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

The extreme f(x)=(x^2-4x)^2 is Minimum(0,0),Maximum(2,16),Minimum(4,0)

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

extreme points

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

Minimum (0, 0), Maximum (2, 16), Minimum (4, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 2

into

(x^{2} - 4 x)^{2} : 16

Maximum (2, 16)

Plug

x = 4

into

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

Minimum (4, 0)

Minimum (0, 0), Maximum (2, 16), Minimum (4, 0)

f (x) = x^{3} + 3

3 x^{4} - 24 x^{3} + 30 x^{2}

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

f (x) = 2 - 13 - e^{4 t}

f (x) = x^{2}, x \geq 0

What is the extreme f(x)=(x^2-4x)^2 ?
The extreme f(x)=(x^2-4x)^2 is Minimum(0,0),Maximum(2,16),Minimum(4,0)