What is the critical f(x)=x^2\sqrt[3]{x^2-4} ?

The critical f(x)=x^2\sqrt[3]{x^2-4} is x=-2,x=-sqrt(3),x=0,x=sqrt(3),x=2

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\sqrt[3]{x^2-4}

critical points

x = - 2, x = - 3, x = 0, x = 3, x = 2

Solution steps

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 0, x = 3, x = - 2, x = 2

Identify critical points not in

f (x)

domain

Domain of

All critical points are in domain

x = - 2, x = - 3, x = 0, x = 3, x = 2

y = cos (x) + sin (x)

y = x^{3} e^{x}

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

(x - 6)^{2} (x + 9)

x^{3} - 3 x^{2} - 24 x - 2

What is the critical f(x)=x^2\sqrt[3]{x^2-4} ?
The critical f(x)=x^2\sqrt[3]{x^2-4} is x=-2,x=-sqrt(3),x=0,x=sqrt(3),x=2