What is the extreme f(x)=x^3(x-8) ?

The extreme f(x)=x^3(x-8) is Saddle(0,0),Minimum(6,-432)

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extreme f(x)=x^3(x-8)

extreme points

f (x) = x^{3} (x - 8)

Saddle (0, 0), Minimum (6, - 432)

Solution steps

f^{'} (x) = 4 x^{2} (x - 6)

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

x^{3} (x - 8) : 0

Saddle (0, 0)

Plug

x = 6

into

x^{3} (x - 8) : - 432

Minimum (6, - 432)

Saddle (0, 0), Minimum (6, - 432)

f (x) = x^{2} - 4 x + 1

x^{2} - 4 x + 5

f (x) = \frac{2 x + 3}{3 x - 1}

f (x) = x^{2} - 2 x - 9

f (x) = \frac{x - 1}{2 x - 1}

What is the extreme f(x)=x^3(x-8) ?
The extreme f(x)=x^3(x-8) is Saddle(0,0),Minimum(6,-432)