What is the extreme f(x)=2x^3+3x^2-12x+1 ?

The extreme f(x)=2x^3+3x^2-12x+1 is Maximum(-2,21),Minimum(1,-6)

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extreme f(x)=2x^3+3x^2-12x+1

extreme points

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 1

Maximum (- 2, 21), Minimum (1, - 6)

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 + 1 : 21

Maximum (- 2, 21)

Plug

x = 1

into

2 x^{3} + 3 x^{2} - 12 x + 1 : - 6

Minimum (1, - 6)

Maximum (- 2, 21), Minimum (1, - 6)

f (x) = x + 5 y = 10

- \frac{1}{6} sin (6 x)

f (x) = - \frac{1}{2 5 - x}

f (x) = \frac{x ^{2} - 9}{x + 3}

\frac{1}{2} (x - 2)^{2} - 3

What is the extreme f(x)=2x^3+3x^2-12x+1 ?
The extreme f(x)=2x^3+3x^2-12x+1 is Maximum(-2,21),Minimum(1,-6)