Find all the relative maxima, relative minima and saddle points for
f(x,y) = x^3+y^2-6x^2+y-1 classify each using the second partials test
f(x,y) = x^3+y^2-6x^2+y-1 classify each using the second partials test
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Critical points:
f_x = 3x^2 - 12x = 3x(x - 4)
f_y = 2y + 1.
Setting these equal to 0 yields (x, y) = (0, -1/2), (4, -1/2).
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Classifying these with the Second Derivative Test:
f_xx = 6x - 12, f_yy = 2, f_xy = 0
==> D = (f_xx)(f_yy) - (f_xy)^2 = 2(6x - 12).
Since D(0, -1/2) < 0, we have a saddle point at (0, -1/2).
Since D(4, -1/2) > 0 and f_xx (4, -1/2) > 0, we have a local minimum at (4, -1/2).
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I hope this helps!
f_x = 3x^2 - 12x = 3x(x - 4)
f_y = 2y + 1.
Setting these equal to 0 yields (x, y) = (0, -1/2), (4, -1/2).
----------------
Classifying these with the Second Derivative Test:
f_xx = 6x - 12, f_yy = 2, f_xy = 0
==> D = (f_xx)(f_yy) - (f_xy)^2 = 2(6x - 12).
Since D(0, -1/2) < 0, we have a saddle point at (0, -1/2).
Since D(4, -1/2) > 0 and f_xx (4, -1/2) > 0, we have a local minimum at (4, -1/2).
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I hope this helps!