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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = y + x y^{4}

f (x) = x^{4} + 32 x^{2}

f (x) = \frac{x ^{4} - 32 x ^{2}}{9}

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

f (x) = [0.2 π]

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