Answer
Over the specified interval, the function has an absolute maximum equal to $2$ and an absolute minimum equal to $-2$.
Work Step by Step
$g'(x)=\dfrac{1}{3\sqrt[3]{x^2}}$
$g'(x)$ 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 $-2$.
