F (x) = e ^ x + e ^ (- x), the calculation process of inverse function

F (x) = e ^ x + e ^ (- x), the calculation process of inverse function

Y = e ^ x + e ^ (- x) is even function, and there is no inverse function on the straight line
On [0, + infinity),
e^x =t
Then t + 1 / T = y, T ^ 2 - y, t + 1 = 0
The solution is: T = (y + - sqrt (y ^ 2-1)) / 2
Because x > 0, t = (y + sqrt (y ^ 2-1)) / 2
Then on [0, + infinity], x = LN, t = ln [(y + sqrt (y ^ 2-1)) / 2]
Similarly, on (- infinity, 0), the inverse function is:
x=ln [(y-sqrt(y^2-1))/2]

Let f (x) = (e ^ x + 1) / (e ^ X - 1) and find its inverse function

y=f(x)=(e^x+1)/(e^x-1)
y-1=(e^x+1)/(e^x-1)-1=2/(e^x-1)
e^x-1=2/(y-1)
e^x=2/(y-1)+1=(y+1)/(y-1)
The inverse function e ^ y = (x + 1) / (x-1)
So y = ln [(x + 1) / (x-1)], (x1)

Inverse function of function f (x) = e ^ X-1 / (e ^ x)

Let f (x) = y, then y = (e ^ x-1) / e ^ X
y*e^x=e^x-1
e^x(y-1)=-1
e^x=-1/(y-1)
ln[-1/(y-1)]=x
And then we swap X and y,
ln[-1/(x-1)]=y
y=ln[-1/(x-1)] x

The inverse function f ^ - 1 (x) =?

y+1=e^x
x=ln(y+1)
So f ^ - 1 (x) = ln (x + 1), x > - 1