### Instruction: Find the domain of the function $f(x)=\sqrt{\sqrt{x^2-16}-3}$.### Response: To find the domain of a function, I need to determine what values of x make the function well-defined and real.
I know that the square root of a negative number is not real, so I need to avoid that situation in both the inner and outer square roots.
For the inner square root, I need $\sqrt{x^2-16}$ to be non-negative, or $x^2-16\geq 0$.
To solve this inequality, I can factor the left side as $(x-4)(x+4)$ and use a sign chart or a number line to find where the product is positive or zero.
I see that the critical points are $x=-4$ and $x=4$, and the sign of the product changes at each point.
So the solution is $x\leq -4$ or $x\geq 4$.
However, this is not the final answer, because I still need to consider the outer square root.
For that, I need $\sqrt{x^2-16}-3$ to be non-negative as well, or $\sqrt{x^2-16}\geq 3$.
To solve this inequality, I can isolate the square root and square both sides, remembering to check for extraneous solutions later.
I get $x^2-16\geq 9$, or $x^2\geq 25$, or $|x|\geq 5$.
This means that $x\leq -5$ or $x\geq 5$, which is a stricter condition than the previous one.
So this is the final domain of the function.