What is the extreme y=4x^3-x ?

The extreme y=4x^3-x is Maximum(-1/(2sqrt(3)), 1/(3sqrt(3))),Minimum(1/(2sqrt(3)),-1/(3sqrt(3)))

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extreme y=4x^3-x

extreme points

y = 4 x^{3} - x

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

Solution steps

f^{'} (x) = 12 x^{2} - 1

Find intervals:

Increasing

: - \infty < x < - \frac{1}{2 3},

Decreasing

: - \frac{1}{2 3} < x < \frac{1}{2 3},

Increasing

: \frac{1}{2 3} < x < \infty

Plug

x = - \frac{1}{2 3}

into

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

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

Plug

x = \frac{1}{2 3}

into

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

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

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

(\frac{x}{4})^{4} - \frac{4}{3} x^{3} + \frac{1}{2} x^{2} + 6 x + 2

\frac{1}{3} x^{3} - 4 x^{2} + 12 x + 200

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

f (x) = g (t) = 6 t^{4} - 8 t^{3} + 5

f (x) = - \frac{x ^{2}}{8} + \frac{y ^{2}}{78.13} + 8

What is the extreme y=4x^3-x ?
The extreme y=4x^3-x is Maximum(-1/(2sqrt(3)), 1/(3sqrt(3))),Minimum(1/(2sqrt(3)),-1/(3sqrt(3)))