## Calculus 8th Edition

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