What is the extreme f(x)=x^4-4x^3+4x^2 ?

The extreme f(x)=x^4-4x^3+4x^2 is Minimum(0,0),Maximum(1,1),Minimum(2,0)

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extreme f(x)=x^4-4x^3+4x^2

extreme points

f (x) = x^{4} - 4 x^{3} + 4 x^{2}

Minimum (0, 0), Maximum (1, 1), Minimum (2, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 12 x^{2} + 8 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x^{4} - 4 x^{3} + 4 x^{2} : 0

Minimum (0, 0)

Plug

x = 1

into

x^{4} - 4 x^{3} + 4 x^{2} : 1

Maximum (1, 1)

Plug

x = 2

into

x^{4} - 4 x^{3} + 4 x^{2} : 0

Minimum (2, 0)

Minimum (0, 0), Maximum (1, 1), Minimum (2, 0)

f (x) = \frac{x}{x + 8}

f (x) = \frac{x + 2}{6 x ^{2} + x - 2}

f (x) = x^{2} - x - 30

(- 1, 3), (6, 2)

f (x) = \frac{x ^{2} - 9 x + 11}{x - 3}

What is the extreme f(x)=x^4-4x^3+4x^2 ?
The extreme f(x)=x^4-4x^3+4x^2 is Minimum(0,0),Maximum(1,1),Minimum(2,0)