### Instruction: Compute the domain of the real-valued function \[f(x)=\sqrt{1-\sqrt{2-\sqrt{x}}}.\].### Response: To find the domain of a function, I need to identify the values of x that make the function well-defined and real.
The first thing I notice is that the function involves square roots, which means I need to avoid negative radicands (the expressions under the square root symbols).
Working from the inside out, I see that the innermost square root is \(\sqrt{x}\), which means I need to have x \(\geq 0\).
Next, I look at the second square root, which is \(\sqrt{2-\sqrt{x}}\).
I need this expression to be nonnegative as well, so I set up an inequality: \[2-\sqrt{x} \geq 0.\]
Solving for x, I get \[\sqrt{x} \leq 2\] and then \[x \leq 4.\]
So far, I have two conditions for x: x \(\geq 0\) and x \(\leq 4\).
Combining them, I get \[0 \leq x \leq 4.\]
Finally, I look at the outermost square root, which is \(\sqrt{1-\sqrt{2-\sqrt{x}}}\).
I need this expression to be nonnegative as well, so I set up another inequality: \[1-\sqrt{2-\sqrt{x}} \geq 0.\]
Solving for x, I get \[\sqrt{2-\sqrt{x}} \leq 1\] and then \[2-\sqrt{x} \leq 1\] and then \[\sqrt{x} \geq 1\] and then \[x \geq 1.\]
This is another condition for x: x \(\geq 1\).
Combining it with the previous one, I get \[1 \leq x \leq 4.\]
This is the final answer for the domain of the function.
It is the set of all x values that satisfy all three inequalities.