Question

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

$\begin{array}{l}f\left(x\right)=6x-5\\ g\left(x\right)=\frac{x}{2}\end{array}$

Short answer

$\begin{array}{l}f\circ g=3x-5\\ g\circ f=3x-\frac{5}{2}\\ f\circ f=36x-35\\ g\circ g=\frac{x}{4}\end{array}$

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,

$\begin{array}{l}f\left(x\right)=6x-5\\ g\left(x\right)=\frac{x}{2}\end{array}$

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

$\begin{array}{l}f\circ g=f\left(g\left(x\right)\right)\\ \text{=}f\left(\frac{x}{2}\right)\\ \text{=6}\left(\frac{x}{2}\right)-5\\ \text{=}\left(\frac{6}{2}\right)x-5\\ \text{=}3x-5\\ \Rightarrow f\circ g=3x-5\end{array}$

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

$\begin{array}{l}g\circ f=g\left(f\left(x\right)\right)\\ \text{=g}\left(6x-5\right)\\ \text{=}\frac{6x-5}{2}\\ \text{=}\frac{6x}{2}-\frac{5}{2}\\ \text{=}3x-\frac{5}{2}\\ \Rightarrow g\circ f=3x-\frac{5}{2}\end{array}$

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

$\begin{array}{l}f\circ f=f\left(f\left(x\right)\right)\\ \text{=}f\left(6x-5\right)\\ \text{=6}\left(6x-5\right)-5\\ \text{=6}\left(6x\right)-6\left(5\right)-5\\ \text{=36}x-30-5\\ \text{=36}x-35\\ \Rightarrow f\circ f=\text{36}x-35\end{array}$

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

$\begin{array}{l}g\circ g=g\left(g\left(x\right)\right)\\ \text{=g}\left(\frac{x}{2}\right)\\ \text{=}\frac{\frac{x}{2}}{2}\\ \text{=}\frac{x}{2}\times \frac{1}{2}\\ \text{=}\frac{x}{4}\\ \Rightarrow g\circ g=\frac{x}{4}\end{array}$

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