### 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 figure out what values of x make the function well-defined and real-valued.
I know that I cannot take the square root of a negative number and get a real number, so I need to avoid that.
That means that the expression inside the square root sign, $\sqrt{x^2-16}-3$, must be non-negative.
To simplify this condition, I can add 3 to both sides and then square both sides, since squaring preserves the order of non-negative numbers.
This gives me $x^2-16 \geq 9$, or $x^2 \geq 25$.
To solve this inequality, I can take the square root of both sides again, but I have to remember to include both the positive and negative roots.
This gives me $|x| \geq 5$, or $x \leq -5$ or $x \geq 5$.
This is the domain of the function, the set of all x-values that make the function real-valued.