# Inverting Rational Functions

Consider these two rational functions

$$

f(x)=\frac{2x+9}{x+2}\quad\quad g(x)=\frac{9-2x}{x-2}

$$

Show that they are inverses of each other, in that

$$

g(f(x))=f(g(x))=x

$$

What happens for the values $x=\pm 2$?

Can you invert the rational function

$$

h(x)=\frac{x-7}{2x+1}

$$

Do rational functions always have inverse functions? Why?

In the examples given here, the inverses of our rational functions were also rational functions. Will this be the case more generally? Why not explore more generally or try to find inverse pairs of rational functions?

As you consider these rational functions, many questions might emerge in your mind such as: "do rational functions have fixed points?" or "Is there a relationship between the asymptotes in a function and the zeroes of its inverse?". Why not make a note of these questions and ask your teacher, yourself or your friends to try to solve them?

One of our most prolific solvers, Patrick from Woodbridge School, sent in his thoughts on this problem

To invert a function, $f(x)$, the following procedure is used: say

$$f(x) = \frac{2x+9}{ x+2}$$

then the graph is

$$y = \frac{2x+9}{x+2}$$

It is inverted by replacing $x$ with $y$, and $y$ with $x$:

$$

x = \frac{2y + 9}{y + 2}

$$

and rearranging gives

$$

y = \frac{9-2x}{x-2}$$

Thus

$$g(x) = \frac{9-2x}{x-2}$$

However, for values $x = \pm 2$, we have a denominator $0$ in one of the fractions, so these must be excluded from the domain of the functions.

For

$$

y= \frac{x-7}{2x+1}

$$

To invert this put

$$

x = \frac{y-7}{2y+1}

$$

Rearrange:

$$

2xy + x - y = -7\,\quad y(2x-1) = -7-x\, \quad y = -\frac{x+7}{2x-1}

$$

Thus, the inverse of $h$ is

$$

k(x) = -\frac{x+7}{2x-1}

$$

The procedure used by Patrick can be used to invert more general rational functions as follows

$$

f(x)= \frac{ax+b}{cx+d}\quad\mbox{has inverse}\quad g(x)=\frac{b-dx}{cx-a}

$$

To check this properly, consider

$$

f(g(x))=\frac{a\left(\frac{b-dx}{cx-a}\right)+b}{c\left(\frac{b-dx}{cx-a}\right)+d}

$$

This reduces to

$$

f(g(x)) = \frac{a(b-dx)+b(cx-a)}{c(b-dx)+d(cx-a)}=\frac{(bc-ad)x}{bc-ad}

$$

Which cancels to $x$, provided that $bc-ad \neq 0$. The same holds for $g(f(x))$.

We can also consider the generalisation to rational functions involving quadratics.

Some progress can be made, in that if

$$

y = f(x) = \frac{P(x)}{Q(x)}

$$

then you might try to construct the inverse using the idea that $x\leftrightarrow y$, corresponding to reflecting the graph in the line $x=y$ so that

$$

x = \frac{P(y)}{Q(y)}

$$

Rearranging gives

$$

Q(y)x = P(y)

$$

This is a polynomial equation in $y$ of degree equal to the maximum of the degree of $P$ and $Q$.

However, a unique solution will only typically follow if $P(y)$ and $Q(y)$ are linear, and in general no algebraic solution will exist. Moreover, for the inverse to be unique, the function $f(x)$ must be one-to-one, which will not be the case for anything but linear rational functions.