Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.9 - Antiderivatives - 4.9 Exercises: 40

Answer

$f(x) = \frac{2}{5}x^5 + \frac{5}{2} x^2 + x -3.9$

Work Step by Step

$f'(x) = 2x^4 + 5x + C$ $f'(1) = 2(1)^4 + 5(1) + C$ Given $f'(1) = 8$ $8 = 2 + 5 + C$ $C = 1$ $f'(x) = 2x^4 + 5x + 1$ Find $f$: $f(x) = 2 \times \frac{x^{4+1}}{4+1} + 5 \times \frac{x^{1+1}}{1+1} + 1(x) $ $f(x) = 2 \times \frac{x^5}{5} + 5 \times \frac{x^2}{2} + x $ $f(x) = \frac{2x^5}{5} + \frac{5x^2}{2} + x + C$ $f(1) = \frac{2(1)^5}{5} + \frac{5(1)^2}{2} + 1 + C$ $f(1) = \frac{2}{5} + \frac{5}{2} + 1 + C$ Given $f(1) = 0$: $0 = \frac{39}{10} + C$ $C = -3.9$ $f(x) = \frac{2}{5}x^5 + \frac{5}{2} x^2 + x -3.9$
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