Inverse of linear fractional function

Functions: Fractional functions

Inverse of linear fractional function

We have seen that determining the inverse function is the same as isolating the variable $x$ in a formula of the form $y=\ldots$ . Now we will investigate how to do that for linear fractional functions.

	Procedure We determine the inverse function of the linear fractional function $\green{y}=\frac{a\blue{x}+b}{c\blue{x}+d}$ with $a$ , $b$ , $c$ and $d$ as numbers.	Example $\green{y}=\frac{2\blue{x}-5}{3\blue{x}+2}$
Step 1	Multiply by the denominator of the fraction: $c\blue{x}+d$ .	$\green{y} \left(3\blue{x}+2\right)=2\blue{x}-5$
Step 2	Expand the brackets.	$3\blue{x}\green{y}+2 \green{y}=2\blue{x}-5$
Step 3	By means of reduction move the terms without $x$ to the right and the terms with a $x$ to the left hand side.	$3\blue{x}\green{y}-2\blue{x}=-2 \green{y}-5$
Step 4	Move $x$ outside brackets.	$\blue x \left(3 \green{y}-2\right)=-2 \green{y}-5$
Step 5	Divide by what's in between the brackets, so that we only have $x$ at the left hand side.	$\blue x=\frac{-2 \green{y}-5}{3 \green{y}-2}$
Step 6	Swap the $\blue x$ into a $\green y$ and the $\green y$ into a $\blue x$ to get the inverse function.	$\green y=\frac{-2 \blue{x}-5}{3 \blue{x}-2}$

If the question is to isolate the variable $x$ in the formula, we don't have to execute step 6 of the procedure. We are ready after step 5.

On the other hand, if we are asked to determine the inverse of a function of the form $f(x)=\tfrac{a\blue{x}+b}{c\blue{x}+d}$ we first replace $f(x)$ with $\green y$ and then execute the procedure. In the end, after step 6, we can replace $\green y$ again with $f^{-1}(x)$ .

Isolate $x$ in
$y={{-2\cdot x-4}\over{2\cdot x+6}}$

$x={{-3\cdot y-2}\over{y+1}}$

$\begin{array}{rcl} y&=&{{-2\cdot x-4}\over{2\cdot x+6}} \\ &&\phantom{xxx}\blue{\text{the original function }}\\ y \cdot \left(2\cdot x+6\right)&=& -2\cdot x-4 \\ &&\phantom{xxx}\blue{\text{both sides divided by }2\cdot x+6}\\ 2\cdot x\cdot y+6\cdot y&=&-2\cdot x-4 \\ &&\phantom{xxx}\blue{\text{brackets expanded}}\\ 2\cdot x\cdot y+2\cdot x &=&-6\cdot y-4 \\&&\phantom{xxx}\blue{\text{ter‍ms with } x \text{ to the left hand side, terms without }x \text{ to the right hand side }}\\ x\cdot \left(2\cdot y+2\right) &=& -6\cdot y-4 \\ &&\phantom{xxx}\blue{x \text{ moved outside brackets}}\\ x&=&{{-3\cdot y-2}\over{y+1}} \\ &&\phantom{xxx}\blue{\text{divided by }2\cdot y+2}\\ \end{array}$

New example