## Calculus 10th Edition

Over the specified interval, the function has an absolute maximum equal to $5$ and an absolute minimum equal to $0$
$y'=\dfrac{2}{\sqrt[3]{x}}-2=\dfrac{2-2\sqrt[3]{x}}{\sqrt[3]{x}}.$ $y'=0\to2-2\sqrt[3]{x}=0\to x=1.$ $y'$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 $0$.