# Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals

Language: English

Pages: 531

ISBN: B00722S6M8

Format: PDF / Kindle (mobi) / ePub

Student Solutions Manual for Stewart's Single Variable Calculus: Early Transcendentals [Paperback] [Jan 01, 2008] Daniel Anderson, Jeffery Cole, Daniel Drucker

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f (x) = (2x − 1)(x + 1) 2x2 + x − 1 = , so lim f(x) = ∞, x2 + x − 2 (x + 2)(x − 1) x→−2− lim f (x) = −∞, lim f (x) = −∞, and lim f (x) = ∞. Thus, x = −2 x→1− x→−2+ x→1+ and x = 1 are vertical asymptotes. The graph confirms our work. 43. y = f (x) = x2 x(x2 − 1) x(x + 1)(x − 1) x(x + 1) x3 − x = = = = g(x) for x 6= 1. − 6x + 5 (x − 1)(x − 5) (x − 1)(x − 5) x−5 The graph of g is the same as the graph of f with the exception of a hole in the graph of f at x = 1. By long division, g(x) = x2

given, then −1/x < ε Take N = −1/ε. Then x < N ⇔ x < −1/ε. ⇒ x < −1/ε ⇒ |(1/x) − 0| = −1/x < ε, so lim (1/x) = 0. x→−∞ 69. Given M > 0, we need N > 0 such that x > N ⇒ ex > M . Now ex > M ⇔ x > ln M , so take N = max(1, ln M ). (This ensures that N > 0.) Then x > N = max(1, ln M) ⇒ ex > max(e, M) ≥ M , so lim ex = ∞. x→∞ 71. Suppose that lim f (x) = L. Then for every ε > 0 there is a corresponding positive number N such that |f (x) − L| < ε x→∞ whenever x > N . If t = 1/x, then x > N ⇔

+ (ln 2 + ln u) − ln u 1 + ln 2 = = [1 + ln(2u)]2 u[1 + ln(2u)]2 u[1 + ln(2u)]2 1 −10x − 1 10x + 1 · (−1 − 10x) = or 2 − x − 5x2 2 − x − 5x2 5x2 + x − 2 19. y = ln(e−x + xe−x ) = ln(e−x (1 + x)) = ln(e−x ) + ln(1 + x) = −x + ln(1 + x) y 0 = −1 + −1 − x + 1 x 1 = =− 1+x 1+x 1+x ⇒ SECTION 3.6 21. y = 2x log10 Note: √ x = 2x log10 x1/2 = 2x · 1 2 DERIVATIVES OF LOGARITHMIC FUNCTIONS log10 x = x log10 x ⇒ y 0 = x · 1 1 + log10 x · 1 = + log10 x x ln 10 ln 10 ln e 1 1 = = log10 e, so

= 200 7 ≈ 28.57 and t2 = 600 7 ≈ 85.71. At t1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at t2 minutes, the rate of decrease is the greatest. 67. f (x) = ax3 + bx2 + cx + d ⇒ f 0 (x) = 3ax2 + 2bx + c. We are given that f(1) = 0 and f (−2) = 3, so f (1) = a + b + c + d = 0 and f (−2) = −8a + 4b − 2c + d = 3. Also f 0 (1) = 3a + 2b + c = 0 and f 0 (−2) = 12a − 4b + c = 0 by Fermat’s Theorem. Solving these four equations, we get a = 29 ,

− 3x ⇒ y = 10 − 3x (y ≥ 0) ⇒ y 2 = 10 − 3x ⇒ 3x = 10 − y 2 Interchange x and y: y = − 13 x2 + 23. y = f (x) = ex 3 ⇒ ln y = x3 25. y = f (x) = ln (x + 3) 10 3 . So f −1 (x) = − 13 x2 + ⇒ x= ⇒ x + 3 = ey 10 3 . 10 . 3 Note that the domain of f −1 is x ≥ 0. √ √ √ 3 ln y. Interchange x and y: y = 3 ln x. So f −1 (x) = 3 ln x. ⇒ x = ey − 3. Interchange x and y: y = ex − 3. So f −1 (x) = ex − 3. √ ⇒ y − 1 = x4 ⇒ x = 4 y − 1 (not ± since √ √ x ≥ 0). Interchange x and y: y = 4 x − 1. So f