Question

1. f(x) = \frac{8}{x - 3} 14. f(x) = \frac{12}{x} + 9

Explanation:

Assuming we need to find the domain of these functions (a common calculus - related task).

Step1: Recall the domain rule for rational functions

For a rational function \(y = \frac{f(x)}{g(x)}\), the denominator \(g(x)
eq0\) because division by zero is undefined.

Step2: Find the domain of \(f(x)=\frac{8}{x - 3}\)

Set the denominator equal to zero: \(x-3 = 0\).
Solve for \(x\): \(x=3\).
So the domain of \(f(x)=\frac{8}{x - 3}\) is all real numbers except \(x = 3\), which can be written in interval notation as \((-\infty,3)\cup(3,\infty)\).

Step3: Find the domain of \(f(x)=\frac{12}{x}+9\)

Set the denominator equal to zero: \(x = 0\).
So the domain of \(f(x)=\frac{12}{x}+9\) is all real numbers except \(x=0\), which can be written in interval notation as \((-\infty,0)\cup(0,\infty)\).

Answer:

For \(f(x)=\frac{8}{x - 3}\), domain is \((-\infty,3)\cup(3,\infty)\); for \(f(x)=\frac{12}{x}+9\), domain is \((-\infty,0)\cup(0,\infty)\)