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ame

  • 2 years ago

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

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

  2. ame
    • 2 years ago
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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.

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

  4. ZeHanz
    • 2 years ago
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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

  5. ZeHanz
    • 2 years ago
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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...

  6. ame
    • 2 years ago
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    so, there is only one critical number (0)..

  7. ZeHanz
    • 2 years ago
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    Yes, we can get no more from this one...

  8. ame
    • 2 years ago
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    so what is critical point is sqrt 6 !

  9. ame
    • 2 years ago
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    (0, sqrt6)

  10. ZeHanz
    • 2 years ago
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    Here is the graph, it confirms what you found:

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  11. ame
    • 2 years ago
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    its webwork solution and it is not getting right..

  12. ZeHanz
    • 2 years ago
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    Thanks!

  13. ame
    • 2 years ago
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    what would be the critical point ?

  14. ZeHanz
    • 2 years ago
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    (0, sqrt(6))

  15. ZeHanz
    • 2 years ago
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    There is even a maximum in that point.

  16. ame
    • 2 years ago
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    the answer (0, sqrt(6)) cameout wrong in the webwork..

  17. ZeHanz
    • 2 years ago
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    But the real question now is: are you convinced that it is true?

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