What is the extreme 2x^3-3x^2-12x+8 ?

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

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extreme 2x^3-3x^2-12x+8

extreme points

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, y) = x y + e^{x y}

g (x, y) = ln (x^{2} + y^{2} - 1)

f (x) = x^{3} - 3 x - 1

f (x, y) = - 3 x^{2} - 2 y^{2} + 3 x - 4 y + 5

P (x, y) = 36 x^{2} - 9 y^{2}

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