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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = (x - 1) (x + 2) (x + 5)

f (x) = - \frac{x ^{2}}{2} - 3 x - \frac{11}{2}

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

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

f (x) = \frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 1

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