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!

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;

$\displaystyle \\2x = e^y - e^{-y}\\2xe^y=e^{2y}-1\\e^{2y}-2xe^y-1=0\\\\\text{So }e^y=\frac{2x+\sqrt{4x^2+4}}{2}$

Re: f ^-1 (x) = (e^x - e^-x) / 2

When Plato go to $\displaystyle e^{2y}- 2xe^y- 1= 0$, he thought of that as a **quadratic** equation, $\displaystyle (e^y)^2- 2x e^y- 1= 0$ of the form $\displaystyle au^2+ bu+ c= 0$ with a= 1, $\displaystyle u= e^y$, b= -2x, and c= -1. The quadratic formula says that the solutions to $\displaystyle au^2+ bu+ c= 0$ are of the form $\displaystyle u= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$. Setting a= 1, b= -2x, c= -1 that gives $\displaystyle u= e^y= \frac{-(-2x)\pm\sqrt{(-2x)^2- 4(1)(-1)}}{2(1)}= \frac{2x\pm\sqrt{4x^2+ 4}}{2}$.

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

(It can be simplified a little more: $\displaystyle 4x^2+ 4= 4(x^2+ 1)$ and $\displaystyle \sqrt{4(x^2+ 1)}= 2\sqrt{x^2+ 1}$ so that $\displaystyle \frac{2x+ \sqrt{4x^2+ 4}}{2}= x+ \sqrt{x^2+ 1}$.)

Re: f ^-1 (x) = (e^x - e^-x) / 2