What is the extreme f(x)=x^6(1-x)^{18} ?

The extreme f(x)=x^6(1-x)^{18} is Minimum(0,0),Maximum(1/4 ,(387420489)/(4^{24)}),Minimum(1,0)

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extreme f(x)=x^6(1-x)^{18}

extreme

Minimum (0, 0), Maximum (\frac{1}{4}, \frac{387420489}{4 ^{24}}), Minimum (1, 0)

Solution steps

f^{'} (x) = 6 x^{5} (x - 1)^{17} (4 x - 1)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{4},

Decreasing

: \frac{1}{4} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

x^{6} (1 - x)^{18} : 0

Minimum (0, 0)

Plug

x = \frac{1}{4}

into

x^{6} (1 - x)^{18} : \frac{387420489}{4 ^{24}}

Maximum (\frac{1}{4}, \frac{387420489}{4 ^{24}})

Plug

x = 1

into

x^{6} (1 - x)^{18} : 0

Minimum (1, 0)

Minimum (0, 0), Maximum (\frac{1}{4}, \frac{387420489}{4 ^{24}}), Minimum (1, 0)

What is the extreme f(x)=x^6(1-x)^{18} ?
The extreme f(x)=x^6(1-x)^{18} is Minimum(0,0),Maximum(1/4 ,(387420489)/(4^{24)}),Minimum(1,0)