What is the critical 2x^5+4x^3-16x ?

The critical 2x^5+4x^3-16x is x=-2/(sqrt(5)),x= 2/(sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

critical 2x^5+4x^3-16x

critical points

2 x^{5} + 4 x^{3} - 16 x

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

Solution steps

2 x^{5} + 4 x^{3} - 16 x

Find where

f^{'} (x)

is equal to zero or undefined

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

Identify critical points not in

f (x)

domain

Domain of

2 x^{5} + 4 x^{3} - 16 x : - \infty < x < \infty

All critical points are in domain

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

f (x) = 2 x^{3} - \frac{5}{2} x^{2} - 4 x

7 x^{5} - x^{4} + 2 x^{3} + 2 x^{2} - x + 4

f (x) = 5 x^{2} + 3 x - 1

f (x) = 6 x y - 2 x^{2} y - 3 x y^{2}

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

What is the critical 2x^5+4x^3-16x ?
The critical 2x^5+4x^3-16x is x=-2/(sqrt(5)),x= 2/(sqrt(5))