1. ## inverse of x-1/x

Its been quite a few years since my calculus days so I'm probably missing something obvious. Somebody can probably do this in a couple seconds...

f(x)=y-(1/y)

solve for y.

many thanks

2. first of all you have a typo

you probably mean f(y)=y-1/y

in any case to solve y=x-1/x, you multiply by x giving you a quadractic

$xy=x^2-1$ so use the quadratic equation and solve for x.

3. Originally Posted by scottboston
Its been quite a few years since my calculus days so I'm probably missing something obvious. Somebody can do this a a couple seconds...

f(x)=y-(1/y)

solve for y.

many thanks
Not sure if you were looking for a more elegant technique here, but the quadratic formula should do the job:

$y = x - \frac{1}{x}$

$y = \frac{x^2-1}{x}$

$xy = x^2-1$

$x^2-yx - 1 = 0$

$x = \frac{y\pm \sqrt{y^2+4}}{2}$

Good luck!

4. Thanks for the replies

Something is up using the quadratic here because I'm getting 0=4 when I try to use the result $

0 = (x + \frac{y + \sqrt{y^2+4}}{2})(x + \frac{y - \sqrt{y^2+4}}{2})
$

Any thoughts? thanks

5. Originally Posted by scottboston
Thanks for the replies

Something is up using the quadratic here because I'm getting 0=4 when I try to use the result $

0 = (x + \frac{y + \sqrt{y^2+4}}{2})(x + \frac{y - \sqrt{y^2+4}}{2})
$

Any thoughts? thanks
why are you taking x plus these roots?

6. I'm just trying to solve one of them at zero

Souldn't I be trying to solve this?
$
x = \frac{y + \sqrt{y^2+4}}{2}
$

And then this?
$
x = \frac{y - \sqrt{y^2+4}}{2}
$

I know I'm missing something here...something tells me I'll be doing and infinite loop of quadratic equations.

7. try subtracting

8. ok

$

0 = (x - \frac{y + \sqrt{y^2+4}}{2})(x - \frac{y - \sqrt{y^2+4}}{2})
$

try the first one....

$
0 = x - \frac{y + \sqrt{y^2+4}}{2}
$

$
-2x = -y + \sqrt{y^2+4}
$

This is where I end. I don't think I want any xy multiple so I'm not sure what to do with the square root. How am I doing?