What is the extreme f(x)=-5x^{5/3}+6\sqrt[3]{x} ?

The extreme f(x)=-5x^{5/3}+6\sqrt[3]{x} is

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extreme f(x)=-5x^{5/3}+6\sqrt[3]{x}

extreme points

Solution steps

f^{'} (x) = - \frac{25 x ^{\frac{2}{3}}}{3} + \frac{2}{x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - \frac{6 ^{\frac{3}{4}}}{5 \cdot 5 ^{\frac{1}{2}}},

Increasing

: - \frac{6 ^{\frac{3}{4}}}{5 \cdot 5 ^{\frac{1}{2}}} < x < 0,

Increasing

: 0 < x < \frac{6 ^{\frac{3}{4}}}{5 \cdot 5 ^{\frac{1}{2}}},

Decreasing

: \frac{6 ^{\frac{3}{4}}}{5 \cdot 5 ^{\frac{1}{2}}} < x < \infty

Plug

x = - \frac{6 ^{\frac{3}{4}}}{5 5}

into

Minimum - \frac{6 ^{\frac{3}{4}}}{5 5}, - 5 (- \frac{6 ^{\frac{3}{4}}}{5 5})^{\frac{5}{3}} - 6 (\frac{6 ^{\frac{3}{4}}}{5 5})^{\frac{1}{3}}

Plug

x = \frac{6 ^{\frac{3}{4}}}{5 5}

into

f (x, y) = 2 x^{2} + 2 x y + 2 y^{2} - 6 x

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

f (x, y) = x y (1 + 2 x - 3 y)

f (x) = \frac{x}{2} - sin (x)

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

What is the extreme f(x)=-5x^{5/3}+6\sqrt[3]{x} ?
The extreme f(x)=-5x^{5/3}+6\sqrt[3]{x} is