What is the extreme f(x,y)=x^3+3xy^2-3y^2-3x^2+11 ?

The extreme f(x,y)=x^3+3xy^2-3y^2-3x^2+11 is Minimum(2,0),Maximum(0,0),Saddle(1,1),Saddle(1,-1)

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extreme f(x,y)=x^3+3xy^2-3y^2-3x^2+11

extreme points

f (x, y) = x^{3} + 3 x y^{2} - 3 y^{2} - 3 x^{2} + 11

Minimum (2, 0), Maximum (0, 0), Saddle (1, 1), Saddle (1, - 1)

Solution steps

Find the critical points:

(2, 0), (0, 0), (1, 1), (1, - 1)

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

D (x, y) = (6 x - 6)^{2} - 36 y^{2}

Check the sign of at each critical point

Check critical point

(2, 0) :

Minimum

Check critical point

(0, 0) :

Maximum

Check critical point

(1, 1) :

Saddle

Check critical point

(1, - 1) :

Saddle

Minimum (2, 0), Maximum (0, 0), Saddle (1, 1), Saddle (1, - 1)

What is the extreme f(x,y)=x^3+3xy^2-3y^2-3x^2+11 ?
The extreme f(x,y)=x^3+3xy^2-3y^2-3x^2+11 is Minimum(2,0),Maximum(0,0),Saddle(1,1),Saddle(1,-1)