Q.An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is __________.

We need a function that is continuous for all $x$ but has exactly two "sharp corners." Each absolute-value term contributes one corner, so adding two of them at different places gives exactly two non-differentiable points.

Take $f(x)=|x|+|x-1|$.

• It is continuous everywhere (a sum of continuous functions).
• $|x|$ has a corner only at $x=0$; $|x-1|$ has a corner only at $x=1$. Away from those points both pieces are smooth, so $f$ is differentiable there. …

The idea

Differentiability can fail where a continuous graph has a sharp corner: at a corner the slope coming from the left and the slope coming from the right disagree, so there is no single tangent. The absolute-value function $|x|$ is the classic one-corner example (its corner is at $x=0$). To get exactly two corners, we add two absolute-value terms whose corners sit at two different places.

Step 1 — Build the function

Let

$|x|$ has its only corner at $x=0$; $|x-1|$ has its only corner at $x=1$.

Step 2 — It is continuous everywhere

Each absolute-value function is continuous on all of $\mathbb{R}$, and the sum of continuous functions is continuous. So $f$ is continuous for every real $x$ — the graph has no breaks or jumps.

Step 3 — Where differentiability fails

Write $f$ piecewise:

• At $x=0$: the left slope is $-2$, the right slope is $0$. They differ, so $f'(0)$ does not exist. …

Method: Building a Function That Is Continuous Everywhere but Non-Differentiable at Chosen Points

This method solves "construct an example" questions that ask for a function with a prescribed continuity/differentiability profile — continuous on all of $\mathbb{R}$ but with corners (non-differentiable points) at exactly the locations you choose.

Steps

Step 1: Know your building block

The function $|x-c|$ is continuous everywhere and has exactly one corner, at $x=c$. Its left-hand derivative there is $-1$ and its right-hand derivative is $+1$; away from $x=c$ it is a straight line (slope $-1$ or $+1$), so it is differentiable everywhere except that single point.

Step 2: Stack one building block per required corner

To get corners at $n$ prescribed points $c_1,c_2,\dots,c_n$, add one absolute-value term per point:

Step 3: Confirm continuity

Each $|x-c_i|$ is continuous on $\mathbb{R}$, and a finite sum of continuous functions is continuous (algebra of continuous functions). So $f$ is automatically continuous everywhere — no extra work needed.

Step 4: Confirm the corners land exactly where intended …

Common Mistakes

Mistake 1: Reaching for a single absolute-value term

A student often writes $f(x)=|x|$ or $f(x)=|x-1|$ and thinks shifting it satisfies "exactly two points." Why it's wrong: a lone term like $|x-a|$ has exactly ONE corner, at $x=a$ — one term can never produce two non-differentiable points on its own. Correct approach: you need two separate absolute-value terms, each contributing its own corner at a different location, e.g. $|x|+|x-1|$.

Mistake 2: Forgetting to justify continuity, not just assume it

Students state "it's continuous" without reasoning. Why it's wrong: continuity has to be argued (here, as a sum of continuous absolute-value functions) — with a different construction (e.g. involving a quotient), continuity can actually fail and must be checked, not assumed. Correct approach: always name why the sum is continuous (sum of continuous functions is continuous) rather than treating it as automatic.

Mistake 3: Sign errors in the piecewise slopes …