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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = 32 x (x - 1) (2 x - 1)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

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

Decreasing

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

Increasing

: 1 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = \frac{1}{2}

into

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

Maximum (\frac{1}{2}, 1)

Plug

x = 1

into

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

Minimum (1, 0)

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

f (x) = 3 x^{2} - 24 x + 9, 0 \leq x \leq 9

f (x) = 2 x^{3} - 9 x^{2} + 12 x

f (x) = 2 - cos (3 x)

4 x + 3 y = 21

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

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