## Denise 5 years ago Find the critical vales of 12x^3 -24x

1. anonymous

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

2. anonymous

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

3. Denise

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

4. anonymous

yep, now derive it

5. Denise

12x(x^2-12)

6. anonymous

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

7. anonymous

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

8. Denise

that is where I get stuck.

9. anonymous

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

10. anonymous

so your critical values are: x = 0 and $x = \pm \sqrt(2)$ ^_^

11. anonymous

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

12. Denise

yep

13. Denise

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