## anonymous one year ago Anyone good at finding critical points???

1. anonymous

$y=3x^2-96\sqrt{x}$

2. ChillOut

Take it's derivative y'. Do you know how?

3. anonymous

I know i have to take y' but that's the thing, would it be $6x-48(x)^{-1/2}$ ??

4. ChillOut

Correct.

5. anonymous

then?

6. anonymous

umm

7. anonymous

I'm confused :(

8. ChillOut

9. anonymous

Don't i have to solve for the independent variable?

10. ChillOut

You have to solve f'(x)=0.

11. anonymous

yeah that's where i got stuck, because i'm used to questions where i favtor (x+y)(x+y)

12. anonymous

Factor*

13. ChillOut

Well, you just need to solve f'(x)=0 really. Those points can be extrema points

14. anonymous

So $6x-48(x)^{-1/2} = 0 ?$

15. ChillOut

Yes.

16. ChillOut

It has only one solution.

17. anonymous

Square the whole thing?

18. ChillOut

Might do. Try it

19. ChillOut

Well, I have to leave soon, do you want me to give you the x coordinate?

20. anonymous

sure

21. ChillOut

x=4. It shouldn't be hard to find it.

22. anonymous

okay i got it, thank you so much

23. ChillOut

Now you just plug it in the original function and find its y coordinate.

24. anonymous

i plug the 4 instead of x?

25. anonymous

nvm that was a dumb question, i got it haha thanks

26. anonymous

y=-30?

27. anonymous

-144**