Checking MVT hypotheses for an absolute value function

The function $f(x) = |x-2|$ can be written piecewise as $f(x) = -(x-2)$ for $x<2$ and $f(x) = x-2$ for $x \ge 2$. This function is continuous everywhere, including on $[0,5]$, since the two pieces meet at $x=2$ with the same value $f(2)=0$. So the continuity hypothesis of the MVT is satisfied.

However, differentiability fails at $x=2$: the left-hand derivative is $-1$ and the right-hand derivative is $+1$, so $f'(2)$ does not exist. Since $2$ lies in the open interval $(0,5)$, $f$ is not differentiable on all of $(0,5)$. Because this hypothesis of the MVT fails, the theorem simply does not apply here — it makes no claim about existence of $c$, and indeed no such $c$ works, since $f'(x) = \pm 1$ everywhere it's defined but the secant slope is $\frac{f(5)-f(0)}{5-0} = \frac{3-2}{5} = \frac{1}{5} \ne \pm 1$.

The answer is x equals two. The function f of x equals the absolute value of x minus two has a sharp corner exactly at x equals two, where the left-hand derivative is negative one and the right-hand derivative is positive one. Since these one-sided derivatives disagree, the derivative does not exist at that point. This is the specific location inside the interval where the differentiability hypothesis of the Mean Value Theorem breaks down.