What is the extreme f(x)=x^4-12x^3 ?

The extreme f(x)=x^4-12x^3 is Saddle(0,0),Minimum(9,-2187)

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extreme f(x)=x^4-12x^3

extreme points

f (x) = x^{4} - 12 x^{3}

Saddle (0, 0), Minimum (9, - 2187)

Solution steps

f^{'} (x) = 4 x^{3} - 36 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 9,

Increasing

: 9 < x < \infty

Plug

x = 0

into

x^{4} - 12 x^{3} : 0

Saddle (0, 0)

Plug

x = 9

into

x^{4} - 12 x^{3} : - 2187

Minimum (9, - 2187)

Saddle (0, 0), Minimum (9, - 2187)

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

f (x) = 200 x + 500 yy = 40000

f (x) = 78

f (x) = \frac{3 + 2 x}{1 - 4 x}

x^{2} - 6 x + 8

What is the extreme f(x)=x^4-12x^3 ?
The extreme f(x)=x^4-12x^3 is Saddle(0,0),Minimum(9,-2187)