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

The extreme f(x)=9x^4-4x^3 is Saddle(0,0),Minimum(1/3 ,-1/27)

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

extreme points

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

Saddle (0, 0), Minimum (\frac{1}{3}, - \frac{1}{27})

Solution steps

f^{'} (x) = 36 x^{3} - 12 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{1}{3},

Increasing

: \frac{1}{3} < x < \infty

Plug

x = 0

into

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

Saddle (0, 0)

Plug

x = \frac{1}{3}

into

9 x^{4} - 4 x^{3} : - \frac{1}{27}

Minimum (\frac{1}{3}, - \frac{1}{27})

Saddle (0, 0), Minimum (\frac{1}{3}, - \frac{1}{27})

f (x, y) = 4 x^{2} y - 2 x - y^{2}

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

f (x) = (x - 3) (x + 1) (x + 5)

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

f (x) = 3 x^{2} - 7 x + 2

What is the extreme f(x)=9x^4-4x^3 ?
The extreme f(x)=9x^4-4x^3 is Saddle(0,0),Minimum(1/3 ,-1/27)