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

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

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

extreme points

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

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) = 4 sin (x) + 2

f (x) = 9 x ln (x)

f (x) = 8 x + 9 x^{\frac{8}{9}}

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

f (x) = - 5 x + 3 ln (4 x)