What is the extreme f(x,y)=x^3-y^3+6xy ?

The extreme f(x,y)=x^3-y^3+6xy is Saddle(0,0),Minimum(2,-2)

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extreme f(x,y)=x^3-y^3+6xy

extreme points

f (x, y) = x^{3} - y^{3} + 6 x y

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, - 2) :

Minimum

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

f (x, y) = 288 y^{2} + x^{2} - x^{2} y

f (x, y) = x (x - 4) y (y - 4)

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

f (x) = 2 x^{3} - 3 x^{2} - 12 x - 3

f (x) = 5 (5 x)^{x}, 0.05 \leq x \leq 1

What is the extreme f(x,y)=x^3-y^3+6xy ?
The extreme f(x,y)=x^3-y^3+6xy is Saddle(0,0),Minimum(2,-2)