# Find inverse of a function

• Sep 18th 2011, 01:57 PM
freestar
Find inverse of a function
$\displaystyle y=\sqrt{x^2-2x}$

I figure the restriction on the domain would be $\displaystyle x\geq2$ or $\displaystyle x\leq0$

$\displaystyle x = \sqrt{y^2-2y}$

$\displaystyle x^2 = y(y-2)$

It's driving me insane. How do I isolate the 'y'?!?
• Sep 18th 2011, 02:03 PM
TheChaz
Re: Find inverse of a function
y^ - 2y - x^2 = 0
a = 1
b = -2
c = -x^2
• Sep 18th 2011, 02:11 PM
skeeter
Re: Find inverse of a function
Quote:

Originally Posted by freestar
$\displaystyle y=\sqrt{x^2-2x}$

I figure the restriction on the domain would be $\displaystyle x\geq2$ or $\displaystyle x\leq0$

$\displaystyle x = \sqrt{y^2-2y}$

$\displaystyle x^2 = y(y-2)$

It's driving me insane. How do I isolate the 'y'?!?

$\displaystyle x^2 = y^2 - 2y$

$\displaystyle x^2 + 1 = y^2 - 2y + 1$

$\displaystyle x^2 + 1 = (y-1)^2$

$\displaystyle \pm \sqrt{x^2+1} = y - 1$

$\displaystyle y = 1 \pm \sqrt{x^2+1}$

now all you have to do is figure out if its plus, or minus
• Sep 18th 2011, 02:52 PM
freestar
Re: Find inverse of a function
ahh completing the square. Thanks a lot!
• Sep 18th 2011, 03:04 PM
TheChaz
Re: Find inverse of a function
Quote:

Originally Posted by freestar
ahh completing the square. And the quadratic formula works too! You guys are geniuses.
Thanks a lot!

I couldn't have said it better myself. You're welcome!(Cool)