What is the extreme f(x)=x^{1/3}(x+8) ?

The extreme f(x)=x^{1/3}(x+8) is Minimum(-2,-6*2^{1/3})

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extreme f(x)=x^{1/3}(x+8)

extreme points

f (x) = x^{\frac{1}{3}} (x + 8)

Minimum (- 2, - 6 \cdot 2^{\frac{1}{3}})

Solution steps

f^{'} (x) = \frac{4 ( x + 2 )}{3 x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

x^{\frac{1}{3}} (x + 8) : - 6 \cdot 2^{\frac{1}{3}}

Minimum (- 2, - 6 \cdot 2^{\frac{1}{3}})

am pl i t u d e y = \frac{1}{2} cos (x)

in v erse \frac{( x + 1 )}{( x - 1 )}

e x t re m e f (x) = - x^{3} + 5 x^{2} + 8 x + 3

s im pl i f y (- 3.1) (- 8.8)

d i s t an ce (8, 2), (14, 3)

What is the extreme f(x)=x^{1/3}(x+8) ?
The extreme f(x)=x^{1/3}(x+8) is Minimum(-2,-6*2^{1/3})