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

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

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extreme f(x,y)=x^3-12x-3y+y^3

extreme points

f (x, y) = x^{3} - 12 x - 3 y + y^{3}

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

Solution steps

Find the critical points:

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

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

D (x, y) = 36 x y

Check the sign of at each critical point

Check critical point

(- 2, 1) :

Saddle

Check critical point

(- 2, - 1) :

Maximum

Check critical point

(2, 1) :

Minimum

Check critical point

(2, - 1) :

Saddle

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

f (x) = - x^{3} + 3 x + 4

f (x) = \frac{x ^{2} - 8}{x + 3}

f (x, y) = 3 x y + 2 x y^{2}

\frac{1}{3} (x - 2.1)^{2} + 7.9

f (x, y) = 120 x + 120 y - x y - x^{2} - y^{2}

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