Answer
$G(x)=f\circ g(x)$ for
$g(x)=\displaystyle \frac{x}{1+x},\qquad f(x)=\sqrt[3]{x}$
Work Step by Step
$f\circ g(x)=f[g(x)]$
Rule of thumb: Ask yourself
what would be the last operation if we used a calculator, step by step?
(Answer: If R was the current result, we would calculate $\sqrt[3]{R}$)
$f(x)=\sqrt[3]{x}$
$G(x)=f(\displaystyle \frac{x}{1+x})$
So, if
$g(x)=\displaystyle \frac{x}{1+x}$ and $f(x)=\sqrt[3]{x}$,
then
$G(x)=f\circ g(x)$
