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

The critical y=x(1-x^2)^2 is x=-1,x=-1/(sqrt(5)),x= 1/(sqrt(5)),x=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical y=x(1-x^2)^2

critical points

y = x (1 - x^{2})^{2}

x = - 1, x = - \frac{1}{5}, x = \frac{1}{5}, x = 1

Solution steps

x (1 - x^{2})^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = - \frac{1}{5}, x = \frac{1}{5}, x = 1

Identify critical points not in

f (x)

domain

Domain of

x (1 - x^{2})^{2} : - \infty < x < \infty

All critical points are in domain

x = - 1, x = - \frac{1}{5}, x = \frac{1}{5}, x = 1

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

f (x) = - 2 x^{2} - 12 x - 13

f (x, y) = y x - y^{2} - x + 6 y

f (x, y) = x^{3} + 12 x y^{2} - 15 x - 24 y

x^{\frac{2}{3}}

What is the critical y=x(1-x^2)^2 ?
The critical y=x(1-x^2)^2 is x=-1,x=-1/(sqrt(5)),x= 1/(sqrt(5)),x=1