Question

24 find the domain of the function
$g(x) = \frac{3x}{x^2 - 9}$
a. $(-\infty, 0) \cup (0, \infty)$
b. $(-\infty, \infty)$
c. $(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$
d. $(9, \infty)$

Explanation:

Step1: Identify the restriction

For a rational function \( g(x)=\frac{3x}{x^2 - 9} \), the denominator cannot be zero. So we set \( x^2-9
eq0 \).

Step2: Solve the equation \( x^2 - 9=0 \)

Factor the left side: \( (x - 3)(x + 3)=0 \).
Set each factor equal to zero: \( x - 3 = 0 \) gives \( x = 3 \), and \( x + 3 = 0 \) gives \( x=-3 \).

Step3: Determine the domain

The domain is all real numbers except \( x = 3 \) and \( x=-3 \). In interval notation, this is \( (-\infty, - 3)\cup(-3,3)\cup(3,\infty) \).

Answer:

C. \( (-\infty, - 3)\cup(-3,3)\cup(3,\infty) \)