Q.Find the values of $a$ and $b$ such that the function defined by $f(x) = \begin{cases} 5, & \text{if } x \leq 2 \\ ax+b, & \text{if } 2 < x < 10 \\ 21, & \text{if } x \geq 10 \end{cases}$ is a continuous function.

Concept: Continuity Condition — For a piecewise function to be continuous at a junction, the left-hand limit, right-hand limit, and the function value must all be equal.

Step 1: At $x = 2$

Left limit: $\lim_{x \to 2^-} f(x) = 5$

Right limit: $\lim_{x \to 2^+} f(x) = a(2) + b = 2a + b$

Continuity requires $2a + b = 5$.

Step 2: At $x = 10$

Left limit: $\lim_{x \to 10^-} f(x) = a(10) + b = 10a + b$

Right limit: $\lim_{x \to 10^+} f(x) = 21$

Continuity requires $10a + b = 21$. …

Why continuity at the boundaries is the key

A piecewise function is continuous if it has no jumps, breaks, or holes. Since each piece ($5$, $ax+b$, $21$) is itself continuous on its own interval, the only places where continuity could fail are the boundary points $x=2$ and $x=10$. At each boundary, the value coming from the left must match the value coming from the right — and both must equal the function's defined value at that point.

This gives us two conditions, which become two equations in $a$ and $b$.

Step-by-step solution

1. Continuity at $x = 2$

For $x \leq 2$, $f(x) = 5$, so $f(2) = 5$.

For $2 < x < 10$, $f(x) = ax + b$. As $x$ approaches $2$ from the right, the value approaches $a(2) + b = 2a + b$.

Continuity at $x=2$ requires:

2. Continuity at $x = 10$

For $2 < x < 10$, $f(x) = ax + b$. As $x$ approaches $10$ from the left, the value approaches $a(10) + b = 10a + b$.

For $x \geq 10$, $f(x) = 21$, so $f(10) = 21$.

Continuity at $x=10$ requires:

$10a + b = 21 \quad \text{(Equation 2)}$ …

Method: Two Unknowns from Two Junction Points

When a three-piece function has an unknown linear middle piece ($ax+b$) sandwiched between two known pieces, continuity must hold at BOTH junction points — this gives two equations in two unknowns.

Steps

Step 1: Identify both junction points from the piecewise conditions

There is one junction wherever the defining inequality changes — typically two junctions for a three-piece function.

Step 2: At the first junction, equate the left-hand and right-hand limits

Evaluate the piece just before the junction and the middle piece $ax+b$ at that junction's $x$-value; set them equal. This gives your first linear equation in $a$ and $b$.

Step 3: At the second junction, do the same

Evaluate the middle piece $ax+b$ and the piece just after the second junction at that $x$-value; set them equal. This is your second equation. …

Common Mistakes

Mistake 1: Using $ax+b$ instead of the actual defined value at $x=10$

Why it's wrong: the function's own definition says $f(10) = 21$ (from the third piece, since $x \geq 10$) — the middle piece $ax+b$ only tells you the left-hand limit approaching $10$, not the function's value there. Confusing the two gives an equation using the wrong quantity. Correct approach: always write $f(10)$ from the piece whose inequality actually contains $10$, and treat the middle piece's value at $10$ purely as the left-hand limit.

Mistake 2: Sign or subtraction error when eliminating $b$ from the system …