Inversion in one to one region of a polynomial

Question

Suppose I have a polynomial: $P(x)=x-x^3$ defined for $x\in[0,1]$. Plot of $P(x)$ over this range of $x$ shows that each value of $P$ corresponds to two different values of $x$ (the curve is concave downward, beginning and ending at zero), except at the peak value of the curve, $\hat{P}$, which occurs at $x=1/\sqrt{3}$. Now if I limit $x$ to the interval $[0,1/\sqrt{3}]$, I have a one-to-one correspondence between $P$ and $x$. I was wondering if I can invert the polynomial in this limited region to obtain $x(P)$. My ultimate aim is to obtain $\frac{dx}{dP}$ in this region. Any help is appreciated. Thanks in advance.

Answer 1

$x-x^3=y$ is a cubic equation. It can be solved by algebraic means, but the answer does not look pretty: $$ x=-\frac{\sqrt[3]{2} \left(\sqrt{81 y^2-12}+9 y\right)^{2/3}+2 \sqrt[3]{3}}{6^{2/3} \sqrt[3]{\sqrt{81 y^2-12}+9 y}}. $$ You can find the derivative by implicit differentiation: $$ (3\,x^2-1)\,\frac{dx}{dy}=1. $$