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

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

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extreme y=9x^4-4x^3,(-3,3)

extreme points

y = 9 x^{4} - 4 x^{3}, (- 3, 3)

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

Solution steps

Find the critical points:

x = 0, x = \frac{1}{3}

Domain of

9 x^{4} - 4 x^{3}, (- 3, 3) : - 3 < x < 3

Find intervals:

Decreasing

: - 3 < x < 0,

Decreasing

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

Increasing

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

Plug the extreme point

x = 0

into

9 x^{4} - 4 x^{3} \Rightarrow y = 0

Saddle (0, 0)

Plug the extreme point

x = \frac{1}{3}

into

9 x^{4} - 4 x^{3} \Rightarrow y = - \frac{1}{27}

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

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

f (x) = \frac{x ^{2}}{x ^{2} - 108}

y = 2 x^{2} - 64 x

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

f (x) = 3 x - 2 y

f (x) = - \frac{12}{x ^{4}} + \frac{14}{x ^{3}}

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