What is the critical f(x)=x^4+8x^3+18x^2-8 ?

The critical f(x)=x^4+8x^3+18x^2-8 is x=-3,x=0

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+8x^3+18x^2-8

critical points

f (x) = x^{4} + 8 x^{3} + 18 x^{2} - 8

x = - 3, x = 0

Solution steps

x^{4} + 8 x^{3} + 18 x^{2} - 8

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{4} + 8 x^{3} + 18 x^{2} - 8 : - \infty < x < \infty

All critical points are in domain

x = - 3, x = 0

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

h (x) = x^{3} - 5 x^{2} + 4 x - 20

f (x) = \frac{x + 6}{x ( x + 11 )}

f (x) = \frac{x - 8}{x + 5}

tan (\frac{π}{9} x)

What is the critical f(x)=x^4+8x^3+18x^2-8 ?
The critical f(x)=x^4+8x^3+18x^2-8 is x=-3,x=0