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

The extreme f(x,y)=x^2+y^2-4y+4 is Minimum(0,2)

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

extreme points

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

Minimum (0, 2)

Solution steps

Find the critical points:

(0, 2)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 2) :

Minimum

Minimum (0, 2)

f (x) = x^{3} - 29 x^{2} - 40 x

f (x) = - 4 x^{2} - 8 x - 1.5

f (x, y) = 3 x - 4 x y + 5 y

f (x) = x^{5} + 243

f (x) = 6 - \frac{x ^{3}}{3} + 15 x - x^{2}

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