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

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

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

extreme points

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

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)

e^{- 3.5 x^{2}}

\frac{x ^{2}}{x ^{2} - 4}

y = x^{2} e^{- x}

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

f (x) = x^{3} - 5 x^{2} + 3 x + 10

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