#### Answer

$F(x)=f\circ g(x)$ for
$g(x)=\sqrt[3]{x},\displaystyle \qquad f(x)=\frac{x}{1+x}$

#### Work Step by Step

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