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
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
Not sure if you were looking for a more elegant technique here, but the quadratic formula should do the job:
$\displaystyle y = x - \frac{1}{x}$
$\displaystyle y = \frac{x^2-1}{x}$
$\displaystyle xy = x^2-1$
$\displaystyle x^2-yx - 1 = 0$
Apply quadratic formula:
$\displaystyle x = \frac{y\pm \sqrt{y^2+4}}{2}$
Good luck!
I'm just trying to solve one of them at zero
Souldn't I be trying to solve this?
$\displaystyle
x = \frac{y + \sqrt{y^2+4}}{2}
$
And then this?
$\displaystyle
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.
ok
$\displaystyle
0 = (x - \frac{y + \sqrt{y^2+4}}{2})(x - \frac{y - \sqrt{y^2+4}}{2})
$
try the first one....
$\displaystyle
0 = x - \frac{y + \sqrt{y^2+4}}{2}
$
$\displaystyle
-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?