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

The extreme f(x)=2x^3-3x^2-12x+2 is Maximum(-1,9),Minimum(2,-18)

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

extreme points

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

Maximum (- 1, 9), Minimum (2, - 18)

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 + 2 : 9

Maximum (- 1, 9)

Plug

x = 2

into

2 x^{3} - 3 x^{2} - 12 x + 2 : - 18

Minimum (2, - 18)

Maximum (- 1, 9), Minimum (2, - 18)

\frac{5}{9} v + w

f (x) = (x + 1) e^{- x}

f (x) = x^{4} - 3 x^{3}

f (x) = \frac{x ^{3}}{3} - 2 x^{2} + 3 x

f (x) = x + \frac{64}{x}

What is the extreme f(x)=2x^3-3x^2-12x+2 ?
The extreme f(x)=2x^3-3x^2-12x+2 is Maximum(-1,9),Minimum(2,-18)