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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = 6 x^{2} - 3

Find intervals:

Increasing

: - \infty < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{1}{2}

into

2 x^{3} - 3 x : 2

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

Plug

x = \frac{1}{2}

into

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

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

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

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

f (x) = \frac{x + 2}{x + 10}

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

f (x) = \frac{x - 4}{x - 6}

f (x) = x^{4} + x

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