Chapter 2.6, Problem 44E

To determine

To find:The functions $fog$ , $gof$ , $fof$ , $gog$ and their domains for $f\left(x\right)=\frac{2}{x}$ and $g\left(x\right)=\frac{x}{x+2}$ .

Answer to Problem 44E

The function $fog$ is equal to $\frac{2\left(x+2\right)}{x}$ with domain $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ , $gof$ is equal to $\frac{1}{1+x}$ with domain set of all the real numbers except $x=-2$ and $x=-1$ , $fof$ is equal to $x$ with domain set of all the real numbers except $x=0$ , $gog$ is equal to $\left(gog\right)\left(x\right)=\frac{x}{3x+4}$ with domain set of real numbers except $x=-2$ and $x=-\frac{4}{3}$ . for $f\left(x\right)=\frac{2}{x}$ and $g\left(x\right)=\frac{x}{x+2}$ .

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\frac{2}{x}$ and $g\left(x\right)=\frac{x}{x+2}$ .

Calculation:

Simplify the function $fog$ .

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

The function $f\left(x\right)=\frac{2}{x}$ is defined for all the real numbers of $x$ except $x=0$ and the function $\left(fog\right)\left(x\right)=\frac{2\left(x+2\right)}{x}$ is defined only if $x\ne 0$ . Combine both the conditions.

So, the domain of the function $\left(fog\right)\left(x\right)=\frac{2\left(x+2\right)}{x}$ is equal to $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ .

Simplify the function $gof$ .

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

Further simplify,

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

The function $g\left(x\right)=\frac{x}{x+2}$ is defined for all the real values of $x$ except $x=-2$ and the function $\left(gof\right)\left(x\right)=\frac{1}{1+x}$ is defined only if $1+x\ne 0$ i.e. $x\ne -1$ . Combine both the conditions.

So, the domain of the function $\left(gof\right)\left(x\right)=\frac{1}{1+x}$ is set of all the real numbers except $x=-2$ and $x=-1$ .

Simplify the function $fof$ .

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

The function $f\left(x\right)=\frac{2}{x}$ is defined for all the real numbers of $x$ except $x=0$ and the function $\left(fof\right)\left(x\right)=x$ is defined for all real values of $x$ . Combine both the conditions.

So, the domain of the function $\left(fof\right)\left(x\right)=x$ is set of all the real numbers except $x=0$ .

Simplify the function $gog$ .

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

Further simplify,

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

The function $g\left(x\right)=\frac{x}{x+2}$ is defined for all the real values of $x$ except $x=-2$ and the function $\left(gog\right)\left(x\right)=\frac{x}{3x+4}$ is defined only if $3x+4\ne 0$ i.e. $x\ne -\frac{4}{3}$ .

So, the domain of the function $\left(gog\right)\left(x\right)=\frac{x}{3x+4}$ is set of real numbers except $x=-2$ and $x=-\frac{4}{3}$ .

Therefore, the function $fog$ is equal to $\frac{2\left(x+2\right)}{x}$ with domain $\left(-\infty ,0\right)\cup \left(0,\infty \right)$ , $gof$ is equal to $\frac{1}{1+x}$ with domain set of all the real numbers except $x=-2$ and $x=-1$ , $fof$ is equal to $x$ with domain set of all the real numbers except $x=0$ , $gog$ is equal to $\left(gog\right)\left(x\right)=\frac{x}{3x+4}$ with domain set of real numbers except $x=-2$ and $x=-\frac{4}{3}$ . for $f\left(x\right)=\frac{2}{x}$ and $g\left(x\right)=\frac{x}{x+2}$ .