What is the critical y=x^2-5x ?

The critical y=x^2-5x is x= 5/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical y=x^2-5x

critical points

y = x^{2} - 5 x

x = \frac{5}{2}

Solution steps

x^{2} - 5 x

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{5}{2}

Identify critical points not in

f (x)

domain

Domain of

x^{2} - 5 x : - \infty < x < \infty

All critical points are in domain

x = \frac{5}{2}

y = 6 cos (3 x + \frac{π}{2})

f (x) = \frac{6 x + 7}{2 x - 9}

y = 4 x + 1

f (x) = \frac{4 x + 3}{1 - 9 x}

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