What is the critical f(x)=(x^2-9)/(2x-4) ?

The critical f(x)=(x^2-9)/(2x-4) is None

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical f(x)=(x^2-9)/(2x-4)

critical points

f (x) = \frac{x ^{2} - 9}{2 x - 4}

None

Solution steps

\frac{x ^{2} - 9}{2 x - 4}

Find where

f^{'} (x)

is equal to zero or undefined

x = 2

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2} - 9}{2 x - 4} : x < 2 or x > 2

\frac{x ^{2} - 9}{2 x - 4}

is not defined at

x = 2

, therefore:

None

h (x) = sin^{2} (x) + cos (x), 0 < x < 2 π

3 x + 4 y

7 x^{5} - 3 x^{4}

f (x) = x^{3} - 6 x^{2} + 12 x + 10

x^{3} + y^{3} + 3 x^{2} - 18 y^{2} + 81 y + 5

What is the critical f(x)=(x^2-9)/(2x-4) ?
The critical f(x)=(x^2-9)/(2x-4) is None