(switch input/output names method)

by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
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)

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 switchedthe 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!

  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 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 $\,f^{-1}(x)\,$:

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


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
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
1 2 3 4 5

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