Chapter 1.9, Problem 72E

To determine

Whether the given function $f(x)=\frac{5x-3}{2x+5}$ has an inverse function and if it does, find the inverse function.

Answer to Problem 72E

The inverse function of given function is $f^{-1}(x)=\frac{5x+3}{5-2x}forx\ne \frac{5}{2}$

Explanation of Solution

Given information :

Given function $f(x)=\frac{5x-3}{2x+5}$

A function $f(x)$ has an inverse function, if it is a one-one function.

Since $f(x)=\frac{5x-3}{2x+5}$ is one-one for $x\ne -\frac{5}{2}$ as each value of the dependent variable corresponds to exactly one value of the independent variable.

Thus, the function $f(x)=\frac{5x-3}{2x+5},x\ne -\frac{5}{2}$ has an inverse function.

To find inverse function:

Step 1: Replace $y=f(x)$ in $f(x)=\frac{5x-3}{2x+5},x\ne -\frac{5}{2}$ , we get

  $y=\frac{5x-3}{2x+5},x\ne -\frac{5}{2}$

Step 2: interchange role of $x\text{and}y$ . And solve for $y$ .

  $\begin{array}{l}x=\frac{5y-3}{2y+5},x\ne -\frac{5}{2}\\ \Rightarrow x=\frac{5}{2}-\frac{31}{2\left(2y+5\right)}\end{array}$

  $\begin{array}{l}\Rightarrow \frac{31}{2\left(2y+5\right)}=\frac{5}{2}-x\\ \Rightarrow \frac{31}{2\left(2y+5\right)}=\frac{5-2x}{2}\\ \Rightarrow \frac{\left(2y+5\right)}{31}=\frac{1}{5-2x}\\ \Rightarrow \left(2y+5\right)=\frac{31}{(5-2x)}\\ \Rightarrow 2y=\frac{31}{(5-2x)}-5\\ \Rightarrow 2y=\frac{31-25+10x}{5-2x}\\ \Rightarrow 2y=\frac{10x+6}{5-2x}\\ \therefore y=\frac{5x+3}{5-2x}\end{array}$

Step 3: replace $y$ with $f^{-1}(x)$ in new equation, we get

  $f^{-1}(x)=\frac{5x+3}{5-2x}forx\ne \frac{5}{2}$

Therefore, the inverse function of given function is $f^{-1}(x)=\frac{5x+3}{5-2x}forx\ne \frac{5}{2}$