Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 3 - Applications of Differentiation - 3.1 Maximum and Minimum Values - 3.1 Execises: 29

Answer

$x=\frac{1}{3}$

Work Step by Step

Find the critical numbers of $f(x) = 4+ \frac{1}{3}x - \frac{1}{2}x^2$ Differentiate and set the derivative $= 0$ to find the critical numbers. $f'(x) = \frac{1}{3} - x$ $0=\frac{1}{3} - x$ $x=\frac{1}{3}$
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