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

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

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extreme g(x,y)=x^3+y^3-3x-3y

extreme points

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

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 (x) = - x^{6} - 3 x^{3} + x + 1

x^{2} = - 4 a y

f (x) = 3 sin (x) cos (x)

f (x) = 2 - ∣ x ∣

f (x) = 4 sin^{2} (x)

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