### 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 I notice is that the function has nested square roots, which means I have to be careful about negative values inside any of the radicals.
To avoid negative values, I need to make sure that each expression under a square root is greater than or equal to zero.
So I have to impose some inequalities.
Starting from the innermost radical, I need to have \[3 - x \ge 0.\]
This means that \[x \le 3.\]
Next, I need to have \[2 - \sqrt{3 - x} \ge 0.\]
This means that \[\sqrt{3 - x} \le 2.\]
Squaring both sides, I get \[3 - x \le 4.\]
This means that \[x \ge -1.\]
Finally, I need to have \[1 - \sqrt{2 - \sqrt{3 - x}} \ge 0.\]
This means that \[\sqrt{2 - \sqrt{3 - x}} \le 1.\]
Squaring both sides, I get \[2 - \sqrt{3 - x} \le 1.\]
This means that \[\sqrt{3 - x} \ge 1.\]
Squaring both sides again, I get \[3 - x \ge 1.\]
This means that \[x \le 2.\]
So I have three inequalities that x must satisfy: \[x \le 3, \quad x \ge -1, \quad x \le 2.\]
The common values of x that satisfy all three inequalities are those in the closed interval \[-1, 2\].
This is the domain of the function.