What is the extreme f(x)=2x^3-3x^2-12x+8 ?

The extreme f(x)=2x^3-3x^2-12x+8 is Maximum(-1,15),Minimum(2,-12)

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extreme f(x)=2x^3-3x^2-12x+8

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 8

Maximum (- 1, 15), Minimum (2, - 12)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

2 x^{3} - 3 x^{2} - 12 x + 8 : 15

Maximum (- 1, 15)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 8 : - 12

Minimum (2, - 12)

Maximum (- 1, 15), Minimum (2, - 12)

f (x) = 3 x^{2} - x^{3}, 1 \leq x \leq 5

f (x, y) = x^{2} + y^{3} - 4 x - 3 y

f (x) = cos (3 x) - 2

f (x) = - sin (x) - 4

f (x) = 2 x^{3} - x^{2} - 4 x + 8 (- 1)

What is the extreme f(x)=2x^3-3x^2-12x+8 ?
The extreme f(x)=2x^3-3x^2-12x+8 is Maximum(-1,15),Minimum(2,-12)