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

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

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

extreme points

f (x) = 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) = 2 x + 2 y

f (x) = \frac{ln ( x )}{2 x}

f (x, y) = 23 - 2 x^{3} - y^{2} - 6 x y

2 x + \frac{1}{x}

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

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