What is the extreme f(x)=-x^3+12x-19 ?

The extreme f(x)=-x^3+12x-19 is Minimum(-2,-35),Maximum(2,-3)

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extreme f(x)=-x^3+12x-19

extreme points

f (x) = - x^{3} + 12 x - 19

Minimum (- 2, - 35), Maximum (2, - 3)

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 - 19 : - 35

Minimum (- 2, - 35)

Plug

x = 2

into

- x^{3} + 12 x - 19 : - 3

Maximum (2, - 3)

Minimum (- 2, - 35), Maximum (2, - 3)

f (x) = x - 9

\frac{4 x - 2 y}{3} = x + 1

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

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

What is the extreme f(x)=-x^3+12x-19 ?
The extreme f(x)=-x^3+12x-19 is Minimum(-2,-35),Maximum(2,-3)