Chapter 2.6, Problem 43E

To determine

To find:The functions $fog$ , $gof$ , $fof$ , $gog$ and their domains for $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x}$ .

Answer to Problem 43E

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

Explanation of Solution

Given information:

The functions are $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=\frac{1}{x}$ .

Calculation:

Simplify the function $fog$ .

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

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

So, the domain of the function $\left(fog\right)\left(x\right)=\frac{1}{1+x}$ is set of real numbers except $x=-1$ and $x=0$ .

Simplify the function $gof$ .

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

The function $f\left(x\right)=\frac{x}{x+1}$ is defined for all the real values of $x$ except $x=-1$ as the denominator becomes zero and the function $\left(gof\right)\left(x\right)=\frac{x+1}{x}$ is defined only if $x\ne 0$ .

So, the domain of the function $\left(gof\right)\left(x\right)=\frac{x+1}{x}$ is set of real numbers except $x=-1$ and $x=0$ .

Simplify the function $fof$ .

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

Further simplify,

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

The function $f\left(x\right)=\frac{x}{x+1}$ is defined for all the real values of $x$ except $x=-1$ and the function $\left(fof\right)\left(x\right)=\frac{x}{2x+1}$ is defined only if $2x+1\ne 0$ .

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

Simplify the function $gog$ .

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

The function $g\left(x\right)=\frac{1}{x}$ is defined for all the real values of $x$ except $x=0$ and the function $\left(gog\right)\left(x\right)=x$ is defined for all the real values of $x$ . Combine both the conditions.

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

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