Finding Inverse Functions (Switch Input/Output Names Method)

Index card: 23b

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)

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 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 (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!

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:

4. 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!