Find the inverse of $xe^{1-2x^2}$, $x\geqslant1$

Question

Find the inverse of $xe^{1-2x^2}$, $x\geqslant1$

This question is killing me and this is my last resort. Thank you in advance I've found great help on here before.

Answer 1

$$y=xe^{1-2x^2}$$ $-4x^2e^{-4x^2}= -4y^2e^{-2}\quad\to\quad We^W=X\quad \begin{cases} W=-4x^2 \\ X=-4y^2e^{-2} \end{cases}$

A closed form requires the special function $W(X)$ namely the Lambert W-function : http://mathworld.wolfram.com/LambertW-Function.html $$-4x^2=W(-4y^2e^{-2})$$ $$x=\pm\frac{1}{2}\sqrt{-W(-4y^2e^{-2})}$$ $x(y)$ isn't bijective on the whole range. But if the range is limited to $x\geq \frac{1}{2}$ the relationship is one-to-one : $$x=\frac{1}{2}\sqrt{-W_{-1}( -4y^2e^{-2} )}$$ where $W_{-1}$ denotes the negative branch of the Lambert W-function, where $W_{-1}(X)<-1$ on the range $-e^{-1}<X<0$

$ -e^{-1}<-4y^2e^{-2}<0 \quad\to\quad 0<y<\frac{\sqrt{e}}{2}$