What is the extreme f(x,y)=x^3+3xy^2-3x^2-3y^2+10 ?

The extreme f(x,y)=x^3+3xy^2-3x^2-3y^2+10 is Minimum(2,0),Maximum(0,0),Saddle(1,1),Saddle(1,-1)

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extreme f(x,y)=x^3+3xy^2-3x^2-3y^2+10

extreme points

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

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)

f (x) = - \frac{2}{3} x^{3} - \frac{11}{2} x^{2} - 5 x + 1

f (x, y) = x^{3} + y^{3} - 3 x - 12 y + 13 e

f (x) = - \frac{2}{3} x^{3} - \frac{11}{2} x^{2} - 5 x + 9

f (x) = x^{3} - 9 x^{2} + 27 x - 26

f (x) = \frac{x ^{2} - 10 x + 25}{x - 8}

What is the extreme f(x,y)=x^3+3xy^2-3x^2-3y^2+10 ?
The extreme f(x,y)=x^3+3xy^2-3x^2-3y^2+10 is Minimum(2,0),Maximum(0,0),Saddle(1,1),Saddle(1,-1)