What is the extreme f(x)=x^3-12x^2+36x-1 ?

The extreme f(x)=x^3-12x^2+36x-1 is Maximum(2,31),Minimum(6,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^3-12x^2+36x-1

extreme points

f (x) = x^{3} - 12 x^{2} + 36 x - 1

Maximum (2, 31), Minimum (6, - 1)

Solution steps

f^{'} (x) = 3 x^{2} - 24 x + 36

Find intervals:

Increasing

: - \infty < x < 2,

Decreasing

: 2 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 2

into

x^{3} - 12 x^{2} + 36 x - 1 : 31

Maximum (2, 31)

Plug

x = 6

into

x^{3} - 12 x^{2} + 36 x - 1 : - 1

Minimum (6, - 1)

Maximum (2, 31), Minimum (6, - 1)

What is the extreme f(x)=x^3-12x^2+36x-1 ?
The extreme f(x)=x^3-12x^2+36x-1 is Maximum(2,31),Minimum(6,-1)