Unformatted text preview: ˆ’ 9
ln
(d)(d) t ln t= 9= 9ln 2ln 2
ln
âˆ’âˆ’
=âˆ’
1
1 2
2 1 2 2 5. 5. 5lny 1â‰ˆ 2x603210678;yln/y = 1lim (1 + magnitudeeof error is < 0< 0.0063
ln
. 1. , lim (1 ln 2 = .609437912;1/y = 2 .
(b) â‰ˆ5 =603210678; + 5)=51.609437912; y ) magnitude of error is .0063
y â†’0 y â†’0 6. 6. 3ln 31â‰ˆ 1.098242635;3ln 31= 1.098612289; magnitude error is < 0< 0.0004
ln â‰ˆ .098242635; ln = .098612289; magnitude of of error is .0004
13. g (x) = x2 âˆ’ x.
7. 7. (a) âˆ’1 , âˆ’1 , x0> 0
(a) x âˆ’1 x >
(b)(b) 2 , x2 , x0ï¿¿= 0
x 2 x ï¿¿=
x
122
1
3
ln x
2
15. (a)(c)3x âˆ’x2 , = < .x < +âˆž âˆž e
) âˆ’âˆž < x <(b)
= 1.
(c) âˆ’ (3,xâˆ’âˆž
(d)(d) x, âˆ’âˆž < x < +âˆž âˆž
âˆ’ âˆ’x, âˆ’âˆž < x < +
+
x
x
x
33
(e)(e) 3, x , x0> 0
x x>
(f ) (f ln xln xx+x , x0> 0
) + , x>
xx
sin âˆš
x
(x2 + 1) cos x âˆ’ 2x sin x
x
3
17. F (x)x âˆ’ âˆšx,3 âˆ’âˆž <x) < +âˆž
,
. (h) e ,ex > 0
3
(g)(g)= x 2 + 1x,F ( x = < +âˆž (x2 + 1)2
âˆ’
âˆ’âˆž < x
(h)x
,x>0
x
âˆ’ ln 2
ln(1/9)
âˆ’âˆ’
/9)
8. 8. (a) (ln 3) =(b)2eln 3 ln 3e= e(c)/9)1=91/9
(a) f
(a) 0 f (ln 3)e=2 0 3 = ln(1ln(1 =1 / 19. True; both integrals are equal to âˆ’ ln a.
21. False; the integral does not exist.
23. (a) d
dx
x2 (b)
1 x2 âˆš
t 1 + tdt = x2 1 + x2 (2x) = 2x3 1 + x2 . 1 âˆš
22
22
42
3/2
5/2
t 1 + tdt = âˆ’ (x + 1) + (x + 1) âˆ’
.
3
5
15
âˆš 25. (a) âˆ’ cos x3 (b) âˆ’ tan2 x
sec2 x = âˆ’ tan2 x.
1 + tan2 x Chapter 5 so d
dx ï¿¿ g (x) f (t)dt = âˆ’f (h(x))hï¿¿ (x) + f (g (x))g ï¿¿ (x) h(x) Exercise Set 5.103 )(3x2 ) âˆ’ sin2 (x2 )(2x) = 3x2 sin2 (x3 ) âˆ’ 2x sin2 (x2 )
29. (a) sin2 (x 153 1
1
2
(b)3x âˆ’ 1 (1) âˆ’ 2 âˆ’ 1 (âˆ’1) =
1 + x+ 2x x1 âˆ’ x.
1 âˆ’ x2
27. âˆ’3 2
9x + 1
x4 + 1
1
1
ï¿¿
30. (a)(xsin2 (x3 )(3x2 ) âˆ’ sin2 (xso Fx) ) is constant 3 ) âˆ’ 2,x sin2 (x2 ). = ln 5 so F (x) = ln 5 for all x > 0.
29. F ) = 5x (5) âˆ’ x (1) = 0 2 )(2 (x = 3x2 sin2 (x on (0 +âˆž). F (1)
1
1
2
ï¿¿
ï¿¿
(b)
(1) âˆ’ ï¿¿ 3 (âˆ’1) = ï¿¿ 5 2 (for âˆ’1 < x7< 1).
1+x
1 âˆ’fx t)dt = 0, âˆ’ xf (t)dt = 6,
1
31. from geometry,
(
f (t)dt = 0; and
0 3 3 31. From geometry,
0 5 7
ï¿¿ 10 5
f
(t dt / dt
f (t)dt = 0, (4f âˆ’)37)=36, = âˆ’3(t)dt = 0; and
=
t
7 5 3 10 f (t)dt 7
10 10 f (t)dt =
7 7 (4t âˆ’ 37)/3dt = âˆ’3. (a) FF (0) = 0,FF (3) = 0,FF (5) = 6,FF (7) = 6,FF (10) = 3
(a) (0) = 0, (3) = 0, (5) = 6, (7) = 6, (10) = 3.
(b) F is increasing where ï¿¿ =
(b) F is increasing where FF = f fisis positive, so on [3/2, 6] and [37/4, 10], decreasing on ,[0, 3/2] [6, 37/4].
positive, so on [3/2, 6] and [37/4, 10], decreasing on [0 3/2] and
and [6, 37/4]
(c) Critical points when F ï¿¿ (xx)= f (x) == 0, so= 3=2, 6,237/4; maximum 15/215/2 = 6, minimum
when F ( ) = f (x) 0, so x x / 3/ , 6, 37/4; maximum at x at x = 6, minimum âˆ’9/4 at
(c) critical
x = 3/2. /4 at x = 3/2F (0) = 0 and F (10) = 3.)
âˆ’9 (Endpoints:
F(x) (d)
6
4
2 x
2 4 6 8 10 â€“2 (d)
x x
1
32. F ï¿¿ is increasing (resp. decreasing) where1 f is increasing (resp. decreasing), namely on (0, 3) and
33. x < 0 : F (x) =
(âˆ’t)dt = âˆ’ t2
= (1 âˆ’ x2 ),
(7, 10) (resp. (5, âˆ’1 The only 2 âˆ’1 common to two of these intervals is x = 7, and that is the
7)).
endpoint 2
only point of inï¬‚ection of F . x
0
(1 âˆ’ x2 )/2, x < 0
11
ï¿¿ = + x2 ; F (x) =
x â‰¥ 0 : F (x) = ï¿¿ x(âˆ’t)dt +
t dt x
2 12
(1 + x2 )/2, x â‰¥ 0
01
33. x < 0 : F (x) = âˆ’1 (âˆ’t)dt = âˆ’ t2
= (1 âˆ’ x2 ),
2 âˆ’1
2
âˆ’1
x
x
ï¿¿
ï¿¿
ï¿¿x2
2t20+ 1
2
35. y (x) = 2 +
dt = 2 + (t + ln 1) 1 2 2 + ln x + 1.(1 âˆ’ x )/2, x < 0
t
=x
x â‰¥ 0 : F (x)1= t (âˆ’t)dt +
t dt = + x ; F (x) =
2 12
(1 + x2 )/2, x â‰¥ 0
âˆ’1
0
x
âˆš
37. y (x) = 1 +
(sec2 t âˆ’ sin t)dt = tan x + cos x âˆ’ 2/2.
Ï€/4 x 39. P (x) = P0 + r(t)dt individuals.
0 41. II has a minimum at x = 12, and I has a zero there, so I could be the derivative of II; on the other hand I has a
minimum near x = 1/3, but II is not zero there, so II could not be the derivative of I, so I is the graph of f (x)
x
and II is the graph of 0 f (t) dt.
43. (a) Where f (t) = 0; by the First Derivative Test, at t = 3.
(b) Where f (t) = 0; by the First Derivative Test, at t = 1, 5.
(c) At t = 0, 1 or 5; from the graph it is evident that it is at t = 5.
(d) At t = 0, 3 or 5; from the graph it is evident that it is at t = 3.
(e) F is concave up when F = f is positive, i.e. where f is increasing, so on (0, 1/2) and (2, 4); it is concave
down on (1/2, 2) and (4, 5). (b) where ,f2,t. .= 0; by the) Firstrelative maxima when= x, 52 = (4k + 1)Ï€ /2, x = Â±âˆš4k + 1,
k = 1 ( ) ., and C (x has Derivative Test, at t Ï€ 12 /
k = 0 , or
(c) at t = ,01,1. . .. 5; from the graph it is evident that it is at t = 5
âˆš
(b) sin = 0, 3 or sign at t graph so evident inï¬‚ection t = 3
(d) at t t changes5; from the = k Ï€ , it isC (x) hasthat it is atpoints at Ï€ x2 /2 = k Ï€ , x = Â± 2k ,
154 (e) F is concave; up when F ï¿¿ï¿¿ = f ï¿¿ is positive, i.e. where f is of x in C ï¿¿ï¿¿ (x),on (0x1changes sign at
k = 1, 2, . . . the case k = 0 is distinct due to the factor increasing, so but , /2) and (2, 4);
x = 0 and sin(Ï€...
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 Spring '14

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