## ame 2 years ago Find critical points h(x)= root over(6-3x^2)

1. ZeHanz

To find critical points, you have to differentiate h. Then solve h'(x)=0. The solutions are the critical points.

2. ame

HI , i did and sort of stuck at the end -3x(6-3x^2)^-1/2=0, x=0, and (6-3x^2)^(-1/2)=0, so how do i do at the second part... thanks for the help.

3. dpaInc

can you retype the function? as written, this is what you have: h(x)= root over(6-3x^2)

4. ZeHanz

OK, so you've got your derivative! Solving h'(x) = 0: $-\frac{ 3x }{ \sqrt{6-3x^2} }=0 \Leftrightarrow 3x=0$ So x=0

5. ZeHanz

Remember: If a/b=0 this means: a=0 (and b is not 0). So there is an easy solution here...

6. ame

so, there is only one critical number (0)..

7. ZeHanz

Yes, we can get no more from this one...

8. ame

so what is critical point is sqrt 6 !

9. ame

(0, sqrt6)

10. ZeHanz

Here is the graph, it confirms what you found:

11. ame

its webwork solution and it is not getting right..

12. ZeHanz

Thanks!

13. ame

what would be the critical point ?

14. ZeHanz

(0, sqrt(6))

15. ZeHanz

There is even a maximum in that point.

16. ame

the answer (0, sqrt(6)) cameout wrong in the webwork..

17. ZeHanz

But the real question now is: are you convinced that it is true?