﻿ Finding Inverse Functions (Switch Input/Output Names Method)

# Finding Inverse Functions (Switch Input/Output Names Method)

This lesson will be more meaningful if you fully understand the concepts in these prior lessons:

Every one-to-one function $\,f\,$ has an inverse, denoted by $\,f^{-1}\,,$ that ‘undoes’ what $\,f\,$ does.

In this lesson and the previous one, we look at two common techniques for getting a formula for $\,f^{-1}\,.$

This author strongly prefers the mapping diagram method of the previous lesson, because it emphasizes the fact that $\,f\,$ does something, and $\,f^{-1}\,$ undoes it. That method, however, only works when the formula for $\,f\,$ contains exactly one appearance of the input variable.

The method discussed in this lesson, dubbed the ‘Switch Input/Output Names’ method, is more widely applicable. However, it tends to be quite mechanicalif you're not careful, you can just ‘go through the motions’ and forget the underlying idea!

## Input/Output Roles for a Function and its Inverse are Switched

The input/output roles for a function and its inverse are switched: the inputs to one are the outputs from the other.

If a (one-to-one) function $\,f\,$ takes $\,x\,$ to $\,y\,,$ then $\,f^{-1}\,$ takes $\,y\,$ back to $\,x\,.$

In other words, if $\,y = f(x)\,,$ then $\,f^{-1}(y) = x\,.$

This is the reason we ‘switch the names’ in the method discussed next!

‘SWITCH INPUT/OUTPUT NAMES’ METHOD FOR FINDING $\,f^{-1}\,$
1. Replace the function notation $\,f(x)\,$ by the variable $\,y\,.$ This is the equation $\,y = f(x)\,.$
2. Switch $\,x\,$ and $\,y\ .$ This new equation is $\,x = f(y)\,.$
3. Solve this new equation for $\,y\,.$ This yields the equation $\,y = f^{-1}(x)\,.$
4. Switch to function notation by replacing $\,y\,$ by $\,f^{-1}(x)\,.$

## Example: the ‘Switch Input/Output Names’ Method

In this example, the ‘switch input/output names’ method for finding the inverse is applied to the function: $$f(x) = \frac{1-3x}{5+2x}$$

Note that the mapping diagram method cannot be used for this function, since it contains two appearances of the input variable $\,x\,.$

1. Start with: $$f(x) = \frac{1-3x}{5+2x}$$

Replace the function notation $\,f(x)\,$ with $\,y\,,$ giving: $$\cssId{s37}{y = \frac{1-3x}{5+2x}}$$

In this equation, $\,x\,$ is the input to $\,f\,$ and $\,y\,$ is the output from $\,f\,.$

2. Switch the names $\,x\,$ and $\,y\,$ to get:

$$\cssId{s40}{x = \frac{1-3y}{5+2y}}$$

Now, $\,x\,$ represents an output from $\,f\,,$ which is an input to $\,f^{-1}\,.$ Now, $\,y\,$ represents an input to $\,f\,,$ which is an output from $\,f^{-1}\,.$

3. Solve this new equation for $\,y\,.$ This is the part that requires some work:
$\displaystyle x = \frac{1-3y}{5+2y}$
You must get all the variables $\,y\,$ ‘upstairs’, on the same side of the equation.
$x(5+2y) = 1-3y$
Start by clearing fractions.
$5x + 2xy = 1 - 3y$
Multiply out.
$2xy + 3y = 1 - 5x$
Rearrange: get all terms containing $\,y\,$ on the same side; move other terms to the other side.
$y(2x + 3) = 1 - 5x$
Factor out $\,y\,.$
$\displaystyle y = \frac{1-5x}{2x+3}$
Solve for $\,y\,.$
1. Switch to function notation, by renaming $\,y\,$ as $\,f^{-1}(x)\,$:

$$\cssId{s59}{f^{-1}(x) = \frac{1-5x}{2x+3}}$$

Done!

## Checking: Great Practice with Function Composition

It's fantastic practice to check that $\,f(f^{-1}(x)) = x\,$ and $\,f^{-1}(f(x)) = x\,.$

Along the way you end up with ‘complex fractions’—fractions within fractions. Note the multiply-by-one technique used to turn these complex fractions into ‘simple’ fractions!

\begin{align} &\cssId{s65}{f(f^{-1}(x))} \cr\cr &\quad \cssId{s66}{=\ f\left(\frac{1-5x}{2x+3}\right)}\cr\cr &\quad \cssId{s67}{=\ \frac{1-3\cdot\frac{1-5x}{2x+3}}{5 + 2\cdot\frac{1-5x}{2x+3}}}\cr\cr &\quad \cssId{s68}{=\ \frac{\left(1-3\cdot\frac{1-5x}{2x+3}\right)}{\left(5 + 2\cdot\frac{1-5x}{2x+3}\right)}\cdot\frac{(2x+3)}{(2x+3)}}\cr\cr &\quad \cssId{s69}{=\ \frac{2x + 3 - 3(1-5x)}{5(2x+3) + 2(1-5x)}}\cr\cr &\quad \cssId{s70}{=\ \frac{2x + 3 - 3 + 15x}{10x + 15 + 2 - 10x}}\cr\cr &\quad \cssId{s71}{=\ \frac{17x}{17}}\cr\cr &\quad \cssId{s72}{=\ x} \end{align}
\begin{align} &\cssId{s73}{f^{-1}(f(x))} \cr\cr &\quad \cssId{s74}{=\ f^{-1}\left(\frac{1-3x}{5+2x}\right)}\cr\cr &\quad \cssId{s75}{=\ \frac{1-5\cdot \frac{1-3x}{5+2x}}{2\cdot \frac{1-3x}{5+2x} + 3}}\cr\cr &\quad \cssId{s76}{=\ \frac{1-5\cdot \frac{1-3x}{5+2x}}{2\cdot \frac{1-3x}{5+2x} + 3}\cdot\frac{(5+2x)}{(5+2x)}}\cr\cr &\quad \cssId{s77}{=\ \frac{1(5+2x) - 5(1-3x)}{2(1-3x) + 3(5+2x)}}\cr\cr &\quad \cssId{s78}{=\ \frac{5 + 2x - 5 + 15x}{2 - 6x + 15 + 6x}}\cr\cr &\quad \cssId{s79}{=\ \frac{17x}{17}}\cr\cr &\quad \cssId{s80}{=\ x} \end{align}