anonymous
  • anonymous
Can someone please help me with this.. exam day 2moro.. "Find extrema y=y(x) x^3+y^3-3x^2y-3=0 thnx..
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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ggrree
  • ggrree
Have you done derivatives yet? if so, this is a problem with implicit differentiation.
anonymous
  • anonymous
yes we've done derivatives but i'm getting kind of lost..
ggrree
  • ggrree
if we take the derivative of x^3+y^3-3x^2y-3=0 you get: (using the product rule) 3x^2 + 3y^2*y'-6x*y-3x^2y'-0= 0 rearranging for y': 3y^2*y' - 3x^2y' = -3x^2 +6x*y y' (3y^2-3x^2) = -3x^2 +6x*y y' = (-3x^2 +6x*y)/(3y^2-3x^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")

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anonymous
  • anonymous
it's the extreme values for the eq. that really helped. thanx ggrree

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