Q.Find the value of $k$ so that the function $f$ is continuous at the indicated point: $f(x) = \begin{cases} 3x - 8, & x \le 5 \\ 2k, & x > 5 \end{cases}$ at $x = 5$.

Concept: Continuity Condition — For $f$ to be continuous at $x = 5$, the left-hand limit, right-hand limit, and $f(5)$ must all be equal.

Step 1: Compute $f(5)$ using the first piece:

$f(5) = 3(5) - 8 = 15 - 8 = 7$.

Step 2: Compute the left-hand limit as $x \to 5^-$:

$\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (3x - 8) = 3(5) - 8 = 7$.

Step 3: Compute the right-hand limit as $x \to 5^+$: …

The idea of continuity at a point is simple: the function should not "jump" there. If you approach $x=5$ from the left, the function follows $3x-8$; from the right, it is constant $2k$. For the graph to be unbroken at $x=5$, these two pieces must meet at the same height. That meeting height must also equal $f(5)$ itself, which is given by the left-side rule (since $x \le 5$ includes $x=5$).

Let’s work through the condition step by step.

1. Write the continuity condition at $x=5$. A function $f$ is continuous at $x=a$ if

Here $a=5$.

2. Compute the left-hand limit. For $x \le 5$, $f(x) = 3x - 8$. As $x$ approaches $5$ from the left,

3. Compute the right-hand limit. For $x > 5$, $f(x) = 2k$, a constant. So as $x$ approaches $5$ from the right,

4. Set the two limits equal. Continuity requires

5. Solve for $k$. …

Method: Finding an Unknown Constant That Makes a Piecewise Function Continuous

This is the general technique whenever a problem gives a piecewise function containing an unknown constant (commonly called $k$) and asks for the value that makes the function continuous at the junction point — the continuity condition itself becomes the equation to solve.

Steps

Step 1: Identify which piece defines the function's actual value at the junction point, and compute it (it may or may not involve $k$).

Step 2: Compute the left-hand limit at the junction using the formula that applies immediately to its left.

Step 3: Compute the right-hand limit at the junction using the formula that applies immediately to its right.

One of Steps 2–3 will typically produce an expression in terms of $k$ (whichever side's formula contains it).

Step 4: Set up the continuity equation and solve for $k$. …

Common Mistakes

Mistake 1: Computing $f(5)$ using the wrong piece of the definition.

Why it's wrong: the point $x=5$ satisfies $x \le 5$, so $f(5)$ must be computed from the first piece ($3x-8$), not the second; using $2k$ for $f(5)$ would introduce $k$ into a quantity that is actually already numerically fixed. Correct approach: carefully check which inequality includes the boundary point itself before deciding which formula gives $f(a)$.

Mistake 2: Setting up the continuity equation with the wrong two limits equated. …