Chapter 1.4, Problem 16E

Solution

To find: The composite function of two functions and find its domain.

The required functions are $f\left(g\left(x\right)\right)=\frac{2x-x^2}{{\left(x-1\right)}^2}$ , domain and $g\left(f\left(x\right)\right)=\frac{1}{x^2-2}$ , domain.

Given information: The two functions are $f\left(x\right)=x^2-1$ and $g\left(x\right)=\frac{1}{x-1}$ . The required composite functions are $f\left(g\left(x\right)\right)$ and $g\left(f\left(x\right)\right)$ .

Method/Formula used:

The composite functions of two functions $f\left(x\right)$ and $g\left(x\right)$ are denoted by $\left(fog\right)\left(x\right)$ and $\left(gof\right)\left(x\right)$ , and are given by $\left(fog\right)\left(x\right)=f\left(g\left(x\right)\right)$ , and $\left(gof\right)\left(x\right)=g\left(f\left(x\right)\right)$ .

  

Calculations:

The given functions are; $f\left(x\right)=x^2-1$ and $g\left(x\right)=\frac{1}{x-1}$ . Therefore,

  $\begin{array}{c}f\left(g\left(x\right)\right)=f\left(g\left(x\right)\right)\leftarrow \text{substitute}\frac{1}{\left(x-1\right)}\text{for}g\left(x\right)\\ =f\left(\frac{1}{x-1}\right)\leftarrow \text{use}\text{definition}\text{of}f\left(x\right)=x^2-1\\ ={\left(\frac{1}{x-1}\right)}^2-1\leftarrow \text{simplify}\\ =\frac{1}{{\left(x-1\right)}^2}-1\\ =\frac{1-1{\left(x-1\right)}^2}{{\left(x-1\right)}^2}\\ =\frac{1-\left(x^2-2x+1\right)}{{\left(x-1\right)}^2}\\ =\frac{1-x^2+2x-1}{{\left(x-1\right)}^2}\\ =\frac{2x-x^2}{{\left(x-1\right)}^2}\end{array}$

Since, the function $f\left(g\left(x\right)\right)=\frac{2x-x^2}{{\left(x-1\right)}^2}$ is defined for all $x\in ℝ$ (set of real numbers), except $x=1$ , therefore or $\left(-\infty ,\infty \right)-\left\{1\right\}$ is the domain of the function $f\left(g\left(x\right)\right)$ .

Further, $f\left(x\right)=x^2-1$ and $g\left(x\right)=\frac{1}{x-1}$ . Therefore,

  $\begin{array}{c}g\left(f\left(x\right)\right)=g\left(f\left(x\right)\right)\leftarrow \text{substitute}\left(x^2-1\right)\text{for}f\left(x\right)\\ =g\left(x^2-1\right)\leftarrow \text{use}\text{definition}\text{of}g\left(x\right)=\frac{1}{x-1}\\ =\frac{1}{\left(x^2-1\right)-1}\leftarrow \text{simplify}\\ =\frac{1}{x^2-1-1}\\ =\frac{1}{x^2-2}\end{array}$

Since, the function $g\left(f\left(x\right)\right)=\frac{1}{x^2-2}$ is t defined for all $x\in ℝ$ (set of real numbers), except $x=\pm \sqrt{2}$ , therefore or $\left(-\infty ,\infty \right)-\left(-\sqrt{2},\sqrt{2}\right)$ is the domain of the function $g\left(f\left(x\right)\right)$ .