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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = \frac{e ^{x}}{1 + x ^{2}}

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

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

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

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

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