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)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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, y) = - [x^{2} - y^{2}] e^{- x^{2} - y^{2}}

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

f (x) = 6 x^{5} - 2 x^{4}

f (x) = - x^{3} + 3 x^{2} + 2880 x - 500

x^{3} - 3 x^{2} - 9 x + 10

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)