What is the extreme f(x)=x^3-3x^2+5 ?

The extreme f(x)=x^3-3x^2+5 is Maximum(0,5),Minimum(2,1)

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+5

extreme points

f (x) = x^{3} - 3 x^{2} + 5

Maximum (0, 5), Minimum (2, 1)

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} + 5 : 5

Maximum (0, 5)

Plug

x = 2

into

x^{3} - 3 x^{2} + 5 : 1

Minimum (2, 1)

Maximum (0, 5), Minimum (2, 1)

g (x) = 9 - x

f (x) = x^{2} + 2 x - 5

f (x) = e^{- x}

f (x) = ln (1 + x)

\frac{5 x + 1}{x - 3}

What is the extreme f(x)=x^3-3x^2+5 ?
The extreme f(x)=x^3-3x^2+5 is Maximum(0,5),Minimum(2,1)