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

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

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extreme y=-4x^3+x

extreme points

y = - 4 x^{3} + x

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

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Decreasing

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

Plug

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

into

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

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

Plug

x = \frac{1}{2 3}

into

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

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

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

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

y = x 4 - x

f (x) = (3 x + 1)^{5} (2 x + 3)^{2} (2 - x)^{3}

y = 2 x^{3} - 12 x^{2} - 270 x + 2

y = 0.0004 x^{2} - 0.0118 x + 2.6821

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