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: 42

Answer

The critical numbers are $x=0, \pm 1$

Work Step by Step

Find the critical numbers of $g(x) = \sqrt {1-x^2}$ Differentiate and set the derivative $=0$ $g'(x) = \frac{-2x}{2\sqrt{1-x^2}}$ $0=\frac{-x}{\sqrt{1-x^2}}$ When $x=0$, $g'(x) = 0$. Therefore $x=0$ is a critical number. There are also critical numbers when $g'(x)$ does not exist. This occurs when $x=\pm 1$ Therefore the critical numbers are $x=0, \pm 1$
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