1. **Find the domain of the function** $f(x) = \frac{1}{x^2 - 9}$. 2. The domain of a function is the set of all $x$ values for which the function is defined. 3. For $f(x) = \frac{1}{x^2 - 9}$, the denominator cannot be zero because division by zero is undefined. 4. Set the denominator equal to zero and solve: $$x^2 - 9 = 0$$ $$x^2 = 9$$ $$x = \pm 3$$ 5. Therefore, $x = 3$ and $x = -3$ are excluded from the domain. 6. The domain is all real numbers except $x = 3$ and $x = -3$. 7. In interval notation, the domain is: $$(-\infty, -3) \cup (-3, 3) \cup (3, \infty)$$