Answer
Rolle's theorem does not apply because $f(x)$ is not differentiable over the interval $[-1,1]$
Work Step by Step
We first look at $f(x)$
$f(x)=\sqrt((2-x^{\frac{2}{3}})^{3})$
Before we differentiate, we can rewrite $f(x)$ to make it easier to differentiate by combining the square root and the cube exponent
$f(x)=(2-x^{\frac{2}{3}})^{\frac{3}{2}}$
$f′(x)=\frac{3}{2}(2-x^{\frac{2}{3}})^{\frac{1}{2}}*(-\frac{2}{3}x^{-\frac{1}{3}})$
$f′(x)=\frac{3\sqrt (2-x^{\frac{2}{3}})}{2}*-\frac{2}{3\sqrt[3] x}$
$f′(x)=-\frac{\sqrt (2-x\frac{2}{3})}{\sqrt[3] x}$
At $x=0$, $f′(x)$ is undefined, which makes $f(x)$ not differentiable, and Rolle's Theorem does not apply/.
