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

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

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

extreme points

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

Saddle (0, 0), Maximum (- 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) :

Maximum

Saddle (0, 0), Maximum (- 2, - 2)

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

x^{4} - 6 x^{2}

f (x) = x^{2} - 8 x + 12

f (x) = x^{3} - 3 x^{2} - 9 x + 7

f (x) = x^{3} - 3 x^{2} - 9 x + 4

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