Chapter 1.5, Problem 23E

To determine

To find: The inverse of the given function and also verify that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Answer to Problem 23E

The inverse of the given function is $f^{-1}\left(x\right)=\frac{1-3x}{x-2}$ and it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{2x+1}{x+3}$ .

Calculation:

Substitute $y$ for $f\left(x\right)$ in the equation $f\left(x\right)=\frac{2x+1}{x+3}$ and solve for $x$ in terms of $y$ .

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

Simplify further.

  $x=\frac{1-3y}{y-2}$

Interchange $x$ and $y$ , and substitute $f^{-1}\left(x\right)$ for $y$ .

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

Verify:

The composite function $\left(f\circ f^{-1}\right)\left(x\right)$ can be written as:

  $\left(f\circ f^{-1}\right)\left(x\right)=f\left(f^{-1}\left(x\right)\right)$

Substitute $\frac{1-3x}{x-2}$ for $f^{-1}\left(x\right)$ to find $\left(f\circ f^{-1}\right)\left(x\right)$ for $x>0$ .

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

Simplify further.

  $\begin{array}{c}\left(f\circ f^{-1}\right)\left(x\right)=\frac{2-6x+x-2}{1-3x+3x-6}\\ =\frac{-5x}{-5}\\ =x\end{array}$

The composite function $\left(f^{-1}\circ f\right)\left(x\right)$ can be written as:

  $\left(f^{-1}\circ f\right)\left(x\right)=f^{-1}\left(f\left(x\right)\right)$

Substitute $\frac{2x+1}{x+3}$ for $f\left(x\right)$ to find $\left(f^{-1}\circ f\right)\left(x\right)$ for $x>0$ .

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

Simplify further.

  $\begin{array}{c}\left(f^{-1}\circ f\right)\left(x\right)=\frac{x+3-6x-3}{2x+1-2x-6}\\ =\frac{-5x}{-5}\\ =x\end{array}$

It is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Therefore, the inverse of the given function is $f^{-1}\left(x\right)=\frac{1-3x}{x-2}$ and it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .