Q.Does there exist a function which is continuous everywhere but not differentiable at exactly two points? Justify your answer.

Concept: Continuity Differentiability Relationship — a function can be continuous everywhere yet fail to be differentiable at isolated corner points.

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

1. It's continuous everywhere, being a sum of the continuous functions $|x|$ and $|x-1|$.
2. Piecewise: $f(x)=1-2x$ for $x<0$, $f(x)=1$ for $0\le x\le1$, $f(x)=2x-1$ for $x>1$.
3. At $x=0$: left derivative $-2 \ne$ right derivative $0$ — not differentiable. …

Differentiability is a stricter requirement than continuity: a function can be unbroken (continuous) yet still have a sharp corner at isolated points, where no single tangent line exists. The absolute value function $|x|$ is the standard example of one corner, at $x=0$. To get exactly two non-differentiable points, add together two absolute-value functions with corners at two different locations.

Step 1 — Construct the function.

Let $f(x) = |x| + |x-1|$, which has potential corners at $x=0$ (from $|x|$) and $x=1$ (from $|x-1|$).

Step 2 — Continuity.

Each of $|x|$ and $|x-1|$ is continuous on $\mathbb{R}$ (absolute value of a continuous function is continuous), so their sum $f$ is continuous everywhere, by the algebra of continuous functions.

Step 3 — Write $f$ piecewise.

Step 4 — Check differentiability at $x=0$.

Left-hand derivative (from the $x<0$ piece $1-2x$): slope $-2$. Right-hand derivative (from the $0\le x\le 1$ piece, constant $1$): slope $0$. Since $-2 \ne 0$, $f$ is not differentiable at $x=0$.

Step 5 — Check differentiability at $x=1$. …

Method: Constructing a Function with Exactly $n$ Non-Differentiable Points

To build a function that is continuous everywhere but fails to be differentiable at exactly a chosen finite set of points, sum shifted copies of $|x|$ — each copy contributes exactly one corner, at the point it's shifted to.

Steps

Step 1: Recall that $g(x)=|x|$ is continuous everywhere but has one non-differentiable point (a corner at $x=0$, where the left and right derivatives disagree)

Step 2: Shift the corner to any desired point $c$ by using $|x-c|$ instead

Step 3: Sum several shifted copies, one for each desired non-differentiable point

A sum of continuous functions is always continuous, so the sum stays continuous everywhere. Differentiability fails only where at least one individual term has a corner.

Step 4: Verify each corner point explicitly by computing the left-hand and right-hand derivatives there …

Common Mistakes

Mistake 1: Only checking that the corners "look like" they should be non-differentiable, without computing the one-sided derivatives explicitly.

Why it's wrong: a "show that" / "justify your answer" question requires the explicit left-hand and right-hand derivative computation at each claimed corner — an intuitive appeal to the shape of $|x-c|$ is not a complete justification on its own. Correct approach: write out the piecewise formula near each corner and compute both one-sided derivatives.

Mistake 2: Assuming continuity is "automatic" without stating why the sum of two absolute-value functions is continuous. …