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

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

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

extreme points

f (x) = 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)

ln (x + 2) + \frac{1}{x}, 1 \leq x \leq 5

\frac{( x ^{2} - 3 )}{x - 2}

f (x) = - 2 x^{2} + 68 x - 60

y = 2 x^{2} + 8 x + 8

- \frac{6 x}{6 x ^{2} + 8}

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