What is the extreme f(x)=3x^4+8x^3-6x^2-24x ?

The extreme f(x)=3x^4+8x^3-6x^2-24x is Minimum(-2,8),Maximum(-1,13),Minimum(1,-19)

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extreme f(x)=3x^4+8x^3-6x^2-24x

extreme points

f (x) = 3 x^{4} + 8 x^{3} - 6 x^{2} - 24 x

Minimum (- 2, 8), Maximum (- 1, 13), Minimum (1, - 19)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

3 x^{4} + 8 x^{3} - 6 x^{2} - 24 x : 8

Minimum (- 2, 8)

Plug

x = - 1

into

3 x^{4} + 8 x^{3} - 6 x^{2} - 24 x : 13

Maximum (- 1, 13)

Plug

x = 1

into

3 x^{4} + 8 x^{3} - 6 x^{2} - 24 x : - 19

Minimum (1, - 19)

Minimum (- 2, 8), Maximum (- 1, 13), Minimum (1, - 19)

x^{3} - 3 x^{2} - 9 x + 1

f (x) = x^{2} + x y + y^{2} - 2 x + 2 y - 8

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

f (x) = 2 x^{3} - 9 x^{2} + 12 x + 6, 0 \leq x \leq 3

y = \frac{x}{x ^{2} - x + 4}

What is the extreme f(x)=3x^4+8x^3-6x^2-24x ?
The extreme f(x)=3x^4+8x^3-6x^2-24x is Minimum(-2,8),Maximum(-1,13),Minimum(1,-19)