Answer
The derivative does not exist when $x = \frac{1}{2}$. The function is not differentiable on the interval $(0,3)$. The hypotheses of the Mean Value Theorem are not satisfied.
Work Step by Step
$f(x) = 2-\vert 2x-1 \vert$
$f'(x) = 2~~$ if $x \lt \frac{1}{2}$
$f'(x) = -2~~$ if $x \gt \frac{1}{2}$
We can find $c$ such that $f(3)- f(0) = f'(c)(3-0)$:
$f(3)- f(0) = f'(c)(3-0)$
$(-3)- (1) = f'(c)(3)$
$f'(c) = -\frac{4}{3}$
However, there is no number $c$ such that $f'(c) = -\frac{4}{3}$ since either $f'(x) = -2$ or $f'(x) = 2$
Thus there is no number $c$ in the interval $(0,3)$ such that $f(3)- f(0) = f'(c)(3-0)$
The derivative does not exist when $x = \frac{1}{2}$. Thus, the function is not differentiable on the interval $(0,3)$. The hypotheses of the Mean Value Theorem are not satisfied.
Therefore, the fact that there is no number $c$ in the interval $(0,3)$ such that $f(3)- f(0) = f'(c)(3-0)$ does not contradict the Mean Value Theorem.