What is the extreme sqrt(x^2+y^2+4x+20) ?

The extreme sqrt(x^2+y^2+4x+20) is Minimum(-2,0)

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extreme sqrt(x^2+y^2+4x+20)

extreme

Minimum (- 2, 0)

Solution steps

Find the critical points:

(- 2, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} + 4 x + 20}{( x ^{2} + y ^{2} + 4 x + 20 ) x ^{2} + y ^{2} + 4 x + 20}

D (x, y) = \frac{16}{( x ^{2} + y ^{2} + 20 + 4 x ) ^{2}}

Check the sign of at each critical point

Check critical point

(- 2, 0) :

Minimum

Minimum (- 2, 0)

What is the extreme sqrt(x^2+y^2+4x+20) ?
The extreme sqrt(x^2+y^2+4x+20) is Minimum(-2,0)