FINDING INVERSE FUNCTIONS
(switch input/output names method)

Thanks for your support!

 

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

The method discussed in this lesson, 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!

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 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 $\,y\,$

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!

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

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