What is the extreme f(x)=sqrt(x)((x^2)/5-4) ?

The extreme f(x)=sqrt(x)((x^2)/5-4) is Maximum(0,0),Minimum(2,-(16sqrt(2))/5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=sqrt(x)((x^2)/5-4)

extreme

Maximum (0, 0), Minimum (2, - \frac{16 2}{5})

Solution steps

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

Find intervals:

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x (\frac{x ^{2}}{5} - 4) : 0

Maximum (0, 0)

Plug

x = 2

into

x (\frac{x ^{2}}{5} - 4) : - \frac{16 2}{5}

Minimum (2, - \frac{16 2}{5})

Maximum (0, 0), Minimum (2, - \frac{16 2}{5})

What is the extreme f(x)=sqrt(x)((x^2)/5-4) ?
The extreme f(x)=sqrt(x)((x^2)/5-4) is Maximum(0,0),Minimum(2,-(16sqrt(2))/5)