Chapter 0.5, Problem 7E

To determine

Whether the function has inverse.

Answer to Problem 7E

The required inverse of $f\left(x\right)$ is $f^{-1}\left(x\right)=\frac{2x+5}{x+1}$ .

Explanation of Solution

Given data:

The function is $f\left(x\right)=\frac{3}{x-2}-1$ .

Method/Formula used:

Substitute y for $f\left(x\right)$ and find x in terms of y , write x as $f^{-1}\left(y\right)$ , then replace y by x throughout.

Calculation:

The given function is $f\left(x\right)=\frac{3}{x-2}-1$ .

Substitute y for $f\left(x\right)$ and solve for x as:

  $\begin{array}{c}y=\frac{3}{x-2}-1\\ \left(y+1\right)\left(x-2\right)=3\\ yx-2y+x-2=3\\ yx+x=3+2+2y\end{array}$

Solve further,

  $\begin{array}{c}yx+x=3+2+2y\\ x\left(y+1\right)=2y+5\\ x=\frac{2y+5}{y+1}\end{array}$

Substitute, $f^{-1}\left(y\right)$ for x ,

  $f^{-1}\left(y\right)=\frac{2y+5}{y+1}$

Change variable − replace y by customary variable x or another,

  $f^{-1}\left(x\right)=\frac{2x+5}{x+1}$ .

Now, verify $f^{-1}\left(f\left(x\right)\right)=x$

  $\begin{array}{c}{f}^{-1}\left(f\left(x\right)\right)=\frac{2f\left(x\right)+5}{f\left(x\right)+1}\\ =\frac{2\left[\frac{3}{x-2}-1\right]+5}{\left[\frac{3}{x-2}-1\right]+1}\\ =\frac{2\left[\frac{3-x+2}{x-2}\right]+5}{\left[\frac{3-x+2}{x-2}\right]+1}\end{array}$

Solve further,

  $\begin{array}{c}{f}^{-1}\left(f\left(x\right)\right)=\frac{2\left[\frac{5-x}{x-2}\right]+5}{\left[\frac{5-x}{x-2}\right]+1}\\ =\frac{\left[\frac{10-2x+5x-10}{x-2}\right]}{\left[\frac{5-x+x-2}{x-2}\right]}\\ =\frac{\left[\frac{3x}{x-2}\right]}{\left[\frac{3}{x-2}\right]}\\ =x\end{array}$

For given $f\left(x\right)$ and computed $f^{-1}\left(x\right)$ , there exists $f^{-1}\left(f\left(x\right)\right)$ equals x, therefore, $f^{-1}\left(x\right)=$

  $f^{-1}\left(x\right)=\frac{2x+5}{x+1}$ is the inverse of $f\left(x\right)=\frac{3}{x-2}-1$ .