What is the extreme f(x)=-2x^3-3x^2+12x+11 ?

The extreme f(x)=-2x^3-3x^2+12x+11 is Minimum(-2,-9),Maximum(1,18)

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extreme f(x)=-2x^3-3x^2+12x+11

extreme points

f (x) = - 2 x^{3} - 3 x^{2} + 12 x + 11

Minimum (- 2, - 9), Maximum (1, 18)

Solution steps

f^{'} (x) = - 6 x^{2} - 6 x + 12

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 2

into

- 2 x^{3} - 3 x^{2} + 12 x + 11 : - 9

Minimum (- 2, - 9)

Plug

x = 1

into

- 2 x^{3} - 3 x^{2} + 12 x + 11 : 18

Maximum (1, 18)

Minimum (- 2, - 9), Maximum (1, 18)

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

f (x) = \frac{4 x - 8}{3}

h (x) = 4^{x}

y = \frac{1}{3} x - 3

8 x - 5 y + 13 = 0

What is the extreme f(x)=-2x^3-3x^2+12x+11 ?
The extreme f(x)=-2x^3-3x^2+12x+11 is Minimum(-2,-9),Maximum(1,18)