Q.Find the values of $k$ so that the function $f$ is continuous at the indicated point, where $f$ is defined by $f(x) = \begin{cases} kx^2, & \text{if } x \leq 2 \\ 3, & \text{if } x > 2 \end{cases}$ at $x = 2$

The key idea is continuity at a point: a function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For a piecewise function, we check the left-hand limit, right-hand limit, and the function value at the point.

Step 1: Compute the left-hand limit as $x \to 2^{-}$.

For $x \leq 2$, $f(x) = kx^2$, so

Step 2: Compute the right-hand limit as $x \to 2^{+}$.

For $x > 2$, $f(x) = 3$, so

Step 3: The function value at $x = 2$ is $f(2) = k(2)^2 = 4k$ (since $x \leq 2$ includes $x = 2$). …

We are checking continuity at $x = 2$ — the point where the definition of $f$ changes. The function is given by two different rules on either side of 2, so the standard three-part test applies:

If even one of these equalities fails, the function is discontinuous at that point.

1. Find the left-hand limit ($x \to 2^-$). For $x \leq 2$, the rule is $f(x) = kx^2$. As $x$ approaches 2 from the left, we substitute directly (since $kx^2$ is a polynomial, hence continuous everywhere):

2. Find the right-hand limit ($x \to 2^+$). For $x > 2$, the rule is $f(x) = 3$, a constant function. So:

3. Find the function value at $x = 2$. The definition says: for $x \leq 2$, use $kx^2$. Since $2$ satisfies $x \leq 2$, we have:

4. Apply the continuity condition. We need:

That gives:

Both conditions are the same equation. Solving:

$4k = 3 \quad\Rightarrow\quad k = \frac{3}{4}.$ …

Method: Finding an Unknown Constant So a Piecewise Polynomial Matches a Constant Piece

Use this when one piece of a piecewise function is a polynomial containing an unknown constant, the other piece is a plain number (or another simple expression), and you must find the constant that makes the two pieces meet without a jump at the boundary.

Steps

Step 1: Identify which piece actually defines the function value at the boundary

Look carefully at whether the boundary inequality is $\le$/$\ge$ (includes the point) on the polynomial side or the other side — this decides which formula to use for $f(a)$ itself.

Step 2: Compute the left-hand limit using the polynomial piece

Substitute the boundary value into the polynomial containing the unknown constant; since polynomials are continuous everywhere, this limit equals direct substitution.

Step 3: Compute the right-hand limit using the other piece …

Common Mistakes

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

Why it's wrong: since the condition is $x \le 2$ (not $x<2$), the polynomial piece $kx^2$ — not the constant piece — defines $f(2)$; using the constant piece here is a common slip that silently changes which equation gets solved. Correct approach: read the inequality symbols carefully ($\le$ vs $<$) before deciding which formula supplies the boundary point's value.

Mistake 2: Not realizing continuity here reduces to a single equation

Why it's wrong: because $f(2)$ equals the same expression as the left-hand limit ($4k$), a student might think there are two independent equations to solve, when in fact two of the three required quantities are automatically equal, leaving just one genuine condition ($4k=3$). Correct approach: write out all three quantities (LHL, RHL, $f(2)$) explicitly first, then see how many genuinely independent equations remain before solving. …