Q. 97 - Precalculus Enhanced with Graphing Utilities

Question

Given

$f\left(x\right)=\frac{ax+b}{cx+d}$

find $f^{-1}$$\left(x\right)$. If $c\ne 0$, under what conditions on $a, b, c,$and

$d$ is $f=f^{-1}\left(x\right)$?

Step-by-Step Solution

Answer

Part (a). The inverse of the function is

$f^{-1}\left(x\right)=\frac{-dx+b}{cx-a}$.

Part (b). If $f^{-1}\left(x\right)=f\left(x\right)$ then $a=-d$.

Step 1. Find f - 1 x

The given function is $f\left(x\right)=\frac{ax+b}{cx+d}$.

Substituting $f\left(x\right)=y$ and replace $y$ by $x$

$x=\frac{ay+b}{cy+d}$

Finding the inverse, we get:

$$\begin{gathered} cxy+dx=ay+b \\ cxy-ay=b-dx \\ y\left(cx-a\right)=b-dx \\ y=\frac{b-dx}{cx-a} \end{gathered}$$

Replace $y=f^{-1}\left(x\right)$, we get:

$f^{-1}\left(x\right)=\frac{-dx+b}{cx-a}$

Step 2. Equate f x = f - 1 x

Equating $f^{-1}\left(x\right)=f\left(x\right)$, we get:

$\frac{ax+b}{cx+d}=\frac{-dx+b}{cx-a}$

From above equation, we can observe that, $a=-d$