How to "undo" a function defined by a rational expresssion in order to find its inverse function, say $f(x)=\frac{x-1}{x+5}$

Question

The method I want to apply is this one.

For example, in order to find the inverse function of $f(x)=3x + 4$ , I analyze function $f$ as follows:

x $\rightarrow_{\times3} (3x) \rightarrow_{+4} (3x+4)$

Then, I "undo" function $f$ by following the same path backwards

$x \rightarrow_{-4} (x-4) \rightarrow_{/3} (\frac{x-4}{3})$.

So the inverse function of $f$ is function $g$ such that : $ g(x)= \frac{x-4}{3}$.

But it seems more difficult to apply this method to a function defined by a rational expression. The reason is that the " path" here is not linear. Is there a way to cope with this difficulty and to adapt the " method" I showed to this kind of function. For example:

$f(x)= \frac {x-1} {x+5}$.

Source : Larson's College Algebra, Section $2.7$, Exercice $23$.

Answer 1

For the given example, if we write $$\frac{x-1}{x+5}=\frac{x+5-6}{x+5}=1-\frac6{x+5}$$ then the way to obtaining the inverse becomes clear. In general obtaining an explicit inverse relation is extremely hard, but for linear-over-linear functions this is very much doable.

Answer 2

Write $y=f(x)$ and solve the equation for $x$.

E.g.

$$y=\frac{x-1}{x+5}\to yx+5y=x-1\to5y+1=x(1-y)\to x=\frac{5y+1}{1-y}.$$

If you have learnt the quadratic equation, you will be able to invert functions like

$$y=\frac{ax^2+bx+c}{dx^2+ex+f}$$ (where some of the coefficients can be zero).