## ame Group Title Find critical points h(x)= root over(6-3x^2) one year ago one year ago

1. ZeHanz Group Title

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

2. ame Group Title

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 Group Title

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

4. ZeHanz Group Title

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 Group Title

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

6. ame Group Title

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

7. ZeHanz Group Title

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

8. ame Group Title

so what is critical point is sqrt 6 !

9. ame Group Title

(0, sqrt6)

10. ZeHanz Group Title

Here is the graph, it confirms what you found:

11. ame Group Title

its webwork solution and it is not getting right..

12. ZeHanz Group Title

Thanks!

13. ame Group Title

what would be the critical point ?

14. ZeHanz Group Title

(0, sqrt(6))

15. ZeHanz Group Title

There is even a maximum in that point.

16. ame Group Title

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

17. ZeHanz Group Title

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