Question

What is the inverse of `f(x) = e^x-2e^(-x)` ?

Answer 1

`f^(-1)(y) = ln(1/2(y+sqrt(y^2+8)))`

Explanation:

Given:

`f(x) = e^x-2e^(-x)`

Let `y = f(x)`

Then multiplying by `4e^x` and rearranging slightly, we have:

`0 = 4(e^x)^2-4y(e^x)-8`

`color(white)(0) = (2e^x)^2-2(2e^x)y+y^2-y^2-8`

`color(white)(0) = (2e^x-y)^2-sqrt(y^2+8)`

`color(white)(0) = (2e^x-y-sqrt(y^2+8))(2e^x-y+sqrt(y^2+8))`

So:

`2e^x = y+-sqrt(y^2+8)`

For real valued solutions `e^x > 0` and hence we need the `+` sign here to find:

`2e^x = y+sqrt(y^2+8)`

Hence:

`x = ln(1/2(y+sqrt(y^2+8)))`

That is:

`f^(-1)(y) = ln(1/2(y+sqrt(y^2+8)))`