What is the extreme f(x,y)=2x^2-4x+y^2-4y+6 ?

The extreme f(x,y)=2x^2-4x+y^2-4y+6 is Minimum(1,2)

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extreme f(x,y)=2x^2-4x+y^2-4y+6

extreme points

f (x, y) = 2 x^{2} - 4 x + y^{2} - 4 y + 6

Minimum (1, 2)

Solution steps

Find the critical points:

(1, 2)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(1, 2) :

Minimum

Minimum (1, 2)

p (x) = 2 x^{3} - 11 x^{2} + a x - 40

\frac{x ^{4} - 98 x ^{2}}{8}

2 x e^{3 x}

f (x, y) = 1 + x^{2} + x y + y^{2}

f (x, y) = 11 - x^{2} - y^{2}

What is the extreme f(x,y)=2x^2-4x+y^2-4y+6 ?
The extreme f(x,y)=2x^2-4x+y^2-4y+6 is Minimum(1,2)