# f ^-1 (x) = (e^x - e^-x) / 2

• Jul 18th 2013, 01:44 PM
kjtruitt
f ^-1 (x) = (e^x - e^-x) / 2
That is, the inverse of (e^x - e^-x) / 2. The answer is x + sqrt(x^2 + 1). If you start with x = (e^y - e^-y ) / 2, what intermediary steps lead to the answer?

You can clear the denominator:
2x = e^y - e^-y;

you can re-write e^-y as 1/e^y:
2x = e^y - 1/e^y;

you can then clear that denominator:
2xe^y = e^2y - 1;

I'm basically out of ideas at this point. Hmm term on the right is a difference of two squares. I would appreciate any comments!
• Jul 18th 2013, 02:01 PM
Plato
Re: f ^-1 (x) = (e^x - e^-x) / 2
Quote:

Originally Posted by kjtruitt
That is, the inverse of (e^x - e^-x) / 2. The answer is x + sqrt(x^2 + 1). If you start with x = (e^y - e^-y ) / 2, what intermediary steps lead to the answer?
You can clear the denominator:
2x = e^y - e^-y;

$\\2x = e^y - e^{-y}\\2xe^y=e^{2y}-1\\e^{2y}-2xe^y-1=0\\\\\text{So }e^y=\frac{2x+\sqrt{4x^2+4}}{2}$
• Jul 18th 2013, 02:32 PM
HallsofIvy
Re: f ^-1 (x) = (e^x - e^-x) / 2
When Plato go to $e^{2y}- 2xe^y- 1= 0$, he thought of that as a quadratic equation, $(e^y)^2- 2x e^y- 1= 0$ of the form $au^2+ bu+ c= 0$ with a= 1, $u= e^y$, b= -2x, and c= -1. The quadratic formula says that the solutions to $au^2+ bu+ c= 0$ are of the form $u= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$. Setting a= 1, b= -2x, c= -1 that gives $u= e^y= \frac{-(-2x)\pm\sqrt{(-2x)^2- 4(1)(-1)}}{2(1)}= \frac{2x\pm\sqrt{4x^2+ 4}}{2}$.

But notice that $4x^2+ 4> 4x^2$ so that $2x- \sqrt{4x^2+ 4}< 0$ and, or course, $e^y$ is never negative so we can drop the "-" sign to get exactly what Plato gave.

(It can be simplified a little more: $4x^2+ 4= 4(x^2+ 1)$ and $\sqrt{4(x^2+ 1)}= 2\sqrt{x^2+ 1}$ so that $\frac{2x+ \sqrt{4x^2+ 4}}{2}= x+ \sqrt{x^2+ 1}$.)
• Jul 18th 2013, 05:55 PM
kjtruitt
Re: f ^-1 (x) = (e^x - e^-x) / 2
Got it thanks