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anonymous
 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...
anonymous
 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0damn i got cut off!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here is my guess start with \[x ^{2}12x+18\ln(x)+11\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0take the derivative, show that it is positive on \(x>1\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That should do it, thanks!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually at 1 it is zero but that doesn't change the solution
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