Question

Find f^(-1) when f(x)=tanhx

Answer 1

You know that\[\sinh(x)=\frac{e^{x}-e^{-x}}{2}, \cosh(x)=\frac{e^{x}+e^{-x}}{2}\]
So\[f(x)=\frac{e^{2x}-1}{e^{2x}+1}\] here I denote
\[s=\frac{e^{2x}-1}{e^{2x}+1}\]
How do you isolate x though? I'm stuck. :(