C x cosx 2 c x x sinx 2 1 48 a the area

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Unformatted text preview: 9 ln (d)(d) t ln t= 9= 9ln 2ln 2 ln −− =− 1 1 2 2 1 2 2 5. 5. 5lny 1≈ 2x603210678;yln/y = 1lim (1 + magnitudeeof error is < 0< 0.0063 ln . 1. , lim (1 ln 2 = .609437912;1/y = 2 . (b) ≈5 =603210678; + 5)=51.609437912; y ) magnitude of error is .0063 y →0 y →0 6. 6. 3ln 31≈ 1.098242635;3ln 31= 1.098612289; magnitude error is < 0< 0.0004 ln ≈ .098242635; ln = .098612289; magnitude of of error is .0004 13. g (x) = x2 − x. 7. 7. (a) −1 , −1 , x0> 0 (a) x −1 x > (b)(b) 2 , x2 , x0￿= 0 x 2 x ￿= x 122 1 3 ln x 2 15. (a)(c)3x −x2 , = < .x < +∞ ∞ e ) −∞ < x <(b) = 1. (c) − (3,x−∞ (d)(d) x, −∞ < x < +∞ ∞ − −x, −∞ < x < + + x x x 33 (e)(e) 3, x , x0> 0 x x> (f ) (f ln xln xx+x , x0> 0 ) + , x> xx sin √ x (x2 + 1) cos x − 2x sin x x 3 17. F (x)x − √x,3 −∞ <x) < +∞ , . (h) e ,ex > 0 3 (g)(g)= x 2 + 1x,F ( x = < +∞ (x2 + 1)2 − −∞ < x (h)x ,x>0 x − ln 2 ln(1/9) −− /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 − ln a. 21. False; the integral does not exist. 23. (a) d dx x2 (b) 1 x2 √ t 1 + tdt = x2 1 + x2 (2x) = 2x3 1 + x2 . 1 √ 22 22 42 3/2 5/2 t 1 + tdt = − (x + 1) + (x + 1) − . 3 5 15 √ 25. (a) − cos x3 (b) − tan2 x sec2 x = − tan2 x. 1 + tan2 x Chapter 5 so d dx ￿ g (x) f (t)dt = −f (h(x))h￿ (x) + f (g (x))g ￿ (x) h(x) Exercise Set 5.103 )(3x2 ) − sin2 (x2 )(2x) = 3x2 sin2 (x3 ) − 2x sin2 (x2 ) 29. (a) sin2 (x 153 1 1 2 (b)3x − 1 (1) − 2 − 1 (−1) = 1 + x+ 2x x1 − x. 1 − x2 27. −3 2 9x + 1 x4 + 1 1 1 ￿ 30. (a)(xsin2 (x3 )(3x2 ) − sin2 (xso Fx) ) is constant 3 ) − 2,x sin2 (x2 ). = ln 5 so F (x) = ln 5 for all x > 0. 29. F ) = 5x (5) − x (1) = 0 2 )(2 (x = 3x2 sin2 (x on (0 +∞). F (1) 1 1 2 ￿ ￿ (b) (1) − ￿ 3 (−1) = ￿ 5 2 (for −1 < x7< 1). 1+x 1 −fx t)dt = 0, − xf (t)dt = 6, 1 31. from geometry, ( f (t)dt = 0; and 0 3 3 31. From geometry, 0 5 7 ￿ 10 5 f (t dt / dt f (t)dt = 0, (4f −)37)=36, = −3(t)dt = 0; and = t 7 5 3 10 f (t)dt 7 10 10 f (t)dt = 7 7 (4t − 37)/3dt = −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 ￿ = (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 ￿ (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 −9/4 at (c) critical x = 3/2. /4 at x = 3/2F (0) = 0 and F (10) = 3.) −9 (Endpoints: F(x) (d) 6 4 2 x 2 4 6 8 10 –2 (d) x x 1 32. F ￿ is increasing (resp. decreasing) where1 f is increasing (resp. decreasing), namely on (0, 3) and 33. x < 0 : F (x) = (−t)dt = − t2 = (1 − x2 ), (7, 10) (resp. (5, −1 The only 2 −1 common to two of these intervals is x = 7, and that is the 7)). endpoint 2 only point of inflection of F . x 0 (1 − x2 )/2, x < 0 11 ￿ = + x2 ; F (x) = x ≥ 0 : F (x) = ￿ x(−t)dt + t dt x 2 12 (1 + x2 )/2, x ≥ 0 01 33. x < 0 : F (x) = −1 (−t)dt = − t2 = (1 − x2 ), 2 −1 2 −1 x x ￿ ￿ ￿x2 2t20+ 1 2 35. y (x) = 2 + dt = 2 + (t + ln 1) 1 2 2 + ln x + 1.(1 − x )/2, x < 0 t =x x ≥ 0 : F (x)1= t (−t)dt + t dt = + x ; F (x) = 2 12 (1 + x2 )/2, x ≥ 0 −1 0 x √ 37. y (x) = 1 + (sec2 t − sin t)dt = tan x + cos x − 2/2. π/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)π /2, x = ±√4k + 1, k = 1 ( ) ., and C (x has Derivative Test, at t π 12 / k = 0 , or (c) at t = ,01,1. . .. 5; from the graph it is evident that it is at t = 5 √ (b) sin = 0, 3 or sign at t graph so evident inflection t = 3 (d) at t t changes5; from the = k π , it isC (x) hasthat it is atpoints at π x2 /2 = k π , x = ± 2k , 154 (e) F is concave; up when F ￿￿ = f ￿ is positive, i.e. where f is of x in C ￿￿ (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(π...
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