What is the extreme f(x,y)=x^2-4x+2y^2+4y+7 ?

The extreme f(x,y)=x^2-4x+2y^2+4y+7 is Minimum(2,-1)

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extreme f(x,y)=x^2-4x+2y^2+4y+7

extreme points

f (x, y) = x^{2} - 4 x + 2 y^{2} + 4 y + 7

Minimum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(2, - 1) :

Minimum

Minimum (2, - 1)

y = - 2 cos (x)

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

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

f (x) = e^{6 x^{2} + 5 y^{2} + 7}

f (x) = x^{2} + x y + y^{2} - 9 x + 7

What is the extreme f(x,y)=x^2-4x+2y^2+4y+7 ?
The extreme f(x,y)=x^2-4x+2y^2+4y+7 is Minimum(2,-1)