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

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

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

extreme points

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

Minimum (0, 1)

Solution steps

Find the critical points:

(0, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 1) :

Minimum

Minimum (0, 1)

36 - x^{2} - x + 3

f (x, y) = (x^{2} + y) e^{\frac{y}{2}}

h (t, v) = 5 t^{2} + v t

x 5 - x

- x^{3} + 6 x^{2} - 9 x

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