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

The extreme f(x,y)=2x^2+2xy+y^2+4x-2y+1 is Minimum(-3,4)

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extreme f(x,y)=2x^2+2xy+y^2+4x-2y+1

extreme points

f (x, y) = 2 x^{2} + 2 x y + y^{2} + 4 x - 2 y + 1

Minimum (- 3, 4)

Solution steps

Find the critical points:

(- 3, 4)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 3, 4) :

Minimum

Minimum (- 3, 4)

x + cos (x)

f (x) = \frac{900 x}{10 + 45 x}

\frac{5 x}{x ^{2} - 1}

f (x) = x^{3} + 2 x^{2} - 4 x + 2

\frac{d}{d x} (8 x^{4} - 24 x^{2} + r - 1)

What is the extreme f(x,y)=2x^2+2xy+y^2+4x-2y+1 ?
The extreme f(x,y)=2x^2+2xy+y^2+4x-2y+1 is Minimum(-3,4)