What is the critical (16)/(x^2(x-4)) ?

The critical (16)/(x^2(x-4)) is x= 8/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical (16)/(x^2(x-4))

critical points

\frac{16}{x ^{2} ( x - 4 )}

x = \frac{8}{3}

Solution steps

\frac{16}{x ^{2} ( x - 4 )}

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{8}{3}, x = 0, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{16}{x ^{2} ( x - 4 )} : x < 0 or 0 < x < 4 or x > 4

\frac{16}{x ^{2} ( x - 4 )}

is not defined at

x = 0, x = 4

, therefore:

x = \frac{8}{3}

f (x) = 2 x ln (16 x^{2})

\frac{3 x ^{2}}{x ^{2} - 4}

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

f (x) = \frac{x}{x ^{2} + 81}

f (x, y) = (x - 1)^{2} + (x - y)^{4}