What is the extreme f(x)=2x^3-3x^2-12x+4 ?

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

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extreme f(x)=2x^3-3x^2-12x+4

extreme points

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 4

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

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 + 4 : 11

Maximum (- 1, 11)

Plug

x = 2

into

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

Minimum (2, - 16)

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

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

(3, 6), (5.5, 5)

y = \frac{x ^{3}}{3} - 2 x^{2} - 12 x

(0, 2), (1, 4)

g (x) = x^{2} + 3

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