Identifying the critical number at a non-differentiable point

The absolute value function $|x-2|$ has a corner at $x = 2$, where the left-hand and right-hand derivatives disagree, so $f'(2)$ does not exist. This makes $x = 2$ a critical number of $f$ on $[-1,5]$.

For $x < 2$, $f(x) = -(x-2)+3 = -x+5$, so $f'(x) = -1 < 0$. For $x > 2$, $f(x) = (x-2)+3 = x+1$, so $f'(x) = 1 > 0$. Since $f'$ changes from negative to positive at $x = 2$, the First Derivative Test says $f$ has a local minimum there, with $f(2) = 3$.

At the left endpoint x equals negative one, f equals six, since f is negative x plus five there. The local minimum found earlier occurs at x equals two, where f equals three. Since six is greater than three, the function value at the left endpoint is bigger than the local minimum value.