What is the extreme f(x,y)=sqrt(x^2+y^2+9) ?

The extreme f(x,y)=sqrt(x^2+y^2+9) is Minimum(0,0)

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extreme f(x,y)=sqrt(x^2+y^2+9)

extreme points

f (x, y) = x^{2} + y^{2} + 9

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} + 9}{( x ^{2} + y ^{2} + 9 ) x ^{2} + y ^{2} + 9}

D (x, y) = \frac{9}{( x ^{2} + y ^{2} + 9 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

f (x) = 2 + 9 x + 3 x^{2} - x^{3}, - 2 \leq x \leq 6

f (x, y) = 2 x^{2} + 3 y^{2} - 4 x - 6 y + 19

f (x) = x^{3} - 24 x^{2} + 135 x

f (x) = x^{3} + y^{3} - 30 x y

What is the extreme f(x,y)=sqrt(x^2+y^2+9) ?
The extreme f(x,y)=sqrt(x^2+y^2+9) is Minimum(0,0)