# Isolating a Variable

• Apr 10th 2010, 06:18 AM
saurabh91
Isolating 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 AM
macosxnerd101
Are you sure factoring doesn't work? I factor $x^2$ out to get $x^2(2-x)$. So the roots are at x = 0, 2.
• Apr 10th 2010, 07:14 AM
Neverquit
solve for x?
Quote:

Originally Posted by saurabh91
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.

Hi do you mean solve for x? eg. x=2?
• Apr 10th 2010, 07:20 AM
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 AM
Sudharaka
Quote:

Originally Posted by saurabh91
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.

Dear saurabh91,

I think you want to subject x in this equation is'nt?? I don't have a hint of how to get the answer but I will let you know when I do. Anyway I will give you the answer which I obtained by using the "Microsoft math".

• Apr 10th 2010, 08:31 AM
saurabh91
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 AM