What is the extreme f(x)=6x^{4/3}-3x^{1/3} ?

The extreme f(x)=6x^{4/3}-3x^{1/3} is Minimum(1/8 ,-9/8)

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extreme f(x)=6x^{4/3}-3x^{1/3}

extreme points

f (x) = 6 x^{\frac{4}{3}} - 3 x^{\frac{1}{3}}

Minimum (\frac{1}{8}, - \frac{9}{8})

Solution steps

f^{'} (x) = \frac{- 1 + 8 x}{x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{1}{8},

Increasing

: \frac{1}{8} < x < \infty

Plug

x = \frac{1}{8}

into

6 x^{\frac{4}{3}} - 3 x^{\frac{1}{3}} : - \frac{9}{8}

Minimum (\frac{1}{8}, - \frac{9}{8})

f (x) = ln (x^{2} + 1) (x^{2} - 1)

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

f (x, y) = 4 - 2 x - 3 y

f (x) = 7 - x

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

What is the extreme f(x)=6x^{4/3}-3x^{1/3} ?
The extreme f(x)=6x^{4/3}-3x^{1/3} is Minimum(1/8 ,-9/8)