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

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

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

extreme points

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

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)

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

f (x) = e^{x^{2} - 3 x - 1}, - 3 \leq x \leq 3

f (x, y) = 4 - x^{2} + 9 - y^{2}

f (x, y) = (5 x + 7 y + 25) e^{- (x^{2} + x y + y^{2})}

x^{3} - 3 x^{2} + 2

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