Is a function that is continuous at a point always differentiable there?

The click

Differentiability implies continuity, but not the other way round. Continuity at a point means no break: the limit of f(x) as x → a equals f(a). Differentiability is stronger: the graph must have a single, well-defined slope there, meaning the left-hand and right-hand derivatives both exist and are equal.

f(x) = |x| at x = 0 is continuous: both sides approach 0, and f(0) = 0. But the right-hand derivative is the limit of (|h| − 0)/h as h → 0⁺, which is h/h = 1, and the left-hand derivative is the limit of |h|/h as h → 0⁻, which is −h/h = −1. Since 1 ≠ −1, f′(0) does not exist. The graph has a sharp corner at the origin.

A continuous function can fail to be differentiable in other ways too: at a vertical tangent, as with f(x) = x^(1/3) at x = 0, where the slope becomes infinite, or at a cusp, as with x^(2/3) at x = 0. There are even functions, such as the Weierstrass function, that are continuous everywhere and differentiable nowhere.

The direction that is true: if f is differentiable at a, it must be continuous at a. So a break in the graph at a point immediately rules out differentiability there. For JEE questions with |x − 2|, the greatest integer function [x] or piecewise definitions, check continuity first, then compare the left and right derivatives at each joining point.