# 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)

= 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)]

We'll re-write f(x)=[e^x-(1/e^x)]/[e^x+(1/e^x)]

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

2/(e^2x+1)<1, f(x)>0, so f(x) is strictly increasing, so f is bijective.

f(x) = y

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

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

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

We'll apply, to the right side, the quotient property of the logarithms:

ln(1-y)=ln2-ln(e^2x+1)

We'll subtract ln2 both sides:

ln(1-y)-ln2 = -ln(e^2x+1)

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

We'll apply one to one property:

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

We'll cross multiply:

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

e^2x+1-y*e^2x-y=2

We'll group the terms in x:

e^2x-y*e^2x - (1+y) = 0

We'll factorize e^2x:

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

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

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

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

We'll divide by 2 both sides:

**x = ln [(1+y)/(1-y)]/2**