What is the extreme f(x,y)=x^3+3x^2+6xy+y^2 ?

The extreme f(x,y)=x^3+3x^2+6xy+y^2 is Saddle(0,0),Minimum(4,-12)

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extreme f(x,y)=x^3+3x^2+6xy+y^2

extreme points

f (x, y) = x^{3} + 3 x^{2} + 6 x y + y^{2}

Saddle (0, 0), Minimum (4, - 12)

Solution steps

Find the critical points:

(0, 0), (4, - 12)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 12 x - 24

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, - 12) :

Minimum

Saddle (0, 0), Minimum (4, - 12)

\frac{( 2 x + 1 )}{( e ^{x^{2}} )}

f (y) = 3 y^{2} - 3 x^{2} - 6 y

f (x, y) = x^{3} y + 12 y x^{2} - 8 y

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

f (x) = \frac{5 - x ^{2}}{x}

What is the extreme f(x,y)=x^3+3x^2+6xy+y^2 ?
The extreme f(x,y)=x^3+3x^2+6xy+y^2 is Saddle(0,0),Minimum(4,-12)