)=10x0.001x2 b. R(x)=100.002xR(x)=100.002x c. $6 per item, $0 per item

163.

a. C(x)=65C(x)=65 b. R(x)=143x0.03x2,R(x)=1430.06xR(x)=143x0.03x2,R(x)=1430.06x c. 83,−97.83,−97. At a production level of 1000 cordless drills, revenue is increasing at a rate of $83 per drill; at a production level of 4000 cordless drills, revenue is decreasing at a rate of $97 per drill. d. P(x)=−0.03x2+78x75000,P(x)=−0.06x+78P(x)=−0.03x2+78x75000,P(x)=−0.06x+78 e. 18,−162.18,−162. At a production level of 1000 cordless drills, profit is increasing at a rate of $18 per drill; at a production level of 4000 cordless drills, profit is decreasing at a rate of $162 per drill.

165.

a. N(t)=3000(−4t2+400(t2+100)2)N(t)=3000(−4t2+400(t2+100)2) b. 120,0,−14.4,−9.6120,0,−14.4,−9.6 c. The bacteria population increases from time 0 to 10 hours; afterwards, the bacteria population decreases. d. 0,−6,0.384,0.432.0,−6,0.384,0.432. The rate at which the bacteria is increasing is decreasing during the first 10 hours. Afterwards, the bacteria population is decreasing at a decreasing rate.

167.

a. P(t)=0.03983+0.4280P(t)=0.03983+0.4280 b. P(t)=0.03983.P(t)=0.03983. The population is increasing. c. P(t)=0.P(t)=0. The rate at which the population is increasing is constant.

169.

a. p(t)=−0.6071x2+0.4357x0.3571p(t)=−0.6071x2+0.4357x0.3571 b. p(t)=−1.214x+0.4357.p(t)=−1.214x+0.4357. This is the velocity of the sensor. c. p(t)=−1.214.p(t)=−1.214. This is the acceleration of the sensor; it is a constant acceleration downward.

171.

a.

The graph is a straight line drawn through the origin with slope 1/2.


b. f(x)=a.f(x)=a. The more increase in prey, the more growth for predators. c. f(x)=0.f(x)=0. As the amount of prey increases, the rate at which the predator population growth increases is constant. d. This equation assumes that if there is more prey, the predator is able to increase consumption linearly. This assumption is unphysical because we would expect there to be some saturation point at which there is too much prey for the predator to consume adequately.

173.

a.

The graph increases from the origin quickly at first and then slowly to (10, 0.4).


b. f(x)=2axn2(n2+x2)2.f(x)=2axn2(n2+x2)2. When the amount of prey increases, the predator growth increases. c. f(x)=2an2(n23x2)(n2+x2)3.f(x)=2an2(n23x2)(n2+x2)3. When the amount of prey is extremely small, the rate at which predator growth is increasing is increasing, but when the amount of prey reaches above a certain threshold, the rate at which predator growth is increasing begins to decrease. d. At lower levels of prey, the prey is more easily able to avoid detection by the predator, so fewer prey individuals are consumed, resulting in less predator growth.

Section 3.5 Exercises

175.

d y d x = 2 x sec x tan x d y d x = 2 x sec x tan x

177.

d y d x = 2 x cot x x 2 csc 2 x d y d x = 2 x cot x x 2 csc 2 x

179.

d y d x = x sec x tan x sec x x 2 d y d x = x sec x tan x sec x x 2

181.

d y d x = ( 1 sin x ) ( 1 sin x ) cos x ( x + cos x ) d y d x = ( 1 sin x ) ( 1 sin x ) cos x ( x + cos x )

183.

d y d x = 2 csc 2 x ( 1 + cot x ) 2 d y d x = 2 csc 2 x ( 1 + cot x ) 2

185.

y=xy=x

The graph shows negative sin(x) and the straight line T(x) with slope −1 and y intercept 0.
187.

y=x+23π2y=x+23π2

The graph shows the cosine function shifted up one and has the straight line T(x) with slope 1 and y intercept (2 – 3π)/2.
189.

y=xy=x

The graph shows the function as starting at (−1, 3), decreasing to the origin, continuing to slowly decrease to about (1, −0.5), at which point it decreases very quickly.
191.

3 cos x x sin x 3 cos x x sin x

193.

1 2 sin x 1 2 sin x

195.

2 csc x ( csc 2 x + cot 2 x ) 2 csc x ( csc 2 x + cot 2 x )

197.

( 2 n + 1 ) π 4 , where n is an integer ( 2 n + 1 ) π 4 , where n is an integer

199.

( π 4 ,   1 ) ,   ( 3 π 4 ,   −1 ) ,   ( 5 π 4 ,   1 ) ,   ( 7 π 4 ,   - 1 ) ( π 4 ,   1 ) ,   ( 3 π 4 ,   −1 ) ,   ( 5 π 4 ,   1 ) ,   ( 7 π 4 ,   - 1 )

201.

a = 0 , b = 3 a = 0 , b = 3

203.

y=5cos(x),y=5cos(x), increasing on (0,π2),(3π2,5π2),(0,π2),(3π2,5π2), and (7π2,12)(7π2,12)

209.

3 sin x 3 sin x

211.

5 cos x 5 cos x

213.

720 x 7 5 tan ( x ) sec 3 ( x ) tan 3 ( x ) sec ( x ) 720 x 7 5 tan ( x ) sec 3 ( x ) tan 3 ( x ) sec ( x )

Section 3.6 Exercises

215.

18 u 2 · 7 = 18 ( 7 x 4 ) 2 · 7 18 u 2 · 7 = 18 ( 7 x 4 ) 2 · 7

217.

sin u · −1 8 = sin ( x 8 ) · −1 8 sin u · −1 8 = sin ( x 8 ) · −1 8

219.

8 x 24 2 4 u + 3 = 4 x 12 4 x 2 24 x + 3 8 x 24 2 4 u + 3 = 4 x 12 4 x 2 24 x + 3

221.

a. u=3x2+1;u=3x2+1; b. 18x(3x2+1)218x(3x2+1)2

223.

a. f(u)=u7,u=x7+7x;f(u)=u7,u=x7+7x; b. 7(x7+7x)6·(177x2)7(x7+7x)6·(177x2)

225.

a. f(u)=cscu,u=πx+1;f(u)=cscu,u=πx+1; b. πcsc(πx+1)·cot(πx+1)πcsc(πx+1)·cot(πx+1)

227.

a. f(u)=−6u−3,u=sinx,f(u)=−6u−3,u=sinx, b. 18(sin)−4x·cosx18(sin)−4x·cosx

229.

4 ( 5 2 x ) 3 4 ( 5 2 x ) 3

231.

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

233.

−3 ( tan x + sin x ) −4 · ( sec 2 x + cos x ) −3 ( tan x + sin x ) −4 · ( sec 2 x + cos x )

235.

−7 cos ( cos 7 x ) · sin 7 x −7 cos ( cos 7 x ) · sin 7 x

237.

−12 cot 2 ( 4 x + 1 ) · csc 2 ( 4 x + 1 ) −12 cot 2 ( 4 x + 1 ) · csc 2 ( 4 x + 1 )

239.

10 3 4 10 3 4

241.

y = −1 2 x y = −1 2 x

243.

x = ± 6 x = ± 6

245.

10

247.

1 8 1 8

249.

−4 −4

251.

−12 −12

253.

a. 200343200343 m/s, b. 60024016002401 m/s2, c. The train is slowing down since velocity and acceleration have opposite signs.

255.

a. C(x)=0.0003x20.04x+3C(x)=0.0003x20.04x+3 b. dCdt=100·(0.0003x20.04x+3)dCdt=100·(0.0003x20.04x+3) c. Approximately $90,300 per week

257.

a. dSdt=8πr2(t+1)3dSdt=8πr2(t+1)3 b. The volume is decreasing at a rate of π36π36 ft3/min.

259.

~2.3~2.3 ft/hr

Section 3.7 Exercises

261.

a.

A curved line starting at (−3, 0) and passing through (−2, 1) and (1, 2). There is another curved line that is symmetric with this about the line x = y. That is, it starts at (0, −3) and passes through (1, −2) and (2, 1).


b. (f−1)(1)~2(f−1)(1)~2

263.

a.

A quarter circle starting at (0, 4) and ending at (4, 0).


b. y'=-x16-x2y'1=-116-1=-115y'=-x16-x2y'1=-116-1=-115

265.

a. 6, b. x=f−1(y)=(y+32)1/3,x=f−1(y)=(y+32)1/3, c. 1616

267.

a. 1,1, b. x=f−1(y)=sin−1y,x=f−1(y)=sin−1y, c. 11

269.

1 5 1 5

271.

1 3 1 3

273.

1 1

275.

a. 4,4, b. y=4xy=4x

277.

a. 113,113, b. y=113x+1813y=113x+1813

279.

2 x 1 x 4 2 x 1 x 4

281.

−1 1 x 2 −1 1 x 2

283.

3 ( 1 + tan −1 x ) 2 1 + x 2 3 ( 1 + tan −1 x ) 2 1 + x 2

285.

−1 ( 1 + x 2 ) ( tan −1 x ) 2 −1 ( 1 + x 2 ) ( tan −1 x ) 2

287.

x ( 5 x 2 ) 4 x 2 x ( 5 x 2 ) 4 x 2

289.

−1 −1

291.

1 2 1 2

293.

1 10 1 10

295.

a. v(t)=11+t2v(t)=11+t2 b. a(t)=−2t(1+t2)2a(t)=−2t(1+t2)2 c. (a)0.2,0.06,0.03;(b)0.16,−0.028,−0.0088(a)0.2,0.06,0.03;(b)0.16,−0.028,−0.0088 d. The hockey puck is decelerating/slowing down at 2, 4, and 6 seconds.

297.

−0.0168−0.0168 radians per foot

299.

a. dθdx=10100+x2401600+x2dθdx=10100+x2401600+x2 b. 18325,9340,424745,018325,9340,424745,0 c. As a person moves farther away from the screen, the viewing angle is increasing, which implies that as he or she moves farther away, his or her screen vision is widening. d. 5412905,3500,19829945,913605412905,3500,19829945,91360 e. As the person moves beyond 20 feet from the screen, the viewing angle is decreasing. The optimal distance the person should stand for maximizing the viewing angle is 20 feet.

Section 3.8 Exercises

301.

d y d x = −2 x y d y d x = −2 x y

303.

d y d x = x 3 y y 2 x d y d x = x 3 y y 2 x

305.

d y d x = y y 2 x + 4 x + 4 x d y d x = y y 2 x + 4 x + 4 x

307.

d y d x = y 2 cos ( x y ) 2 y sin ( x y ) x y cos x y d y d x = y 2 cos ( x y ) 2 y sin ( x y ) x y cos x y

309.

d y d x = −3 x 2 y y 3 x 3 + 3 x y 2 d y d x = −3 x 2 y y 3 x 3 + 3 x y 2

311.


The graph has a crescent in each of the four quadrants. There is a straight line marked T(x) with slope −1/2 and y intercept 2.


y=−12x+2y=−12x+2

313.


The graph has two curves, one in the first quadrant and one in the fourth quadrant. They are symmetric about the x axis. The curve in the first quadrant goes from (0.3, 5) to (1.5, 3.5) to (5, 4). There is a straight line marked T(x) with slope 1/(π + 12) and y intercept −(3π + 38)/(π + 12).


y=1π+12x3π+38π+12y=1π+12x3π+38π+12

315.


The graph starts in the third quadrant near (−5, 0), remains near 0 until x = −4, at which point it decreases until it reaches near (0, −5). There is an asymptote at x = 0. The graph begins again near (0, 5) decreases to (1, 0) and then increases a little bit before decreasing to be near (5, 0). There is a straight line marked T(x) that coincides with y = 0.


y=0y=0

317.

a. y=x+2y=x+2 b. (3,−1)(3,−1)

319.

a. (±7,0)(±7,0) b. −2−2 c. They are parallel since the slope is the same at both intercepts.

321.

y = x + 1 y = x + 1

323.

a. −0.5926−0.5926 b. When $81 is spent on labor and $16 is spent on capital, the amount spent on capital is decreasing by $0.5926 per $1 spent on labor.

325.

−8 −8

327.

−2.67 −2.67

329.

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

Section 3.9 Exercises

331.

2 x e x + x 2 e x 2 x e x + x 2 e x

333.

e x 3 ln x ( 3 x 2 ln x + x 2 ) e x 3 ln x ( 3 x 2 ln x + x 2 )

335.

4 ( e x + e x ) 2 4 ( e x + e x ) 2

337.

2 4 x + 2 · ln 2 + 8 x 2 4 x + 2 · ln 2 + 8 x

339.

π x π 1 · π x + x π · π x ln π π x π 1 · π x + x π · π x ln π

341.

5 2 ( 5 x 7 ) 5 2 ( 5 x 7 )

343.

tan x ln 10 tan x ln 10

345.

2 x · ln 2 · log 3 7 x 2 4 + 2 x · 2 x ln 7 ln 3 2 x · ln 2 · log 3 7 x 2 4 + 2 x · 2 x ln 7 ln 3

347.

( sin 2 x ) 4 x [ 4 · ln ( sin 2 x ) + 8 x · cot 2 x ] ( sin 2 x ) 4 x [ 4 · ln ( sin 2 x ) + 8 x · cot 2 x ]

349.

x log 2 x · 2 ln x x ln 2 x log 2 x · 2 ln x x ln 2

351.

x cot x · [ csc 2 x · ln x + cot x x ] x cot x · [ csc 2 x · ln x + cot x x ]

353.

x −1 / 2 ( x 2 + 3 ) 2 / 3 ( 3 x 4 ) 4 · [ −1 2 x + 4 x 3 ( x 2 + 3 ) + 12 3 x 4 ] x −1 / 2 ( x 2 + 3 ) 2 / 3 ( 3 x 4 ) 4 · [ −1 2 x + 4 x 3 ( x 2 + 3 ) + 12 3 x 4 ]

355.


The function starts at (−3, 0), decreases slightly and then increases through the origin and increases to (1.25, 10). There is a straight line marked T(x) with slope −1/(5 + 5 ln 5) and y intercept 5 + 1/(5 + 5 ln 5).


y=−15+5ln5x+(5+15+5ln5)y=−15+5ln5x+(5+15+5ln5)

357.

a. x=e~2.718x=e~2.718 b. y' > 0 on (0, e); y' < 0 on (e, )y' > 0 on (0, e); y' < 0 on (e, )

359.

a. P=500,000(1.05)tP=500,000(1.05)t individuals b. P(t)=24395·(1.05)tP(t)=24395·(1.05)t individuals per year c. 39,73739,737 individuals per year

361.

a. At the beginning of 1960 there were 5.3 thousand cases of the disease in New York City. At the beginning of 1964 there were approximately 723 cases of the disease in the United States. b. At the beginning of 1960 the number of cases of the disease was decreasing at rate of −4.611−4.611 thousand per year; at the beginning of 1963, the number of cases of the disease was decreasing at a rate of −0.2808−0.2808 thousand per year.

363.

p = 35741 ( 1.045 ) t p = 35741 ( 1.045 ) t

365.
Years since 1790 PP
0 69.25
10 107.5
20 167.0
30 259.4
40 402.8
50 625.5
60 971.4
70 1508.5

Review Exercises

367.

False.

369.

False

371.

1 2 x + 4 1 2 x + 4

373.

9 x 2 + 8 x 3 9 x 2 + 8 x 3

375.

e sin x cos x e sin x cos x

377.

x sec 2 ( x ) + 2 x cos ( x ) + tan ( x ) x 2 sin ( x ) x sec 2 ( x ) + 2 x cos ( x ) + tan ( x ) x 2 sin ( x )

379.

1 4 ( x 1 x 2 + sin −1 ( x ) ) 1 4 ( x 1 x 2 + sin −1 ( x ) )

381.

cos x · ( ln x + 1 ) x ln ( x ) sin x cos x · ( ln x + 1 ) x ln ( x ) sin x

383.

4 x ( ln 4 ) 2 + 2 sin x + 4 x cos x x 2 sin x 4 x ( ln 4 ) 2 + 2 sin x + 4 x cos x x 2 sin x

385.

T = ( 2 + e ) x 2 T = ( 2 + e ) x 2

387.


The function is the straight line y = −4 until x = 0, at which point it becomes a straight line starting at the origin with slope 2. There is no value assigned for this function at x = 0.
389.

w(3)=2.9π6.w(3)=2.9π6. At 3 a.m. the tide is decreasing at a rate of 1.518 ft/hr.

391.

−7.5.−7.5. The wind speed is decreasing at a rate of 7.5 mph/hr

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