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

The extreme f(x)=(x^3)/3-2x^2+4 is Maximum(0,4),Minimum(4,-20/3)

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)/3-2x^2+4

extreme points

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

Maximum (0, 4), Minimum (4, - \frac{20}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

\frac{x ^{3}}{3} - 2 x^{2} + 4 : 4

Maximum (0, 4)

Plug

x = 4

into

\frac{x ^{3}}{3} - 2 x^{2} + 4 : - \frac{20}{3}

Minimum (4, - \frac{20}{3})

Maximum (0, 4), Minimum (4, - \frac{20}{3})

f (x) = x^{3} + 3 x^{2} - 28 x

y = \frac{1 - x ^{2}}{x}

x^{4} - 5 x^{2} - 4

F (X, Y) = 2 X + 3 Y X^{2}

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

What is the extreme f(x)=(x^3)/3-2x^2+4 ?
The extreme f(x)=(x^3)/3-2x^2+4 is Maximum(0,4),Minimum(4,-20/3)