What is the extreme f(x)=3\sqrt[3]{x}-4x ?

The extreme f(x)=3\sqrt[3]{x}-4x is Minimum(-1/8 ,-1),Minimum(1/8 ,1)

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extreme f(x)=3\sqrt[3]{x}-4x

extreme points

Minimum (- \frac{1}{8}, - 1), Minimum (\frac{1}{8}, 1)

Solution steps

f^{'} (x) = \frac{1}{x ^{\frac{2}{3}}} - 4

Find intervals:

Increasing

: - \infty < x < - 8,

Decreasing

: - 8 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = - \frac{1}{8}

into

Minimum (- \frac{1}{8}, - 1)

Plug

x = \frac{1}{8}

into

Minimum (\frac{1}{8}, 1)

Minimum (- \frac{1}{8}, - 1), Minimum (\frac{1}{8}, 1)

What is the extreme f(x)=3\sqrt[3]{x}-4x ?
The extreme f(x)=3\sqrt[3]{x}-4x is Minimum(-1/8 ,-1),Minimum(1/8 ,1)