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anonymous
 4 years ago
Can someone please help me with this.. exam day 2moro.. "Find extrema y=y(x) x^3+y^33x^2y3=0 thnx..
anonymous
 4 years ago
Can someone please help me with this.. exam day 2moro.. "Find extrema y=y(x) x^3+y^33x^2y3=0 thnx..

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Have you done derivatives yet? if so, this is a problem with implicit differentiation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes we've done derivatives but i'm getting kind of lost..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if we take the derivative of x^3+y^33x^2y3=0 you get: (using the product rule) 3x^2 + 3y^2*y'6x*y3x^2y'0= 0 rearranging for y': 3y^2*y'  3x^2y' = 3x^2 +6x*y y' (3y^23x^2) = 3x^2 +6x*y y' = (3x^2 +6x*y)/(3y^23x^2) now you have the expression for y'. you should know that when the derivative of a function is 0, it is either a max or a min (which I'm guessing are "extrema")

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's the extreme values for the eq. that really helped. thanx ggrree
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