What is the extreme f(x)=x^2+y^2-6x+2 ?

The extreme f(x)=x^2+y^2-6x+2 is Minimum(3,0)

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extreme f(x)=x^2+y^2-6x+2

extreme points

f (x) = x^{2} + y^{2} - 6 x + 2

Minimum (3, 0)

Solution steps

Find the critical points:

(3, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(3, 0) :

Minimum

Minimum (3, 0)

Q (x, y) = x y

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

f (x) = (8 - 4 x)^{8}

f (x) = 2 x^{2} - 8 x^{4}

f (x) = - x^{\frac{2}{3}} (x - 3), - 3 \leq x \leq 3

What is the extreme f(x)=x^2+y^2-6x+2 ?
The extreme f(x)=x^2+y^2-6x+2 is Minimum(3,0)