Chapter 1.3, Problem 43E

To determine

To express: The function $F\left(x\right)=\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}$ in the form of $\left(f\circ g\right)\left(x\right)$.

Answer to Problem 43E

The function $F\left(x\right)=\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}$ can be expressed as $F\left(x\right)=f\left(g\left(x\right)\right),$ where $f\left(x\right)=\frac{x}{1+x}$ and $g\left(x\right)=\sqrt[3]{x}$.

Explanation of Solution

The given composite function is $F\left(x\right)=\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}$.

Name the expression $\sqrt[3]{x}$ as $g\left(x\right)=\sqrt[3]{x}$.

Thus, $F\left(x\right)=\frac{g\left(x\right)}{1+g\left(x\right)}$

It can be expressed as $F\left(x\right)=f\left(g\left(x\right)\right)$ as F(x) is a composition of f and g.

Replace $g\left(x\right)$ by x to obtain the expression for $f\left(x\right)$.

Therefore, $f\left(x\right)=\frac{x}{1+x}$.

Hence, it can be concluded that $F\left(x\right)=\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}$ can be expressed as $F\left(x\right)=f\left(g\left(x\right)\right),$ where $f\left(x\right)=\frac{x}{1+x}$ and $g\left(x\right)=\sqrt[3]{x}$.

Verification:

The composite function $\left(f\circ g\right)\left(x\right)$ is defined as follows.

$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substitute $\sqrt[3]{x}$ for x in $f\left(x\right)$.

$\begin{array}{c}f\left(\sqrt[3]{x}\right)=\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}\\ =F\left(x\right)\end{array}$

Thus, it is verified that $F\left(x\right)=\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}$ can be expressed as $F\left(x\right)=f\left(g\left(x\right)\right),$ where $f\left(x\right)=\frac{x}{1+x}$ and $g\left(x\right)=\sqrt[3]{x}$.