# Inverse function of a polynomial

What is the inverse function of $f(x) = x^5 + 2x^3 + x - 1?$ I have no idea how to find the inverse of a polynomial, so I would greatly appreciate it if someone could show me the steps to solving this problem. Thank you in advance!

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please enclose your math in \$\$ to have it typeset nicely – Valentin Oct 10 '12 at 21:41
Even Mathematica can't find inverse function, but you can be confident - inverse function does exist – Norbert Oct 10 '12 at 21:42
Your polynomial is increasing, and its range is all reals, so there is an inverse. Finding a pleasant expression for the inverse is another matter. But one can find information about the derivative of the inverse without knowing a formula. – André Nicolas Oct 10 '12 at 22:05
I think, it's very hard. Is it homework? Or where is this problem from? – Berci Oct 10 '12 at 22:21
This question has been asked before, albeit in greater detail. Most likely we're having trouble answering it because we're missing this information: math.stackexchange.com/questions/85226/… – Gyu Eun Lee Dec 15 '12 at 3:42

Generally, you say $y=$ your polynomial and solve for $x$. Fifth degree polynomials are generally not solvable. The general approach for a quadratic would be essentially the quadratic formula. Given $y=ax^2+bx+c$, you find $x=\frac {-b \pm \sqrt{b^2-4ac}}{2a}$. You need to pick one sign to get a function.

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Thank you! I will try it again. – Jaden M. Oct 10 '12 at 23:47

Maybe try the Lagrange inversion theorem

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As others indicated, there is no algebraic formula for the inverse function $f^{-1}$. The inverse functions exists (since $f$ is increasing), but there are serious algebraic obstructions to solving $y=x^5 + 2x^3 + x - 1$ for $x$.

But we can find particular values of $f^{-1}$ and of its derivative. For example, to find $f^{-1}(3)$ we would just have to note that $f(1)=3$. Therefore, $f^{-1}(3)=1$. (The number $3$ is lucky here; if asked about $f^{-1}(4)$, one would need a numerical method, e.g., a calculator.)

Also, the inverse function theorem can be used to find the derivative of $f^{-1}$ at $3$: $$(f^{-1})'(3) = \frac{1}{f'(1)} = \frac{1}{5+6+1} = \frac{1}{12}$$

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switch $f(x)$ with $x$ and you get $x= y^5 + 2y^3 +y-1$ and solve for $y$.

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How is this different from the original question? – leo Dec 15 '12 at 3:54
How do you solve for $y$ then? – Haskell Curry Jan 15 '13 at 6:32