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 T O P I C    R E V I E W Kevitzinn 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. 3   L A T E S T    R E P L I E S    (Newest First) Kevitzinn 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. Kevitzinn 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. skeeter 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.

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