Answer
$f(x)$ satisfies the hypotheses because
$f(x)$ is continuous on $[0,1]$
$f'(x)$ is defined on $(0,1)$
Work Step by Step
The Mean Value Theorem requires
- that $f(x)$ is continuous on $[a,b]$
- that $f'(x)$ is defined on $(a,b)$
$x^{4/5}=\sqrt[5]{x^{4}}$, and by Theorem 8 (section 2.5),
roots and positive integer powers are continuous functions over their domain, so f is continuous over $[0,1].$
$f'(x)=\displaystyle \frac{4}{5}x^{-1/5}=\frac{4}{5\sqrt[5]{x}},\quad $
which is not defined at $x=0$, but it needs not be, as the theorem requires it to be defined on $(0,1)$, $(0,1)$ does not contain x=0.
