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.

$\begin{array}{l}f\left(x\right)=\frac{1}{x}\\ g\left(x\right)=2x+4\end{array}$

Short answer

$\begin{array}{l}f\circ g=\frac{1}{2x+4}\\ g\circ f=\frac{2+4x}{x}\\ f\circ f=x\\ g\circ g=4x+12\end{array}$

The domain$f\circ g$ is$\left\{x|x\ne 2\right\}$

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

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

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,

$$\begin{gathered} f\left(x\right)=\frac{1}{x} \\ g\left(x\right)=2x+4 \end{gathered}$$

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

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

We know that, the domain of the function is where that function is defined

As the function$f\circ g$is not defined when the denominator is zero

So,

$\begin{array}{l}2x+4=0\\ \Rightarrow 2x=-4\\ \Rightarrow x=-\frac{4}{2}\\ \Rightarrow x=-2\end{array}$

Therefore, we get

The domain$f\circ g$is all real numbers except$x=-2$

i.e.

The domain of $f\circ g$is$\left\{x|x\ne 2\right\}$

Step 4. To find g∘f.

By using Step 2, we have

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

We know that, the domain of the function is where that function is defined

As the function$g\circ f$is not defined when the denominator is zero

So,

$x=0$

Therefore, we get

The domain$g\circ f$is all real numbers except$x=0$

i.e.

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

Step 5. To find f∘f.

By using Step 2, we have

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

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

$\begin{array}{l}g\circ g=g\left(g\left(x\right)\right)\\ \text{=g}\left(2x+4\right)\\ \text{=2}\left(2x+4\right)+4\\ \text{=2}\left(2x\right)+2\left(4\right)+4\\ \text{=}4x+8+4\\ \text{=}4x+12\\ \Rightarrow g\circ g=4x+12\end{array}$

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