What is the extreme f(x)=4x^{2/3}-5x^{1/3} ?

The extreme f(x)=4x^{2/3}-5x^{1/3} is Minimum(125/512 ,-25/16)

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extreme f(x)=4x^{2/3}-5x^{1/3}

extreme points

f (x) = 4 x^{\frac{2}{3}} - 5 x^{\frac{1}{3}}

Minimum (\frac{125}{512}, - \frac{25}{16})

Solution steps

f^{'} (x) = \frac{8}{3 x ^{\frac{1}{3}}} - \frac{5}{3 x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{125}{512},

Increasing

: \frac{125}{512} < x < \infty

Plug

x = \frac{125}{512}

into

4 x^{\frac{2}{3}} - 5 x^{\frac{1}{3}} : - \frac{25}{16}

Minimum (\frac{125}{512}, - \frac{25}{16})

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

f (x) = x^{5} e^{- x}

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 2 x + 10 y

f (x) = 1 - y + x

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

What is the extreme f(x)=4x^{2/3}-5x^{1/3} ?
The extreme f(x)=4x^{2/3}-5x^{1/3} is Minimum(125/512 ,-25/16)