What is the extreme f(x)=sqrt(x^2+y^2+1) ?

The extreme f(x)=sqrt(x^2+y^2+1) is Minimum(0,0)

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extreme f(x)=sqrt(x^2+y^2+1)

extreme points

f (x) = x^{2} + y^{2} + 1

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} + 1}{( x ^{2} + y ^{2} + 1 ) x ^{2} + y ^{2} + 1}

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

f (x) = - 5 x^{2} + 50 x - 80

f (x) = 1 - 7 x^{2}

f (x) = - 12 x^{2} + 24 x

f (x) = 4 x + 4 sin (x)

What is the extreme f(x)=sqrt(x^2+y^2+1) ?
The extreme f(x)=sqrt(x^2+y^2+1) is Minimum(0,0)