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If rational functions have the form $f(x)=\frac{P(x)}{Q(x)}$, where $P(x)$ and $Q(x)$ are polynomials, how do you find the inverse of $f(x)$ when $P(x)$ and/or $Q(x)$ are polynomials of degree $2$ or higher?

for example: $$\frac {3x+2}{x^2-4}, \frac {x^2-4}{3x+2}, \frac {x^2+2x-3}{x^3-4x}$$

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It's tough, whenever even possible, to get an explicit expression. You have to solve $P(x)-yQ(x)=0$ for $x$ in terms of $y$ (assuming $P$ and $Q$ have no common root), and there's no closed-form ways of doing that for degree larger than $4$ (short of hypergeometric functions). If the degree of both $P$ and $Q$ are $2$ or less then you're safe with the quadratic formula. (And in case it isn't clear enough, there is generally no unique inverse.) – anon Sep 28 '11 at 5:56
@anon if $P(x) - yQ(x) = 0$, then that would mean that $P(x) = yQ(x)$? After you solved y, then what would you do? – Matt Munson Sep 28 '11 at 6:37
After you solved for $y$ you'd have your inverse. Do you understand that $y=P(x)/Q(x)$ means asking for the inverse is equivalent to asking to solve for $x$ in terms of $y$...? – anon Sep 28 '11 at 6:47
@anon yeah I get that. I was just a bit sidetracked by the $=0$ notation. What would be the closed form methods for degree 2 and 3. how would you reliably get $P(x)-yQ(x)$ to a form that could be plugged into the quadratic formula? – Matt Munson Sep 28 '11 at 7:08
If $P$ and $Q$ are of degree 2 then $P(x)-yQ(x)$ can be rewritten as $()x^2+()x+()$, then you just plug the expressions in where necessary. For degree 3 you'll need to be able to solve a cubic - which generally requires a reduction method and really big formula. – anon Sep 28 '11 at 7:24

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