MATH10560 Midterm: Math10560Practice Exam1 Solutions

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31 Jan 2019
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Solutions to practice exam 1, math 10560: simplify the following expression for x x = log3 81 + log3. = log3 9 = log3 32 = 2 log3 3 = 2 : the function f (x) = x3 + 3x + e2x is one-to-one. By trial-and-error we determine that f 1(1) = 0. f (x) = 3x2 + 3 + 2e2x. Hence f (f 1(1)) = f (0) = 5. 5 : di erentiate the function f (x) = (x2. Solution: use logarithmic di erentiation. (take logarithm of both sides of equation, then do implicit di erentiation. ) ln f = 4 ln(x2. 1)4 f f x(x2 f (x) : compute the integral. 2 ln(x2 + 1) x x2 + 1. Solution: make the substitution u = ln x. At x = 2e, have u = 1 and at x = 2e2 have u = 2.

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