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.

$\begin{array}{l}f\left(x\right)=\frac{x}{x+1}\\ g\left(x\right)=2x-1\end{array}$

Short answer

$\begin{array}{l}f\circ g=\frac{2x-1}{2x}\\ g\circ f=\frac{x-1}{x+1}\\ f\circ f=\frac{x}{2x+1}\\ g\circ g=4x-3\end{array}$

The domain$f\circ g$ is$\left\{x|x\ne 0\right\}$

The domain$g\circ f$ is$\left\{x|x\ne -1\right\}$

The domain$f\circ f$ is$\left\{x|x\ne -\frac{1}{2}\right\}$

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

Step-by-step solution

Step 1. Given Information.

Given that,

$\begin{array}{l}f\left(x\right)=\frac{x}{x+1}\\ g\left(x\right)=2x-1\end{array}$

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

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

The domain$f\circ g$ is where the function is defined.

i.e., domain of$f\circ g$ is defined where the denominator is not zero.

So,

$\begin{array}{l}2x=0\\ \Rightarrow x=0\end{array}$

Therefore, we get

The domain of $f\circ g$is$\left\{x|x\ne 0\right\}$

Step 4. To find g∘f.

By using Step 2, we have

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

The domain$g\circ f$ is where the function is defined.

i.e., domain of$g\circ f$ is defined where the denominator is not zero.

So,

$\begin{array}{l}x+1=0\\ \Rightarrow x=-1\end{array}$

Therefore, we get

The domain of $g\circ f$is$\left\{x|x\ne -1\right\}$

Step 5. To find f∘f.

By using Step 2, we have

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

The domain$f\circ f$ is where the function is defined.

i.e., domain of$f\circ f$ is defined where the denominator is not zero.

So,

$\begin{array}{l}2x+1=0\\ \Rightarrow 2x=-1\\ \Rightarrow x=-\frac{1}{2}\end{array}$

Therefore, we get

Domain of$f\circ f$is $\left\{x|x\ne -\frac{1}{2}\right\}$

Step 6. To find g∘g.

By using Step 2, we have

$\begin{array}{l}g\circ g=g\left(g\left(x\right)\right)\\ \text{=g}\left(2x-1\right)\\ \text{=2}\left(2x-1\right)-1\\ \text{=2}\left(2x\right)-2\left(1\right)-1\\ \text{=}4x-2-1\\ \text{=}4x-3\\ \Rightarrow g\circ g=4x-3\end{array}$

So,

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