What is the extreme f(x)=-x^{2/3}(x-2) ?

The extreme f(x)=-x^{2/3}(x-2) is Minimum(0,0),Maximum(4/5 ,(6*4^{2/3})/(5*5^{2/3)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=-x^{2/3}(x-2)

extreme points

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

Minimum (0, 0), Maximum (\frac{4}{5}, \frac{6 \cdot 4 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Solution steps

f^{'} (x) = - \frac{5 x - 4}{3 x ^{\frac{1}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{4}{5},

Decreasing

: \frac{4}{5} < x < \infty

Plug

x = 0

into

- x^{\frac{2}{3}} (x - 2) : 0

Minimum (0, 0)

Plug

x = \frac{4}{5}

into

- x^{\frac{2}{3}} (x - 2) : \frac{6 \cdot 4 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}}

Maximum (\frac{4}{5}, \frac{6 \cdot 4 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

Minimum (0, 0), Maximum (\frac{4}{5}, \frac{6 \cdot 4 ^{\frac{2}{3}}}{5 \cdot 5 ^{\frac{2}{3}}})

What is the extreme f(x)=-x^{2/3}(x-2) ?
The extreme f(x)=-x^{2/3}(x-2) is Minimum(0,0),Maximum(4/5 ,(6*4^{2/3})/(5*5^{2/3)})