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

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

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

extreme points

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

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

\frac{x - 7}{x ^{2}}

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

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

\frac{2 x + 1}{x ^{2} + 2 x - 3}

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