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ame Group Title

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

  • one year ago
  • one year ago

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

    • one year ago
  2. ame Group Title
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    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.

    • one year ago
  3. dpaInc Group Title
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    can you retype the function? as written, this is what you have: h(x)= root over(6-3x^2)

    • one year ago
  4. ZeHanz Group Title
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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

    • one year ago
  5. ZeHanz Group Title
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    Remember: If a/b=0 this means: a=0 (and b is not 0). So there is an easy solution here...

    • one year ago
  6. ame Group Title
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    so, there is only one critical number (0)..

    • one year ago
  7. ZeHanz Group Title
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    Yes, we can get no more from this one...

    • one year ago
  8. ame Group Title
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    so what is critical point is sqrt 6 !

    • one year ago
  9. ame Group Title
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    (0, sqrt6)

    • one year ago
  10. ZeHanz Group Title
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    Here is the graph, it confirms what you found:

    • one year ago
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  11. ame Group Title
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    its webwork solution and it is not getting right..

    • one year ago
  12. ZeHanz Group Title
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    Thanks!

    • one year ago
  13. ame Group Title
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    what would be the critical point ?

    • one year ago
  14. ZeHanz Group Title
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    (0, sqrt(6))

    • one year ago
  15. ZeHanz Group Title
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    There is even a maximum in that point.

    • one year ago
  16. ame Group Title
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    the answer (0, sqrt(6)) cameout wrong in the webwork..

    • one year ago
  17. ZeHanz Group Title
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    But the real question now is: are you convinced that it is true?

    • one year ago
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