⎡ ( 2 x 4 1) ∫ ⎤ + cos ( ⎡ 2 x + 1) 4 ⎤ ( 2 + 1) 3 x dx 3.9 Concepts Review

sin 3 ⎡ ( 2 x 4 1) ∫ ⎤ + cos ( ⎡ 2 x + 1) 4 ⎤ ( 2 + 1) 3 x dx 3.9 Concepts Review

1. = differential equation ∫

u ⋅ du = ∫ u du = ⎜ + ⎜ C 1

2. function

sin 4 ⎡ ( x 2 + 1) 4 ⎤

3. separate variables

32 4. − 32 t + v 0 ; 16 − t 2 + vt 0 + s 0

51. If x ≥

0, then x = x and ∫ x dx = 2 x + C .

Problem Set 3.9

If x < 0, then

52. Using sin 2 u

1 1 ∫ dx

sin x dx = ∫

53. Different software may produce different, but

dx

equivalent answers. These answers were

dy 2

produced by Mathematica.

=− C 1 sin xC − 2 cos x

dx 2

2 a. ∫ 6sin 3( (

x − 2) ) dx =− 2 cos 3( ( x − 2) ) + C dy

b. ∫ sin ⎜⎟ dx = cos ⎜⎟ − cos ⎜⎟ + C 1 2 1 2

=− ( C sin xC − cos ) ( x + C sin x + C cos ) x = 0

⎝⎠ 6 2 ⎝⎠ 2 2 ⎝⎠ 6 x 2 sin 2 x

c. ∫ ( x 2 cos 2 x + x sin 2 ) x dx =

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dy x

4. For y = sin(x + C),

= cos( x + C ) 7. =

1 =±+ 1 CC ; = and the square root is 0

∫ = ∫ ( 2 dy x + 1) dx dy x

∫ dy = − 3 ∫ ( x + 2) dx z − ∫ 2 dz = 2 ∫ t dt

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ds = (16 2 t +− 4 t 1) dt − ∫ y − 2 dy = 2( xx 2 2) ∫ 4 + dx ∫ ∫ 2

At t = 0, s = 100:C = 100

∫ 10 u du = ∫ ( t − t dt )

y = ∫ 3 x dx = 2 x + C

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dx

− y 2 dy

∫ 4/3 = 3 ∫ dx 3 = (2 t + 1) + C 1

s = ∫ (2 t + 1) 4/3 dt − 1 C x dt

At (1, 2):

∫ (2 t + 1) 2 dt − ∫ 1 dt 2 = 16 8

(2 t + 1) 73 − 1111 t +

∫ 3 t dt = + v 0 3

At t = 2: v = (5) 43 −≈ 2.83

s = ∫ ⎜ + 3 ⎟ dt = ++ 3 t s

6 20. v = ∫ (3 t + 1) − dt = (3 t + 1) − 3 3 dt

++= 3 t 0 + 3 t 1

6 6 =− (3 t + 1) − 2 + C 1 6

At t = 2: v = 5 cm/s

18. v = ∫ (1 + t ) − 4 dt =−

3 + C s =− ∫ (3 t + 1) − 2 dt + dt 3(1

=− ∫ (3 t + 1) 3 dt + ∫ dt

3 6 1 At t = 2: 25 =− (7) 2 v − + ≈ 4.16

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21. v = –32t + 96,

27. v esc = 2 gR s =− 16 t 2 + 96 t + s =− 16 t 0 2 + 96 t For the Moon, v

esc ≈ 2(0.165)(32)(1080 5280) ⋅ v = 0 at t = 3

≈ 7760 ft/s ≈ 1.470 mi/s.

At t = 3, s =− 16(3 ) 96(3) 2 + = 144 ft

For Venus, v esc ≈ 2(0.85)(32)(3800 5280) ⋅

≈ 33,038 ft/s ≈ 6.257 mi/s.

22. a =

dv

For Jupiter, v esc ≈ 194,369 ft/s ≈ 36.812 mi/s.

dt

For the Sun, v esc ≈ 2,021,752 ft/s

ds

v = ∫ k dt =+ kt v 0 = ; ≈ 382.908 mi/s.

dt

s = ∫ ( kt + v dt 0 ) = t + vt 0 + s 0 = 2 t + vt 28. 0 0 60 mi/h = 88 ft/s

2 2 v = 0 = –11t + 88; t = 8 sec

v = 0 when t =− 0 . Then

st () =− t 2 + 88 t

⎟ =− 0 . 2 s () 8 =− () 8 + () 88 8 = 352 feet

2 2 ⎝ ⎠ k The shortest distance in which the car can be

braked to a halt is 352 feet.

= 1.5 mi/h/s = 2.2 ft/s 2

∫ Δ dv =− ∫ 5.28 dt dt t 10

ds

=− 5.28 t + v 0 = –5.28t + 56

30. 75 = 2 (3.75) + v 0 (3.75) 0; 5 + v 0 = ft/s

dt

∫ ds =− ∫ ( 5.28 t + 56) dt

s 2 =− 2.64 t + 56 t + s

0 =− 2.64 t 2 + 56 t + 1000 31. For the first 10 s, a = = tv 6, 3 = t , and

dv

When t = 4.5, v = 32.24 ft/s and s = 1198.54 ft s 3 = . So v(10) = 300 and s(10) = 1000. After t

24. v = 0 when t =

≈ 10.6061 . Then

10 s, a =

=− 10 , v = –10(t – 10) + 300, and

dt

+ 10) 1000. v = 0 at ≈ 1296.97 ft

s ≈− 2.64(10.6061) + 56(10.6061) 1000 + s =− 5( t − 10) + 300( t −

t = 40, at which time s = 50 m.

dV

32. a. After accelerating for 8 seconds, the velocity

25. =− kS is 8 · 3 = 24 m/s.

dt

Since V =π r 3 and S =π 4 r 2 , b. Since acceleration and deceleration are

constant, the average velocity during those times is

12 m/s . Solve 0 = –4t + 24 to get the

dr =− ∫ k dt

24 r = –kt + C

time spent decelerating. t = = 6 s; 4

2 = –k(0) + C and 0.5 = –k(10) + C, so

d = (12)(8) (24)(100) (12)(6) + + = 2568 m.

C = 2 and k =

. Then, r =− t + 2 .

26. Solving v = –136 = –32t yields t = .

Then s ==− 0 16 ⎛ ⎞ ⎜ ⎟ + (0) ⎛ ⎞ ⎜ ⎟ + s 0 , so

s 0 = 289 ft.

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3.10 Chapter Review

Concepts Test

x →∞ 1– x 2 x →∞ 1 2 x –1

1. True:

Max-Min Existence Theorem

Since c is an interior point and f is

differentiable ( fc ′

() exists), by the

lim

= lim

Critical Point Theorem, c is a

x →∞ – 1– x 2 x →∞ – 1

stationary point ( fc ′ () = 0).

3. True:

For example, let f(x) = sin x.

2 4. False: fx () = x 1/ 3 is continuous and 3 x + 2 x + sin x

() x does

increasing for all x, but f ' x

sin x

sin x

not exist at x = 0.

fx ′ ( ) 18 = x 5 + 16 x 3 + 4; x 17. True:

The function is differentiable on

f ′′ () x = 90 x 4 + 48 x 2 + 4 , which is

greater than zero for all x.

18. False:

fx ′ () =

so f′ (0) does not exist.

6. False:

For example, fx () = x 3 is increasing

on [–1, 1] but f ′ (0) = 0.

19. False:

There are two points: x =− , .

7. True:

When ( ) fx ′ > fx 0, ( ) is increasing.

8. False:

If ( ) f ′′ c = 0 , c is a candidate, but not

20. True:

Let g(x) = D where D is any number. Then gx ′ ()

= and so, by Theorem B 0 example, if fx () = x 4 , (0) of Section 3.6, P ′′ = 0 but f (x) = g(x) + C = D + C, which is a

necessarily an inflection point. For

x = 0 is not an inflection point.

constant, for all x in (a, b).

9. True:

fx () = ax 2 + bx + c ; 21. False:

For example if fx () = x 4 , fx ′ () = 2 ax + bf ; () ′′ x = 2 a f ′ (0) = f ′′ (0) = but f has a 0

minimum at x = 0.

10. True:

If f(x) is increasing for all x in [a, b], the maximum occurs at b.

= cos ; sin ; x =− x –sin x = 0

11. False:

dx

tan x has a minimum value of 0. 2 dx

This occurs whenever x = k π where k

has infinitely many solutions.

is an integer.

23. False:

The rectangle will have minimum perimeter if it is a square.

12. True:

lim (2 x 3 + x ) = ∞ while

++ x tan ) x = ∞ while

when , x = Ky = K .

x →− π

24. True:

By the Mean Value Theorem, the

14. False:

At x = 3 there is a removable

derivative must be zero between each

discontinuity.

pair of distinct x-intercepts.

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25. True:

If fx () 1 < fx ( 2 ) and gx () 1 < gx ( 2 ) is an inflection point while if

for x 1 < x 2 , b 2 –3 ac < there are no critical 0

fx () 1 + gx () 1 < fx ( 2 ) + gx ( 2 ), so

points.) On an open interval, no local maxima

f + g is increasing. can come from endpoints, so there can

26. False: Let f(x) = g(x) = 2x, fx ′

() > and 0

be at most one local maximum in an

open interval.

gx ′ () > for all x, but 0

fx ′ () = ≠ so f(x) has no local a 0 (– ∞ , 0).

fxgx ()() = 4 x 2 is decreasing on

34. True:

minima or maxima. On an open interval, no local minima or maxima

27. True:

Since ( ) f ′′ x > fx ′ 0, ( ) is increasing

can come from endpoints, so f(x) has no local minima.

for x ≥

0. Therefore, fx ′ () > 0 for x

in [0, ∞ ), so f(x) is increasing.

35. True:

Intermediate Value Theorem

28. False: If f(3) = 4, the Mean Value Theorem

The Bisection Method can be very requires that at some point c in [0, 3],

36. False:

slow to converge.

does not contradict that fx ′ () ≤ 2 for

Newton’s method can fail to exist for several reasons (e.g. if f’(x) is 0 at or

29. True: If the function is nondecreasing,

near r). It may be possible to achieve fx () must be greater than or equal to convergence by selecting a different zero, and if fx ′ () ≥ 0,

f is

starting value.

nondecreasing. This can be seen using the Mean Value Theorem.

39. True:

From the Fixed-point Theorem, if g is

continuous on [] ab , and

30. True: However, if the constant is 0, the

functions are the same. ≤≤, x b

a ≤ gx () ≤ b whenever a

then there is at least one fixed point

For example, let fx () = e x . on [] ab , . The given conditions satisfy

31. False:

lim e x = 0, so y = 0 is a horizontal

these criteria.

The Bisection Method always converges as long as the function is

32. True: If f(c) is a global maximum then f(c) continuous and the values of the is the maximum value of f on

function at the endpoints are of (a, b) ↔ S where (a, b) is any interval

opposite sign.

containing c and S is the domain of f. Hence, f(c) is a local maximum value.

41. True:

Theorem 3.8.C

42. True:

fx ′ () = 3 ax 2 + 2 bx + c ; fx ′ () = 0 the Product Rule

Obtained by integrating both sides of

33. True:

– b ± b 2 –3 ac

when x =

3 a Quadratic Formula. f ′′ () x = 6 ax + 2 b

44. True:

If ( ) Fx = ∫ () f x dx f x , () is a

so

′′ ⎜ – b ± b 2 –3 ac f ⎞ ⎟ =± 2 b 2 –3. ac

derivative of F(x).

⎝ 2 ⎠ 45. False: fx () = x + 2 x + and 1

gx () = x 2 + 7 x − are a counter- 5 point is a local maximum and the

Thus, if b 2 –3 ac > 0, one critical

example.

other is a local minimum.

b (If 2 –3 ac = the only critical point 0

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46. False:

The two sides will in general differ by

5. fx ′ () = ; () fx ′ does not exist at x = 0.

a constant term.

47. True:

At any given height, speed on the

Critical points: – , 0, 1

downward trip is the negative of

speed on the upward.

f ⎜ – ⎞= ⎟ , (0) f =

f 0, (1) 1 =

⎝ 2 ⎠ 2 Global minimum f(0) = 0;

Sample Test Problems

global maximum f(1) = 1

6. fs ′ ()1 =+ ; () fs ′ does not exist when s = 0. Critical points: 0, 1, 4

1. fx ′ () = x 2 – 2; 2x – 2 = 0 when x = 1.

For s < 0, s = – s so f(s) = s – s = 0 and Global minimum f(1) = –1;

f (0) = 0, f(1) = –1, f(4) = 8

global maximum f(4) = 8

fs ′ ( ) 1 – 1 0. = = Critical points: 1 and all s in [–1, 0]

2. ft ′

() = – 2 ; – 2 is never 0.

1 1 f (1) = 2, f(s) = 0 for s in [–1, 0]

Global minimum f(s) = 0, –1 ≤ s ≤ 0; Critical points: 1, 4

global maximum f(1) = 2.

f (1) = 1, f (4) =

7. fx ′ ( ) 12 = x 3 – 12 x 2 = 12 x 2 ( – 1); ( ) x fx ′ = 0

1 when x = 0, 1 Global minimum f (4) = ; Critical points: –2, 0, 1, 3

4 f (–2) = 80, f(0) = 0, f(1) = –1, f(3) = 135 global maximum f(1) = 1.

Global minimum f(1) = –1; global maximum f(3) = 135

3. fz ′ () = – 3 ; –

is never 0.

fu 2/3 ′ ; () = 0 when u = 0,

Critical points: –2, – 3( – 2) u

2 f′ (2) does not exist.

f (–2) ⎞ = ,– f ⎜ ⎟ = 4 12

Critical points: –1, 0,

Global minimum f (–2) = ; f (–1) = 3 –3 ≈ –1.44, (0) f = 0,

global maximum f ⎜ – ⎞= 4. f ⎜ ⎞=

⎟ 3 – ≈ –1.94, f(2) = 0, f(3) = 9

⎛ 12 Global minimum f ⎜ ⎞≈ ⎟ –1.94;

4. fx ′ () = – 3 ; –

is never 0.

global maximum f(3) = 9

Critical point: –2

9. fx ′ ( ) 10 = x 4 – 20 x 3 = 10 x 3 ( – 2); x

f (–2) =

fx ′ () = 0 when x = 0, 2

fx ′ () > 0 for x < 0, so f is increasing.

Critical points: –1, 0, 2, 3

Global minimum f (–2) = ; no global

1 f (–1) = 0, f(0) = 7, f(2) = –9, f(3) = 88

4 Global minimum f(2) = –9;

maximum.

global maximum f(3) = 88

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10. fx ′ () = 3( – 1) ( x 2 x + 2) 2 + 2( – 1) ( x 3 x + 2) 14. fx ′ () = 9 x 8 ; () fx ′ > 0 for all x ≠ 0.

= ( – 1) ( x 2 x + 2)(5 x + 4); ( ) fx ′ = 0 when

f ′′ () x = 72 x 7 ; () f ′′ x < 0 when x < 0.

f (x) is increasing on (– ∞ , ∞ ) and concave down x =

on (– ∞ , 0).

15. fx ′ () = 3 x 2 – 3 3( = x 2 – 1); ( ) fx ′ > 0 when

Critical points: –2, – , 1 ,2

f () x = 6; () xf ′′ x < 0 when x < 0.

f (–2) = 0, f

f (x) is increasing on (– ∞ , –1] ∪ [1, ∞ ) and

f (1) = 0, f(2) = 16

concave down on (– ∞ , 0).

Global minimum f ⎜ – ⎞≈ ⎟ –8.40; 16. fx ′ () =− 6 x 2 −

6 x + 12 = –6(x + 2)(x – 1);

fx ′ () > 0 when –2 < x < 1.

global maximum f(2) = 16

f ′′ () x =− 12 x − 6 = –6(2x + 1); f ′′ () x < 0 when

11. f ′ () θ = cos ; ( ) θ f ′ θ = 0 when θ =

f (x) is increasing on [–2, 1] and concave down on ⎣ 4 3

Critical points: ,, ⎝

1 when 0 << x .

⎟ ⎞≈ –0.87; when x

Global minimum f

f (x) is increasing on ⎡ 0, 1 ⎤ and concave down on global maximum f ⎛⎞= ⎜⎟ 1 ⎣ 5 ⎦

12. f ′ () θ = 2sin cos – cos θ θ θ = cos (2sin – 1); θ θ ⎝ 20 ⎠

() θ = 0 when θ = ,, in [0, π ]

Critical points:

f ′′ () x = x 6 – 24 x = x 6 (1 – 4 x 2 ); ( ) f ′′ x < 0 when

f (0) = 0, f ⎜⎟ = – , 0, f ⎜⎟ = ⎝⎠ 6 4 ⎝⎠ 2 1 – 1 << x 0 or x > .

f ⎜ ⎞= ⎟ –, f( π ) = 0 ⎝ 6 ⎠

f (x) is increasing on ⎢ –

⎥ and concave

Global minimum f ⎜⎟ = – or f ⎜ ⎟ = –;

4 ⎛ 1 ⎞⎛ ∪ 1 down on ⎞

global maximum f(0) = 0, f ⎛⎞= ⎜⎟ 0, or ⎝⎠ 2

f (π)=0

13. fx ′ () = 3–2; () xfx ′ > 0 when x < .

f ′′ () x = –2; ( ) f ′′ x is always negative.

f (x) is increasing on ⎜ –, ∞

and concave down

on (– ∞ , ∞ ).

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19. fx ′ () = 3 x 2 –4 x 3 = x 2 (3 – 4 ); ( ) x fx ′ > 0 when

f () x = x 6 – 8; ( ) f ′′ x > 0 when x > .

f (x) is concave up on ⎜ , ∞ ⎟ and concave down

f ′′ () x = x 6 – 12 x 2 = x 6 (1 – 2 ); ( ) x f ′′ x < 0 when

on ⎜ –,; ⎟ inflection point ⎜ , x < 0 or x > . ⎟

f (x) is increasing on ⎜ –, ∞ 3 ⎤ and concave down

on (– , 0) ∞ ∪ ⎜ , ∞ ⎞ ⎟ .

20. gt ′ () = 2 3 1 t – ; () gt ′ > when 2 2 1 0 3t > or

t 4 > , so t < – 1/ 4 or t > 1/ 4 .

3 3 3 8 22. x fx ′ () = – ; () fx ′

2 2 = 0 when x = 0.

() is increasing on ⎜ –,– ∞

; (0) f ′′ = –8, so f(0) = 6 is a

( x 2 + 1) 3

and decreasing on ⎢ –

local maximum. fx ′ () > 0 for x < 0 and

fx ′ () < 0 for x > 0 so

Local minimum g =

f(0) = 6 is a global maximum value. f(x) has no local maximum

minimum value.

23. 3 ′ fx () = 4 x 3 – 2; ( ) fx ′ = 0 when x = 3 . 2

gt () = 6 t + ; () gt ′′ > 0 when t > 0. g(t) has no

f ′′ ( ) 12 x x 3 2 = ; () f ′′ x = 0 when x = 0.

inflection point since g(0) does not exist.

is a global

f ′′ () x > 0 for all x ≠ 0; no inflection points No horizontal or vertical asymptotes

21. fx ′ () = xx 2 ( – 4) + x 2 = 3 x 2 –8 x = xx (3 – 8);

fx ′ () > 0 when x < 0 or x >

f (x) is increasing on (– , 0] ∞ ∪ ⎡

, ∞⎟ ⎢⎣ and

decreasing on 0,

Local minimum f ⎛⎞= ⎜⎟ –

local maximum f(0) = 0

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24. fx ′ () = 2( x 2 – 1)(2 ) x = 4( xx 2 – 1) = 4 x 3 –4; x Vertical asymptote x = 3

fx ′ () = 0 when x = –1, 0, 1.

f ′′ ( ) 12 x = x 2 –4 = 4(3 x 2 – 1); ( ) f ′′ x = 0 when

x =± 1 . 3

f ′′ (–1) = f 8, (0) ′′ = –4, (1) f ′′ = 8

Global minima f(–1) = 0, f(1) = 0; local maximum f(0) = 1

Inflection points ±

27. fx

x 3 ⎜ 2 ⎟ ′ ( ) 12 = – 12 x = 12 x 2 ( – 1); ( ) x fx ′ = 0

when x = 0, 1.

No horizontal or vertical asymptotes

f ′′ () x = 36 x 2 – 24 x = xx 12 (3 – 2); ( ) f ′′ x = 0

2 when x = 0, .

3 (1) 12, f ′′ = so f(1) = –1 is a minimum. Global minimum f(1) = –1; no local maxima

Inflection points (0, 0), ⎜ ,

No horizontal or vertical asymptotes.

25. fx ′ () =

3–6 x

; () fx ′ = 0 when x = 2, but x = 2

2 x –3 is not in the domain of f(x). fx ′ () does not exist when x = 3.

f ′′ () x =

3( – 4) x

; () f ′′ x = when x = 4. 0

4( – 3) x 3/2

Global minimum f(3) = 0; no local maxima Inflection point (4, 4) No horizontal or vertical asymptotes.

f () x – 3 ; () f ′′ x > 0 when x < 0 and

f ′′ () x < 0 when x > 0. No local minima or maxima

No inflection points

2 ; () fx ′ < 0 for all x ≠ 3. oblique asymptote.

( – 3) x

Vertical asymptote x = 0

3 ; () f ′′ x > when x > 3. ( – 3) 0 x No local minima or maxima

f ′′

2 () x =

No inflection points x 2 –2 1– lim

= lim

x = 1 x →∞ x –3 x →∞ 1– 3 x

Horizontal asymptote y = 1

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1 31. fx ′ () = – sin – cos ; ( ) x xfx ′ = 0 when

3 x > 0 x when x < 0 and

; () f ′′

f ′′ () x = – cos x + sin ; ( ) xf ′′ x = 0 when

f ′′ () x < 0 when x > 0

No local minima or maxima

No inflection points

Global minimum f ⎜ ⎟ ⎞= – 2;

lim [ ( ) 3 ] fx − x = lim ⎜ − ⎟ = 0 and y = 3x is an

oblique asymptote. global maximum f ⎜ – ⎞= ⎟ 2

x →∞

x →∞ ⎝ x ⎠

Vertical asymptote x = 0 ⎛ 3 π ⎞⎛ π ⎞ Inflection points ⎜ − ,0, ⎟⎜ ,0 ⎟

3 ; () fx ′ > 0 when x <− 1 and

( x + 1)

32. fx fx ′ () < 0 when x > –1. () = cos – sec x 2 xfx ; () ′ = 0 when x = 0

f ′′ () x = – sin – 2sec x 2 x tan x

= – sin (1 2sec x + 3 x )

No local minima or maxima

f ′′ () x = 0 when x = 0

No inflection points

No local minima or maxima

Inflection point f(0) = 0

horizontal asymptote. ππ

Vertical asymptotes x = –, Vertical asymptote x = –1

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33. fx ′ () = x sec 2 x + tan ; ( ) xfx ′ = 0 when x = 0

f ′′ () x = 2sec 2 x (1 + x tan ); ( ) x f ′′ x is never 0 on

⎝ 22 ⎠ (0) f ′′ = 2 Global minimum f(0) = 0

36. fx ′ () = –2sin – 2 cos ; ( ) x xfx ′ = 0 when

ππ 3 x = –,

f ′′ () x = –2 cos x + 2sin ; ( ) xf ′′ x = 0 when

2 () ′′ = –2 cot csc x xf ; () ′′ x = 0 when f ⎜ – ⎟ = –2 2, 2 2 f ′′ ⎜ ⎟ =

x = ; f ′′ () x > 0 on

Global minimum f ⎞=

Inflection point ⎜ , π ⎟ ⎛ global maximum π f – ⎞= 22

⎛ 3 π ⎞⎛ π ⎞ Inflection points ⎜ − ,0, ⎟⎜ ,0 ⎟

35. fx ′ () = cos – 2 cos sin x x x = cos (1 – 2sin ); x x

ππππ 5 fx ′ () = 0 when – , , , x =

f ′′ () x = – sin x + 2sin 2 x – 2 cos 2

xf ;() ′′ x = 0

when x ≈ –2.51, –0.63, 1.00, 2.14

Global minimum f ⎜ – ⎞=

local minimum f ⎜⎟ ⎛⎞= 0; ⎝⎠ 2

global maxima f ⎜⎟ = , f =

Inflection points (–2.51, –0.94), (–0.63, –0.94), (1.00, 0.13), (2.14, 0.13)

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=+ ⎛ 1 When x = 4, ⎞ p ⎜ 1 ⎟

+ 16 64 ≈ 11.18 ft.

42. Let x be the width and y the height of a page.

A = xy. Because of the margins, 27

(y – 4)(x – 3) = 27 or y =

27 A x = + 4; x x − 3

40. Let x be the length of a turned up side and let l be

dA ( x − 3)(27) 27 − x

+=− the (fixed) length of the sheet of metal. 81 4 + 4

= 0 when x =− 3 15 ,

dV

= 16 l − 4;0 xl V ′= when 4 x = dx

0 when x =

2 =− 4; l

4 inches should be turned up for

each side.

; y = 10

41. Let p be the length of the plank and let x be the distance from the fence to where the plank

43. π rh 2 = 128 π

touches the ground.

See the figure below.

= h 256

Let S be the surface area of the trough.

2 2 256 S π =π+π=π+ r rh r

0; 3 r = 128, r = 42 3

r 2 dS 2

Since

> 0 when r = 42 3 , r = 42 3

dr 2 minimizes S.

By properties of similar triangles,

x + 64

p =+ ⎜ 1 ⎟ x 2 + 64

Minimize p:

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2 –1; 1 c =± 2 ( c − 1)

2 = 0; 2, x = − which is not in the domain.

3 Only c 1 2 is in the interval (2, 3). Critical points: x = –2, 0, 2

f (–2) = 0, f(0) = 2, f(2) = 0 Minima f(–2) = 0, f(2) = 0, maximum f(0) = 2. ⎧ 1

Concave up on (–2, 0), concave down on (0, 2)

2 = 12 x 2 − 36 x + 24; 12( x 2 − 3 x + 2) = when dx 0 x = 1, 2

Inflection points: x = 1, y = 5 and x = 2, y = 11

Slope at x = 1: dy

dx x = 1

45. a.

fx ′ () = x 2 Tangent line: y – 5 = 7(x – 1); y = 7x – 2

f (3) − f ( 3) − 99 + = dy =

3 Slope at x = 2:

dx x = 2

c 2 = 3; 3, c =− 3 Tangent line: y – 11 = 5(x – 2); y = 5x + 1

b. The Mean Value Theorem does not apply because F′ (0) does not exist.

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49. Let fx () = 3 x − cos 2 x ; a 1 =, 0 b 1 =. 1 cos 2 x

51. x

f () 0 =−; 1 f () 1 ≈ 3.4161468 3

fm () n

50. f(x) = 3x – cos 2x, fx ′ () =+ 3 2sin 2 x

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53. x 3

3 x 2 + 3 x dx

sin x dx ) or

58. Let u cos x ; then du

− du = sin x dx .

3 2 ∫ 1/ 2 ( x − 3 x + 3 x ) dx

4 ∫ 4 cos x sin x dx = ∫ ( cos x ) sin x dx

x x 3 +⋅ 3 x 3/ 2 + C

3 4 ∫ u ⋅− du

= 1 x 4 − x 3 + 2 x 3/ 2 + C

∫ u du

2 x 2 − ∫ 2 ( −+ 3 x

) 5 dx

2 2 = 2 x 3 − 3 x − x 1 + C 59. u = tan(3 x + x du 6 ), (6 = x + 6) sec (3 x + 6) x 3

∫ ( x + ) 1 tan ( 3 x + 6 x )( sec 3 x + 6 2 x dx 1 2 9 3 )

− 3 x −+ C or + C

= 1 ∫ 1 u du 2 = u 3 + C

y 3 − y 9 sin y + y − 1 26

dy = 1 tan 3 3 x 2 + 6 x + C

9sin y + 26 ) dy

56. Let u = y 2 − ; then 4 =

1 du 2 ydy or du = ydy .

4 ∫ u du

∫ 2 yy −

4 dy = ∫ u ⋅ du

=⋅ 2 u + C

u ∫ 1/ 2 du

=+ 5 = 4 1 1 2 3/ 2 61. Let u t 5 ; then du 5 t dt or du = t dt 4 .

t 4 t 5 ∫ 2/3 ( +

5 ) 2/3 dt = ∫ u du

57. Let u = 2 z 2 − ; then 3 du = 4 zdz or du = zdz .

4 = 1 u 2/3 du

2 − 3 ) dz = u 1/ 3 ∫ ⋅ du

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62. Let u x + ; then 4 du 2 x dx or du = xdx .

66. Let u = y 3 − 3 y ; then

) dy . ∫

2 du =

( 3 y − 3 ) dy = 3 ( y

= dx 1 ∫

du

y x 2 + 4 2 u − 1 1 ∫ du

2 dy 3 = 3 ∫ u 2

∫ u du

∫ − u du 2

63. Let u = x 3 + ; then 9 du = 3 x dx 2 or du = x dx 2 .

x 2 1 du

dx = ∫

∫ u du = (2 y + 3 y 2 + 6) y 4/5 + C =⋅ 2 u + C 6 24

x 3 ++ 9 C 68. ∫ dy = ∫ sin x dx

y =− cos x + C

64. Let u =+ y 1 ; then du = dy .

y = –cos x + 3

dy

du

∫ 1 2 ∫ 2 69. dy = dx

= − ∫ u du 2 y = 2 x ++ 1 C

=− u − 1 + C y = 2 x ++ 1 14 =− 1 + C 70. sin y dy = dx

− cos y =+ x C

65. Let u = 2 y − 1 ; then du = 2 dy .

x = –1 – cos y

2 ∫ du

= ∫ u du

22 ( y − 1 )

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73. 3 7. ∫ A 2 y dy = ∫ (6 x − x dx ) region = ( 0.5 1 1.5 2 2.5 + ++ ) = 3.5

y 2 = 3 x 2 − 1 x 4 + C 8. A region =

4 9. A region = A rect +

A tri = 1 x + xx ⋅= x + x

10. 1 1 1 A 2 region = bh = x xt ⋅= xt

74. ∫ cos y dy = ∫ x dx

2 11. y =− 5 xA ; region = A rect + A tri x sin y =

y = sin − 1 ⎜ ⎜

12. A region = A rect + A tri

75. st () =− 16 t 2 + 48 t + 448; s = 0 at t = 7;

vt () = st ' () =− 32 t + 48

when t = 7, v = –32(7) + 48 = –176 ft/s

Review and Preview Problems

1. A region = bh = aa o

sin 60 2 = a

2. A region = 6 ⎜ base height × ⎟ = 6 ⎜ a ⎟⎜ ⎜

= 33 a 2 2

3. A region = 10 ⎜ base height × ⎟ = 5 cot 36

a 2 cot 36

4. A region = A rect + A tri = () 17 8.5 + 17

5. A region = A rect + A semic. = ⋅ 3.6 5.8 + π () 1.8

6. A region = A #5 + 2 A tri =

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33. a.

b. − ∫ 1/ 2 10 V dV = ∫ 1 C dt ; 20 V = Ct 1 + C 2 ;

V (0) = 1600: C 2 = ⋅ 20 40 = 800;

V (40) = 0: C 1 =−

( 20 − t + 800) = ( 40 − t )

2 c. (10) = − 40 10 = 900 cm 3

b. Since the trip that involves 1 min more travel

time at speed v m is 0.6 mi longer, v = 0.6 mi/min

dP

1 P , P(0) = 1000, P(10) = 1700 = 36 mi/h.

36. a.

dt

where t is the number of years since 1980.

c. From part b, v m = 0.6 mi/min. Note that the

b.

average speed during acceleration and

P − 1/ 3

dP = ∫ 1 C dt ; P 2/3 = Ct 1 + C 2

deceleration is m = 0.3 mi/min. Let t be the

= 150 time spent between stop C and stop D at the

2 P (0) = 1000: C =⋅

− 150 0.6t + 0.3(4 – t)= 2 miles. Therefore,

constant speed v m , so

P (10) = 1700: C = 2

2 t = 2 min and the time spent accelerating

P = (4.2440 t + 100) 3/2

42 − 2 is

min.

2 3 c. 4000 = (4.2440 t + 100) 3/2

− 0.6 0 a =

0.9 mi/min . 2 4000 2/3 − 2 100

t ≈ 36 years, so the population will reach

dh

4000 by 2016.

34. For the balloon,

= 4 , so ht () = 4 t + C 1 . Set

dt

t 2 = 0 at the time when Victoria threw the ball, and 37. Initially, v = –32t and s =− 16 t + 16 . s = 0 when height 0 at the ground, then h(t) = 4t + 64. The

t = 1. Later, the ball falls 9 ft in a time given by

0 =− 16 t 2 + , or 9 s, and on impact has a since s 0 = 0 . The maximum height of the ball is

height of the ball is given by st () =− 16 t 2 + vt 0 ,

v velocity of − 32 =− 24 ft/s. By symmetry,

when t = 0 , since then st ′ () = 0 . At this time

2 24 ft/s must be the velocity right after the first

h (t) = s(t) or 4 ⎜ ⎟ + 64 =− 16 0 + v

bounce. So

Solve this for v 0 to get v 0 ≈ 68.125 feet per

a. vt () =⎨

second. ⎩ − 32( t −+ 1) 24 for 1 <≤ t 2.5

dV

b. 9 =− 16 t 2 + 16 ⇒≈ t

0.66 sec; s also equals 9

35. a.

= C 1 h where h is the depth of the

dt at the apex of the first rebound at t = 1.75 sec.

water. Here, V =π rh 2 = 100 h , so h =

= C 1 , V(0) = 1600,

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The Definite Integral