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

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

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

extreme points

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

Minimum (- 1, 1)

Solution steps

Find the critical points:

(- 1, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 1, 1) :

Minimum

Minimum (- 1, 1)

K (y, z) = 32 yz

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

f (x) = - x^{4} - 8 x^{3} - 18 x^{2} + 6

f (x) = x^{3} - 6 x^{2} + 9 x - 2

f (x) = \frac{x}{x ^{2} - x + 9}

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