Question

Composition of functions- Find the functions$f\circ g,g\circ f,f\circ f,\text{\hspace{0.33em}and}g\circ g$and their domains.

50.$\begin{array}{l}f\left(x\right)=x^3+2\\ g\left(x\right)=\sqrt[3]{x}\end{array}$

Short answer

$\begin{array}{l}f\circ g=x+2\\ g\circ f=\sqrt[3]{x^3+2}\\ f\circ f={\left(x^3+2\right)}^3+2\\ g\circ g=\sqrt[9]{x}\end{array}$

The domain$f\circ g$ is all real numbers$\left(-\infty ,\infty \right)$

The domain$g\circ f$ is all real numbers$\left(-\infty ,\infty \right)$

The domain$f\circ f$ is all real numbers$\left(-\infty ,\infty \right)$

The domain $g\circ g$is all real numbers$\left(-\infty ,\infty \right)$

Step-by-step solution

Step 1- Given Information

Given that,

$\begin{array}{l}f\left(x\right)=x^3+2\\ g\left(x\right)=\sqrt[3]{x}=x^{\frac{1}{3}}\end{array}$

Step 2- Concept used

The Compositions are defined as

$\begin{array}{l}f\circ g=f\left(g\left(x\right)\right)\\ g\circ f=g\left(f\left(x\right)\right)\\ f\circ f=f\left(f\left(x\right)\right)\\ g\circ g=g\left(g\left(x\right)\right)\end{array}$

Step 3- To find f∘g

By using Step 2, we have

$\begin{array}{l}f\circ g=f\left(g\left(x\right)\right)\\ \text{=}f\left(x^{\frac{1}{3}}\right)\\ \text{=}{\left(x^{\frac{1}{3}}\right)}^3+2\\ \text{=}{x}^{\frac{3}{3}}+2\\ \text{=}x+2\\ \Rightarrow f\circ g=x+2\end{array}$

Also, the domain $f\circ g$is all real numbers$\left(-\infty ,\infty \right)$

Step 4- To find g∘f

By using Step 2, we have

$\begin{array}{l}g\circ f=g\left(f\left(x\right)\right)\\ \text{=g}\left(x^3+2\right)\\ \text{=}{\left(x^3+2\right)}^{\frac{1}{3}}\\ \text{=}\sqrt[3]{x^3+2}\\ \Rightarrow g\circ f=\sqrt[3]{x^3+2}\end{array}$

Also, the domain $g\circ f$is all real numbers$\left(-\infty ,\infty \right)$

Step 5- To find f∘f

By using Step 2, we have

$\begin{array}{l}f\circ f=f\left(f\left(x\right)\right)\\ \text{=}f\left(x^3+2\right)\\ \text{=}{\left(x^3+2\right)}^3+2\\ \Rightarrow f\circ f={\left(x^3+2\right)}^3+2\end{array}$

Also, the domain $f\circ f$is all real numbers$\left(-\infty ,\infty \right)$

Step 6- To find g∘g

By using Step 2, we have

$\begin{array}{l}g\circ g=g\left(g\left(x\right)\right)\\ \text{=g}\left(x^{\frac{1}{3}}\right)\\ \text{=}{\left(x^{\frac{1}{3}}\right)}^{\frac{1}{3}}\\ \text{=}{x}^{\frac{1}{3}\times \frac{1}{3}}\\ \text{=}{x}^{\frac{1}{9}}\\ \text{=}\sqrt[9]{x}\\ \Rightarrow g\circ g=\sqrt[9]{x}\end{array}$

Also, the domain $g\circ g$is all real numbers$\left(-\infty ,\infty \right)$