anonymous
  • anonymous
how to find critical point
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
critical points are where the derivative of the equation equal zero. These points are potential relative maximums and relative minimum depending on if the derivative is before and after the point.
anonymous
  • anonymous
2y^3-x^3+147x-54y+12 find the critical point and relative max or min
anonymous
  • anonymous
so if you take a derivative (with respect to x) you will have to use implicit differentiation so you get: \[6y ^{2}dy/dx-3x ^{2} + 147-54dy/dx\], if you put zero in for dy/dx this simplifies to -3x^2+147 so we have 147 = 3x^2. x = ±7 for our critical points (starting with x^3 we expect to get 2 points)

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