Chapter 1.4, Problem 14E

Solution

To find: The functions $fog\left(x\right)$ and $gof\left(x\right)$ from the functions $f\left(x\right)$ and $g\left(x\right)$ .

The required functions values are $fog\left(3\right)=0$ and $gof\left(-2\right)=5$ .

Given data:

The functions $f\left(x\right)$ and $g\left(x\right)$ are; $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=9-x^2$ . The required functions are $fog\left(3\right)$ and $gof\left(-2\right)$ .

Method/Formula used:

The composition function of two functions $f\left(x\right)$ and $g\left(x\right)$ is denoted by $\left(f\text{ο}g\right)\left(x\right)$ is also a function defined by $\left(f\text{ο}g\right)\left(x\right)=f\left(g\left(x\right)\right)$ . Similarly, $\left(g\text{ο}f\right)\left(x\right)=g\left(f\left(x\right)\right)$

Calculations:

The given functions $f\left(x\right)$ and $g\left(x\right)$ are $f\left(x\right)=\frac{x}{x+1}$ and $g\left(x\right)=9-x^2$ . Then by the definition of the function $fog\left(x\right)$ ,

  $\begin{array}{c}fog\left(x\right)=f\left(g\left(x\right)\right)\leftarrow \text{use}\text{definition}\text{of}g\left(x\right)\text{, i}\text{.e}\text{., rereplace}g\left(x\right)\text{by}9-x^2\\ =f\left(9-x^2\right)\leftarrow \text{use}\text{definition}\text{of}f\left(x\right),\text{replace}x\text{by}9-x^2\text{in}f\left(x\right)=\frac{x}{x+1}\\ =\frac{9-x^2}{9-x^2+1}\\ =\frac{9-x^2}{10-x^2}\end{array}$

Now, substitute 3 for x in $fog\left(x\right)=\frac{9-x^2}{10-x^2}$

  $\begin{array}{c}fog\left(3\right)=\frac{9-{\left(3\right)}^2}{10-{\left(3\right)}^2}\\ =\frac{9-9}{10-9}\\ =0\end{array}$

Further, by the definition of $gof\left(x\right)$

  $\begin{array}{c}gof\left(x\right)=g\left(f\left(x\right)\right)\leftarrow \text{use}\text{definition}\text{of}f\left(x\right)\text{, rereplace}f\left(x\right)\text{by}f\left(x\right)=\frac{x}{x+1}\\ =g\left(\frac{x}{x+1}\right)\leftarrow \text{use}\text{definition}\text{of}g\left(x\right),\text{replace}x\text{by}\frac{x}{x+1}\text{in}g\left(x\right)=9-x^2\\ =9-{\left(\frac{x}{x+1}\right)}^2\end{array}$

Now, substitute -2 for x in $gof\left(x\right)=9-{\left(\frac{x}{x+1}\right)}^2$ ,

  $\begin{array}{c}gof\left(-2\right)=9-{\left(\frac{-2}{-2+1}\right)}^2\\ =9-{\left(\frac{-2}{-1}\right)}^2\\ =5\end{array}$

Therefore, $fog\left(3\right)=0$ and $gof\left(-2\right)=5$ .