What is the extreme f(x)=-x^3+12x-16 ?

The extreme f(x)=-x^3+12x-16 is Minimum(-2,-32),Maximum(2,0)

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extreme f(x)=-x^3+12x-16

extreme points

f (x) = - x^{3} + 12 x - 16

Minimum (- 2, - 32), Maximum (2, 0)

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 : - 32

Minimum (- 2, - 32)

Plug

x = 2

into

- x^{3} + 12 x - 16 : 0

Maximum (2, 0)

Minimum (- 2, - 32), Maximum (2, 0)

- x + 3

f (x) = \frac{4}{3} - 2 cos (2 + \frac{1}{3} x)

\frac{x}{2 x + 1}

f (x) = - 2 49 x + 49

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

What is the extreme f(x)=-x^3+12x-16 ?
The extreme f(x)=-x^3+12x-16 is Minimum(-2,-32),Maximum(2,0)