Chapter 1 - Functions and Limits - 1.3 New Functions from Old Functions - 1.3 Exercises - Page 44: 45

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

$f\circ g(x)=f[g(x)]$ 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 $F(x)=f\circ g(x)$