Solução_Calculo_Stewart_6e

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F. TX.10 SECTION 4.9 ANTIDERIVATIVES ¤ 211 35. f 0 (x) =x −1/3 has domain (−∞, 0) ∪ (0, ∞) ⇒ f(x) = 3 2 x2/3 + C 1 if x>0 3 2 x2/3 + C 2 if x0 2 Thus, f(x) = 3 2 x2/3 − 5 if x0 ⇒ f 0 (x) =−1/x + C ⇒ f(x) =− ln |x| + Cx + D = − ln x + Cx + D [since x>0]. f(1) = 0 ⇒ C + D =0and f(2) = 0 ⇒ −ln 2 + 2C + D =0 ⇒ −ln 2 + 2C − C =0 [since D = −C] ⇒ − ln 2 + C =0 ⇒ C =ln2and D = − ln 2. Sof(x) =− ln x +(ln2)x − ln 2. 47. Given f 0 (x) =2x +1,wehavef(x) =x 2 + x + C. Sincef passes through (1, 6), f(1) = 6 ⇒ 1 2 +1+C =6 ⇒ C =4. Therefore, f(x) =x 2 + x +4and f(2) = 2 2 +2+4=10. 49. b is the antiderivative of f. Forsmallx, f is negative, so the graph of its antiderivative must be decreasing. But both a and c are increasing for small x,soonlyb can be f’s antiderivative. Also, f is positive where b is increasing, which supports our conclusion. 51. The graph of F must start at (0, 1). Where the given graph, y = f(x),hasa local minimum or maximum, the graph of F will have an inflection point. Where f is negative (positive), F is decreasing (increasing). Where f changes from negative to positive, F will have a minimum. Where f changes from positive to negative, F will have a maximum. Where f is decreasing (increasing), F is concave downward (upward).
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Solução_Calculo_Stewart_6e

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