Answer
$ x=0$ and $ x= 4$
Work Step by Step
Given $$ f(x) =6 x^2-x^3$$
Then to find the critical points
\begin{align*}
f'(x)&=0\\
12x-3x^2&=0
\end{align*}
Hence the critical points are $ x=0$ and $ x= 4$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 4: Applications of Derivatives - Section 4.1 - Extreme Values of Functions - Exercises 4.1 - Page 192: 42
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