What is the extreme f(x)=-2x^{5/3}+4\sqrt[3]{x} ?

The extreme f(x)=-2x^{5/3}+4\sqrt[3]{x} is

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extreme f(x)=-2x^{5/3}+4\sqrt[3]{x}

extreme points

f (x) = - 2 x^{\frac{5}{3}} + 4 3 x

Minimum - (\frac{8}{125})^{\frac{1}{4}}, - 2 (- (\frac{8}{125})^{\frac{1}{4}})^{\frac{5}{3}} - 4 (\frac{8}{125})^{\frac{1}{12}}, Maximum (\frac{8}{125})^{\frac{1}{4}}, - 2 (\frac{8}{125})^{\frac{5}{12}} + 4 3 (\frac{8}{125})^{\frac{1}{4}}

Solution steps

f^{'} (x) = - \frac{10 x ^{\frac{2}{3}}}{3} + \frac{4}{3 x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - (\frac{2}{5})^{\frac{3}{4}},

Increasing

: - (\frac{2}{5})^{\frac{3}{4}} < x < 0,

Increasing

: 0 < x < (\frac{2}{5})^{\frac{3}{4}},

Decreasing

: (\frac{2}{5})^{\frac{3}{4}} < x < \infty

Plug

x = - (\frac{8}{125})^{\frac{1}{4}}

into

- 2 x^{\frac{5}{3}} + 4 3 x : - 2 (- (\frac{8}{125})^{\frac{1}{4}})^{\frac{5}{3}} - 4 (\frac{8}{125})^{\frac{1}{12}}

Minimum - (\frac{8}{125})^{\frac{1}{4}}, - 2 (- (\frac{8}{125})^{\frac{1}{4}})^{\frac{5}{3}} - 4 (\frac{8}{125})^{\frac{1}{12}}

Plug

x = (\frac{8}{125})^{\frac{1}{4}}

into

- 2 x^{\frac{5}{3}} + 4 3 x : - 2 (\frac{8}{125})^{\frac{5}{12}} + 4 3 (\frac{8}{125})^{\frac{1}{4}}

Maximum (\frac{8}{125})^{\frac{1}{4}}, - 2 (\frac{8}{125})^{\frac{5}{12}} + 4 3 (\frac{8}{125})^{\frac{1}{4}}

Minimum - (\frac{8}{125})^{\frac{1}{4}}, - 2 (- (\frac{8}{125})^{\frac{1}{4}})^{\frac{5}{3}} - 4 (\frac{8}{125})^{\frac{1}{12}}, Maximum (\frac{8}{125})^{\frac{1}{4}}, - 2 (\frac{8}{125})^{\frac{5}{12}} + 4 3 (\frac{8}{125})^{\frac{1}{4}}

f (x) = \frac{( x ^{2} - 5 x + 4 )}{x - 5}

f (x) = 2 x^{3} - 3 x^{2} - 72 x + 15

- x^{3} - 9 x^{2}

y = \frac{x ^{2} + 2}{x ^{2} - 25}

f (x) = - sin (x) - 4, (0, 2 π)

What is the extreme f(x)=-2x^{5/3}+4\sqrt[3]{x} ?
The extreme f(x)=-2x^{5/3}+4\sqrt[3]{x} is