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frx Group Title

Show that \[x ^{2}+18\ln(x)+11>12x \] for x>1, I have no idea how to do the 18ln(x) messes things up for me...

  • one year ago
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  1. satellite73 Group Title
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    damn i got cut off!!!

    • one year ago
  2. satellite73 Group Title
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    here is my guess start with \[x ^{2}-12x+18\ln(x)+11\]

    • one year ago
  3. satellite73 Group Title
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    take the derivative, show that it is positive on \(x>1\)

    • one year ago
  4. satellite73 Group Title
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    that will mean your function is increasing, and since it is greater than zero at 1 (i am assuming it is, i didn't check) that means it will always be greater than zero

    • one year ago
  5. frx Group Title
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    That should do it, thanks!

    • one year ago
  6. satellite73 Group Title
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    actually at 1 it is zero but that doesn't change the solution

    • one year ago
  7. satellite73 Group Title
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    yw

    • one year ago
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