# Thread: Find inverse of a function

1. ## 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'?!?

2. ## Re: Find inverse of a function

y^ - 2y - x^2 = 0
a = 1
b = -2
c = -x^2

3. ## Re: Find inverse of a function

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

4. ## Re: Find inverse of a function

ahh completing the square. Thanks a lot!

5. ## Re: Find inverse of a function

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!