What is the extreme f(x,y)=2x^3-2y^3+12xy+3 ?

The extreme f(x,y)=2x^3-2y^3+12xy+3 is Saddle(0,0),Minimum(2,-2)

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extreme f(x,y)=2x^3-2y^3+12xy+3

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 12 y

D (x, y) = - 144 x y - 144

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)

2 x^{3} - 3 x^{2} - 12 x + 8

f (x, y) = x y + e^{x y}

g (x, y) = ln (x^{2} + y^{2} - 1)

f (x) = x^{3} - 3 x - 1

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

What is the extreme f(x,y)=2x^3-2y^3+12xy+3 ?
The extreme f(x,y)=2x^3-2y^3+12xy+3 is Saddle(0,0),Minimum(2,-2)