What is the extreme f(x)=sqrt(x^2+y^2+5) ?

The extreme f(x)=sqrt(x^2+y^2+5) is Minimum(0,0)

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extreme f(x)=sqrt(x^2+y^2+5)

extreme points

f (x) = x^{2} + y^{2} + 5

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} + 5}{( x ^{2} + y ^{2} + 5 ) x ^{2} + y ^{2} + 5}

D (x, y) = \frac{5}{( x ^{2} + y ^{2} + 5 ) ^{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = 5 x^{2} - 4 x y + y^{2} - 16 x - 12 y - 4

f (x) = x^{4} + 20 x^{3}

f (x) = 7 x^{3} + 6 x^{2} - 7 x - 6

f (x) = 3^{\frac{3}{4}} - 4 x^{\frac{1}{4}}

\frac{x}{x ^{2} - x + 16}, 0 \leq x \leq 12

What is the extreme f(x)=sqrt(x^2+y^2+5) ?
The extreme f(x)=sqrt(x^2+y^2+5) is Minimum(0,0)