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

The extreme f(x)=x^3+4x^2 is Maximum(-8/3 , 256/27),Minimum(0,0)

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+4x^2

extreme points

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

Maximum (- \frac{8}{3}, \frac{256}{27}), Minimum (0, 0)

Solution steps

f^{'} (x) = 3 x^{2} + 8 x

Find intervals:

Increasing

: - \infty < x < - \frac{8}{3},

Decreasing

: - \frac{8}{3} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{8}{3}

into

x^{3} + 4 x^{2} : \frac{256}{27}

Maximum (- \frac{8}{3}, \frac{256}{27})

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- \frac{8}{3}, \frac{256}{27}), Minimum (0, 0)

\frac{x ^{2} + 8}{x ^{2} - 4}

f (x) = \frac{8}{x ^{2} - x - 6}

ln (ln (x))

f (x) = \frac{5}{( x - 2 ) ^{4}}

f (x) = \frac{( 3 x + 2 )}{x + 5}

What is the extreme f(x)=x^3+4x^2 ?
The extreme f(x)=x^3+4x^2 is Maximum(-8/3 , 256/27),Minimum(0,0)