What is the extreme f(x)=3x^4-12x^2+9 ?

The extreme f(x)=3x^4-12x^2+9 is Minimum(-sqrt(2),-3),Maximum(0,9),Minimum(sqrt(2),-3)

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extreme f(x)=3x^4-12x^2+9

extreme points

f (x) = 3 x^{4} - 12 x^{2} + 9

Minimum (- 2, - 3), Maximum (0, 9), Minimum (2, - 3)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

3 x^{4} - 12 x^{2} + 9 : - 3

Minimum (- 2, - 3)

Plug

x = 0

into

3 x^{4} - 12 x^{2} + 9 : 9

Maximum (0, 9)

Plug

x = 2

into

3 x^{4} - 12 x^{2} + 9 : - 3

Minimum (2, - 3)

Minimum (- 2, - 3), Maximum (0, 9), Minimum (2, - 3)

(8, - 6), (7, - 8)

(- 2.5) (10)

(- 1.4) (6.1)

f (x) = 8 x^{3} - 10

What is the extreme f(x)=3x^4-12x^2+9 ?
The extreme f(x)=3x^4-12x^2+9 is Minimum(-sqrt(2),-3),Maximum(0,9),Minimum(sqrt(2),-3)