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

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

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

extreme points

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

Minimum (0, 0), Maximum (\frac{2}{5}, \frac{108}{3125}), Saddle (1, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{2}{5},

Decreasing

: \frac{2}{5} < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = \frac{2}{5}

into

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

Maximum (\frac{2}{5}, \frac{108}{3125})

Plug

x = 1

into

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

Saddle (1, 0)

Minimum (0, 0), Maximum (\frac{2}{5}, \frac{108}{3125}), Saddle (1, 0)

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

C (t) = 6.5 (e^{- a t} - e^{- b t})

f (x) = - x^{3} + 3 x^{2} + 9 x + 2

f (x) = 9 x^{3} - 7 x^{2} + 3 x + 10 [- 5.6]

f (x) = x^{3} - 5 x^{2} + 8 x

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