The question asks to determine the domain of the function $f(x) = \sqrt{3x - 9}$. To find the domain, we must ensure that the expression under the square root is non-negative, as the square root function is only defined for non-negative numbers (i.e., the radicand must be greater than or equal to zero).

So, we set up the inequality:

$3x - 9 \geq 0$

Now, solve for $x$:

$3x \geq 9$

$x \geq \frac{9}{3} = 3$

Therefore, the domain of the function is:

$x \geq 3$

The domain in interval notation is:

$[3, \infty)$

Would you like more details or have any further questions on this?

Here are five related questions for further exploration:

1. How would the domain change if the function were $f(x) = \sqrt{5x - 2}$?
2. What is the range of $f(x) = \sqrt{3x - 9}$?
3. How would you graph $f(x) = \sqrt{3x - 9}$?
4. What happens to the domain if a negative square root is involved, e.g., $f(x) = -\sqrt{3x - 9}$?
5. How do you determine the domain of a rational function involving a square root?

Tip: For any function involving a square root, always check that the expression inside the root is non-negative to find the domain.