## University Calculus: Early Transcendentals (3rd Edition)

$c=\dfrac{8}{27}$
Here, the function $f(x)=x^{\frac{2}{3}}$ is continuous on $[0,1]$ and differentiable on $(0,1)$ when $f'(x)=\dfrac{2}{3}x^{\frac{-1}{3}}$ . The Mean value Theorem states that there is a point $c \in (0,1)$ such that $f'(c)=\dfrac{f(1)-f(0)}{1-0}=\dfrac{1-0}{1-0}=1$ Then $f'(c)=\dfrac{2}{3}c^{\frac{-1}{3}}=1$ This implies $c=\dfrac{8}{27}$