If (x) = [(e^x) - ( e^-x)]/[(e^x) + (e^-x)] bijective, calculate the n order of it's inverse.

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f(x) = [(e^x) - (e^-x)]/ (e^x)+(2^-x)]

     = [(e^x)-(1/e^x)]/[(e^x)+(1/e^x)]

     = (e^2x -1)/ (e^2x) +1)

     = (e^2x +1 -2)/ (e^2x)+1)

    =  1 -[2/(e^2x)+1]

y= 1- [2/(e^2x)+1)]

y-1= -2/(e^2x+1)

==> e^2x +1 = -2/(y-1)

==> e^2x = -2/(y-1)  -1

==> 2x= ln [(-2/(y-1)  -1]

        = ln (-2-(y-1)/(y-1)

        =...

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f(x) = [(e^x) - (e^-x)]/ (e^x)+(2^-x)]

     = [(e^x)-(1/e^x)]/[(e^x)+(1/e^x)]

     = (e^2x -1)/ (e^2x) +1)

     = (e^2x +1 -2)/ (e^2x)+1)

    =  1 -[2/(e^2x)+1]

y= 1- [2/(e^2x)+1)]

y-1= -2/(e^2x+1)

==> e^2x +1 = -2/(y-1)

==> e^2x = -2/(y-1)  -1

==> 2x= ln [(-2/(y-1)  -1]

        = ln (-2-(y-1)/(y-1)

        = ln (-1-y)(y-1)

==> x=(1/2)*ln (-1-y)/(y-1)

The the inverse is:

f(x)^-1= (1/2)*ln [(-1-x)/(x-1)]

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