### 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 identify the values of x that make the function well-defined and real.
One condition is that the expression under any square root must be nonnegative, since otherwise the function would be undefined or complex.
Starting from the innermost square root, I see that $x$ must be at least 0, so $x \geq 0$.
Next, I look at the expression under the second square root, which is $5-\sqrt{x}$.
This must also be nonnegative, so $5-\sqrt{x} \geq 0$.
Solving for x, I get $\sqrt{x} \leq 5$, or $x \leq 25$.
Finally, I look at the expression under the outermost square root, which is $3-\sqrt{5-\sqrt{x}}$.
This must also be nonnegative, so $3-\sqrt{5-\sqrt{x}} \geq 0$.
Solving for x, I get $\sqrt{5-\sqrt{x}} \leq 3$, or $5-\sqrt{x} \leq 9$, or $\sqrt{x} \geq -4$.
Since $x$ is nonnegative, the last inequality is always true, so it does not affect the domain.
Therefore, the domain of the function is the intersection of the three inequalities I found, which is $0 \leq x \leq 25$.