Question

Find the critical vales of 12x^3 -24x

Answer 1

to find the critical numbers, you have to derive the function once and take 2 conditions for f'(x):

1) f'(x) = 0
2 f'(x) = UND (undefined)

in your case, it'll be, f'(x) = 0 <-- to find the critical numbers ^_^ give it a try now

Answer 2

set f'(x) to zero after deriving and find the zeros ^_^

Answer 3

The original equation was f(x)=3x^4-12x^2+4

Answer 4

yep, now derive it

Answer 5

12x(x^2-12)

Answer 6

now set this equal to 0 and find the zeros normally ^_^

Answer 7

the zeros that you'll find = will be your critical values :)

Answer 8

that is where I get stuck.

Answer 9

my dear:

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

Answer 10

so your critical values are:

x = 0

and \[x = \pm \sqrt(2)\]

^_^

Answer 11

you had a lil mistake in finding the derivative ^_^

Answer 12

yep

Answer 13

so to find the relative min and max I need to do ???