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

The extreme \sqrt[3]{x^2}*(x-2)^2 is Minimum(0,0),Maximum(1/2 , 9/(4\sqrt[3]{4)}),Minimum(2,0)

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extreme \sqrt[3]{x^2}*(x-2)^2

extreme points

Solution steps

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

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

Decreasing

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

Increasing

: 2 < x < \infty

Plug

x = 0

into

Minimum (0, 0)

Plug

x = \frac{1}{2}

into

Plug

x = 2

into

Minimum (2, 0)

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

f (x, y) = 12 - x^{2} - y^{2}

f (x, y) = x + y + x y

f (x) = 1 + \frac{5}{x} - \frac{8}{x ^{2}}

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

What is the extreme \sqrt[3]{x^2}*(x-2)^2 ?
The extreme \sqrt[3]{x^2}*(x-2)^2 is Minimum(0,0),Maximum(1/2 , 9/(4\sqrt[3]{4)}),Minimum(2,0)