### 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 look for any values of x that would make the function undefined or nonsensical.
One thing that could cause trouble is taking the square root of a negative number, since that would give an imaginary result.
So I want to avoid any x that would make $\sqrt{x^2-16}$ or $\sqrt{\sqrt{x^2-16}-3}$ negative.
To ensure that $\sqrt{x^2-16}$ is not negative, I need to have $x^2-16\geq 0$, or equivalently, $x^2\geq 16$.
This means that $x\leq -4$ or $x\geq 4$, since squaring preserves the inequality for positive numbers and reverses it for negative numbers.
But this is not enough, because I also need to ensure that $\sqrt{\sqrt{x^2-16}-3}$ is not negative, which means that $\sqrt{x^2-16}-3\geq 0$, or equivalently, $\sqrt{x^2-16}\geq 3$.
Squaring both sides, I get $x^2-16\geq 9$, or $x^2\geq 25$, or $x\leq -5$ or $x\geq 5$.
So this is a stricter condition than the previous one, and it narrows down the possible values of x even further.
Therefore, the domain of the function is the set of all x such that $x\leq -5$ or $x\geq 5$.