Let's work through the problem step-by-step.
### Part (a)
First, let's find [tex]\( f(x) \)[/tex] for [tex]\( x = -2 \)[/tex] and [tex]\( x = p \)[/tex], if possible.
1. For [tex]\( x = -2 \)[/tex]:
Plugging [tex]\( x = -2 \)[/tex] into the function [tex]\( f(x) = \frac{3x - 18}{x + 6} \)[/tex],
[tex]\[
f(-2) = \frac{3(-2) - 18}{-2 + 6} = \frac{-6 - 18}{4} = \frac{-24}{4} = -6
\][/tex]
Therefore, [tex]\( f(-2) = -6 \)[/tex].
2. For [tex]\( x = p \)[/tex]:
Plugging [tex]\( x = p \)[/tex] into the function [tex]\( f(x) = \frac{3x - 18}{x + 6} \)[/tex],
[tex]\[
f(p) = \frac{3p - 18}{p + 6}
\][/tex]
Hence, [tex]\( f(p) = \frac{3p - 18}{p + 6} \)[/tex].
### Part (b)
Next, let's find the domain of the function [tex]\( f \)[/tex].
The function [tex]\( f(x) = \frac{3x - 18}{x + 6} \)[/tex] will be defined for all real numbers except where the denominator is zero, as division by zero is undefined.
1. Set the denominator equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
x + 6 = 0 \\
x = -6
\][/tex]
Thus, the function is undefined at [tex]\( x = -6 \)[/tex].
2. Therefore, the domain of the function [tex]\( f(x) \)[/tex] is all real numbers except [tex]\( x = -6 \)[/tex].
The domain can be written in interval notation as:
[tex]\[
(-\infty, -6) \cup (-6, \infty)
\][/tex]
### Summary
- For [tex]\( x = -2 \)[/tex], [tex]\( f(x) = -6 \)[/tex].
- For [tex]\( x = p \)[/tex], [tex]\( f(x) = \frac{3p - 18}{p + 6} \)[/tex].
- The domain of [tex]\( f \)[/tex] is all real numbers except [tex]\( x = -6 \)[/tex], represented as [tex]\( (-\infty, -6) \cup (-6, \infty) \)[/tex]. The restriction that makes the function undefined is [tex]\( x = -6 \)[/tex].
