What is the extreme y=2x^3-3x^2-12x+7 ?

The extreme y=2x^3-3x^2-12x+7 is Maximum(-1,14),Minimum(2,-13)

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extreme y=2x^3-3x^2-12x+7

extreme points

y = 2 x^{3} - 3 x^{2} - 12 x + 7

Maximum (- 1, 14), Minimum (2, - 13)

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 + 7 : 14

Maximum (- 1, 14)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 7 : - 13

Minimum (2, - 13)

Maximum (- 1, 14), Minimum (2, - 13)

- x + 12

x^{3}

2 x + 4 - 5

f (x) = \frac{e ^{x}}{1 + e ^{2 x}}

What is the extreme y=2x^3-3x^2-12x+7 ?
The extreme y=2x^3-3x^2-12x+7 is Maximum(-1,14),Minimum(2,-13)