Find the domain of the function \[f(x)=\sqrt{1-\sqrt{2-\sqrt{3 - \sqrt{x}}}}.\]
Since all the square roots in the expression must be real, we must have \[3 - \sqrt{x} \ge 0,\]so $\sqrt{x} \le 3,$ or $x \le 9.$ Also, we must the expression inside the square root to be nonnegative, so \[2 - \sqrt{3 - \sqrt{x}} \ge 0,\]which turns into $\sqrt{3 - \sqrt{x}} \le 2.$ Squaring, we get $3 - \sqrt{x} \le 4,$ so $\sqrt{x} \ge -1.$ Since we must also have $\sqrt{x} \le 3,$ we must have $\sqrt{x} \le 3$ and $\sqrt{x} \ge 0.$ Hence, $0 \le x \le 9.$ The domain of $f(x)$ is $\boxed{[0,9]}.$
To find the domain of the function \[f(x)=\sqrt{1-\sqrt{2-\sqrt{3 - \sqrt{x}}}},\] we need to determine the values of \(x\) for which the function is defined.
Starting from the innermost radical, we have \(\sqrt{x}\). For this expression to be defined, \(x\) must be greater than or equal to zero (i.e., \(x \geq 0\)).
Next, we have \(3 - \sqrt{x}\). To guarantee that this expression is defined, we need \(x \leq 3\) since \(\sqrt{x}\) must be nonnegative.
Moving outward, we have \(2 - \sqrt{3-\sqrt{x}}\). This expression requires \(3 - \sqrt{x} \geq 2\). To simplify, let's add \(\sqrt{x}\) to both sides of the inequality to get \(\sqrt{x} \leq 1\). By squaring both sides, we have \(x \leq 1\).
Finally, we have \(1 - \sqrt{2-\sqrt{3 - \sqrt{x}}}\). This expression is defined as long as \(2 - \sqrt{3 - \sqrt{x}} \geq 0\). Again, we'll add \(\sqrt{3 - \sqrt{x}}\) to both sides of the inequality: \(\sqrt{3 - \sqrt{x}} \leq 2\). By squaring both sides, we obtain \(3 - \sqrt{x} \leq 4\), which simplifies to \(\sqrt{x} \geq -1\). Since \(\sqrt{x}\) is always nonnegative, we have \(x \geq 0\).
Combining all the conditions, the domain of the function is \(\boxed{0 \leq x \leq 1}\).
To find the domain of the function \[f(x)=\sqrt{1-\sqrt{2-\sqrt{3 - \sqrt{x}}}},\] we need to consider the restriction on its input value.
The square root function is defined only for non-negative numbers since the square root of a negative number is not a real number. Therefore, for our function to be defined, every square root within it must yield a non-negative value.
Let's start from the innermost square root: \[\sqrt{x}.\] To ensure that this expression is defined, we need \(x\) to be non-negative or \(x \geq 0\).
Next, we have \[\sqrt{3 - \sqrt{x}}.\] For this expression to be defined, the value inside the square root must be non-negative. Therefore, we need \(3 - \sqrt{x} \geq 0\). Rearranging this inequality, we find \(\sqrt{x} \leq 3\), which implies \(x \leq 9\). Furthermore, since \(\sqrt{x}\) must be non-negative, we also have \(x \geq 0\).
Moving on, we have \[\sqrt{2-\sqrt{3 - \sqrt{x}}}.\] For this expression to be defined, the value inside the square root must be non-negative. Therefore, we need \(2 - \sqrt{3 - \sqrt{x}} \geq 0\). Rearranging this inequality, we find \(\sqrt{3 - \sqrt{x}} \leq 2\), which implies \(\sqrt{x} \geq 1\). Furthermore, considering the previous restrictions, we also have \(x \geq 1\).
Finally, we have \[f(x) = \sqrt{1-\sqrt{2-\sqrt{3 - \sqrt{x}}}}.\] For this expression to be defined, the value inside the outermost square root must be non-negative. Therefore, we need \(1 - \sqrt{2 - \sqrt{3 - \sqrt{x}}} \geq 0\). Rearranging this inequality, we find \(\sqrt{2 - \sqrt{3 - \sqrt{x}}} \leq 1\). Since the square root must yield a non-negative value, we have \(\sqrt{3 - \sqrt{x}} \leq 1\), which implies \(\sqrt{x} \geq 2\). Considering the previous restrictions, we also have \(x \geq 4\).
Combining all the restrictions, the domain of the function \(f(x)\) is \(\boxed{[4, 9]}\), or equivalently, \(4 \leq x \leq 9\).