What is the extreme f(x)=x^2+y^2-4x-1 ?

The extreme f(x)=x^2+y^2-4x-1 is Minimum(2,0)

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extreme f(x)=x^2+y^2-4x-1

extreme points

f (x) = x^{2} + y^{2} - 4 x - 1

Minimum (2, 0)

Solution steps

Find the critical points:

(2, 0)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(2, 0) :

Minimum

Minimum (2, 0)

f (x) = \frac{x ^{2} + 7 x + 3}{x + 2}

(5 + x)^{5}

f (x) = (x - 6) (x - 4)

f (x) = \frac{1}{4} x - ln (x)

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

What is the extreme f(x)=x^2+y^2-4x-1 ?
The extreme f(x)=x^2+y^2-4x-1 is Minimum(2,0)