To find the domain of the function $f(x) = \sqrt{x^2 - 3 + 2}$, follow these steps:

1. Simplify the expression inside the square root: $f(x) = \sqrt{x^2 - 3 + 2} = \sqrt{x^2 - 1}$

2. Determine the condition for the square root to be defined: The square root function is only defined for non-negative values, so the expression inside the square root must be greater than or equal to zero: $x^2 - 1 \geq 0$

3. Solve the inequality $x^2 - 1 \geq 0$: $x^2 \geq 1$ This inequality holds when: $x \leq -1 \quad \text{or} \quad x \geq 1$

4. Express the domain: The domain of the function is the set of all $x$ values that satisfy the inequality. In interval notation, this is: $(-\infty, -1] \cup [1, \infty)$

So, the domain of the function $f(x) = \sqrt{x^2 - 1}$ is $(-\infty, -1] \cup [1, \infty)$.

Would you like more details or have any questions? Here are five related questions you might explore:

1. What is the domain of $\sqrt{x^2 + 4}$?
2. How would the domain change if the function were $\sqrt{-x^2 + 4}$?
3. What is the range of the function $f(x) = \sqrt{x^2 - 1}$?
4. How do you find the domain of a rational function?
5. What happens to the domain if you add or subtract a constant inside the square root?

Tip: When finding the domain of functions involving square roots, always set the expression inside the square root to be greater than or equal to zero.