sech(x) inverse for x< 0

Question

I know this is probably a basic question but I spent about an hour googling it and can't find any answer actually dressing this.

I have a function $f(x) = sech(x)$ for $x<0 $

I got the log form of the inverse as

$\log{((1- \sqrt{1-X^2} /x))} $

which seems correct.

I'm just confused because the range of the inverse function doesn't seem to equal the domain of the original function

Thanks

Answer 1

Draw graphs

sec(x) always $>0$, $\le 1$

interchange $x,y$ axes for inverse function

$\sec^{-1}(x) $always $x>0$ as a real function. It does not exist real for $x<0$.