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

The extreme f(x)=x^3-3x^2+1 is Maximum(0,1),Minimum(2,-3)

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+1

extreme points

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

Maximum (0, 1), Minimum (2, - 3)

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} + 1 : 1

Maximum (0, 1)

Plug

x = 2

into

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

Minimum (2, - 3)

Maximum (0, 1), Minimum (2, - 3)

y = \frac{1}{- 4 - 3}

y = - \frac{1}{6} x + 4

f (x) = \frac{x ^{2} + 7}{x ^{2}}

y = - x^{2} + 4 x

y = 4^{x} + 8

What is the extreme f(x)=x^3-3x^2+1 ?
The extreme f(x)=x^3-3x^2+1 is Maximum(0,1),Minimum(2,-3)