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 $x=0, \pm 1$