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

The critical f(x)=x^4(x-2)^3 is x=0,x= 8/7 ,x=2

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(x-2)^3

critical points

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

x = 0, x = \frac{8}{7}, x = 2

Solution steps

x^{4} (x - 2)^{3}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = \frac{8}{7}, x = 2

Identify critical points not in

f (x)

domain

Domain of

x^{4} (x - 2)^{3} : - \infty < x < \infty

All critical points are in domain

x = 0, x = \frac{8}{7}, x = 2

x \cdot ln (x)

h (t) = 50 + 40 t^{3} - 5 t^{4} - 4 t^{5}

R (t) = 1 + 80 t^{3} + 5 t^{4} - 2 t^{5}

f (x) = \frac{x ^{3} - 4 x ^{2} + 7 x}{x}

f (x) = x^{3} - 6 x + 1

What is the critical f(x)=x^4(x-2)^3 ?
The critical f(x)=x^4(x-2)^3 is x=0,x= 8/7 ,x=2