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

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

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

extreme points

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

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)

f (x) = 2 x^{3} - 3 x^{2} - 12 x - 3

f (x) = 5 (5 x)^{x}, 0.05 \leq x \leq 1

f (x) = 3 x^{3} e^{- x}, - 1 \leq x \leq 6

y = x^{4} - 4 x^{2} = x^{2} (x^{2} - 4)

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 7

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