Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
Have you done derivatives yet? if so, this is a problem with implicit differentiation.
yes we've done derivatives but i'm getting kind of lost..
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")

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question