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

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

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

extreme points

f (x) = 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)

P (y, x) = - y^{3} - 2 y^{2} - x y

x 2 - x

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

f (x) = - e^{- x}

f (x) = \frac{4}{3} x^{3} - 12 x^{2} + 27 x - 6

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