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

The extreme f(x)=-x^3+12x is Minimum(-2,-16),Maximum(2,16)

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

extreme points

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

Minimum (- 2, - 16), Maximum (2, 16)

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

- x^{3} + 12 x : - 16

Minimum (- 2, - 16)

Plug

x = 2

into

- x^{3} + 12 x : 16

Maximum (2, 16)

Minimum (- 2, - 16), Maximum (2, 16)

f (x) = 5 - 7 x^{2}

f (x) = x^{3} - 6 x^{2} + 6

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

f (x) = - 2 x^{3} + 33 x^{2} - 108 x + 1

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

What is the extreme f(x)=-x^3+12x ?
The extreme f(x)=-x^3+12x is Minimum(-2,-16),Maximum(2,16)