What is the extreme f(x)=(x-1)\sqrt[3]{x^2} ?

The extreme f(x)=(x-1)\sqrt[3]{x^2} is Maximum(0,0),Minimum(2/5 ,-(3\sqrt[3]{4})/(5\sqrt[3]{25)})

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extreme f(x)=(x-1)\sqrt[3]{x^2}

extreme points

Solution steps

Find the critical points:

x = 0, x = \frac{2}{5}

Domain of

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{2}{5},

Increasing

: \frac{2}{5} < x < \infty

Plug the extreme point

x = 0

into

Maximum (0, 0)

Plug the extreme point

x = \frac{2}{5}

into

f (x) = \frac{x + 3}{x ^{2} - 4 x - 21}

2.5 x^{2} + 75 x + 25000

f (x) = (100 - 2 x) \cdot (75 - 2 x) \cdot x

f (x) = 5 x x - x^{2}

f (x) = x^{2} - 72 ln (x)

What is the extreme f(x)=(x-1)\sqrt[3]{x^2} ?
The extreme f(x)=(x-1)\sqrt[3]{x^2} is Maximum(0,0),Minimum(2/5 ,-(3\sqrt[3]{4})/(5\sqrt[3]{25)})