What is the extreme f(x)= 1/(1-x^2-y^2) ?

The extreme f(x)= 1/(1-x^2-y^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(x)= 1/(1-x^2-y^2)

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{2 ( 3 y ^{2} - x ^{2} + 1 )}{( 1 - x ^{2} - y ^{2} ) ^{3}}

D (x, y) = \frac{- 12 x ^{4} - 24 x ^{2} y ^{2} + 8 x ^{2} - 12 y ^{4} + 8 y ^{2} + 4}{( - x ^{2} - y ^{2} + 1 ) ^{6}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

5 x^{2} ln (x)

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

f (x, y) = \frac{2 x y}{( x - 2 )}

f (x) = - 7 (x + 1)^{2} + 3 x + 7

y = 2 x^{3} - 3 x + 11

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