What is the extreme f(x)=5x^2sqrt(x+1) ?

The extreme f(x)=5x^2sqrt(x+1) is Minimum(-1,0),Maximum(-4/5 ,(16)/(5sqrt(5))),Minimum(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=5x^2sqrt(x+1)

extreme

Minimum (- 1, 0), Maximum (- \frac{4}{5}, \frac{16}{5 5}), Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{5 ( 5 x ^{2} + 4 x )}{2 x + 1}

Find intervals:

Increasing

: - 1 < x < - \frac{4}{5},

Decreasing

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

Increasing

: 0 < x < \infty

Plug

x = - 1

into

5 x^{2} x + 1 : 0

Minimum (- 1, 0)

Plug

x = - \frac{4}{5}

into

5 x^{2} x + 1 : \frac{16}{5 5}

Maximum (- \frac{4}{5}, \frac{16}{5 5})

Plug

x = 0

into

5 x^{2} x + 1 : 0

Minimum (0, 0)

Minimum (- 1, 0), Maximum (- \frac{4}{5}, \frac{16}{5 5}), Minimum (0, 0)

What is the extreme f(x)=5x^2sqrt(x+1) ?
The extreme f(x)=5x^2sqrt(x+1) is Minimum(-1,0),Maximum(-4/5 ,(16)/(5sqrt(5))),Minimum(0,0)