Q.Find the values of $p$ and $q$ so that $f(x) = \begin{cases} x^2 + 3x + p, & x \le 1 \\ qx + 2, & x > 1 \end{cases}$ is differentiable at $x = 1$.

Concept: Continuity and differentiability at a point — for $f$ to be differentiable at $x=1$, it must first be continuous there, and the left-hand derivative must equal the right-hand derivative.

Step 1 — Continuity at $x=1$

Left-hand limit: $\lim_{x\to 1^-} f(x) = 1^2 + 3(1) + p = 4 + p$

Right-hand limit: $\lim_{x\to 1^+} f(x) = q(1) + 2 = q + 2$

Function value: $f(1) = 4 + p$

Continuity requires $4 + p = q + 2$, so

Step 2 — Differentiability at $x=1$ …

The key idea: differentiability at a point is a stronger condition than continuity. A function is differentiable at $x = 1$ only if it is continuous there and its derivative from the left equals its derivative from the right. We'll enforce both conditions step by step.

1. Enforce continuity at $x = 1$

For $f$ to be differentiable at $x = 1$, it must first be continuous there. That means the left-hand limit, right-hand limit, and the function value at $x = 1$ must all be equal.

The left-hand piece is $f(x) = x^2 + 3x + p$ for $x \le 1$, so at $x = 1$:

The left-hand limit as $x \to 1^-$ is the same:

The right-hand piece is $f(x) = qx + 2$ for $x > 1$, so the right-hand limit as $x \to 1^+$ is:

Continuity requires:

which simplifies to:

2. Enforce equal derivatives from both sides

Now we need the left-hand derivative and the right-hand derivative at $x = 1$ to be equal.

Left-hand derivative: For $x \le 1$, $f(x) = x^2 + 3x + p$. Differentiate:

So the left-hand derivative at $x = 1$ is:

Right-hand derivative: For $x > 1$, $f(x) = qx + 2$. Differentiate:

So the right-hand derivative at $x = 1$ is:

For differentiability, we need:

$5 = q$ …

Method: Finding Unknown Constants for Differentiability of a Piecewise Function

Use this whenever a piecewise-defined function contains unknown constants and the question asks for the values that make it differentiable at the junction point. Differentiability at a point is a two-layer condition — you must enforce both layers.

Steps

Step 1: Recognise the two conditions differentiability requires

A function is differentiable at a junction point $x=a$ only if (i) it is continuous there, and (ii) its left-hand derivative equals its right-hand derivative there. Continuity alone is not sufficient — a function can be continuous yet have a sharp corner (unequal one-sided slopes).

Step 2: Enforce continuity at the junction

Compute the left-hand limit (using the piece defined for $x\le a$), the right-hand limit (using the piece for $x>a$), and $f(a)$ itself; set them all equal. This gives one equation relating the unknown constants. …

Common Mistakes

Mistake 1: Checking differentiability without first ensuring continuity

Why it's wrong: Differentiability at a point requires continuity as a prerequisite; a student who jumps straight to equating one-sided derivatives without first equating the one-sided limits (and the function value) can arrive at a value for $q$ that doesn't actually make the pieces meet at $x=1$. Correct approach: always enforce continuity first ($4+p=q+2$) and treat it as a separate equation alongside the derivative-matching equation.

Mistake 2: Using the wrong piece to evaluate $f(1)$

Why it's wrong: Since the domain split is $x\le 1$ versus $x>1$, the point $x=1$ itself belongs to the first piece ($x^2+3x+p$); accidentally using $qx+2$ to evaluate $f(1)$ gives a wrong continuity equation. Correct approach: substitute $x=1$ only into the piece whose inequality includes $x=1$ (here, $x\le 1$). …