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

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

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

extreme points

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

Minimum (- 2, 2)

Solution steps

Find the critical points:

(- 2, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 1

D (x, y) = 1

Check the sign of at each critical point

Check critical point

(- 2, 2) :

Minimum

Minimum (- 2, 2)

F (I, J) = 25 I N + 12 J N

(x - x^{2})^{2}, - 1 \leq x \leq 1

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

f (x, y) = 4 x y

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

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