Q.Prove that the function $f$ given by $f(x) = |x - 1|, x \in \mathbf{R}$ is not differentiable at $x = 1$.

Concept: Differentiability & Left/Right-Hand Derivatives

A function is differentiable at a point if the left-hand derivative equals the right-hand derivative there.

Step 1 – Write the definition for $x=1$

The derivative at $x=1$ is

Step 2 – Evaluate left-hand and right-hand limits

For $h \to 0^+$, $|h| = h$, so the limit is $1$.

For $h \to 0^-$, $|h| = -h$, so the limit is $-1$. …

The Concept: Differentiability Means a Smooth, Unambiguous Slope

For a function to be differentiable at a point, the graph must have a well-defined tangent line there. This means the slope of the secant line from the left must approach the same number as the slope of the secant line from the right. If these two limits disagree, the function has a "corner" or "cusp" — and the derivative does not exist.

The absolute value function $|x-1|$ is the classic example of this. Its graph is a V-shape with the vertex at $x=1$. To the left of 1, the slope is $-1$; to the right, the slope is $+1$. At the vertex itself, there is no single tangent line — the slope changes abruptly.

Let's apply this definition to $f(x) = |x-1|$ at $a=1$.

Step-by-Step Proof

1. Write the function without the absolute value.

The definition of absolute value gives us two cases:

This split is crucial: the rule changes at $x=1$.

2. Compute the left-hand derivative at $x=1$.

We approach 1 from the left, so $h < 0$ and $1+h < 1$. Using the $1-x$ branch:

and $f(1) = |1-1| = 0$.

The left-hand derivative is:

3. Compute the right-hand derivative at $x=1$.

Now approach from the right, so $h > 0$ and $1+h > 1$. Using the $x-1$ branch:

and again $f(1) = 0$.

The right-hand derivative is: …

Method: Testing Differentiability at a Corner Point Using One-Sided Derivatives

Use this method to prove a function is (or isn't) differentiable at a specific point where its formula changes — typically because of an absolute value, a piecewise definition, or a sharp corner in the graph.

Steps

Step 1: Rewrite the function without the absolute value, as a piecewise rule

Split $|E(x)|$ into two cases based on the sign of $E(x)$: one formula for where $E(x)\ge 0$, another (with a flipped sign) for where $E(x)<0$. This tells you which formula applies on each side of the point in question.

Step 2: Write the difference-quotient definition of the derivative at the point $a$

This is the definition to fall back on whenever a function isn't given by a single smooth formula everywhere.

Step 3: Evaluate the limit separately as $h\to 0^+$ and $h\to 0^-$ …

Common Mistakes

Mistake 1: Assuming continuity at $x=1$ is enough to prove differentiability

Why it's wrong: continuity only means the graph has no break or jump — it says nothing about whether the two sides meet at the same slope. $|x-1|$ is perfectly continuous at $x=1$ (both pieces meet at height $0$) but still has a sharp corner there. Correct approach: always remember continuity is necessary but not sufficient — differentiability additionally requires the left- and right-hand derivatives to agree.

Mistake 2: Using the wrong piecewise formula on each side when computing the one-sided limits

Why it's wrong: for $x<1$, $|x-1|=1-x$ (not $x-1$), and mixing these up flips the sign of the resulting derivative. Correct approach: substitute a specific nearby value (like $x=0.9$) into $|x-1|$ mentally to double-check which branch is active before computing the limit. …