What is the extreme f(x,y)=sqrt(x^2+y^2+4) ?

The extreme f(x,y)=sqrt(x^2+y^2+4) is Minimum(0,0)

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extreme f(x,y)=sqrt(x^2+y^2+4)

extreme points

f (x, y) = x^{2} + y^{2} + 4

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = x^{2} - 8 x + 40

f (x) = x^{2} - 8 x + 18

f (x) = \frac{x}{x ^{2} + 4}, - 5 \leq x \leq 5

f (x) = - x^{2} - x - 1

f (x) = 5 (x - 3)^{2} + 2

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