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

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

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

extreme points

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

Minimum (1, 0)

Solution steps

Find the critical points:

(1, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 0) :

Minimum

Minimum (1, 0)

\frac{x + 3}{x - 2}

\frac{3}{4} \frac{2}{5}

f (x) = 3 - 4 x - 5 x^{2} [- 2.4]

f (x) = \frac{x + 2}{( x + 1 ) ( 2 - x )}

4 + \frac{9 x - 36}{( x - 4 ) ^{2} + 65}

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