What is the critical f(x)=(x-3)/(x^2-3x+9) ?

The critical f(x)=(x-3)/(x^2-3x+9) is x=0,x=6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical f(x)=(x-3)/(x^2-3x+9)

critical points

f (x) = \frac{x - 3}{x ^{2} - 3 x + 9}

x = 0, x = 6

Solution steps

\frac{x - 3}{x ^{2} - 3 x + 9}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 6

Identify critical points not in

f (x)

domain

Domain of

\frac{x - 3}{x ^{2} - 3 x + 9} : - \infty < x < \infty

All critical points are in domain

x = 0, x = 6

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

f (x) = \frac{x - 2}{x - 5}

f (x) = - 4

x - 3

f (x) = \frac{6}{x + 9}

What is the critical f(x)=(x-3)/(x^2-3x+9) ?
The critical f(x)=(x-3)/(x^2-3x+9) is x=0,x=6