What is the extreme 2x^3+3x^2-12x-4 ?

The extreme 2x^3+3x^2-12x-4 is Maximum(-2,16),Minimum(1,-11)

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extreme 2x^3+3x^2-12x-4

extreme points

2 x^{3} + 3 x^{2} - 12 x - 4

Maximum (- 2, 16), Minimum (1, - 11)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

2 x^{3} + 3 x^{2} - 12 x - 4 : 16

Maximum (- 2, 16)

Plug

x = 1

into

2 x^{3} + 3 x^{2} - 12 x - 4 : - 11

Minimum (1, - 11)

Maximum (- 2, 16), Minimum (1, - 11)

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

f (x) = x^{2} - 2 x - 1

f (x) = x y + \frac{8}{x} + \frac{8}{y}

f (x, y) = ln (x y - 1)

f (x) = 3 x^{3} - 6 x^{2} + 1

What is the extreme 2x^3+3x^2-12x-4 ?
The extreme 2x^3+3x^2-12x-4 is Maximum(-2,16),Minimum(1,-11)