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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 36 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, - 1) :

Minimum

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

f (x) = 10 x^{2} + 36 x + 36

f (x) = x^{4} - 4 x^{3} + 12

f (x) = 4 x^{3} + 48 x^{2} + 180 x

f (x) = \frac{x ^{2} - 7 x + 26}{x - 5}

f (x) = x^{3} - x^{2} - 8 x + 4

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