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

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

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

extreme points

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

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

g (x, y) = e^{x} - x + e^{y} - y

\frac{1}{3 - 2 x - x ^{2}}

f (x) = x x^{2} + 36

\frac{- 2 x}{x ^{2} + y ^{2} + 1}

f (x) = x^{2} e^{16 x}

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