What is the extreme f(y,z)=ln(1+y^2+z^2) ?

The extreme f(y,z)=ln(1+y^2+z^2) 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(y,z)=ln(1+y^2+z^2)

extreme points

f (y, z) = ln (1 + y^{2} + z^{2})

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (y, z) = - \frac{4 ( y ^{2} + z ^{2} - 1 )}{( y ^{2} + z ^{2} + 1 ) ^{3}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

3 x^{3} - 9 x

f (x) = 2 x^{3} e^{- x}, - 1 \leq x \leq 6

f (x) = (4 x - 3)^{2}

y = x^{2} + z + z^{2}

3 x^{3} - 3 x^{2} - 3 x + 8

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