Chapter 2.7, Problem 58E

Composition of Functions Find the functions f ○ g, g ○ f, f ○ f, and g ○ g and their domains.

58. $f(x)=\frac{2}{x}$, $g(x)=\frac{x}{x+2}$

To determine

To evaluate: The functions $f\circ g$, $g\circ f$, $f\circ f$ and $g\circ g$ and their domains.

Answer to Problem 58E

The function $f\circ g$ is $\left(f\circ g\right)\left(x\right)=\frac{2x+4}{x}$ and its domain is $\left\{x|x\ne -2,0\right\}$. The function $g\circ f$ is $\left(g\circ f\right)\left(x\right)=\frac{1}{1+x}$ and its domain is $\left\{x|x\ne 0,-1\right\}$. The function $f\circ f$ is $\left(f\circ f\right)\left(x\right)=x$ and its domain is $\left\{x|x\ne 0\right\}$. The function $g\circ g$ is $\left(g\circ g\right)\left(x\right)=\frac{x}{3x+4}$ and its domain is $\left\{x|x\ne -\frac{4}{3},-2\right\}$.

Explanation of Solution

Given:

The given functions are,

$f\left(x\right)=\frac{2}{x}$ (1)

$g\left(x\right)=\frac{x}{x+2}$ (2)

Calculation:

The function $f\circ g$ is written as,

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

Substitute $\frac{x}{x+2}$ for $g\left(x\right)$ in above expression,

$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ (3)

Substitute $\frac{x}{x+2}$ for x in equation (1), to find the function $f\left(\frac{x}{x+2}\right)$,

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

Substitute $\frac{2x+4}{x}$ for $f\left(\frac{x}{x+2}\right)$ in equation (3), to find the function $f\circ g$,

$\left(f\circ g\right)\left(x\right)=\frac{2x+4}{x}$

The required function is $\left(f\circ g\right)\left(x\right)=\frac{2x+4}{x}$.

All possible inputs for a function is called the domain of the function.

The function $\left(f\circ g\right)\left(x\right)=\frac{2x+4}{x}$ defined for all real numbers lie except 0 and $-2$ because at $x=-2$ the denominator function $\frac{x}{x+2}$ is not defined, therefore the domain of the function is the set of all real numbers except $-2$ and 0. The domain of the function is $\left\{x|x\ne -2,0\right\}$.

The function $g\circ f$ is written as,

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

Substitute $\frac{2}{x}$ for $f\left(x\right)$ in above expression,

$\left(g\circ f\right)\left(x\right)=g\left(\frac{2}{x}\right)$ (4)

Substitute $\frac{2}{x}$ for x in equation (2), to find the function $g\left(\frac{2}{x}\right)$,

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

Substitute $\frac{1}{1+x}$ for $g\left(\frac{2}{x}\right)$ in equation (4), to find the function $g\circ f$,

$\left(g\circ f\right)\left(x\right)=\frac{1}{1+x}$

The required function is $\left(g\circ f\right)\left(x\right)=\frac{1}{1+x}$.

All possible inputs for a function is called the domain of the function.

The function $\left(g\circ f\right)\left(x\right)=\frac{1}{1+x}$ is defined for all real numbers except $-1$ and 0 because at $x=0$ the denominator function $\frac{2}{x}+2$ is not defined, therefore the domain of the function is the set of all real numbers except $-1$ and 0. The domain of the function is $\left\{x|x\ne 0,-1\right\}$.

Thus, the function $g\circ f$ is $\left(g\circ f\right)\left(x\right)=\frac{1}{1+x}$ and its domain is $\left\{x|x\ne 0,-1\right\}$.

The function $f\circ f$ is written as,

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

Substitute $\frac{2}{x}$ for $f\left(x\right)$ in above expression,

$\left(f\circ f\right)\left(x\right)=f\left(\frac{2}{x}\right)$ (5)

Substitute $\frac{2}{x}$ for x in equation (1), to find the function $f\left(\frac{2}{x}\right)$,

$\begin{array}{c}f\left(\frac{2}{x}\right)=\frac{2}{\left(\frac{2}{x}\right)}\\ =\frac{2x}{2}\\ =x\end{array}$

Substitute x for $f\left(\frac{2}{x}\right)$ in equation (5), to find the function $f\circ f$,

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

The required function is $\left(f\circ f\right)\left(x\right)=x$.

All possible inputs for a function is called the domain of the function.

The function $\left(f\circ f\right)\left(x\right)=x$ is defined for all real numbers except 0 because at $x=0$ the denominator function $\frac{2}{x}$ is not defined, therefore the domain of the function is the set of all real numbers except 0. The domain of the function is $\left\{x|x\ne 0\right\}$.

The function $g\circ g$ is written as shown below,

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

Substitute $\frac{x}{x+2}$ for $g\left(x\right)$ in above expression,

$\left(g\circ g\right)\left(x\right)=g\left(\frac{x}{x+2}\right)$ (6)

Substitute $\frac{x}{x+2}$ for x in equation (2), to find the function $g\left(\frac{x}{x+2}\right)$,

$\begin{array}{c}g\left(\frac{x}{x+2}\right)=\frac{\left(\frac{x}{x+2}\right)}{\left(\frac{x}{x+2}\right)+2}\\ =\frac{\left(\frac{x}{x+2}\right)}{\left(\frac{x+2\left(x+2\right)}{x+2}\right)}\\ =\frac{x}{x+2x+4}\\ =\frac{x}{3x+4}\end{array}$

Substitute $\frac{x}{3x+4}$ for $g\left(\frac{x}{x+2}\right)$ in equation (6), to find the function $g\circ g$,

$\left(g\circ g\right)\left(x\right)=\frac{x}{3x+4}$

The required function is $\left(g\circ g\right)\left(x\right)=\frac{x}{3x+4}$.

All possible inputs for a function is called the domain of the function.

The function $\left(g\circ g\right)\left(x\right)=\frac{x}{3x+4}$ defined for all real numbers except $-\frac{4}{3}$ and $-2$ because at $x=-2$ the denominator function $\left(\frac{x}{x+2}+2\right)$ is not defined, therefore the domain of the function is the set of all real numbers except $-\frac{4}{3}$ and $-2$. The domain of the function is $\left\{x|x\ne -\frac{4}{3},-2\right\}$.