Sure, let's analyze the function [tex]\( f(x) = \frac{x^2 - 36}{x^2 - 9} \)[/tex] to determine its domain.
1. Simplify the Function:
We begin by factoring both the numerator and the denominator.
- The numerator [tex]\( x^2 - 36 \)[/tex] can be factored as a difference of squares:
[tex]\[
x^2 - 36 = (x - 6)(x + 6)
\][/tex]
- The denominator [tex]\( x^2 - 9 \)[/tex] can also be factored as a difference of squares:
[tex]\[
x^2 - 9 = (x - 3)(x + 3)
\][/tex]
So, the function can be simplified to:
[tex]\[
f(x) = \frac{(x - 6)(x + 6)}{(x - 3)(x + 3)}
\][/tex]
2. Identify Points of Discontinuity:
We need to find the values of [tex]\(x\)[/tex] that make the denominator equal to zero, since the function is undefined at those points. To do this, we set the denominator equal to zero and solve for [tex]\(x\)[/tex]:
[tex]\[
(x - 3)(x + 3) = 0
\][/tex]
Solving for [tex]\(x\)[/tex], we get:
[tex]\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\][/tex]
[tex]\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\][/tex]
These are the critical points where the function is not defined because they make the denominator zero.
3. Determine the Domain:
The domain of the function is all real numbers except the critical points [tex]\(x = 3\)[/tex] and [tex]\(x = -3\)[/tex]. In mathematical notation, we can express this as:
[tex]\[
\mathbb{R} \setminus \{-3, 3\}
\][/tex]
or using interval notation:
[tex]\[
(-\infty, -3) \cup (-3, 3) \cup (3, \infty)
\][/tex]
So, the domain of the function [tex]\( f(x) = \frac{x^2 - 36}{x^2 - 9} \)[/tex] is:
[tex]\[
(-\infty, -3) \cup (-3, 3) \cup (3, \infty)
\][/tex]
And the points where the function is not defined are [tex]\( x = -3 \)[/tex] and [tex]\( x = 3 \)[/tex].
