What is the extreme f(x)=2x^3-3x^2-12x+7 ?

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

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extreme f(x)=2x^3-3x^2-12x+7

extreme points

f (x) = 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)

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

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

4 x + 5 y = 16

x^{3} - \frac{15}{2} x^{2} - 18 x - 1

x^{3} - 18 \cdot x^{2} + 105 \cdot x + 9

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