Chapter 1.4, Problem 18E

Solution

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

The required functions are $f\left(g\left(x\right)\right)=x-1$ , domain $\left[0,1\right)\cup \left(1,\infty \right)$ and $g\left(f\left(x\right)\right)=\frac{1}{\sqrt{x-1}}$ , domain $\left(1,\infty \right)$ .

Given information: The two functions are $f\left(x\right)=\frac{1}{x-1}$ and $g\left(x\right)=\sqrt{x}$ . 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; $f\left(x\right)=\frac{1}{x-1}$ and $g\left(x\right)=\sqrt{x}$ . Therefore,

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

Since, the function $f\left(g\left(x\right)\right)=\frac{1}{\sqrt{x}-1}$ is defined for all (set of real numbers) but not defined when $x<0$ and $x=1$ , therefore, $\left[0,1\right)\cup \left(1,\infty \right)$ is the domain of the function $f\left(g\left(x\right)\right)$ .

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

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

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