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

The extreme g(x)=x^{1/3}-x^{-2/3} is Maximum(-2,-2^{1/3}-1/((-2)^{2/3)})

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

extreme points

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

Maximum (- 2, - 2^{\frac{1}{3}} - \frac{1}{( - 2 ) ^{\frac{2}{3}}})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

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

Maximum (- 2, - 2^{\frac{1}{3}} - \frac{1}{( - 2 ) ^{\frac{2}{3}}})

f (x) = \frac{5 x ^{2}}{x ^{2} - 9}

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

32 ln (x) - x^{2}

f (x) = x^{3} - 2 x^{2} - 4 x + 1

f (x, y) = 3 x^{2} + 2 y^{2} - 6 x - 4 y + 16

What is the extreme g(x)=x^{1/3}-x^{-2/3} ?
The extreme g(x)=x^{1/3}-x^{-2/3} is Maximum(-2,-2^{1/3}-1/((-2)^{2/3)})