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Find critical points h(x)= root over(6-3x^2)

Mathematics
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To find critical points, you have to differentiate h. Then solve h'(x)=0. The solutions are the critical points.
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.
can you retype the function? as written, this is what you have: h(x)= root over(6-3x^2)

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OK, so you've got your derivative! Solving h'(x) = 0: \[-\frac{ 3x }{ \sqrt{6-3x^2} }=0 \Leftrightarrow 3x=0\] So x=0
Remember: If a/b=0 this means: a=0 (and b is not 0). So there is an easy solution here...
so, there is only one critical number (0)..
Yes, we can get no more from this one...
so what is critical point is sqrt 6 !
(0, sqrt6)
Here is the graph, it confirms what you found:
1 Attachment
its webwork solution and it is not getting right..
Thanks!
what would be the critical point ?
(0, sqrt(6))
There is even a maximum in that point.
the answer (0, sqrt(6)) cameout wrong in the webwork..
But the real question now is: are you convinced that it is true?

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