#### Answer

Over the specified interval, the function has an absolute maximum equal to $5$ and an absolute minimum equal to $0$

#### Work Step by Step

$y'=\dfrac{2}{\sqrt[3]{x}}-2=\dfrac{2-2\sqrt[3]{x}}{\sqrt[3]{x}}.$
$y'=0\to2-2\sqrt[3]{x}=0\to x=1.$
$y'$is undefined $\to \sqrt[3]{x}=0\to x=0.$
Since both values are in the specified interval, both, along with the interval endpoints, are possible candidates for absolute extrema.
$f(-1)=5.$
$f(0)=0.$
$f(1)=1.$
Over the specified interval, the function has an absolute maximum equal to $5$ and an absolute minimum equal to $0$.