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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

(0, - 2) :

Saddle

Check critical point

(0, 0) :

Maximum

Check critical point

(2, - 2) :

Minimum

Check critical point

(2, 0) :

Saddle

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

f (x) = x^{4} - 2 x^{2} + 3, - 2 \leq x \leq 3

f (x) = 3 cos (3 x), - \frac{π}{2} \leq x \leq \frac{π}{2}

\frac{- 2 ( x ^{2} - 1 )}{x ^{2} - 4}

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

y = x^{4} - 18 x^{2}

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