What is the critical f(x)=x^3-6x^2-15x+40 ?

The critical f(x)=x^3-6x^2-15x+40 is x=-1,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^3-6x^2-15x+40

critical points

f (x) = x^{3} - 6 x^{2} - 15 x + 40

x = - 1, x = 5

Solution steps

x^{3} - 6 x^{2} - 15 x + 40

Find where

f^{'} (x)

is equal to zero or undefined

x = - 1, x = 5

Identify critical points not in

f (x)

domain

Domain of

x^{3} - 6 x^{2} - 15 x + 40 : - \infty < x < \infty

All critical points are in domain

x = - 1, x = 5

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

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

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

y - 27 = \frac{7}{15} (x + 12)

- 2 cos (x - \frac{π}{2})

What is the critical f(x)=x^3-6x^2-15x+40 ?
The critical f(x)=x^3-6x^2-15x+40 is x=-1,x=5