What is the extreme f(x)=380x^2-3800x^3 ?

The extreme f(x)=380x^2-3800x^3 is Minimum(0,0),Maximum(1/15 , 76/135)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=380x^2-3800x^3

extreme

Minimum (0, 0), Maximum (\frac{1}{15}, \frac{76}{135})

Solution steps

f^{'} (x) = 760 x - 11400 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{15},

Decreasing

: \frac{1}{15} < x < \infty

Plug

x = 0

into

380 x^{2} - 3800 x^{3} : 0

Minimum (0, 0)

Plug

x = \frac{1}{15}

into

380 x^{2} - 3800 x^{3} : \frac{76}{135}

Maximum (\frac{1}{15}, \frac{76}{135})

Minimum (0, 0), Maximum (\frac{1}{15}, \frac{76}{135})

What is the extreme f(x)=380x^2-3800x^3 ?
The extreme f(x)=380x^2-3800x^3 is Minimum(0,0),Maximum(1/15 , 76/135)