What is the extreme f(x)=(x^3-3x^2+4) ?

The extreme f(x)=(x^3-3x^2+4) is Maximum(0,4),Minimum(2,0)

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+4)

extreme points

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

Maximum (0, 4), Minimum (2, 0)

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} + 4 : 4

Maximum (0, 4)

Plug

x = 2

into

x^{3} - 3 x^{2} + 4 : 0

Minimum (2, 0)

Maximum (0, 4), Minimum (2, 0)

f (x) = x - sin (- x), 0 \leq x \leq 2 π

f (x) = \frac{( x ^{4} + 1 )}{x ^{2}}

f (x) = x^{\frac{1}{5}} (x + 1)

P (x, y) = 0.4 x + 0.6 y

f (x) = (1 + x)^{2} (6 - x)

What is the extreme f(x)=(x^3-3x^2+4) ?
The extreme f(x)=(x^3-3x^2+4) is Maximum(0,4),Minimum(2,0)