Find the domain of the function $$f(x) = \sqrt{x-2 \sqrt[3]{x-3}}.$$ Express your answer in interval notation.
In order for $f(x)$ to be defined, we must have $x-2\sqrt[3]{x-3} \ge 0.$ To factor this expression, we substitute the substitution $u = \sqrt[3]{x-3}.$ Then $x = u^3 + 3,$ so \begin{align*}
x - 2\sqrt[3]{x-3} &= (u^3 + 3) - 2u \\
&= u^3 - 2u + 3 \\
&= u(u^2 - 2) + 3.
\end{align*}Note that $x-2\sqrt[3]{x-3}$ is an increasing function of $u.$ We can draw a sign chart, which looks like this:
\[
\begin{array}{c|c|c|c}
& u < -\sqrt[3]{2} & -\sqrt[3]{2} < u < 0 & u > 0 \\\hline
u^2 - 2 & - & + & + \\
u(u^2 - 2) & - & - & + \\
u(u^2 - 2) + 3 & - & - & +
\end{array}
\]The expression $x - 2\sqrt[3]{x-3}$ is positive for $-\sqrt[3]{2} < u < 0,$ which corresponds $u$ values $\sqrt[3]{2} < x - 3 < 0.$ Therefore, $f(x)$ is defined on $\boxed{(\sqrt[3]{2} + 3,3) \cup [3,\infty)}.$
To find the domain of a function, we need to determine the set of all possible input values for which the function is defined. In the case of the given function, we need to ensure that the expression inside the square root is non-negative.
The expression inside the square root is $x - 2 \sqrt[3]{x-3}$. For this expression to be non-negative, we need both $x$ and $x-3$ to be greater than or equal to 0.
First, let's consider the condition $x \geq 0$. This ensures that $x - 2 \sqrt[3]{x-3}$ is greater than or equal to 0 because both $x$ and $2 \sqrt[3]{x-3}$ are non-negative.
Second, let's consider the condition $x-3 \geq 0$. This ensures that $\sqrt[3]{x-3}$ is defined and non-negative because the cube root of a non-negative number is always real. Therefore, we have $x \geq 3$.
Combining these conditions, we find that the domain of the function $f(x) = \sqrt{x - 2 \sqrt[3]{x-3}}$ is $[3, \infty)$. This interval notation means that all real numbers $x$ such that $x$ is greater than or equal to 3 are in the domain of the function.
To find the domain of the function, we need to determine for which values of $x$ the function is defined.
Looking at the function, we notice that there are two main considerations:
1. The expression inside the square root must be non-negative.
2. The expression inside the cubic root must be defined.
Let's consider these conditions separately:
1. The expression inside the square root, $x - 2 \sqrt[3]{x-3}$, must be non-negative. This means that $$x - 2 \sqrt[3]{x-3} \geq 0.$$ To simplify this inequality, we can isolate the cubic root term and raise both sides to the power of $3$:
$$(x-3) \geq (2 \sqrt[3]{x-3})^3 = 8(x-3).$$
Expanding the RHS, we have $8(x-3) = 8x-24$. Simplifying the inequality leads to:
$$x - 3 \geq 8x - 24.$$
Solving this inequality gives $x \leq \frac{21}{7} = 3$.
2. The expression inside the cubic root, $x-3$, should be defined. This means that $x-3$ must be greater than or equal to $0$, which gives $x \geq 3$.
Combining these two conditions, we find that the domain of the function is $x \in \left[ 3, \frac{21}{7} \right]$ (interval notation).