What is the extreme f(x)=3x^4-4x^3-12x^2-1 ?

The extreme f(x)=3x^4-4x^3-12x^2-1 is Minimum(-1,-6),Maximum(0,-1),Minimum(2,-33)

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extreme f(x)=3x^4-4x^3-12x^2-1

extreme points

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} - 1

Minimum (- 1, - 6), Maximum (0, - 1), Minimum (2, - 33)

Solution steps

f^{'} (x) = 12 x^{3} - 12 x^{2} - 24 x

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 1

into

3 x^{4} - 4 x^{3} - 12 x^{2} - 1 : - 6

Minimum (- 1, - 6)

Plug

x = 0

into

3 x^{4} - 4 x^{3} - 12 x^{2} - 1 : - 1

Maximum (0, - 1)

Plug

x = 2

into

3 x^{4} - 4 x^{3} - 12 x^{2} - 1 : - 33

Minimum (2, - 33)

Minimum (- 1, - 6), Maximum (0, - 1), Minimum (2, - 33)

What is the extreme f(x)=3x^4-4x^3-12x^2-1 ?
The extreme f(x)=3x^4-4x^3-12x^2-1 is Minimum(-1,-6),Maximum(0,-1),Minimum(2,-33)