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

The extreme f(x)=x^4+4x^3-2x^2-12x is Minimum(-3,-9),Maximum(-1,7),Minimum(1,-9)

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

extreme points

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

Minimum (- 3, - 9), Maximum (- 1, 7), Minimum (1, - 9)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 3

into

x^{4} + 4 x^{3} - 2 x^{2} - 12 x : - 9

Minimum (- 3, - 9)

Plug

x = - 1

into

x^{4} + 4 x^{3} - 2 x^{2} - 12 x : 7

Maximum (- 1, 7)

Plug

x = 1

into

x^{4} + 4 x^{3} - 2 x^{2} - 12 x : - 9

Minimum (1, - 9)

Minimum (- 3, - 9), Maximum (- 1, 7), Minimum (1, - 9)

What is the extreme f(x)=x^4+4x^3-2x^2-12x ?
The extreme f(x)=x^4+4x^3-2x^2-12x is Minimum(-3,-9),Maximum(-1,7),Minimum(1,-9)