What is the extreme f(x)=3x^4-4x^3-12x^2+17 ?

The extreme f(x)=3x^4-4x^3-12x^2+17 is Minimum(-1,12),Maximum(0,17),Minimum(2,-15)

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extreme f(x)=3x^4-4x^3-12x^2+17

extreme points

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 17

Minimum (- 1, 12), Maximum (0, 17), Minimum (2, - 15)

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} + 17 : 12

Minimum (- 1, 12)

Plug

x = 0

into

3 x^{4} - 4 x^{3} - 12 x^{2} + 17 : 17

Maximum (0, 17)

Plug

x = 2

into

3 x^{4} - 4 x^{3} - 12 x^{2} + 17 : - 15

Minimum (2, - 15)

Minimum (- 1, 12), Maximum (0, 17), Minimum (2, - 15)

x^{2} - 6 x + 7

f (x) = x^{3} + 6 x^{2} + 9 x - 8

f (x, y) = x^{3} - y^{3} - 2 x y + 10

f (x) = 2 x^{2} - 8 x, 0 \leq x \leq 6

x^{2} + 2 x - 8

What is the extreme f(x)=3x^4-4x^3-12x^2+17 ?
The extreme f(x)=3x^4-4x^3-12x^2+17 is Minimum(-1,12),Maximum(0,17),Minimum(2,-15)