anonymous
  • anonymous
Find critical points h(x)= root over(6-3x^2)
Mathematics
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anonymous
  • anonymous
Find critical points h(x)= root over(6-3x^2)
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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ZeHanz
  • ZeHanz
To find critical points, you have to differentiate h. Then solve h'(x)=0. The solutions are the critical points.
anonymous
  • 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.
anonymous
  • anonymous
can you retype the function? as written, this is what you have: h(x)= root over(6-3x^2)

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ZeHanz
  • 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
ZeHanz
  • ZeHanz
Remember: If a/b=0 this means: a=0 (and b is not 0). So there is an easy solution here...
anonymous
  • anonymous
so, there is only one critical number (0)..
ZeHanz
  • ZeHanz
Yes, we can get no more from this one...
anonymous
  • anonymous
so what is critical point is sqrt 6 !
anonymous
  • anonymous
(0, sqrt6)
ZeHanz
  • ZeHanz
Here is the graph, it confirms what you found:
1 Attachment
anonymous
  • anonymous
its webwork solution and it is not getting right..
ZeHanz
  • ZeHanz
Thanks!
anonymous
  • anonymous
what would be the critical point ?
ZeHanz
  • ZeHanz
(0, sqrt(6))
ZeHanz
  • ZeHanz
There is even a maximum in that point.
anonymous
  • anonymous
the answer (0, sqrt(6)) cameout wrong in the webwork..
ZeHanz
  • ZeHanz
But the real question now is: are you convinced that it is true?

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