What is the extreme f(x)=x^6(1-x)^3 ?

The extreme f(x)=x^6(1-x)^3 is Minimum(0,0),Maximum(2/3 , 64/19683),Saddle(1,0)

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extreme f(x)=x^6(1-x)^3

extreme points

f (x) = x^{6} (1 - x)^{3}

Minimum (0, 0), Maximum (\frac{2}{3}, \frac{64}{19683}), Saddle (1, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

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

Decreasing

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

Decreasing

: 1 < x < \infty

Plug

x = 0

into

x^{6} (1 - x)^{3} : 0

Minimum (0, 0)

Plug

x = \frac{2}{3}

into

x^{6} (1 - x)^{3} : \frac{64}{19683}

Maximum (\frac{2}{3}, \frac{64}{19683})

Plug

x = 1

into

x^{6} (1 - x)^{3} : 0

Saddle (1, 0)

Minimum (0, 0), Maximum (\frac{2}{3}, \frac{64}{19683}), Saddle (1, 0)

f (x) = 6 x^{2} - 6 x - 12

f (x) = x^{2} + x y + y^{2} - 25 y + 208

f (x, y) = 48 x - 48 y - 6 x^{2} + 6 y^{2}

\frac{x ^{2} y}{2} - x^{2} + x y - 2 x - y^{2}

y = x^{2} - 2 x

What is the extreme f(x)=x^6(1-x)^3 ?
The extreme f(x)=x^6(1-x)^3 is Minimum(0,0),Maximum(2/3 , 64/19683),Saddle(1,0)