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

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

\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}

Solving for critical numbers

\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}

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