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}# |
Isolate #x# in
\[y={{-5\cdot x-6}\over{1-x}}\]
\[y={{-5\cdot x-6}\over{1-x}}\]
#x={{y+6}\over{y-5}}#
#\begin{array}{rcl}
y&=&{{-5\cdot x-6}\over{1-x}} \\ &&\phantom{xxx}\blue{\text{the original function }}\\
y \cdot \left(1-x\right)&=& -5\cdot x-6 \\ &&\phantom{xxx}\blue{\text{both sides divided by }1-x}\\
y-x\cdot y&=&-5\cdot x-6 \\ &&\phantom{xxx}\blue{\text{brackets expanded}}\\
5\cdot x-x\cdot y &=&-y-6 \\&&\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(5-y\right) &=& -y-6 \\ &&\phantom{xxx}\blue{x \text{ moved outside brackets}}\\
x&=&{{y+6}\over{y-5}} \\ &&\phantom{xxx}\blue{\text{divided by }5-y}\\
\end{array}#
#\begin{array}{rcl}
y&=&{{-5\cdot x-6}\over{1-x}} \\ &&\phantom{xxx}\blue{\text{the original function }}\\
y \cdot \left(1-x\right)&=& -5\cdot x-6 \\ &&\phantom{xxx}\blue{\text{both sides divided by }1-x}\\
y-x\cdot y&=&-5\cdot x-6 \\ &&\phantom{xxx}\blue{\text{brackets expanded}}\\
5\cdot x-x\cdot y &=&-y-6 \\&&\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(5-y\right) &=& -y-6 \\ &&\phantom{xxx}\blue{x \text{ moved outside brackets}}\\
x&=&{{y+6}\over{y-5}} \\ &&\phantom{xxx}\blue{\text{divided by }5-y}\\
\end{array}#
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