If f(x) is continuous at x=2 but not differentiable, sketch possible graphs

Question

Give three examples of functions that are continuous at $x = 2$ but not differentiable at $x = 2$ . Sketch the graph for each and explain why differentiability fails.

(JEE Main 2022, similar pattern)

Solution — Step by Step

f(x) = |x - 2|

At $x = 2$ : continuous (both sides approach 0, and $f(2) = 0$ ). But the left derivative is $-1$ and the right derivative is $+1$ . Since left derivative $\neq$ right derivative, $f$ is not differentiable at $x = 2$ .

The graph has a sharp corner (V-shape) at $x = 2$ .

f(x) = (x - 2)^{2/3}

At $x = 2$ : continuous ( $f(2) = 0$ , and $\lim_{x \to 2} f(x) = 0$ ). But:

f'(x) = \frac{2}{3}(x-2)^{-1/3}

As $x \to 2$ , $f'(x) \to \pm\infty$ . The tangent line becomes vertical — the slope is undefined. The graph has a cusp at $x = 2$ .

f(x) = \begin{cases} 2x - 1 & \text{if } x < 2 \\ x + 1 & \text{if } x \geq 2 \end{cases}

At $x = 2$ : from the left, $f(2^-) = 2(2) - 1 = 3$ . From the right, $f(2^+) = 2 + 1 = 3$ . And $f(2) = 3$ . So continuous.

Left derivative = 2, right derivative = 1. Since $2 \neq 1$ , not differentiable. The graph has a kink — two line segments meeting at an angle.

Why This Works

Differentiability at a point requires that the function has a unique, well-defined tangent line there. This fails in three situations:

Sharp corner/kink: The graph suddenly changes direction. The left and right slopes exist but are different.
Cusp/vertical tangent: The slope becomes infinite. The tangent line exists but is vertical — slope is undefined.
Oscillation: Functions like $x \sin(1/(x-2))$ oscillate infinitely fast near the point, so no tangent can be drawn.

Continuity only requires that the function has no “jumps” — the value matches the limit. Differentiability is a stricter requirement: the function must also be “smooth” (no corners, cusps, or wild oscillations).

Key fact for JEE: Differentiability $\Rightarrow$ Continuity, but Continuity $\not\Rightarrow$ Differentiability.

Alternative Method

Instead of sketching, you can verify non-differentiability algebraically by computing the limit definition:

f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}

If $\lim_{h \to 0^+}$ and $\lim_{h \to 0^-}$ give different values (or do not exist), $f$ is not differentiable at $x = 2$ .

In JEE Main, the most common version of this question gives a piecewise function and asks: “At which points is $f$ continuous but not differentiable?” Check the junction points — compute left and right derivatives. If they differ, that point is continuous-but-not-differentiable.

Common Mistake

Students sometimes claim $|x - 2|$ is not continuous at $x = 2$ because of the “break” in the formula. But the absolute value function IS continuous everywhere — there is no gap in the graph. The V-shape is connected. The issue is only with the derivative (slope changes abruptly), not with the function value.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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