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!

  • 2
    $\begingroup$ Even Mathematica can't find inverse function, but you can be confident - inverse function does exist $\endgroup$
    – Norbert
    Oct 10, 2012 at 21:42
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    $\begingroup$ 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. $\endgroup$ Oct 10, 2012 at 22:05
  • $\begingroup$ I think, it's very hard. Is it homework? Or where is this problem from? $\endgroup$
    – Berci
    Oct 10, 2012 at 22:21
  • $\begingroup$ 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/… $\endgroup$ Dec 15, 2012 at 3:42

4 Answers 4


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.

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    $\begingroup$ I don't understand why this isn't the accepted answer. This is brilliant and helped me check the stability of some IIR filters for nonuniformly spaced samples that I have been building. $\endgroup$
    – The Dude
    May 26, 2022 at 17:03

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.

  • $\begingroup$ Thank you! I will try it again. $\endgroup$
    – Jaden M.
    Oct 10, 2012 at 23:47
  • $\begingroup$ The solution that is expected to be x = f(y) does not contain y. Substitute c by c-y ? $\endgroup$
    – h22
    Nov 23, 2018 at 8:02
  • $\begingroup$ @h22: yes, that is correct. Thanks $\endgroup$ Nov 23, 2018 at 15:07

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.


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


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.


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