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

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

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

extreme

Minimum (- 1, 0)

Solution steps

Find the critical points:

(- 1, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} + 2 x + 2}{( x ^{2} + y ^{2} + 2 x + 2 ) x ^{2} + y ^{2} + 2 x + 2}

D (x, y) = \frac{1}{( x ^{2} + y ^{2} + 2 + 2 x ) ^{2}}

Check the sign of at each critical point

Check critical point

(- 1, 0) :

Minimum

Minimum (- 1, 0)

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