What is the extreme f(x)=x^{2/3}-x^{1/3} ?

The extreme f(x)=x^{2/3}-x^{1/3} is Minimum(1/8 ,-1/4)

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extreme f(x)=x^{2/3}-x^{1/3}

extreme points

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

Minimum (\frac{1}{8}, - \frac{1}{4})

Solution steps

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

x^{\frac{2}{3}} - x^{\frac{1}{3}} : - \frac{1}{4}

Minimum (\frac{1}{8}, - \frac{1}{4})

f (x) = 7 (x^{2} + 1)^{2} (x - 5)^{3}

f (x) = 9 x^{4} - 4 x^{3}, - 3 \leq x \leq 3

f (x) = (x^{2} - 5 x + 4) (x)

f (x) = 2600 + 2 x + 0.001 x^{2}

f (x) = x^{3} - 12 x^{2} - 27 x + 9 (- 2)

What is the extreme f(x)=x^{2/3}-x^{1/3} ?
The extreme f(x)=x^{2/3}-x^{1/3} is Minimum(1/8 ,-1/4)