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

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

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

extreme points

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

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)

f (x, y) = 10 - x^{2} - y^{2}

y = \frac{1}{5} x^{5} - \frac{8}{3} x^{3} + 16 x

f (x) = \frac{( x ^{2} - 7 )}{x - 4}

\frac{x ^{2} + 1}{x ^{2} - 1}

f (x) = \frac{x ^{2} - 12}{x - 4}

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