FINDING INVERSE FUNCTIONS
(switch input/output names method)

This lesson will be more meaningful if you fully understand the concepts in these prior web exercises:
using a function box ‘backwards’
one-to-one functions
undoing a one-to-one function; inverse functions
properties of inverse functions
finding inverse functions (when there's only one $x$ in the formula)

Every one-to-one function [beautiful math coming... please be patient]$\,f\,$ has an inverse, denoted by [beautiful math coming... please be patient]$\,f^{-1}\,$, that ‘undoes’ what $\,f\,$ does.
In this section 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 section,
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 section, dubbed the ‘Switch Input/Output Names’ method, is more widely applicable.
However, it tends to be quite mechanical—if 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 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 $\displaystyle\,f(x) = \frac{1-3x}{5+2x}\,$.
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 $\,\displaystyle f(x) = \frac{1-3x}{5+2x}\,$.
    Replace the function notation $\,f(x)\,$ with $\,y\,$, giving:   $\,\displaystyle y = \frac{1-3x}{5+2x}\,$
    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:   $\displaystyle x = \frac{1-3y}{5+2y}$
    Now, $\,x\,$ represents an output from $\,f\,$, which is an input to $\,f^{-1}\,$.
    Now, $\,y\,$ represents an input to $\,f\,$, which is 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 $\,y\,$
  4. Switch to function notation, by renaming $\,y\,$ as $\,f^{-1}(x)\,$:

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

    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!

$$ f(f^{-1}(x)) \ =\ f\left(\frac{1-5x}{2x+3}\right) \ =\ \frac{1-3\cdot\frac{1-5x}{2x+3}}{5 + 2\cdot\frac{1-5x}{2x+3}} \ =\ \frac{\left(1-3\cdot\frac{1-5x}{2x+3}\right)}{\left(5 + 2\cdot\frac{1-5x}{2x+3}\right)}\cdot\frac{(2x+3)}{(2x+3)} \ =\ \frac{2x + 3 - 3(1-5x)}{5(2x+3) + 2(1-5x)} \ =\ \frac{2x + 3 - 3 + 15x}{10x + 15 + 2 - 10x} \ =\ \frac{17x}{17} \ =\ x $$
$$ f^{-1}(f(x)) \ =\ f^{-1}\left(\frac{1-3x}{5+2x}\right) \ =\ \frac{1-5\cdot \frac{1-3x}{5+2x}}{2\cdot \frac{1-3x}{5+2x} + 3} \ =\ \frac{1-5\cdot \frac{1-3x}{5+2x}}{2\cdot \frac{1-3x}{5+2x} + 3}\cdot\frac{(5+2x)}{(5+2x)} \ =\ \frac{1(5+2x) - 5(1-3x)}{2(1-3x) + 3(5+2x)} \ =\ \frac{5 + 2x - 5 + 15x}{2 - 6x + 15 + 6x} \ =\ \frac{17x}{17} \ =\ x $$
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
the graph of an inverse function
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5
AVAILABLE MASTERED IN PROGRESS

(MAX is 5; there are 5 different problem types.)