Hi, I am trying to do this question for a long time now, and i cant figure it out.

F(x) = 2x^2-x^3; isolate x using any way you can think of.

P.S: Factoring, and log/Ln doesn't work. I thought this couldn't be solved but apparently it can.

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- Apr 10th 2010, 06:18 AMsaurabh91Isolating a Variable
Hi, I am trying to do this question for a long time now, and i cant figure it out.

F(x) = 2x^2-x^3; isolate x using any way you can think of.

P.S: Factoring, and log/Ln doesn't work. I thought this couldn't be solved but apparently it can. - Apr 10th 2010, 07:11 AMmacosxnerd101
Are you sure factoring doesn't work? I factor $\displaystyle x^2$ out to get $\displaystyle x^2(2-x)$. So the roots are at x = 0, 2.

- Apr 10th 2010, 07:14 AMNeverquitsolve for x?
- Apr 10th 2010, 07:20 AMmaddas
I believe they mean to solve for x as a function of F(x) i.e. invert the function. It is cubic in x, so the solution is known, but its not pretty, I can tell you that (Wolfram Alpha can do it).

- Apr 10th 2010, 08:19 AMSudharaka
- Apr 10th 2010, 08:31 AMsaurabh91
yes i m trying to find the inverse of the function.. with not much success. I tried taking the integral of the function and finding the area of random point (0,1).... the area of the inverse will also be the same so i tried finding the inverse from there but that leaves me nowhere.

- Apr 10th 2010, 08:35 AMmaddas
- Apr 12th 2010, 06:21 PMsaurabh91
Thank you for the help everyone!