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how do you show that
lnx<(x^1/2) x>0...
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Hence f(x) has a minimum at x = 4. Also, since f(x) → +∞ both as x → 0[sup]+[/sup] and as x → +∞, and f(x) is continuous and has no other critical points, this minimum is a global minimum.
Finally, note that f(4) = 2 − 2ln2 > 0 (since ln2 < 1). Since the minimum value of f(x) is positive, we can conclude that f(x) > 0 for all x > 0.
Theres probably a smarter method to do this. Mathsyperson? Ricky?
Last edited by JaneFairfax (2008-04-28 00:41:52)
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That's the way I'd do it. (Well, my first thought was induction, but that's obviously wrong)
Why did the vector cross the road?
It wanted to be normal.
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