Q.Find the relationship between $a$ and $b$ so that the function $f$ defined by $f(x) = \begin{cases} ax+1, & \text{if } x \le 3 \\ bx+3, & \text{if } x > 3 \end{cases}$ is continuous at $x=3$.

Concept: Continuity At A Point — A function is continuous at $x=3$ if $\lim_{x\to 3^-} f(x) = \lim_{x\to 3^+} f(x) = f(3)$.

Step 1: Compute the left-hand limit and the function value at $x=3$ (since $x \le 3$ uses $ax+1$):

Step 2: Compute the right-hand limit (using $bx+3$ for $x > 3$):

$\lim_{x\to 3^+} f(x) = b(3) + 3 = 3b + 3.$ …

Why continuity at a point works this way

A function is continuous at a point if three things match: the value of the function at that point, the limit as you approach from the left, and the limit as you approach from the right. For a piecewise function like this one, the two pieces meet at $x=3$, but they might not join smoothly — the left piece gives $ax+1$ and the right piece gives $bx+3$. Continuity forces these two expressions to give the same output when $x=3$, even though the right piece technically starts just after $3$.

The key insight: the left-hand limit uses the formula for $x \le 3$, and the right-hand limit uses the formula for $x > 3$. At $x=3$ itself, the function is defined by the first case ($x \le 3$), so $f(3) = a(3)+1 = 3a+1$.

Step-by-step

1. Write the left-hand limit. As $x$ approaches 3 from the left ($x \to 3^-$), we use $f(x) = ax+1$:

2. Write the right-hand limit. As $x$ approaches 3 from the right ($x \to 3^+$), we use $f(x) = bx+3$: $\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (bx+3) = b(3) + 3 = 3b + 3.$ …

Method: Using the Continuity Condition to Find a Relationship Between Unknown Constants

When a piecewise function contains unknown parameters (like $a$ and $b$) and the problem states the function IS continuous at a given point, that continuity condition becomes an equation to solve — this is the reverse of the usual "check whether it's continuous" problem.

Steps

Step 1: Identify which piece defines the function's value at the given point

Match the boundary value to the inequality that includes equality — that piece gives $f(a)$ directly, now written in terms of the unknown constant(s) rather than a number.

Step 2: Write the left-hand and right-hand limits as algebraic expressions

Both sides will contain the unknown constants rather than pure numbers.

Step 3: Set up the continuity equation …

Common Mistakes

Mistake 1: Using $bx+3$ instead of $ax+1$ to compute $f(3)$

Why it's wrong: the boundary condition is $x \leq 3$, so $x=3$ belongs to the first piece, $ax+1$ — using the second piece gives the wrong expression for the function's actual value at the boundary. Correct approach: read the inequality carefully and substitute into the piece that includes equality at the boundary.

Mistake 2: Trying to solve $3a+1 = 3b+3$ for unique numeric values of $a$ and $b$ …