The given function is $f(x) = \sqrt{x - 2}$.

To find the domain of the function:

1. For a square root function, the expression under the square root must be non-negative, since the square root of a negative number is not defined in the set of real numbers.

2. So, we need: $x - 2 \geq 0$

3. Solving for $x$: $x \geq 2$

Therefore, the domain of the function $f(x) = \sqrt{x - 2}$ is $x \geq 2$.

This corresponds to the answer choice that indicates $x \geq 2$.

Would you like more details on how I derived this? Here are five related questions for practice:

1. What is the domain of the function $g(x) = \sqrt{3x - 6}$?
2. How do you determine the domain of a square root function in general?
3. What is the domain of $h(x) = \sqrt{-x + 5}$?
4. If the function is $f(x) = \sqrt{x^2 - 4x + 3}$, how do you find its domain?
5. What is the range of $f(x) = \sqrt{x - 2}$?

Tip: Always remember, for square root functions, the expression inside the root must be non-negative to ensure real number results!