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

The extreme f(x)= x/(x^3-1) is Maximum(-1/(\sqrt[3]{2)},(2^{2/3})/3)

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extreme f(x)= x/(x^3-1)

extreme points

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

Solution steps

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

Find intervals:

Increasing

Decreasing

: 1 < x < \infty

Plug

into

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

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

f (x) = \frac{6 x}{7 x - 3}

15 x^{\frac{2}{3}} - 10 x

7 x^{6} - 4 x^{5}

f (x) = - 2 x^{3} + 18 x^{2} + 49

What is the extreme f(x)= x/(x^3-1) ?
The extreme f(x)= x/(x^3-1) is Maximum(-1/(\sqrt[3]{2)},(2^{2/3})/3)