Answer
$c=\dfrac{8}{27}$
Work Step by Step
The Mean value Theorem states that let there is a point $c$ and $c \in (0,1)$.
This means that that $f'(c)=\dfrac{f(1)-f(0)}{1-0}=\dfrac{1-0}{1-0}=1$
Since, the given function $f(x)=x^{\frac{2}{3}}$ is continuous on $[0,1]$ and differentiable on $(0,1)$ and $f'(x)=\dfrac{2}{3}x^{\frac{-1}{3}}$ .
This implies that $f'(c)=(\dfrac{2}{3})c^{\frac{-1}{3}}=1$
Thus, $c=\dfrac{8}{27}$
