The function given is:

$f(x) = \sqrt{x^2 - x - 6}$

To analyze this function, let's consider the domain and any potential issues with the square root.

Step 1: Domain of the Function

The expression inside the square root, $x^2 - x - 6$, must be non-negative for the square root to be real. So, we set up the inequality:

$x^2 - x - 6 \geq 0$

Step 2: Solve the Quadratic Inequality

First, solve the quadratic equation:

$x^2 - x - 6 = 0$

This can be factored as:

$(x - 3)(x + 2) = 0$

So, the roots are $x = 3$ and $x = -2$.

Now, we test the intervals determined by these roots:

1. For $x < -2$: Choose $x = -3$, $(-3)^2 - (-3) - 6 = 9 + 3 - 6 = 6 \geq 0$

2. For $-2 < x < 3$: Choose $x = 0$, $0^2 - 0 - 6 = -6 \, (\text{which is negative})$

3. For $x > 3$: Choose $x = 4$, $4^2 - 4 - 6 = 16 - 4 - 6 = 6 \geq 0$

Thus, the quadratic is non-negative when $x \leq -2$ or $x \geq 3$.

Step 3: Conclusion

The domain of $f(x)$ is:

$(-\infty, -2] \cup [3, \infty)$

For these values of $x$, the function $f(x)$ is defined and real.

Would you like to explore further details, such as the range, or do you have any specific questions?

---

1. What is the range of the function $f(x)$?
2. How does the function behave as $x$ approaches the boundaries of its domain?
3. Can the function $f(x)$ ever be zero?
4. How would the graph of $f(x)$ look over its domain?
5. What happens if we shift the entire function by adding a constant $c$?

Tip: Always check the domain of a function when dealing with square roots to ensure you're only considering real-valued outputs.