Answer
$ \left (\frac{f}{g} \right )(x)=\frac{x\sqrt [3]{x^2}}{2}$.
$(0,\infty)$.
Work Step by Step
The given functions are
$f(x)=\sqrt[3] {2x^6}$ and $g(x)=\sqrt [3]{16x}$
$\Rightarrow \left (\frac{f}{g} \right )(x)=\frac{f(x)}{g(x)}$
Substitute both functions.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\frac{\sqrt[3] {2x^6}}{\sqrt [3]{16x}}$
Divide the radicands and retain the common index.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\sqrt[3] {\frac{2x^6}{2^4x}}$
Divide factors in the radicand. Subtract exponents on common bases.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\sqrt[3] {2^{1-4}x^{6-1}}$
Simplify.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\sqrt[3] {2^{-3}x^{5}}$
$\Rightarrow \left (\frac{f}{g} \right )(x)=2^{-1}x\sqrt [3]{x^2}$.
$\Rightarrow \left (\frac{f}{g} \right )(x)=\frac{x\sqrt [3]{x^2}}{2}$.
The domain is all positive real numbers.
The interval notation is $(0,\infty)$.
