Answer
$0, ±\sqrt {\frac{2}{3}}, ±1$
Work Step by Step
let $f(x)$ = $x^{2}\sqrt{1-x^{2}}$
$f'(x)$ = $-\frac{x^{3}}{\sqrt{1-x^{2}}}+2x\sqrt{1-x^{2}}$
$f'(x)$ = $\frac{2x-3x^{3}}{\sqrt{1-x^{2}}}$
This derivative is 0
$0$ = $\frac{2x-3x^{3}}{\sqrt{1-x^{2}}}$
$0$ = $2x-3x^{3}$
$0$ = $x(2-3x^{2)}$
$x$ = $0, ±\sqrt {\frac{2}{3}}$
the derivative does not exist when x = ±1
$0, ±\sqrt {\frac{2}{3}}, ±1$ are critical points of f
