1. **State the problem:** Find the domain of the function $$f(x) = \sqrt{x^2 - 9}$$. 2. **Recall the domain rule for square roots:** The expression inside the square root must be greater than or equal to zero because the square root of a negative number is not a real number. 3. **Set the radicand $x^2 - 9$ to be non-negative:** $$x^2 - 9 \geq 0$$ 4. **Solve the inequality:** $$x^2 \geq 9$$ 5. **Rewrite as:** $$x^2 - 3^2 \geq 0$$ 6. **Factor using difference of squares:** $$ (x - 3)(x + 3) \geq 0 $$ 7. **Determine intervals where the product is non-negative:** - The product is zero at $x = 3$ and $x = -3$. - Test intervals: - For $x < -3$, choose $x = -4$: $(-4 - 3)(-4 + 3) = (-7)(-1) = 7 > 0$ - For $-3 < x < 3$, choose $x = 0$: $(0 - 3)(0 + 3) = (-3)(3) = -9 < 0$ - For $x > 3$, choose $x = 4$: $(4 - 3)(4 + 3) = (1)(7) = 7 > 0$ 8. **Conclusion:** The domain is all $x$ such that $x \leq -3$ or $x \geq 3$. **Final answer:** $$\boxed{(-\infty, -3] \cup [3, \infty)}$$