To determine the value of [tex]\( k \)[/tex] that ensures the function [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 3 \)[/tex], we need to satisfy the definition of continuity at that point. Specifically, a function [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 3 \)[/tex] if the following three conditions hold:
1. [tex]\( f(3) \)[/tex] is defined.
2. The limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 3 exists.
3. The limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 3 equals [tex]\( f(3) \)[/tex].
For the given function:
[tex]\[ f(x) = \begin{cases}
\frac{2x^2 - 18}{x - 3} & \text{for } x \neq 3, \\
k & \text{for } x = 3.
\end{cases} \][/tex]
First, we'll find the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 3.
Consider the expression [tex]\(\frac{2x^2 - 18}{x - 3}\)[/tex] for [tex]\( x \neq 3 \)[/tex]. To find this limit, we need to simplify the expression. Notice that the numerator can be factored:
[tex]\[ 2x^2 - 18 = 2(x^2 - 9) = 2(x - 3)(x + 3). \][/tex]
Thus, we can rewrite the function as:
[tex]\[ \frac{2(x - 3)(x + 3)}{x - 3}. \][/tex]
For [tex]\( x \neq 3 \)[/tex], we can cancel out the [tex]\((x - 3)\)[/tex] terms in the numerator and the denominator:
[tex]\[ \frac{2(x - 3)(x + 3)}{x - 3} = 2(x + 3). \][/tex]
Now, we need to find the limit of [tex]\( 2(x + 3) \)[/tex] as [tex]\( x \)[/tex] approaches 3:
[tex]\[ \lim_{x \to 3} 2(x + 3). \][/tex]
Substituting [tex]\( x = 3 \)[/tex] into the expression [tex]\( 2(x + 3) \)[/tex]:
[tex]\[ 2(3 + 3) = 2 \cdot 6 = 12. \][/tex]
Therefore, the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 3 is 12. To ensure continuity at [tex]\( x = 3 \)[/tex], we must have:
[tex]\[ f(3) = \lim_{x \to 3} f(x). \][/tex]
Given that [tex]\( f(3) = k \)[/tex], we set [tex]\( k \)[/tex] equal to the limit we just found:
[tex]\[ k = 12. \][/tex]
So, the value of [tex]\( k \)[/tex] that makes the function [tex]\( f(x) \)[/tex] continuous at [tex]\( x = 3 \)[/tex] is [tex]\( \boxed{12} \)[/tex].
