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 points

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

Minimum (0, 0)

Plug the extreme point

x = 1

into

Maximum (1, 1)

Plug the extreme point

x = 2

into

Minimum (2, 0)

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

(2 x - 1)^{2}

f (x, y) = 5 x^{2} - x y + 3 y^{2}

f (t) = t + (t^{2} + t) \cdot u (t - 3) + (t - t^{2}) u (t - 5)

2 x^{2} y - 4 y^{2} - 8 y - 4

f (x) = (x + 1)^{3} (x - 1)

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)