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!

This is an experimental way of working out the inverse.
We can treat the polynomial like an expansion
\begin{equation}
f(x) = -1 + x + 0x^2 + 2x^3 + 0x^4 + x^5 + 0x^6 + 0x^7 + \cdots
\end{equation}
then we can perform a Series Reversion on this to give the inverse series (as an infinite expansion) \begin{equation}
f^{-1}(x) = (1+x) -2(1+x)^3 +11(1+x)^5-80(1+x)^7+665(1+x)^9-\cdots
\end{equation}
at a glance this doesn't seem to helpful, but if we search the coefficients in the OEIS we seem to get a hit! It would appear (conjecture) that the general coefficient is then \begin{equation}
a(n) = \binom{5n+1}{n}\frac{(-1)^n}{2n+1}
\end{equation}
and we could write that \begin{equation}
f^{-1}(x) = \sum_{n=0}^\infty \binom{5n+1}{n}\frac{(-1)^n}{2n+1}(x+1)^{2n+1}
\end{equation}
if we evaluate that in Mathematica, it gives \begin{equation}
f^{-1}(x)=(1+x)\;_4F_3\left(\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5}\bigg|\frac{3}{4},\frac{5}{4},\frac{6}{4}\bigg|-\frac{5^5}{4^4}(1+x)^2 \right)
\end{equation}
a generalised hypergeometric function. If we plot the composition of these two i.e $f(f^{-1}(x))$ or $f^{-1}(f(x))$ the plot seems to indicate \begin{equation}
f(f^{-1}(x))=f^{-1}(f(x))=x
\end{equation}
of course this is not a proof, and evaluating the inverse function may become numerically unstable if $x$ becomes too large.
