Answer
Critical values at $x=-\sqrt 2, x=0, x=\sqrt 2$
Work Step by Step
In order to find the critical values, we need to find the derivative of $f(x)$
$f(x)=x^{2}(x^{2}-4)$
$f'(x)=((x^{2}\times2x)+((x^{2}-4)\times2x))$
$f'(x)=2x^{3}+2x^{3}-8x=4x^{3}-8x$
$f'(x)=x(4x^{2}-8)$
$x=0$
$4x^{2}-8=0$
$4x^{2}=8$
$x^{2}=\frac{8}{4}=2$
$x=\sqrt 2, x=-\sqrt 2$
Critical values occur at $x=-\sqrt 2, x=0, x=\sqrt 2$
