Chapter 0.5, Problem 13E

To determine

To find: The inverse $f^{-1}$ of a function and to verify $\left(f\text{ο}{f}^{-1}\right)\left(x\right)=\left(f^{-1}\text{ο}f\right)\left(x\right)=x$ .

Answer to Problem 13E

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

Explanation of Solution

Given data:

The function is $f\left(x\right)=2x+3$ .

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. The composite function $\left(f\text{ο}{f}^{-1}\right)\left(x\right)=f\left[f^{-1}\left(x\right)\right]$ and $\left(f^{\text{-1}}\text{ο}f\right)\left(x\right)=f^{-1}\left[f\left(x\right)\right]$ .

Calculation:

The given function is $f\left(x\right)=2x+3$ .

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

  $\begin{array}{c}y=2x+3\\ 2x=y-3\\ x=\frac{y-3}{2}\end{array}$

Replace, x by $f^{-1}\left(y\right)$ ,

  $f^{-1}\left(y\right)=\frac{y-3}{2}$

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

  $f^{-1}\left(x\right)=\frac{x-3}{2}$

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

Now,

  $\begin{array}{c}\left(f\text{ο}{f}^{-1}\right)\left(x\right)=f\left[f^{-1}\left(x\right)\right]\leftarrow \text{}\text{use}\text{definition}{f}^{-1}\left(x\right)=\frac{x-3}{2}\\ =f\left(\frac{x-3}{2}\right)\leftarrow \text{}\text{use}\text{definition}f\left(x\right)=2x+3\\ =\text{2}\left(\frac{x-3}{2}\right)+3\\ =x\end{array}$ .

Also,

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

Thus, the inverse of $f\left(x\right)=2x+3$ is $f^{-1}\left(x\right)=\frac{x-3}{2}$ , and $\left(f\text{ο}{f}^{-1}\right)\left(x\right)=x=\left(f^{-1}\text{ο}f\right)\left(x\right)$ .