y=3x-1/2x
My attempt:
x=
( I get stuck her what do l do with the y in the denominator ? )
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y=3x-1/2x
My attempt:
x=
( I get stuck her what do l do with the y in the denominator ? )
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 "". This function does not have a true inverse.
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 "
I solved it thank you very much for your help
Quote:
Originally Posted by HallsofIvyOnce 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 "
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.