Finding Inverse Functions (Switch Input/Output Names Method)
This lesson will be more meaningful if you fully understand the concepts in these prior lessons:
 Using a Function Box ‘Backwards’
 OnetoOne Functions
 Undoing a OnetoOne Function; Inverse Functions
 Properties of Inverse Functions
 Finding Inverse Functions (when there's only one $x$ in the formula)
Every onetoone 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 (onetoone) 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!
 Replace the function notation $\,f(x)\,$ by the variable $\,y\,.$ This is the equation $\,y = f(x)\,.$
 Switch $\,x\,$ and $\,y\ .$ This new equation is $\,x = f(y)\,.$
 Solve this new equation for $\,y\,.$ This yields the equation $\,y = f^{1}(x)\,.$
 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: $$f(x) = \frac{13x}{5+2x}$$
Note that the mapping diagram method cannot be used for this function, since it contains two appearances of the input variable $\,x\,.$

Start with: $$f(x) = \frac{13x}{5+2x}$$
Replace the function notation $\,f(x)\,$ with $\,y\,,$ giving: $$\cssId{s37}{y = \frac{13x}{5+2x}}$$
In this equation, $\,x\,$ is the input to $\,f\,$ and $\,y\,$ is the output from $\,f\,.$

Switch the names $\,x\,$ and $\,y\,$ to get:
$$\cssId{s40}{x = \frac{13y}{5+2y}}$$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 $\,f^{1}\,.$
 Solve this new equation for $\,y\,.$ This is the part that requires some work:

Switch to function notation, by renaming $\,y\,$ as $\,f^{1}(x)\,$:
$$\cssId{s59}{f^{1}(x) = \frac{15x}{2x+3}}$$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 multiplybyone technique used to turn these complex fractions into ‘simple’ fractions!
$$ \begin{align} &\cssId{s65}{f(f^{1}(x))} \cr\cr &\quad \cssId{s66}{=\ f\left(\frac{15x}{2x+3}\right)}\cr\cr &\quad \cssId{s67}{=\ \frac{13\cdot\frac{15x}{2x+3}}{5 + 2\cdot\frac{15x}{2x+3}}}\cr\cr &\quad \cssId{s68}{=\ \frac{\left(13\cdot\frac{15x}{2x+3}\right)}{\left(5 + 2\cdot\frac{15x}{2x+3}\right)}\cdot\frac{(2x+3)}{(2x+3)}}\cr\cr &\quad \cssId{s69}{=\ \frac{2x + 3  3(15x)}{5(2x+3) + 2(15x)}}\cr\cr &\quad \cssId{s70}{=\ \frac{2x + 3  3 + 15x}{10x + 15 + 2  10x}}\cr\cr &\quad \cssId{s71}{=\ \frac{17x}{17}}\cr\cr &\quad \cssId{s72}{=\ x} \end{align} $$$$ \begin{align} &\cssId{s73}{f^{1}(f(x))} \cr\cr &\quad \cssId{s74}{=\ f^{1}\left(\frac{13x}{5+2x}\right)}\cr\cr &\quad \cssId{s75}{=\ \frac{15\cdot \frac{13x}{5+2x}}{2\cdot \frac{13x}{5+2x} + 3}}\cr\cr &\quad \cssId{s76}{=\ \frac{15\cdot \frac{13x}{5+2x}}{2\cdot \frac{13x}{5+2x} + 3}\cdot\frac{(5+2x)}{(5+2x)}}\cr\cr &\quad \cssId{s77}{=\ \frac{1(5+2x)  5(13x)}{2(13x) + 3(5+2x)}}\cr\cr &\quad \cssId{s78}{=\ \frac{5 + 2x  5 + 15x}{2  6x + 15 + 6x}}\cr\cr &\quad \cssId{s79}{=\ \frac{17x}{17}}\cr\cr &\quad \cssId{s80}{=\ x} \end{align} $$