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 points

x^{2} + y^{2} + 4 x + 20

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)

\frac{- x + 5}{x ^{2}}

f (x) = \frac{x}{x + 3}, - 1 \leq x \leq 6

f (x) = 4 x y^{2} + 10 x - 4 y^{4} + 9

f (x) = x^{3} - 6 x^{2} + 9 x + 10

4 x + 1

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