What is the critical f(x)=(x^2)/(2x-5) ?

The critical f(x)=(x^2)/(2x-5) is x=0,x=5

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)/(2x-5)

critical points

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

x = 0, x = 5

Solution steps

\frac{x ^{2}}{2 x - 5}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 5, x = \frac{5}{2}

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{2}}{2 x - 5} : x < \frac{5}{2} or x > \frac{5}{2}

\frac{x ^{2}}{2 x - 5}

is not defined at

x = \frac{5}{2}

, therefore:

x = 0, x = 5

f (x) = 2 x ln (x)

f (x) = e^{x} (3 - x^{2})

f (x) = x 6 - x

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

x^{4} - 4 x^{3} + 4 x^{2}

What is the critical f(x)=(x^2)/(2x-5) ?
The critical f(x)=(x^2)/(2x-5) is x=0,x=5