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

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

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

extreme points

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

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) = \frac{x ^{2} - 11 x + 82}{x - 9}

x^{3} - 12 x^{2} + 45 x + 1

10

f (x) = 3 x^{4} - 8 x^{3} - 6 x^{2} + 24 x - 3

f (x) = 3 x^{5} - x^{3} + 4 x - 2

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