What is the extreme f(x)=(x^4-4x^3)/2 ?

The extreme f(x)=(x^4-4x^3)/2 is Saddle(0,0),Minimum(3,-27/2)

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extreme f(x)=(x^4-4x^3)/2

extreme points

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

Saddle (0, 0), Minimum (3, - \frac{27}{2})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

\frac{x ^{4} - 4 x ^{3}}{2} : 0

Saddle (0, 0)

Plug

x = 3

into

\frac{x ^{4} - 4 x ^{3}}{2} : - \frac{27}{2}

Minimum (3, - \frac{27}{2})

Saddle (0, 0), Minimum (3, - \frac{27}{2})

f (x) = 0.5 x^{2} + 15 x + 5000

f (x) = \frac{x ^{5}}{20} - 1 x^{3}

\frac{x ^{2} + 4 x}{x ^{3} - 17 x ^{2} + 72 x}

f (x) = 3 x^{3} - 3 x^{2} - 3 x + 10

0.5 x + 15 + \frac{5000}{x}

What is the extreme f(x)=(x^4-4x^3)/2 ?
The extreme f(x)=(x^4-4x^3)/2 is Saddle(0,0),Minimum(3,-27/2)