What is the extreme f(x)=3x^4+20x^3-36x^2-4 ?

The extreme f(x)=3x^4+20x^3-36x^2-4 is Minimum(-6,-1732),Maximum(0,-4),Minimum(1,-17)

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extreme f(x)=3x^4+20x^3-36x^2-4

extreme points

f (x) = 3 x^{4} + 20 x^{3} - 36 x^{2} - 4

Minimum (- 6, - 1732), Maximum (0, - 4), Minimum (1, - 17)

Solution steps

f^{'} (x) = 12 x^{3} + 60 x^{2} - 72 x

Find intervals:

Decreasing

: - \infty < x < - 6,

Increasing

: - 6 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 6

into

3 x^{4} + 20 x^{3} - 36 x^{2} - 4 : - 1732

Minimum (- 6, - 1732)

Plug

x = 0

into

3 x^{4} + 20 x^{3} - 36 x^{2} - 4 : - 4

Maximum (0, - 4)

Plug

x = 1

into

3 x^{4} + 20 x^{3} - 36 x^{2} - 4 : - 17

Minimum (1, - 17)

Minimum (- 6, - 1732), Maximum (0, - 4), Minimum (1, - 17)

2 x^{3} - 30 x^{2}, - 1 \leq x \leq 11

f (x) = x (1 - x)^{\frac{2}{5}}

f (x) = x^{2} - 2 x + 2, - 2 \leq x \leq 2

\frac{6}{- 3 x + 2}

ln (x^{2} - 4)

What is the extreme f(x)=3x^4+20x^3-36x^2-4 ?
The extreme f(x)=3x^4+20x^3-36x^2-4 is Minimum(-6,-1732),Maximum(0,-4),Minimum(1,-17)