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

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

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

extreme points

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

Minimum (2, 1)

Solution steps

Find the critical points:

(2, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Minimum (2, 1)

f (x) = 5 x^{5} - 3 x^{2} + x - 10, - 5 \leq x \leq 6

f (x, y) = 32 y^{2} + x^{2} - x^{2} y

f (x, y) = 3 x^{2} + 6 x y + 7 y^{2} - 2 x + 4 y

f (x) = x^{\frac{1}{7}} - x^{- \frac{6}{7}}

f (x) = (x - 2)^{3} (3 x + 4)^{2}

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