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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 5

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

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

x^{2} y + y^{3} - 27 y

f (x) = \frac{2 x}{x ^{2} - 4}

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

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