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

The extreme f(x)= 1/(x^3-x) is Minimum(-1/(sqrt(3)),(3sqrt(3))/2),Maximum(1/(sqrt(3)),-(3sqrt(3))/2)

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < - \frac{1}{3},

Increasing

: - \frac{1}{3} < x < 0,

Increasing

: 0 < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - \frac{1}{3}

into

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

Minimum (- \frac{1}{3}, \frac{3 3}{2})

Plug

x = \frac{1}{3}

into

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

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

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

y = \frac{x ^{3}}{3} - x^{2} - 8 x

3 x^{2} - 18 x + 24

f (x) = x^{4} - 4 x^{3} + 9, - 5 \leq x \leq 2

f (x, y) = x^{2} y^{2} - 5 x^{2} - 8 x y - 5 y^{2}

\frac{1}{5} x + 1

What is the extreme f(x)= 1/(x^3-x) ?
The extreme f(x)= 1/(x^3-x) is Minimum(-1/(sqrt(3)),(3sqrt(3))/2),Maximum(1/(sqrt(3)),-(3sqrt(3))/2)