What is the extreme f(x)=340x^2-2040x^3 ?

The extreme f(x)=340x^2-2040x^3 is Minimum(0,0),Maximum(1/9 , 340/243)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=340x^2-2040x^3

extreme points

f (x) = 340 x^{2} - 2040 x^{3}

Minimum (0, 0), Maximum (\frac{1}{9}, \frac{340}{243})

Solution steps

f^{'} (x) = 680 x - 6120 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{9},

Decreasing

: \frac{1}{9} < x < \infty

Plug

x = 0

into

340 x^{2} - 2040 x^{3} : 0

Minimum (0, 0)

Plug

x = \frac{1}{9}

into

340 x^{2} - 2040 x^{3} : \frac{340}{243}

Maximum (\frac{1}{9}, \frac{340}{243})

Minimum (0, 0), Maximum (\frac{1}{9}, \frac{340}{243})

f (x) = x (x - 1)

f (x) = - 6 sin^{2} (x)

f (x) = x^{4} - 72 x^{2} + 3

f (x) = x^{4} - 72 x^{2} + 7

f (x, y) = x^{2} + 6 x + y^{3} - 6 y^{2} + 10

What is the extreme f(x)=340x^2-2040x^3 ?
The extreme f(x)=340x^2-2040x^3 is Minimum(0,0),Maximum(1/9 , 340/243)