# 4 272 y a 26 v a 2 b x2 a2 x2 dx a bx a 4b a a2

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Unformatted text preview: t; x < +∞ (f ) ln x + x, x > 0 ex (h) ,x>0 x x2 = lim x→+∞ 1+ y = 2x, lim (1 + y )2/y = lim (1 + y )1/y (a) y = 3x, lim lim (1 + x)1/3x = lim (1 + x)1/x y →+∞ x→0 1 y y /3 = lim (a) 16. (a) 17. F (x) = 1/3 (a) 18. = e2 1 y y 1/3 = lim y →+∞ (b) eln x = e1/3 − 1 x2 sin 1 x cos x −(x2 + 3) sin x − 2x cos x , F (x) = 2+3 x (x2 + 3)2 0 (b) 1/3 √ 3x 3x2 + 1, F (x) = √ 3x2 + 1 F (x) = 0 (b) (a) d dx x2 1 y 1/3 1 y 1 =1 x (b) √ 2x x2 + 1 (b) 1+ 14. g (x) = 1 − cos x 3 1 (3x2 ) = 3 x x (a) 19. 2 ln 2 = e1/3 g (x) = x2 − x 15. 1+ y →+∞ x→0 13. 2 y →0 1+ √ =e = e2 (b) 12. 2 x2 1 x (b) y →0 √ x2 1 √ t 1 + tdt = x2 (c) (c) √ 13 1 + x2 (2x) = 2x3 1 + x2 √ 22 22 42 3/2 5/2 t 1 + tdt = − (x + 1) + (x + 1) − 3 5 15 √ 0 √ 6/ 13 247 20. Chapter 7 (a) (b) a d dx f (t)dt = − x a d dx d dx f (t)dt = − g (x) x f (t)dt = −f (x) a g (x) d dx f (t)dt = −f (g (x))g (x) a tan2 x sec2 x = − tan2 x 1 + tan2 x 21. (a) − sin x2 (b) − 22. (a) −(x2 + 1)40 (b) − cos3 23. −3 1 x − 1 x2 = cos3 (1/x) x2 x2 − 1 3x − 1 + 2x 4 9x2 + 1 x +1 24. If f is continuous on an open interval I and g (x), h(x), and a are in I then g (x) a g (x) f (t)dt = f (t)dt + h(x) so h(x) g (x) d dx h(x) f (t)dt = − a g (x) f (t)dt + a f (t)dt a f (t)dt = −f (h(x))h (x) + f (g (x))g (x) h(x) 25. sin2 (x3 )(3x2 ) − sin2 (x2 )(2x) = 3x2 sin2 (x3 ) − 2x sin2 (x2 ) 1 2 1 (1) − (−1) = (b) 1+x 1−x 1 − x2 26. F (x) = 27. from geometry, (a) 1 1 (3) − (1) = 0 so F (x) is constant on (0, +∞). F (1) = ln 3 so F (x) = ln 3 for all x > 0. 3x x 3 5 f (t)dt = 0, 0 (a) 7 f (t)dt = 6, 3 10 f (t)dt = 0; and 5 10 f (t)dt = 7 (4t − 37)/3dt = −3 7 F (0) = 0, F (3) = 0, F (5) = 6, F (7) = 6, F (10) = 3 (b) F is increasing where F = f is positive, so on [3/2, 6] and [37/4, 10], decreasing on [0, 3/2] and [6, 37/4] (c) (d) critical points when F (x) = f (x) = 0, so x = 3/2, 6, 37/4; maximum 15/2 at x = 6, minimum −9/4 at x = 3/2 F(x) 6 4 2 2 4 6 8 10 x -2 28. fave = 1 10 − 0 10 f (t)dt = 0 1 F (10) = 0.3 10 x 29. x < 0 : F (x) = x ≥ 0 : F (x) = 1 (−t)dt = − t2 2 −1 0 −1 x = −1 x (−t)dt + t dt = 0 1 (1 − x2 ), 2 1 12 + x ; F (x) = 22 (1 − x2 )/2, x < 0 (1 + x2 )/2, x ≥ 0 Exercise Set 7.9 30. 248 x 0 ≤ x ≤ 2 : F (x) = t dt = 0 2 x > 2 : F (x) = x t dt + 0 x 31. y (x) = 2 + 1 x 32. y (x) = 3 t1/3 dt = 2 + t4/3 4 (t1/2 + t−1/2 )dt = x x = 2x − 2, x > 2 1 5 3 4/3 +x 44 2 3/2 2 x − + 2x1/2 − 2 3 3 (sec2 t − sin t)dt = tan x + cos x − y (x) = 1 + x2 /2, 0 ≤ x ≤ 2 2 dt = 2 + 2(x − 2) = 2x − 2; F (x) = 2 1 33. 12 x, 2 √ 2/2 π/4 x 34. 2 tet dt = y (x) = 0 x 1 −x2 1 e − 2 2 35. P (x) = P0 + r(t)dt individuals 0 T 36. s(T ) = s1 + v (t)dt 1 37. II has a minimum at x = 1, 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 38. (b) 39. 1 o lim (xk − 1) = ln x lim xk = ln x by L’Hˆpital’s rule (with respect to k ) k →0 k k →0 (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 (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) (f ) F(x) 1 0.5 1 2 3 4 2 4 5 x -0.5 -1 40. (a) erf (x) 1 -4 -2 x -1 (c) (e) (g) erf (x) > 0 for all x, so there are no relative extrema 2√ erf (x) = −4xe−x / π changes sign only at x = 0 so that is the only point of inﬂection lim erf(x) = +1, lim erf(x) = −1 x→+∞ x→−∞ 249 41. Chapter 7 C (x) = cos(πx2 /2), C (x) = −πx sin(πx2 /2) cos t goes from negative to positive at 2kπ − π/2, and from positive to negative at t = 2kπ + π/2, √ so C (x) has relative minima when πx2 /2 = 2kπ − π/2, x = ± 4k − 1, k = 1, 2, . . ., and C (x) √ has relative maxima when πx2 /2 = (4k + 1)π/2, x = ± 4k + 1, k = 0, 1, . . .. √ (b) sin t changes sign at t = kπ , so C (x) has inﬂection points at πx2 /2 = kπ , x = ± 2k , k = 1, 2, . . .; the case k = 0 is distinct due to the factor of x in C (x), but x changes sign at x = 0 and sin(πx2 /2) does not, so there is also a point of inﬂection at x = 0 (a) F (x + h) − F (x) 1 = lim h→0 h→0 h h x 42. Let F (x) = 1 1 h→0 h ln tdt = ln x x x Diﬀerentiate: f (x) = 3e3x , so 2 + (a) (c) = 2 + e3x − e3a = e3x provided 3e3t dt = 2 + e3t a a The area under 1/t for x ≤ t ≤ x + 1 is less than the area of the rectangle with altitude 1/x and base 1, but greater than the area of the rectangle with altitude 1/(x + 1) and base 1. x+1 (b) x x f (t)dt = 2 + a e3a = 2, a = (ln 2)/3. 44. ln tdt; but F (x) = ln x so x x+h lim 43. x+h ln tdt, F (x) = lim x+1 1 dt = ln t = ln(x + 1) − ln x = ln(1 + 1/x), so t x x 1/(x + 1) < ln(1 + 1/x) < 1/x for...
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