What is the extreme f(x)=-64x^3+12x+4 ?

The extreme f(x)=-64x^3+12x+4 is Minimum(-1/4 ,2),Maximum(1/4 ,6)

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extreme f(x)=-64x^3+12x+4

extreme points

f (x) = - 64 x^{3} + 12 x + 4

Minimum (- \frac{1}{4}, 2), Maximum (\frac{1}{4}, 6)

Solution steps

f^{'} (x) = - 192 x^{2} + 12

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{4},

Increasing

: - \frac{1}{4} < x < \frac{1}{4},

Decreasing

: \frac{1}{4} < x < \infty

Plug

x = - \frac{1}{4}

into

- 64 x^{3} + 12 x + 4 : 2

Minimum (- \frac{1}{4}, 2)

Plug

x = \frac{1}{4}

into

- 64 x^{3} + 12 x + 4 : 6

Maximum (\frac{1}{4}, 6)

Minimum (- \frac{1}{4}, 2), Maximum (\frac{1}{4}, 6)

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

5 - \frac{2 y - 2 x}{2} = 4 x + 4

f (x) = \frac{x ^{2} + 5 x + 4}{x ^{2} - 1}

f (x) = sin (\frac{1}{2} x)

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

What is the extreme f(x)=-64x^3+12x+4 ?
The extreme f(x)=-64x^3+12x+4 is Minimum(-1/4 ,2),Maximum(1/4 ,6)