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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
 one year 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...
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.2
damn i got cut off!!!
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
here is my guess start with \[x ^{2}12x+18\ln(x)+11\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
take the derivative, show that it is positive on \(x>1\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
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

frxBest ResponseYou've already chosen the best response.0
That should do it, thanks!
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
actually at 1 it is zero but that doesn't change the solution
 one year ago
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