Inverse of equation involving e^(-x)

**romjke** · March 19th 2010, 02:57 AM

Hi,

My problem is to find the inverse of $y = f(x) = \frac{1}{2}(e^x - \frac{1}{e^x})$ .

I can get up to using $v$ for $e^x$ , and that $v > 0$

$2yv = v^2 -1$

$0 = v^2 - 2yv - 1$

The using the quadratic formula I get $v = \frac{2y \pm \sqrt{(-2y)^2 + 4}}{2}.$

Here I am stuck

Sorry for bad format.

**Deadstar** · March 19th 2010, 03:25 AM

Following on from where you left off...

$v = \frac{2y}{2} \pm \frac{\sqrt{4(y^2 + 1)}}{2}$ (since $(-2y)^2 + 4 = 4y^2 + 4 = 4(y^2 + 1)$ )

Note that $2 = \sqrt{4}$ so we get

$v = y \pm \sqrt{(y^2 + 1)}$

Now since $v = e^x$ , take natural logs of both sides to get...

$x = \ln(v) =$ ..? Can you finish?

**romjke** · March 19th 2010, 03:43 AM

Yes.

I kept going around in circles about what to do with the square root in the quadratic formula.

All is good now. Thanks

**mr fantastic** · March 19th 2010, 04:11 PM

Originally Posted by Deadstar

Following on from where you left off...

$v = \frac{2y}{2} \pm \frac{\sqrt{4(y^2 + 1)}}{2}$ (since $(-2y)^2 + 4 = 4y^2 + 4 = 4(y^2 + 1)$ )

Note that $2 = \sqrt{4}$ so we get

$v = y \pm \sqrt{(y^2 + 1)}$

Now since $v = e^x$ , take natural logs of both sides to get...

$x = \ln(v) =$ ..? Can you finish?

I'll add a small note: One of the solutions (either the positive or the negative root) must be discarded ....

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