Answer
$f$ is not differentiable in the interval $(-1,1)$, so the theorem does not apply.
Work Step by Step
Rolle's Theorem
Let $f$ be continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ .
If $f(a)=f(b)$ , then there is at least one number $c$ in $(a, b)$ such that $f^{\prime}(c)=0.$
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f(x) is continuous on the interval $[-1,1]$.
Let us check whether it is differentiable on $(-1,1)$.
$f(x)=\sqrt{(2-x^{2/3})^{3}}=[(2-x^{2/3})^{3}]^{1/2}=(2-x^{2/3})^{3/2}$
$f^{\prime}(x)=$ .. chain rule ... $=\displaystyle \frac{3}{2}\cdot[(2-x^{2/3})^{1/2}(-\frac{2}{3})x^{-1/3}$
$=\displaystyle \frac{-\sqrt{(2-x^{2/3})}}{x^{1/3}}$
$f^{\prime}(x)$ is not defined at x=0, which is in the interval $(-1,1)$.
$f$ is not differentiable in the interval $(-1,1)$, so the theorem does not apply.
