What is the extreme f(x)=1+12x-x^3 ?

The extreme f(x)=1+12x-x^3 is Minimum(-2,-15),Maximum(2,17)

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extreme f(x)=1+12x-x^3

extreme points

f (x) = 1 + 12 x - x^{3}

Minimum (- 2, - 15), Maximum (2, 17)

Solution steps

f^{'} (x) = - 3 x^{2} + 12

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

1 + 12 x - x^{3} : - 15

Minimum (- 2, - 15)

Plug

x = 2

into

1 + 12 x - x^{3} : 17

Maximum (2, 17)

Minimum (- 2, - 15), Maximum (2, 17)

f (x) = x^{3} + 3 x^{2} - 9 x - 3

f (x) = - x^{3} - 8 x

f (x) = 2 x^{3} - 27 x^{2} + 108 x + 3

f (x) = - 5 x^{2} + 1630 x

y = \frac{5 - x}{2 x + 6}

What is the extreme f(x)=1+12x-x^3 ?
The extreme f(x)=1+12x-x^3 is Minimum(-2,-15),Maximum(2,17)