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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 6 y (6 x - 6)

Check the sign of at each critical point

Check critical point

(0, 1) :

Saddle

Check critical point

(0, - 1) :

Maximum

Check critical point

(2, 1) :

Minimum

Check critical point

(2, - 1) :

Saddle

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

x^{5} - 5 x^{4} + 8

f (x) = (x^{2} + 12) (4 - x^{2})

4 x^{2} + 6 + \frac{6}{x ^{2} - 4 x + 4}

(x - y) (25 - x y)

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

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