What is the extreme f(x)=3x^4+4x^3-120x^2+120 ?

The extreme f(x)=3x^4+4x^3-120x^2+120 is Minimum(-5,-1505),Maximum(0,120),Minimum(4,-776)

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extreme f(x)=3x^4+4x^3-120x^2+120

extreme points

f (x) = 3 x^{4} + 4 x^{3} - 120 x^{2} + 120

Minimum (- 5, - 1505), Maximum (0, 120), Minimum (4, - 776)

Solution steps

f^{'} (x) = 12 x^{3} + 12 x^{2} - 240 x

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 5

into

3 x^{4} + 4 x^{3} - 120 x^{2} + 120 : - 1505

Minimum (- 5, - 1505)

Plug

x = 0

into

3 x^{4} + 4 x^{3} - 120 x^{2} + 120 : 120

Maximum (0, 120)

Plug

x = 4

into

3 x^{4} + 4 x^{3} - 120 x^{2} + 120 : - 776

Minimum (4, - 776)

Minimum (- 5, - 1505), Maximum (0, 120), Minimum (4, - 776)

What is the extreme f(x)=3x^4+4x^3-120x^2+120 ?
The extreme f(x)=3x^4+4x^3-120x^2+120 is Minimum(-5,-1505),Maximum(0,120),Minimum(4,-776)