### Instruction: Find the domain of the function
\[f(x) = \sqrt{1 - \sqrt{2 - \sqrt{3 - x}}}.\].### Response: To find the domain of a function, I need to think about what values of x make the function undefined or invalid.
In this case, the function involves square roots, so I need to avoid negative values under any of the radicals.
I can start from the innermost radical and work my way outward.
The expression $3 - x$ must be nonnegative, so $3 - x \geq 0$, which implies $x \leq 3$.
Next, the expression $2 - \sqrt{3 - x}$ must also be nonnegative, so $2 - \sqrt{3 - x} \geq 0$, which implies $\sqrt{3 - x} \leq 2$.
Squaring both sides, I get $3 - x \leq 4$, which implies $x \geq -1$.
So far, I have $-1 \leq x \leq 3$ as a condition for the domain.
Finally, the expression $1 - \sqrt{2 - \sqrt{3 - x}}$ must also be nonnegative, so $1 - \sqrt{2 - \sqrt{3 - x}} \geq 0$, which implies $\sqrt{2 - \sqrt{3 - x}} \leq 1$.
Squaring both sides, I get $2 - \sqrt{3 - x} \leq 1$, which implies $\sqrt{3 - x} \geq 1$.
Squaring both sides again, I get $3 - x \geq 1$, which implies $x \leq 2$.