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

The extreme f(x)=x^2(2-4x)^3 is Minimum(0,0),Maximum(1/5 , 216/3125),Saddle(1/2 ,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=x^2(2-4x)^3

extreme

Minimum (0, 0), Maximum (\frac{1}{5}, \frac{216}{3125}), Saddle (\frac{1}{2}, 0)

Solution steps

f^{'} (x) = - 16 x (2 x - 1)^{2} (5 x - 1)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{5},

Decreasing

: \frac{1}{5} < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < \infty

Plug

x = 0

into

x^{2} (2 - 4 x)^{3} : 0

Minimum (0, 0)

Plug

x = \frac{1}{5}

into

x^{2} (2 - 4 x)^{3} : \frac{216}{3125}

Maximum (\frac{1}{5}, \frac{216}{3125})

Plug

x = \frac{1}{2}

into

x^{2} (2 - 4 x)^{3} : 0

Saddle (\frac{1}{2}, 0)

Minimum (0, 0), Maximum (\frac{1}{5}, \frac{216}{3125}), Saddle (\frac{1}{2}, 0)

What is the extreme f(x)=x^2(2-4x)^3 ?
The extreme f(x)=x^2(2-4x)^3 is Minimum(0,0),Maximum(1/5 , 216/3125),Saddle(1/2 ,0)