What is the critical f(x)=((-2(x^2-1)))/(x^2-4) ?

The critical f(x)=((-2(x^2-1)))/(x^2-4) is 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)=((-2(x^2-1)))/(x^2-4)

critical points

f (x) = \frac{( - 2 ( x ^{2} - 1 ) )}{x ^{2} - 4}

x = 0

Solution steps

\frac{- 2 ( x ^{2} - 1 )}{x ^{2} - 4}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

\frac{- 2 ( x ^{2} - 1 )}{x ^{2} - 4} : x < - 2 or - 2 < x < 2 or x > 2

\frac{- 2 ( x ^{2} - 1 )}{x ^{2} - 4}

is not defined at

x = - 2, x = 2

, therefore:

x = 0

x^{\frac{3}{5}} (4 - x)

x^{2} + \frac{1}{x}

f (x) = - 5 - 10 x - 25 x^{2}

f (x) = 16 cos (x) + 8 sin^{2} (x)

f (x) = \frac{16 x}{( x ^{2} + 4 ) ^{2}}

What is the critical f(x)=((-2(x^2-1)))/(x^2-4) ?
The critical f(x)=((-2(x^2-1)))/(x^2-4) is x=0