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

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

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

extreme points

f (x) = 2 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, y) = 5 x^{2} - 3 y^{2} + 4 x y - 32 x + 10 y

f (x, y) = - 3 x^{2} - 2 y^{2} + 3 x - 4 y + 5

f (x) = 5 x + 5 sin (x)

f (x) = 5 - 3 sin (2 x)

\frac{2 x - 4}{3}