If `f: Rvec(-1,1)` defined by `f(x)=(10^x-10^(-x))/(10^x+10^(-x))` is invertible, find `f^(-1)`

Explanation

Injectivity of `f` :


Let `x` and `y` be two elements of domain (R), such that


`f(x)=f(y)`


`Rightarrow frac{10^{x}-10^{-x}}{10^{x}-10^{-x}}=frac{10^{y}-10^{-y}}{10^{y}-10^{-y}}`


`Rightarrow frac{10^{-x}(10^{2 x}-1)}{10^{-y}(10^{2 x}-1)}=frac{10^{-y}(10^{2 y}-1)}{10^{-y}(10^{2 y}-1)}`


`Rightarrow frac{10^{2 x}-1}{10^{2} x+1}=frac{10^{2 y}-1}{10^{2} y+1}`


`Rightarrow(102 {x}-1)(102 {y}+1)=(102 {x}+1)(102 {y}-1)`


`Rightarrow 102 x+2 y+102 x-102 y-1=102 x+2 y-102 x+102 y-1`


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