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

1. anonymous

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

2. anonymous

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. anonymous

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

4. anonymous

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. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

so what is critical point is sqrt 6 !

9. anonymous

(0, sqrt6)

10. anonymous

Here is the graph, it confirms what you found:

11. anonymous

its webwork solution and it is not getting right..

12. anonymous

Thanks!

13. anonymous

what would be the critical point ?

14. anonymous

(0, sqrt(6))

15. anonymous

There is even a maximum in that point.

16. anonymous

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

17. anonymous

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