What is the extreme f(x)=2=x^3-x ?

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

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extreme f(x)=2=x^3-x

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

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

Plug

x = - \frac{1}{3}

into

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

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

Plug

x = \frac{1}{3}

into

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

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

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

\frac{4 x ^{2} - 10}{( x - 8 ) ( x + 6 )}

f (x) = x^{2} - 12 x + 5

4 sin (3 x + π)

f (x) = \frac{( ln ( x ) )}{x}

y = x^{2} - 4 x + 3

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