What is the extreme f(x)=2x^3-3x^2-12x+6 ?

The extreme f(x)=2x^3-3x^2-12x+6 is Maximum(-1,13),Minimum(2,-14)

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extreme f(x)=2x^3-3x^2-12x+6

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 6

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

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 + 6 : 13

Maximum (- 1, 13)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 6 : - 14

Minimum (2, - 14)

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

f (x) = \frac{x}{x ^{2} + 16}

f (x) = \frac{2 + x}{2 - x}

7 x^{2} - 9 x - 5

sin (\frac{2}{x})

\frac{3 x - 3}{2 x - 2}

What is the extreme f(x)=2x^3-3x^2-12x+6 ?
The extreme f(x)=2x^3-3x^2-12x+6 is Maximum(-1,13),Minimum(2,-14)