Question

If $f(x) = \begin{cases} \dfrac{x^{2} - 9}{x - 3}, & x \neq 3 \\ 2x + k, & x = 3 \end{cases}$ is continuous at $x = 3$, then $k$ equals

• A. 3
• B. 0
• C. $-6$
• D. $\dfrac{1}{6}$

Hint:

Factor and cancel first: $\dfrac{x^2-9}{x-3} = x+3$ for $x \neq 3$. The limit at $x=3$ is $6$.

Answer 1

The Correct Option is B

To determine the value of \(k\) that makes the function \(f(x)\) continuous at \(x = 3\), we need to ensure the following condition for continuity:

The function \(f(x)\) is continuous at \(x=3\) if:

• The limit of \(f(x)\) as \(x\) approaches \(3\) exists.
• This limit is equal to \(f(3)\).

Let's first find the limit of \(f(x)\) as \(x\) approaches 3.

For the limit, consider the expression \(\dfrac{x^{2} - 9}{x - 3}\) when \(x \neq 3\):

\(\dfrac{x^{2} - 9}{x - 3} = \dfrac{(x-3)(x+3)}{x-3}\)

This simplifies to \(x+3\) when \(x \neq 3\).

Therefore, the limit as \(x \to 3\) is:

\(\lim_{{x \to 3}} (x+3) = 3+3 = 6\)

Next, we evaluate \(f(x)\) at \(x=3\):

Since \(f(x) = 2x + k\) when \(x = 3\), we have:

\(f(3) = 2(3) + k = 6 + k\)

For \(f(x)\) to be continuous at \(x=3\), the limit must equal the value of the function at that point:

\(6 = 6 + k\)

Solving for \(k\) gives:

\(k = 0\)

Therefore, the correct value of \(k\) is 0.