What is the extreme f(x)=3x^4-4x^3-12x^2+11 ?

The extreme f(x)=3x^4-4x^3-12x^2+11 is Minimum(-1,6),Maximum(0,11),Minimum(2,-21)

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extreme f(x)=3x^4-4x^3-12x^2+11

extreme

Minimum (- 1, 6), Maximum (0, 11), Minimum (2, - 21)

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} + 11 : 6

Minimum (- 1, 6)

Plug

x = 0

into

3 x^{4} - 4 x^{3} - 12 x^{2} + 11 : 11

Maximum (0, 11)

Plug

x = 2

into

3 x^{4} - 4 x^{3} - 12 x^{2} + 11 : - 21

Minimum (2, - 21)

Minimum (- 1, 6), Maximum (0, 11), Minimum (2, - 21)

What is the extreme f(x)=3x^4-4x^3-12x^2+11 ?
The extreme f(x)=3x^4-4x^3-12x^2+11 is Minimum(-1,6),Maximum(0,11),Minimum(2,-21)