Schrödinger operator with a potential defined implicitly

let be the problem

$$-\frac{d^{2}}{dx^{2}}y(x)+f(x)y(x)=E_{n}y(x)$$

however we have a problem, we do not know the potential but its inverse

$$f^{-1}(x)=g(x)$$

we know $g(x)$ but not $f(x)$ what happense then since

a) the function $f(x)$ may be multi-valued

b) the function $f(x)$ may not exists even if we know $g(x)$

howe can we analytically solve $f^{-1}(x)=g(x)$ so we get $f(x)$ ?? in the case $f(x)=f(-x)$ is even.

-
is $f^{-1}=1/f$ or is it the inverse function, i.e., $g(f(x))=x$? – Arnold Neumaier Nov 22 '12 at 13:06
$f^{-1)$ is the inverse function i mean $f^{-1}(f(x))=x$ – Jose Javier Garcia Nov 22 '12 at 13:54

Apart from that, one doesn't need an explicit potential. If one calculates solutions numerically it is enough that one can compute the potential numerically at any given $x$.