Htykuut

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_3left(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.

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.

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.