## Calculus 10th Edition

The Mean Value Theorem can be applied; $c=\dfrac{8}{27}.$
$f(x)$ is continuous over the closed interval and is differentiable on the open interval ( even though $f(x)$ is not differentiable at $x=0$, this is not included in the open interval.) $a=0; b=1\to f(a)=0$ and $f(b)=1.$ Since $f(x)$ is continuous over $[a, b]$ and differentiable over $(a, b)$, applying Mean Value Theorem over the interval $[a , b]$ guarantees the existence of at least one value c such that $a\lt c\lt b$ and $f′(c)=\dfrac{f(b)-f(a)}{b-a}.$ $f'(c)=\dfrac{1-0}{1-0}=1.$ $f'(x)=\dfrac{2}{3\sqrt[3]{x}}\to f'(x)=1\to \dfrac{2}{3\sqrt[3]{x}}=1\to \sqrt[3]{x}=\dfrac{2}{3}\to c=\dfrac{8}{27}.$