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) = 4 cos^{2} (x)

f (x) = 2 x^{3} - 24 x + 1

g (x, y) = e^{x} - x + e^{y} - y

f (x) = (x + 6)^{\frac{2}{3}}

f (x) = (x + 8)^{\frac{2}{3}}

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)