Suppose I have the function
$f(y)=2y^4-5y^3+3y^2,$
with zeroes $y=0$ (2x), $y=1$, $y=3/2$, which I only need on the part of the domain $0\le y\le 1$.
Is there a transformation $y\rightarrow y'$, such that
- $f(y')$ is even
- The transformation is invertible on the specified domain
- The transformation of the derivative $\frac{d^2f}{dy^2}$ still looks 'nice'
- $0$ and $1$ get mapped to $y'_0$ and $-y'_0$ or vice versa.
I am aware that my third requirement is a bit vague, but I don't know a better formulation for now.
I tried some transformations that first maps $3/2$ and $1$ to the same value, but it will not be invertible for sure.