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frx

  • 3 years ago

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...

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  1. anonymous
    • 3 years ago
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    damn i got cut off!!!

  2. anonymous
    • 3 years ago
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    here is my guess start with \[x ^{2}-12x+18\ln(x)+11\]

  3. anonymous
    • 3 years ago
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    take the derivative, show that it is positive on \(x>1\)

  4. anonymous
    • 3 years ago
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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

  5. frx
    • 3 years ago
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    That should do it, thanks!

  6. anonymous
    • 3 years ago
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    actually at 1 it is zero but that doesn't change the solution

  7. anonymous
    • 3 years ago
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    yw

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