What is the critical f(x)= 1/(x^2-9) ?

The critical f(x)= 1/(x^2-9) 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)= 1/(x^2-9)

critical points

f (x) = \frac{1}{x ^{2} - 9}

x = 0

Solution steps

\frac{1}{x ^{2} - 9}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 3, x = 3

Identify critical points not in

f (x)

domain

Domain of

\frac{1}{x ^{2} - 9} : x < - 3 or - 3 < x < 3 or x > 3

\frac{1}{x ^{2} - 9}

is not defined at

x = - 3, x = 3

, therefore:

x = 0

f (x) = e^{x} (8 - x^{2})

\frac{1}{x ^{3} - x}

x^{3} + x^{2} - 8 x + 5

f (x, y) = x y + ln (x) + 32 y^{2}

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