We are given the function $f(x) = \sqrt{x^2 - 9}$ and need to determine its domain, or the set of possible values for $x$.

To ensure the function is defined, the expression under the square root must be non-negative:

$x^2 - 9 \geq 0$

This simplifies to:

$x^2 \geq 9$

Next, solving this inequality:

$x \geq 3 \quad \text{or} \quad x \leq -3$

Thus, the domain of the function $f(x)$ is:

$x \in (-\infty, -3] \cup [3, \infty)$

This corresponds to option e in the provided choices:

$\{ x \,|\, x \leq -9 \, \text{or} \, x \geq 3, x \in \mathbb{R} \}$

Would you like further explanation or have any questions?

Here are 5 related questions to expand on this topic:

1. How do we determine the domain of a function involving square roots?
2. Why is the expression under the square root required to be non-negative?
3. How do you solve inequalities involving squares like $x^2 \geq 9$?
4. What happens to the domain of the function if the square root were replaced by a different function, like $f(x) = x^2 - 9$?
5. How does the range of this function relate to its domain?

Tip: When solving inequalities with squares, always check both positive and negative solutions.