What is the extreme f(x)= 1/9 (x^4-4x^3) ?

The extreme f(x)= 1/9 (x^4-4x^3) is Saddle(0,0),Minimum(3,-3)

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extreme f(x)= 1/9 (x^4-4x^3)

extreme points

f (x) = \frac{1}{9} (x^{4} - 4 x^{3})

Saddle (0, 0), Minimum (3, - 3)

Solution steps

f^{'} (x) = \frac{1}{9} (4 x^{3} - 12 x^{2})

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 0

into

\frac{1}{9} (x^{4} - 4 x^{3}) : 0

Saddle (0, 0)

Plug

x = 3

into

\frac{1}{9} (x^{4} - 4 x^{3}) : - 3

Minimum (3, - 3)

Saddle (0, 0), Minimum (3, - 3)

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

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

f (x) = - 2 x^{2} + 12 x

cos (x) - sin (x)

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

What is the extreme f(x)= 1/9 (x^4-4x^3) ?
The extreme f(x)= 1/9 (x^4-4x^3) is Saddle(0,0),Minimum(3,-3)