### Instruction: Compute the domain of the real-valued function $$f(x)=\sqrt{3-\sqrt{5-\sqrt{x}}}.$$.### Response: To find the domain of a function, I need to think about what values of x make the function well-defined and real-valued.
One thing that I notice is that the function involves square roots, and I know that I can only take the square root of a non-negative number if I want to get a real number as a result.
So, for each square root in the function, I need to make sure that the expression inside it is greater than or equal to zero.
Starting from the innermost square root, I need to have $\sqrt{x}\geq 0$, which means that $x\geq 0$.
Then, moving to the next square root, I need to have $5-\sqrt{x}\geq 0$, which means that $\sqrt{x}\leq 5$, and since $x\geq 0$, I can square both sides and get $x\leq 25$.
Finally, moving to the outermost square root, I need to have $3-\sqrt{5-\sqrt{x}}\geq 0$, which means that $\sqrt{5-\sqrt{x}}\leq 3$, and since $5-\sqrt{x}\geq 0$, I can square both sides and get $5-\sqrt{x}\leq 9$, or equivalently, $\sqrt{x}\geq -4$.
But, I already know that $\sqrt{x}\geq 0$, so this inequality does not add any new restriction on x.
Therefore, the domain of the function is the set of all x that satisfy $0\leq x\leq 25$.