Calculus 10th Edition

Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-2$.
$g'(x)=\dfrac{1}{3\sqrt[3]{x^2}}$ $g'(x)$ is never equal to $0$ but is undefined for $x=0.$ Hence, the endpoints of the interval, along with the critical number $x=0,$ are possible points for absolute extrema. $g(-8)=-2.$ $g(0)=0.$ $g(8)=2.$ Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-2$.