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

The extreme f(x)=3x^4-4x^3-12x^2 is Minimum(-1,-5),Maximum(0,0),Minimum(2,-32)

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

extreme points

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

Minimum (- 1, - 5), Maximum (0, 0), Minimum (2, - 32)

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} : - 5

Minimum (- 1, - 5)

Plug

x = 0

into

3 x^{4} - 4 x^{3} - 12 x^{2} : 0

Maximum (0, 0)

Plug

x = 2

into

3 x^{4} - 4 x^{3} - 12 x^{2} : - 32

Minimum (2, - 32)

Minimum (- 1, - 5), Maximum (0, 0), Minimum (2, - 32)

f (x) = 4 x^{3} - 45 x^{2} + 150 x

f (x) = ∣ 2 x + 5 ∣ - 1

f (x) = y^{6} + 5 y^{4} - 6 y^{2}

5 cos (\frac{x}{6})

f (x) = \frac{x}{4} + 2

What is the extreme f(x)=3x^4-4x^3-12x^2 ?
The extreme f(x)=3x^4-4x^3-12x^2 is Minimum(-1,-5),Maximum(0,0),Minimum(2,-32)