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 $g(f(x))$.

On Ask NRICH, some discussion took place concerning 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.