Single Variable Calculus, Chapter 4, 4.2, Section 4.2, Problem 16

Show that there is no value of $c$ such that $f(3) -f(0) = f'(c)(3-0)$ in the equation $f(x) = 2 - |2x - 1|$. Why does this not contradict the Mean Value Theorem?
Using the definition of absolute value, we can rewrite $f(x)$ as

$
f(x) =
\begin{array}{c}
2 - (2x - 1) & \text{ for } & x \geq \frac{1}{2}\\
2 - (-(2x-1)) & \text{ for } & x < \frac{1}{2}
\end{array}
\qquad \Longrightarrow \qquad f(x)=
$
$
\begin{array}{c}
-2x + 3 & \text{ for } & x \geq \frac{1}{2}\\
2x + 1 & \text{ for } & x < \frac{1}{2}
\end{array}
$


If we evaluate $f(x)$ for $\displaystyle x \geq \frac{1}{2}$. By using Mean Value Theorem, we get $\displaystyle f'(c) = \frac{f(b) - f(a)}{b-a}$

$
\begin{equation}
\begin{aligned}
f'(c) &= \frac{\left[ -2(3) + 3\right] - \left[ -2(0) + 3 \right]}{3-0}\\
\\
f'(c) &= -2 \text{ for } c \geq \frac{1}{2}\\
\\
\text{for } x < \frac{1}{2},\\
\\
f'(c) &= \frac{\left[ -2(3) + 1\right] - \left[ -2(0) + 1 \right]}{3-0}\\
\\
f'(c) &= 2 \text{ for } c < \frac{1}{2}\\
\end{aligned}
\end{equation}
$

We got two different values of $f'(c)$ and it shows that $f(x)$ is not differentiable at $x = \frac{1}{2}$ the function is continuous on the interval $[0,3]$ (See the graph below).



Therefore, it does not satisfy the Mean Value Theorem.