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,-4),Maximum(0,1),Minimum(2,-31)

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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, - 4), Maximum (0, 1), Minimum (2, - 31)

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 : - 4

Minimum (- 1, - 4)

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 : - 31

Minimum (2, - 31)

Minimum (- 1, - 4), Maximum (0, 1), Minimum (2, - 31)

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

f (x) = \frac{x ^{2}}{x + 1}

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

f (x) = x^{4} - 2 x^{2} - 8

f (x, y) = 2 - 1 + x + y

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,-4),Maximum(0,1),Minimum(2,-31)