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Author  Topic

Kevitzinn
Average Member

USA
14 Posts

 Posted - 10/19/2007 :  13:53:53 So I have, y = x(2-x) and I find the derivative is y' = (x)/(3(2-x)^(2/3)) + (2-x)Is there some easy way to find the critical points of that? Thank you very much for your time.

skeeter

USA
5634 Posts

 Posted - 10/19/2007 :  14:47:03 y = x(2 - x)1/3y' = x*(1/3)(2 - x)-2/3*(-1) + (2 - x)1/3y' = (2 - x)-2/3[-(x/3) + (2 - x)]y' = (2 - x)-2/3[2 - (4x/3)]y' = [2 - (4x/3)]/(2 - x)2/3you should be able to almost "see" the critical values now.

Kevitzinn
Average Member

USA
14 Posts

 Posted - 10/19/2007 :  16:26:24 I see you have the right answer and I appreciate your time very much, but I have just one more question.You said:y' = (2 - x)-2/3[-(x/3) + (2 - x)]What I'm confused about is, how did you do that? If -x/3 is to the first power and and (2-x) is to the (1/3) power, how can you pull it out like that? Thanks.

Kevitzinn
Average Member

USA
14 Posts

 Posted - 10/21/2007 :  23:53:39 Ohhhh, that's amazing! Thanks for the help, I would have never thought it could have been pulled out like that! Again, thanks a lot for the help, this site is the best.
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