Question 1208167
<pre>
Here's an easy way to find the inverse when both numerator and denominator are
linear:

{{{y = "f(x)" = (x + 2)/(x - 2)}}}

The inverse of a rational function with a linear numerator and denominator is also a rational 
function with a linear numerator and denominator.

We know that f(x) has a vertical asymptote at x = 2 and a horizontal asymptote at y = 1,

So we know that f<sup>-1</sup>(x) has a vertical asymptote at x=1 and a horizontal asymptote at y = 2,
so we know the denominator could be {{{x-1}}}, and if it were, the numerator would have to be
2x + something, so we have:

{{{f^(-1)}}}{{{"(x)"}}}{{{""=""}}}{{{(2x+A)/(x-1)}}} 

Now since f(x) goes through (0,-1), f<sup>-1</sup> goes through (-1,0), so

{{{0}}}{{{""=""}}}{{{(2(-1)+A)/(-1-1)}}}

and A has to be 2 

So {{{f^(-1)}}}{{{(x)}}}{{{""=""}}}{{{(2x+2)/(x-1)}}}.

You could just about do that in your head.

Edwin</pre>