What is the extreme f(xy)=2x^3+y^3+3x^2-3y-12x-4 ?

The extreme f(xy)=2x^3+y^3+3x^2-3y-12x-4 is Minimum(1,1),Saddle(1,-1),Saddle(-2,1),Maximum(-2,-1)

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extreme f(xy)=2x^3+y^3+3x^2-3y-12x-4

extreme

Minimum (1, 1), Saddle (1, - 1), Saddle (- 2, 1), Maximum (- 2, - 1)

Solution steps

Find the critical points:

(1, 1), (1, - 1), (- 2, 1), (- 2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 6 y (12 x + 6)

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(1, - 1) :

Saddle

Check critical point

(- 2, 1) :

Saddle

Check critical point

(- 2, - 1) :

Maximum

Minimum (1, 1), Saddle (1, - 1), Saddle (- 2, 1), Maximum (- 2, - 1)

What is the extreme f(xy)=2x^3+y^3+3x^2-3y-12x-4 ?
The extreme f(xy)=2x^3+y^3+3x^2-3y-12x-4 is Minimum(1,1),Saddle(1,-1),Saddle(-2,1),Maximum(-2,-1)