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: 25

Answer

$f(x) = x^{5} - x^{4} + x^{3}+Cx+D$

Work Step by Step

$f''(x) = 20x^{3} -12x^{2} +6x$ $f'(x) = 20(\frac{x^{4}}{4}) - 12(\frac{x^{3}}{3}) + 6(\frac{x^{2}}{2})$ $f'(x) = (\frac{20}{4})x^{4} - (\frac{12}{3})x^{3} + (\frac{6}{2})x^{2}$ $f'(x) = 5x^{4} - 4x^{3} + 3x^{2} + C$ $f'(x) = 5x^{4} - 4x^{3} + 3x^{2} + C$ $f(x) = 5(\frac{x^{5}}{5}) - 4(\frac{x^{4}}{4}) + 3(\frac{x^{3}}{3})+Cx$ $f(x) = x^{5} - x^{4} + x^{3}+Cx+D$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.