What is the extreme f(x)=|\sqrt[3]{x^2-2x}| ?

The extreme f(x)=|\sqrt[3]{x^2-2x}| is Minimum(0,0),Maximum(1,1),Minimum(2,0)

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extreme f(x)=|\sqrt[3]{x^2-2x}|

extreme

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

Solution steps

Find the critical points:

x = 0, x = 1, x = 2

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = 0

into

3 x^{2} - 2 x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

3 x^{2} - 2 x \Rightarrow y = 1

Maximum (1, 1)

Plug the extreme point

x = 2

into

3 x^{2} - 2 x \Rightarrow y = 0

Minimum (2, 0)

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

What is the extreme f(x)=|\sqrt[3]{x^2-2x}| ?
The extreme f(x)=|\sqrt[3]{x^2-2x}| is Minimum(0,0),Maximum(1,1),Minimum(2,0)