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!
 A: 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 $3$: 
$$
(f^{-1})'(3) = \frac{1}{f'(1)} = \frac{1}{5+6+1} = \frac{1}{12}
$$

Related question.
A: You need to solve the equation
$$x^5 + 2x^3 + x - 1=y$$ for $x$. Unfortunately, such quintic equations are known to have no closed-form solution in general, and this one does not escape the rule.
Anyway, there is a little backdoor, as a quintic can be (after painful computation) reduced to the form known as Bring Quintic Form
$$x^5-x-a=0.$$
Under this particular form, the solutions of $x$ in terms of $a$ are called the Bring radicals of $a$. So if you accept this special univariate function in your toolbox, then you can invert the quintic polynomials.

The cases of linear, quadratic, cubic and quartic polynomials can be solved with the usual functions, with increasing difficulty.
A: 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.
A: 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 $x=\frac {-b \pm \sqrt{b^2-4a(c-y)}}{2a}$.  You need to pick one sign to get a function.
