What is the extreme f(x)=x^4(x-1)^2 ?

The extreme f(x)=x^4(x-1)^2 is Minimum(0,0),Maximum(2/3 , 16/729),Minimum(1,0)

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extreme f(x)=x^4(x-1)^2

extreme points

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

Minimum (0, 0), Maximum (\frac{2}{3}, \frac{16}{729}), Minimum (1, 0)

Solution steps

f^{'} (x) = 2 x^{3} (x - 1) (3 x - 2)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{2}{3},

Decreasing

: \frac{2}{3} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

x^{4} (x - 1)^{2} : 0

Minimum (0, 0)

Plug

x = \frac{2}{3}

into

x^{4} (x - 1)^{2} : \frac{16}{729}

Maximum (\frac{2}{3}, \frac{16}{729})

Plug

x = 1

into

x^{4} (x - 1)^{2} : 0

Minimum (1, 0)

Minimum (0, 0), Maximum (\frac{2}{3}, \frac{16}{729}), Minimum (1, 0)

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

f (x) = 2 x^{3} + 3 x^{2} - 36 x - 54

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

f (x) = - 3 sin (x) cos (x)

f (x) = \frac{4 x}{x ^{2} + 1}, (- 4, 0)

What is the extreme f(x)=x^4(x-1)^2 ?
The extreme f(x)=x^4(x-1)^2 is Minimum(0,0),Maximum(2/3 , 16/729),Minimum(1,0)