To find the domain of the function $f(x) = \sqrt{-3x - 9}$, we need to ensure that the expression inside the square root is non-negative, since the square root of a negative number is not defined in the set of real numbers.

Step 1: Set the expression inside the square root greater than or equal to zero.

$-3x - 9 \geq 0$

Step 2: Solve for $x$.

First, add 9 to both sides: $-3x \geq 9$

Next, divide both sides by -3. Note that when dividing by a negative number, the inequality sign flips: $x \leq -3$

Step 3: Write the domain in interval notation.

Since $x$ must be less than or equal to $-3$, the domain is: $(-\infty, -3]$

So, the domain of $f(x) = \sqrt{-3x - 9}$ is $(-\infty, -3]$.

Would you like more details or have any questions?

Here are 5 related questions:

1. What is the domain of $f(x) = \sqrt{4x + 12}$?
2. How does the domain change for a function $f(x) = \sqrt{ax + b}$?
3. What is the domain of $f(x) = \frac{1}{\sqrt{-2x - 8}}$?
4. What happens to the domain when you have a cubic root instead of a square root?
5. How do you solve an inequality like $-5x + 7 \geq 0$?

Tip: When working with square roots, always ensure the radicand (the expression inside the square root) is non-negative for real-valued functions.