Find the domain of the function $$f(x) = \sqrt{x-2 \sqrt[3]{x-3}}.$$ Express your answer in interval notation.

Answer 1

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 $\boxed{(\sqrt[3]{2} + 3,3) \cup [3,\infty)}.$