What is the critical (6x^2)/(x^2-16) ?

The critical (6x^2)/(x^2-16) is x=0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical (6x^2)/(x^2-16)

critical points

\frac{6 x ^{2}}{x ^{2} - 16}

x = 0

Solution steps

\frac{6 x ^{2}}{x ^{2} - 16}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 4, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{6 x ^{2}}{x ^{2} - 16} : x < - 4 or - 4 < x < 4 or x > 4

\frac{6 x ^{2}}{x ^{2} - 16}

is not defined at

x = - 4, x = 4

, therefore:

x = 0

y = x^{3} + x^{2} - x

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

f (x) = x^{2} - 6 x + 4

f (x) = \frac{1}{x ^{2} - 3 x + 2}