# FINDING INVERSE FUNCTIONS(switch input/output names method)

• PRACTICE (online exercises and printable worksheets)
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 $\,f\,$ has an inverse, denoted by $\,f^{-1}\,$, 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