Single Variable Calculus, Chapter 4, 4.1, Section 4.1, Problem 38

Determine the critical numbers of the function $g(x) = \sqrt{1 - x^2}$


$
\begin{equation}
\begin{aligned}

g'(x) =& \frac{d}{dx} (\sqrt{1 - x^2})
\\
\\
g'(x) =& \frac{d}{dx} (1 - x^2)^{\frac{1}{2}}
\\
\\
g'(x) =& \frac{1}{2} (1 - x^2)^{- \frac{1}{2}} \frac{d}{dx} (1 - x^2)
\\
\\
g'(x) =& \frac{1}{\cancel{2}} (1 - x^2)^{- \frac{1}{2}} (\cancel{-2}x)
\\
\\
g'(x) =& -x (1 - x^2)^{- \frac{1}{2}}
\\
\\
g'(x) =& \frac{-x}{(1 - x^2)^{\frac{1}{2}}}

\end{aligned}
\end{equation}
$


Solving for critical numbers


$
\begin{equation}
\begin{aligned}

& g'(x) = 0
\\
\\
& 0 = \frac{-x}{(1 - x^2)^{\frac{1}{2}}}
\\
\\
& (1 - x^2)^{\frac{1}{2}} \left[ 0 = \frac{-x}{\cancel{(1 - x^2)^{\frac{1}{2}}}} \right] \cancel{(1 - x^2)^{\frac{1}{2}}}
\\
\\
& 0 = -x
\\
\\
& x = 0
\\
\\
& (1 - x^2)^{\frac{1}{2}} = 0
\\
\\
& \sqrt{(1 - x^2)^{\frac{1}{2}}} = \sqrt{0}
\\
\\
& 1 - x^2 = 0
\\
\\
& x^2 = 1
\\
\\
& \sqrt{x^2} = \pm \sqrt{1}
\\
\\
& x = \pm 1

\end{aligned}
\end{equation}
$


Therefore, the critical numbers are $x = 0, x = 1$ and $x = -1$.