What is the critical f(x)=x^4-162x^2+6561 ?

The critical f(x)=x^4-162x^2+6561 is x=-9,x=0,x=9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical f(x)=x^4-162x^2+6561

critical points

f (x) = x^{4} - 162 x^{2} + 6561

x = - 9, x = 0, x = 9

Solution steps

x^{4} - 162 x^{2} + 6561

Find where

f^{'} (x)

is equal to zero or undefined

x = - 9, x = 0, x = 9

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 162 x^{2} + 6561 : - \infty < x < \infty

All critical points are in domain

x = - 9, x = 0, x = 9

y = 2 x + 3, (1, 3)

f (x) = \frac{8}{( 2 x - 5 ) ^{3}}

f (x) = \frac{x ^{2} + 8 x - 4}{- 2 x - 4}

a

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

What is the critical f(x)=x^4-162x^2+6561 ?
The critical f(x)=x^4-162x^2+6561 is x=-9,x=0,x=9