What is the extreme f(x)=-64x^3+12x+9 ?

The extreme f(x)=-64x^3+12x+9 is Minimum(-1/4 ,7),Maximum(1/4 ,11)

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extreme f(x)=-64x^3+12x+9

extreme points

f (x) = - 64 x^{3} + 12 x + 9

Minimum (- \frac{1}{4}, 7), Maximum (\frac{1}{4}, 11)

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 + 9 : 7

Minimum (- \frac{1}{4}, 7)

Plug

x = \frac{1}{4}

into

- 64 x^{3} + 12 x + 9 : 11

Maximum (\frac{1}{4}, 11)

Minimum (- \frac{1}{4}, 7), Maximum (\frac{1}{4}, 11)

f (x, y) = - 3 x^{2} - 2 y^{2} + 24 x + 20 y + 600

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

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

f (x) = 10 + 5 x - x^{2}

f (x) = 2 x^{2} n x - 19 x^{2}

What is the extreme f(x)=-64x^3+12x+9 ?
The extreme f(x)=-64x^3+12x+9 is Minimum(-1/4 ,7),Maximum(1/4 ,11)