### 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 consider what values of x make the function well-defined and real-valued.
One thing that I notice is that the function involves taking square roots, which means I need to avoid negative numbers under the radicals.
So, I can start by setting up some inequalities to ensure that each expression inside a square root is nonnegative.
The innermost expression is 3 - x, so I need 3 - x >= 0, which implies x <= 3.
The next expression is 2 - the square root of 3 - x, so I need 2 - sqrt(3 - x) >= 0, which implies sqrt(3 - x) <= 2.
Squaring both sides, I get 3 - x <= 4, which implies x >= -1.
This is a weaker condition than x <= 3, so I don't need to change that.
The outermost expression is 1 - the square root of 2 - the square root of 3 - x, so I need 1 - sqrt(2 - sqrt(3 - x)) >= 0, which implies sqrt(2 - sqrt(3 - x)) <= 1.
Squaring both sides, I get 2 - sqrt(3 - x) <= 1, which implies sqrt(3 - x) >= 1.
Squaring both sides again, I get 3 - x >= 1, which implies x <= 2.
This is a stronger condition than x <= 3, so I need to update that.
So, putting it all together, I have -1 <= x <= 2 as the domain of the function.