What is the extreme f(x)=-7x^{5/3}+2\sqrt[3]{x} ?

The extreme f(x)=-7x^{5/3}+2\sqrt[3]{x} is

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extreme f(x)=-7x^{5/3}+2\sqrt[3]{x}

extreme

Minimum - (\frac{8}{42875})^{\frac{1}{4}}, - 7 (- (\frac{8}{42875})^{\frac{1}{4}})^{\frac{5}{3}} - 2 (\frac{8}{42875})^{\frac{1}{12}}, Maximum (\frac{8}{42875})^{\frac{1}{4}}, - 7 (\frac{8}{42875})^{\frac{5}{12}} + 2 3 (\frac{8}{42875})^{\frac{1}{4}}

Solution steps

f^{'} (x) = - \frac{35 x ^{\frac{2}{3}}}{3} + \frac{2}{3 x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - (\frac{2}{35})^{\frac{3}{4}},

Increasing

: - (\frac{2}{35})^{\frac{3}{4}} < x < 0,

Increasing

: 0 < x < (\frac{2}{35})^{\frac{3}{4}},

Decreasing

: (\frac{2}{35})^{\frac{3}{4}} < x < \infty

Plug

x = - (\frac{8}{42875})^{\frac{1}{4}}

into

- 7 x^{\frac{5}{3}} + 2 3 x : - 7 (- (\frac{8}{42875})^{\frac{1}{4}})^{\frac{5}{3}} - 2 (\frac{8}{42875})^{\frac{1}{12}}

Minimum - (\frac{8}{42875})^{\frac{1}{4}}, - 7 (- (\frac{8}{42875})^{\frac{1}{4}})^{\frac{5}{3}} - 2 (\frac{8}{42875})^{\frac{1}{12}}

Plug

x = (\frac{8}{42875})^{\frac{1}{4}}

into

- 7 x^{\frac{5}{3}} + 2 3 x : - 7 (\frac{8}{42875})^{\frac{5}{12}} + 2 3 (\frac{8}{42875})^{\frac{1}{4}}

Maximum (\frac{8}{42875})^{\frac{1}{4}}, - 7 (\frac{8}{42875})^{\frac{5}{12}} + 2 3 (\frac{8}{42875})^{\frac{1}{4}}

Minimum - (\frac{8}{42875})^{\frac{1}{4}}, - 7 (- (\frac{8}{42875})^{\frac{1}{4}})^{\frac{5}{3}} - 2 (\frac{8}{42875})^{\frac{1}{12}}, Maximum (\frac{8}{42875})^{\frac{1}{4}}, - 7 (\frac{8}{42875})^{\frac{5}{12}} + 2 3 (\frac{8}{42875})^{\frac{1}{4}}

What is the extreme f(x)=-7x^{5/3}+2\sqrt[3]{x} ?
The extreme f(x)=-7x^{5/3}+2\sqrt[3]{x} is