What is the extreme f(x)=-4x^{5/3}+7\sqrt[3]{x} ?

The extreme f(x)=-4x^{5/3}+7\sqrt[3]{x} is

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extreme f(x)=-4x^{5/3}+7\sqrt[3]{x}

extreme points

Solution steps

f^{'} (x) = - \frac{20 x ^{\frac{2}{3}}}{3} + \frac{7}{3 x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - (\frac{7}{20})^{\frac{3}{4}},

Increasing

: - (\frac{7}{20})^{\frac{3}{4}} < x < 0,

Increasing

: 0 < x < (\frac{7}{20})^{\frac{3}{4}},

Decreasing

: (\frac{7}{20})^{\frac{3}{4}} < x < \infty

Plug

x = - \frac{7 ^{\frac{3}{4}}}{2 \cdot 2 ^{\frac{1}{2}} \cdot 5 ^{\frac{3}{4}}}

into

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

Plug

x = \frac{7 ^{\frac{3}{4}}}{2 \cdot 2 ^{\frac{1}{2}} \cdot 5 ^{\frac{3}{4}}}

into

f (x, y) = 3 x - 4 y + 5 (1, 1)

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

f (x, y) = \frac{x ^{2} + y ^{2}}{x ^{2} y}

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

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

What is the extreme f(x)=-4x^{5/3}+7\sqrt[3]{x} ?
The extreme f(x)=-4x^{5/3}+7\sqrt[3]{x} is