Continuity & Differentiability: Step-by-Step Worked Examples (10)

Question

Determine if $f(x) = |x - 1| + |x - 2|$ is continuous and differentiable everywhere. Identify the points where it is not differentiable.

Solution — Step by Step

Final answer: $f$ is continuous on $\mathbb{R}$, differentiable on $\mathbb{R} \setminus \{1, 2\}$.

Why This Works

Absolute value functions have “corners” at the points where their argument is zero — these are the only candidates for non-differentiability. Sum of corners is just more corners (or smooth, if they happen to cancel, which they don’t here).

The function is piecewise linear with three pieces: slope $-2$ for $x < 1$, slope $0$ for $1 \leq x \leq 2$, slope $+2$ for $x > 2$. Slopes change discontinuously at $x = 1$ and $x = 2$, so derivatives don’t exist there.

Alternative Method

Sketch the graph. $f(x)$ is V-shaped, with minimum value $1$ on the entire interval $[1, 2]$. Two corners at $x = 1$ and $x = 2$ make it non-differentiable there.

Common Mistake

Concluding non-differentiability from continuity failure. Differentiability is stronger than continuity. If $f$ is continuous, it might or might not be differentiable. Continuity does not guarantee differentiability — corners are continuous but not differentiable.