What is the extreme f(x)=4x^3-4x ?

The extreme f(x)=4x^3-4x is Maximum(-1/(sqrt(3)), 8/(3sqrt(3))),Minimum(1/(sqrt(3)),-8/(3sqrt(3)))

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extreme f(x)=4x^3-4x

extreme points

f (x) = 4 x^{3} - 4 x

Maximum (- \frac{1}{3}, \frac{8}{3 3}), Minimum (\frac{1}{3}, - \frac{8}{3 3})

Solution steps

f^{'} (x) = 12 x^{2} - 4

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

4 x^{3} - 4 x : \frac{8}{3 3}

Maximum (- \frac{1}{3}, \frac{8}{3 3})

Plug

x = \frac{1}{3}

into

4 x^{3} - 4 x : - \frac{8}{3 3}

Minimum (\frac{1}{3}, - \frac{8}{3 3})

Maximum (- \frac{1}{3}, \frac{8}{3 3}), Minimum (\frac{1}{3}, - \frac{8}{3 3})

f (x) = x - 6 x - 1

f (x, y) = x^{3} + y^{3} - 15 x y

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

f (x) = 2 x + 1 - \frac{18}{x}

f (x, y) = y - x

What is the extreme f(x)=4x^3-4x ?
The extreme f(x)=4x^3-4x is Maximum(-1/(sqrt(3)), 8/(3sqrt(3))),Minimum(1/(sqrt(3)),-8/(3sqrt(3)))