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

The extreme f(x)=3x^4+4x^3-12x^2+2 is Minimum(-2,-30),Maximum(0,2),Minimum(1,-3)

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

extreme points

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

Minimum (- 2, - 30), Maximum (0, 2), Minimum (1, - 3)

Solution steps

f^{'} (x) = 12 x^{3} + 12 x^{2} - 24 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

3 x^{4} + 4 x^{3} - 12 x^{2} + 2 : - 30

Minimum (- 2, - 30)

Plug

x = 0

into

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

Maximum (0, 2)

Plug

x = 1

into

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

Minimum (1, - 3)

Minimum (- 2, - 30), Maximum (0, 2), Minimum (1, - 3)

f (x) = 3 x^{4} + 4 x^{3} - 12 x^{2} + 7

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

f (x, y) = x y e^{(- x - y)}

x^{2} - 6 x + 5

\frac{x}{x + 3}

What is the extreme f(x)=3x^4+4x^3-12x^2+2 ?
The extreme f(x)=3x^4+4x^3-12x^2+2 is Minimum(-2,-30),Maximum(0,2),Minimum(1,-3)