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

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)

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)