What is the extreme f(x,y)=sqrt(x^2+y^2+6x+10) ?

The extreme f(x,y)=sqrt(x^2+y^2+6x+10) is Minimum(-3,0)

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extreme f(x,y)=sqrt(x^2+y^2+6x+10)

extreme points

f (x, y) = x^{2} + y^{2} + 6 x + 10

Minimum (- 3, 0)

Solution steps

Find the critical points:

(- 3, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} + 6 x + 10}{( x ^{2} + y ^{2} + 6 x + 10 ) x ^{2} + y ^{2} + 6 x + 10}

D (x, y) = \frac{1}{( x ^{2} + y ^{2} + 10 + 6 x ) ^{2}}

Check the sign of at each critical point

Check critical point

(- 3, 0) :

Minimum

Minimum (- 3, 0)

f (x) = - 34.3 x^{2} + 460.2 x

f (x) = - 3 (5 - 3 x) e^{- 4 x}, (- 1, 4)

f (x) = x^{3} - 6 x^{2} + 18

f (x) = 4 x + \frac{8}{x}

f (x) = - x^{2} + 229 x

What is the extreme f(x,y)=sqrt(x^2+y^2+6x+10) ?
The extreme f(x,y)=sqrt(x^2+y^2+6x+10) is Minimum(-3,0)