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

The extreme f(x)=x^3-3x^2+2 is Maximum(0,2),Minimum(2,-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=x^3-3x^2+2

extreme points

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

Maximum (0, 2), Minimum (2, - 2)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x^{3} - 3 x^{2} + 2 : 2

Maximum (0, 2)

Plug

x = 2

into

x^{3} - 3 x^{2} + 2 : - 2

Minimum (2, - 2)

Maximum (0, 2), Minimum (2, - 2)

f (x) = x^{5} - 4

f (x) = 2 (x - 4)^{2} - 6

f (x) = \frac{x}{x ^{2} + 6}

f (x) = \frac{x}{5}

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

What is the extreme f(x)=x^3-3x^2+2 ?
The extreme f(x)=x^3-3x^2+2 is Maximum(0,2),Minimum(2,-2)