Question

Can someone please help me with this.. exam day 2moro.. "Find extrema y=y(x) x^3+y^3-3x^2y-3=0 thnx..

Answer 1

Have you done derivatives yet? if so, this is a problem with implicit differentiation.

Answer 2

yes we've done derivatives but i'm getting kind of lost..

Answer 3

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")

Answer 4

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