Hi Frank,

thanks for you Answer!

I am not sure if FindRoot is able to do what I want to do. Or I am doing something wrong still, since I am getting cryptic error messages - it doesnt like the starting point term. I will add the code of the concrete function I am given, so it is easier to follow: I have given the relation r = rfunc(R) with

rfunc[R_] := (2 R + M + Sqrt[4 R^2 + 4 M R + 3 M^2])/

4*(((4 + 3 Sqrt[2])*(2 R - 3 M))/(

8 R + 6 M + 3*Sqrt[8 R^2 + 8 M R + 6 M^2]))^(1/Sqrt[2]);

M := 1;

Plot[rfunc[R], {R, 0, 6}]

, and where M = 1.

As one can see from the plot, R goes to 3/2 as r goes to 0, and R goes to infinity as r goes to infinity. Inbetween the function is one-to-one, so it should be possible to invert the function numerically on this domain. So I want an expression R = somefunction.

What I tried before is,

NSolve[rfunc[R]==r,R]

but this does not work.

Can FindRoot do this, and if yes, how? Or can NSolve do this, with specifying additional assumptions on the domain and/or range? If so, how would I do this?

The Mathematica Help to the assumption functions is not really helpfull here for me. : /