What is the extreme f(x,y)=x^2-xy+y^2+6 ?

The extreme f(x,y)=x^2-xy+y^2+6 is Minimum(0,0)

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extreme f(x,y)=x^2-xy+y^2+6

extreme points

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

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)

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

f (x) = \frac{- 4 t ^{2} + 50 t + 16}{( t ^{2} + 4 ) ^{2}}

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

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

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

What is the extreme f(x,y)=x^2-xy+y^2+6 ?
The extreme f(x,y)=x^2-xy+y^2+6 is Minimum(0,0)