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

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

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extreme f(x,y)=x^3+y^3-3x-3y+4

extreme points

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

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

Solution steps

Find the critical points:

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

(- 1, 1) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Check critical point

(1, 1) :

Minimum

Check critical point

(1, - 1) :

Saddle

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

F (m, a) = ma

f (x) = 10 + 12 x - 3 x^{2} - 2 x^{3}

f (x) = 3 x y + 12 y x^{5} - 3

f (x, y) = 6 x^{2} - 2 x^{3} + 3 y^{2} + 6 x y

f (x) = x + y

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