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: 31

Answer

$x=-2$ and $x=3$ are the critical numbers.

Work Step by Step

Find the critical numbers of $f(x) = 2x^3-3x^2-36x$ Differentiate and set the derivative $=0$ $f'(x) = 6x^2-6x-36$ $0=6(x^2-x-6)$ $0=6(x+2)(x-3)$ $x=-2$ and $x=3$ are the critical numbers.
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