Q.Find the values of $k$ so that the function $f$ is continuous at the indicated point, where $f$ is defined by $f(x) = \begin{cases} kx+1, & \text{if } x \leq 5 \\ 3x-5, & \text{if } x > 5 \end{cases}$ at $x = 5$

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

Step 1: Left-hand limit

For $x \leq 5$, $f(x) = kx + 1$.

$\lim_{x \to 5^-} f(x) = k(5) + 1 = 5k + 1$.

Step 2: Right-hand limit

For $x > 5$, $f(x) = 3x - 5$.

$\lim_{x \to 5^+} f(x) = 3(5) - 5 = 15 - 5 = 10$.

Step 3: Value at $x = 5$ …

Why continuity at a join 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 $x=5$. The left piece ($kx+1$) gives us the function value at $x=5$ and the left-hand limit. The right piece ($3x-5$) gives us the right-hand limit. If these three numbers are equal, the function is continuous. If they aren't, there's a jump — a break in the graph.

The question asks us to discuss continuity, which means we need to find the condition on $k$ that makes the function continuous, and then state what happens for other values of $k$.

Step-by-step reasoning

1. Find the function value at $x=5$.

Since $x=5$ falls under the first case ($x \leq 5$), we use $f(x) = kx+1$.

2. Find the left-hand limit as $x \to 5^{-}$.

For $x$ just less than 5, the function is still $kx+1$. So the left-hand limit is the same as the function value:

3. Find the right-hand limit as $x \to 5^{+}$.

For $x$ just greater than 5, the function is $3x-5$. So:

4. Set the three values equal for continuity.

For continuity at $x=5$, we need:

That gives:

5. Solve for $k$.

$5k = 9 \quad \Rightarrow \quad k = \frac{9}{5}$ …

Method: Solving for an Unknown Coefficient at a Piecewise Junction

Any question that gives a piecewise function with one unknown constant and asks for continuity at a named point reduces to the same three-part comparison, every time.

Steps

Step 1: Write down $f(a)$ from the piece that actually owns the junction point

Check the inequality carefully — the piece whose condition includes the equals sign (e.g. $x \leq a$) is the one that defines $f(a)$; the other piece only tells you what the function approaches, not its actual value there.

Step 2: Compute the one-sided limit from the owning side

Since that piece is typically a polynomial (always continuous), its limit as $x \to a$ from that side is just the piece evaluated at $a$ — this automatically matches $f(a)$ from Step 1.

Step 3: Compute the one-sided limit from the other side …

Common Mistakes

Mistake 1: Assuming $f(5)$ comes from the second piece

Why it's wrong: because the condition is $x \leq 5$, $x=5$ belongs to the first piece ($kx+1$), so $f(5) = 5k+1$ — not $3(5)-5=10$. Using the second piece for $f(5)$ can lead to the false conclusion that the function is continuous for every $k$. Correct approach: match the junction value to whichever inequality includes an equals sign, every time, before writing $f(a)$.

Mistake 2: Arithmetic slip when solving $5k+1=10$ …