What is the extreme f(x)=x^3(1-x)^2 ?

The extreme f(x)=x^3(1-x)^2 is Saddle(0,0),Maximum(3/5 , 108/3125),Minimum(1,0)

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extreme f(x)=x^3(1-x)^2

extreme points

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

Saddle (0, 0), Maximum (\frac{3}{5}, \frac{108}{3125}), Minimum (1, 0)

Solution steps

f^{'} (x) = x^{2} (x - 1) (5 x - 3)

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{3}{5},

Decreasing

: \frac{3}{5} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

x^{3} (1 - x)^{2} : 0

Saddle (0, 0)

Plug

x = \frac{3}{5}

into

x^{3} (1 - x)^{2} : \frac{108}{3125}

Maximum (\frac{3}{5}, \frac{108}{3125})

Plug

x = 1

into

x^{3} (1 - x)^{2} : 0

Minimum (1, 0)

Saddle (0, 0), Maximum (\frac{3}{5}, \frac{108}{3125}), Minimum (1, 0)

y = x^{\frac{2}{3}} (6 - x)^{\frac{1}{3}}

f (x) = (x^{2} - 4)^{7}

y = x^{3} - 27 x + 3, - 2 \leq x \leq 4

f (x) = 2 x^{4} - 8 x + 3

2 x^{3} + 3 x^{2} - 12 x + 5, 0 \leq x \leq 2

What is the extreme f(x)=x^3(1-x)^2 ?
The extreme f(x)=x^3(1-x)^2 is Saddle(0,0),Maximum(3/5 , 108/3125),Minimum(1,0)