We are asked to find the domain of the function $f(x) = \sqrt{x^2 + x - 6}$.

1. Problem Description

We are asked to find the domain of the function $f(x) = \sqrt{x^2 + x - 6}$.

2. Solution Steps

The domain of $f(x)$ consists of all real numbers $x$ such that $x^2 + x - 6 \ge 0$.

First, we factor the quadratic expression:

$x^2 + x - 6 = (x+3)(x-2)$.

Now we want to find the values of $x$ for which $(x+3)(x-2) \ge 0$.

We analyze the sign of $(x+3)(x-2)$ by considering the intervals determined by the roots $x=-3$ and $x=2$.

- If $x < -3$, then $x+3 < 0$ and $x-2 < 0$, so $(x+3)(x-2) > 0$.

- If $-3 < x < 2$, then $x+3 > 0$ and $x-2 < 0$, so $(x+3)(x-2) < 0$.

- If $x > 2$, then $x+3 > 0$ and $x-2 > 0$, so $(x+3)(x-2) > 0$.

- If $x=-3$, then $(x+3)(x-2) = 0$.

- If $x=2$, then $(x+3)(x-2) = 0$.

Therefore, the inequality $(x+3)(x-2) \ge 0$ is satisfied when $x \le -3$ or $x \ge 2$.

The domain is $(-\infty, -3] \cup [2, +\infty)$.

3. Final Answer

C. $[-\infty, -3] \cup [2, +\infty]$