What is the critical (2x^2)/(1-x^2) ?

The critical (2x^2)/(1-x^2) is x=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical (2x^2)/(1-x^2)

critical points

\frac{2 x ^{2}}{1 - x ^{2}}

x = 0

Solution steps

\frac{2 x ^{2}}{1 - x ^{2}}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 1, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{2 x ^{2}}{1 - x ^{2}} : x < - 1 or - 1 < x < 1 or x > 1

\frac{2 x ^{2}}{1 - x ^{2}}

is not defined at

x = - 1, x = 1

, therefore:

x = 0

(x^{2} + 10) (9 - x^{2})

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

f (x) = x^{\frac{2}{3}} - \frac{x}{6}

f (x) = x^{- 8} ln (x)

\frac{x ^{2} - 6 x + 12}{x - 4}