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Popular >

extreme f(x)=-8x^{5/3}+\sqrt[3]{x}

Solution

extreme points

Solution

Solution steps

f^{'} (x) = - \frac{40 x ^{\frac{2}{3}}}{3} + \frac{1}{3 x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{4 0 ^{\frac{3}{4}}},

Increasing

: - \frac{1}{4 0 ^{\frac{3}{4}}} < x < 0,

Increasing

: 0 < x < \frac{1}{4 0 ^{\frac{3}{4}}},

Decreasing

: \frac{1}{4 0 ^{\frac{3}{4}}} < x < \infty

Plug

x = - \frac{1}{1 0 ^{\frac{1}{4}} \cdot 4 5}

into

Minimum (- \frac{1}{1 0 ^{\frac{1}{4}} \cdot 4 5}, - 8 (- \frac{1}{4 \cdot 1 0 ^{\frac{1}{4}} 5})^{\frac{5}{3}} - (\frac{1}{4 \cdot 1 0 ^{\frac{1}{4}} 5})^{\frac{1}{3}})

Plug

x = \frac{1}{1 0 ^{\frac{1}{4}} \cdot 4 5}

into

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Frequently Asked Questions (FAQ)

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