To determine the value of [tex]\( n \)[/tex] such that the function [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 6 \)[/tex], we need to ensure that the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( 6 \)[/tex] equals [tex]\( f(6) = n \)[/tex].
The function is given as:
[tex]\[
f(x) =
\begin{cases}
\frac{x^2 - 36}{x - 6} & \text{if } x \neq 6 \\
n & \text{if } x = 6
\end{cases}
\][/tex]
First, we find the limit of [tex]\( \frac{x^2 - 36}{x - 6} \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( 6 \)[/tex].
### Step 1: Simplify the expression
Notice that [tex]\( x^2 - 36 \)[/tex] can be factored using the difference of squares:
[tex]\[
x^2 - 36 = (x + 6)(x - 6)
\][/tex]
Thus, the expression can be rewritten as:
[tex]\[
\frac{x^2 - 36}{x - 6} = \frac{(x + 6)(x - 6)}{x - 6}
\][/tex]
For [tex]\( x \neq 6 \)[/tex], the [tex]\( (x - 6) \)[/tex] in the numerator and denominator cancel each other out:
[tex]\[
\frac{(x + 6)(x - 6)}{x - 6} = x + 6
\][/tex]
Therefore, for [tex]\( x \neq 6 \)[/tex]:
[tex]\[
f(x) = x + 6
\][/tex]
### Step 2: Calculate the limit as [tex]\( x \)[/tex] approaches [tex]\( 6 \)[/tex]
Now we find the limit of [tex]\( f(x) = x + 6 \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( 6 \)[/tex]:
[tex]\[
\lim_{x \to 6} (x + 6) = 6 + 6 = 12
\][/tex]
### Step 3: Ensure continuity at [tex]\( x = 6 \)[/tex]
For [tex]\( f(x) \)[/tex] to be continuous at [tex]\( x = 6 \)[/tex], the following must hold:
[tex]\[
\lim_{x \to 6} f(x) = f(6)
\][/tex]
We have calculated the limit as [tex]\( x \to 6 \)[/tex]:
[tex]\[
\lim_{x \to 6} f(x) = 12
\][/tex]
So, for the function to be continuous at [tex]\( x = 6 \)[/tex], we need:
[tex]\[
f(6) = n = 12
\][/tex]
Therefore, the value of [tex]\( n \)[/tex] that makes [tex]\( f(x) \)[/tex] continuous at [tex]\( x = 6 \)[/tex] is:
[tex]\[
n = 12
\][/tex]
