What is the extreme f(x)=2x^3+3x^2-12x+8 ?

The extreme f(x)=2x^3+3x^2-12x+8 is Maximum(-2,28),Minimum(1,1)

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extreme f(x)=2x^3+3x^2-12x+8

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 8

Maximum (- 2, 28), Minimum (1, 1)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

2 x^{3} + 3 x^{2} - 12 x + 8 : 28

Maximum (- 2, 28)

Plug

x = 1

into

2 x^{3} + 3 x^{2} - 12 x + 8 : 1

Minimum (1, 1)

Maximum (- 2, 28), Minimum (1, 1)

What is the extreme f(x)=2x^3+3x^2-12x+8 ?
The extreme f(x)=2x^3+3x^2-12x+8 is Maximum(-2,28),Minimum(1,1)