To determine the domain of the function $f(x) = \sqrt{x^2 - 49}$, we need to analyze the expression inside the square root.

Step-by-Step Solution

1. Identify the constraint for the square root: Since the square root function is only defined for non-negative values, we require: $x^2 - 49 \geq 0$

2. Solve the inequality:

   • Rewrite the inequality as: $x^2 \geq 49$
   • Taking the square root of both sides, we get: $x \leq -7 \quad \text{or} \quad x \geq 7$

3. Express the domain in interval notation:

   • The values of $x$ for which $f(x)$ is defined are $x \leq -7$ and $x \geq 7$.
   • In interval notation, this is: $(-\infty, -7] \cup [7, \infty)$

Final Answer

The domain of $f(x) = \sqrt{x^2 - 49}$ is: $(-\infty, -7] \cup [7, \infty)$

Would you like further details on any part of this solution or have any questions?

Here are some related questions to deepen your understanding:

1. How would the domain change if the function was $f(x) = \sqrt{x^2 - 25}$?
2. What would be the domain of $f(x) = \sqrt{49 - x^2}$?
3. How do you determine the domain for functions involving higher roots (e.g., cube roots)?
4. If we change the function to $f(x) = \frac{1}{\sqrt{x^2 - 49}}$, what would be the domain?
5. How would the answer change if the function was $f(x) = x^2 - 49$ without the square root?

Tip: When finding the domain of a square root function, ensure the expression under the root is non-negative for real-valued results.