Answer
The function is continuous on $[0,8]$
$f(0) = 1$
$f(8) = 4$
However, the function does not satisfy the conclusion of the Mean Value Theorem on $[0,8]$
Work Step by Step
The function is continuous on $[0,8]$
$f(0) = 1$
$f(8) = 4$
However, there is no number $c$ in the interval $(0,8)$ such that $f'(c) = \frac{f(8)-f(0)}{8-0} = \frac{4-1}{8-0} = \frac{3}{8}$
Therefore, the function does not satisfy the conclusion of the Mean Value Theorem on $[0,8]$