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}#

Isolating a variable or determining the inverseIf 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
\[y={{-x-3}\over{5\cdot x-2}}\]

#x={{2\cdot y-3}\over{5\cdot y+1}}#

#\begin{array}{rcl}
y&=&{{-x-3}\over{5\cdot x-2}} \\ &&\phantom{xxx}\blue{\text{the original function }}\\
y \cdot \left(5\cdot x-2\right)&=& -x-3 \\ &&\phantom{xxx}\blue{\text{both sides divided by }5\cdot x-2}\\
5\cdot y\cdot x-2\cdot y&=&-x-3 \\ &&\phantom{xxx}\blue{\text{brackets expanded}}\\
5\cdot y\cdot x+x &=&2\cdot y-3 \\&&\phantom{xxx}\blue{\text{ter‍ms with } x \text{ to the left hand side, terms without }x \text{ to the right hand side }}\\
\left(5\cdot y+1\right)\cdot x &=& 2\cdot y-3 \\ &&\phantom{xxx}\blue{x \text{ moved outside brackets}}\\
x&=&{{2\cdot y-3}\over{5\cdot y+1}} \\ &&\phantom{xxx}\blue{\text{divided by }5\cdot y+1}\\
\end{array}#

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