# Inverse of function

• Feb 19th 2009, 01:28 PM
heneri
Inverse of function
y=3x-1/2x

My attempt:

$\displaystyle y=(6x^2-1)/(2x)$

x= $\displaystyle (6y^2-1)/(2y)$

$\displaystyle 2yx=6y^2-1$

$\displaystyle 2yx-6y^2=-1$

$\displaystyle 2y(x-3y)=-1$

$\displaystyle 2y=-1/(x-3y)$

( I get stuck her what do l do with the y in the denominator ? )
• Feb 19th 2009, 01:48 PM
HallsofIvy
Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.

Solve that quadratic equation for x you will get a "$\displaystyle \pm$". This function does not have a true inverse.
• Feb 20th 2009, 12:09 AM
nyasha
Quote:

Originally Posted by HallsofIvy
Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.

Solve that quadratic equation for x you will get a "$\displaystyle \pm$". This function does not have a true inverse.

I solved it thank you very much for your help
• Feb 27th 2009, 04:24 AM
nyasha
Quote:

Originally Posted by HallsofIvy
Once you have that 2yx= 6y^2- 1, rewrite it as 6y^2- 2xy- 1 and use the quadratic formula to solve for y.

Solve that quadratic equation for x you will get a "$\displaystyle \pm$". This function does not have a true inverse.

If my restricted domain is D={x|x<0} does that mean for my inverse function is going to be the one with a minus sign on the quadratic equation.