What is the extreme x^2-2xy+2y^2-2x+4y-1 ?

The extreme x^2-2xy+2y^2-2x+4y-1 is Minimum(0,-1)

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extreme x^2-2xy+2y^2-2x+4y-1

extreme points

x^{2} - 2 x y + 2 y^{2} - 2 x + 4 y - 1

Minimum (0, - 1)

Solution steps

Find the critical points:

(0, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, - 1) :

Minimum

Minimum (0, - 1)

f (x) = - 2 sin (x) cos (x)

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

f (x) = 14 x^{2} - x^{3}

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

f (x) = \frac{4 ( x - 10 ) ^{2}}{x - 6}, (0, 6)

What is the extreme x^2-2xy+2y^2-2x+4y-1 ?
The extreme x^2-2xy+2y^2-2x+4y-1 is Minimum(0,-1)