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

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

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

extreme points

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

Minimum (- 1, 2)

Solution steps

Find the critical points:

(- 1, 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Minimum

Minimum (- 1, 2)

f (x) = 3 x^{4} + 8 x^{3} - 14 x^{2} - 24 x + 17

f (x, y) = \frac{x + y}{x + y}

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

f (x) = 5 x^{2} - 3 x + 1, - 3 \leq x \leq 3

P (a, b) = 6 a^{2} b + 2 a b^{2}

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