## Calculus 10th Edition

$g(x)$ has three critical numbers : {$-2, 0, 2$}.
$g'(x)=4x^3-16x=4x(x^2-4)=4x(x-2)(x+2).$ $g'(x)=0\to 4x(x-2)(x+2)=0\to x=-2, x=0$ or $x=2.$ $g(x)$ has three critical numbers : {$-2, 0, 2$}.