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July 28th, 2012, 09:42 PM   #1
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Can anyone prove this?

We know:
(1) f(x)/x is strictly decreasing in x;
(2) f(0)=0 and f(3)=0;
(3) f(x) is continuous in x;

Can we prove that f(x) is concave in x on [0, 3]? Or f(x) is single-peaked on [0, 3]?

Thanks.
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July 29th, 2012, 02:23 PM   #2
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Re: Can anyone prove this?

[color=#000000]No, take as a counter example.[/color]
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July 29th, 2012, 02:40 PM   #3
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Re: Can anyone prove this?

Surely, is not "strictly decreasing"?
HallsofIvy is offline  
July 30th, 2012, 01:04 AM   #4
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Re: Can anyone prove this?

Quote:
Originally Posted by HallsofIvy
Surely, is not "strictly decreasing"?
[color=#000000]I made a typo and I cannot understand the ironic tone of your post (unfortounately there is no real time contact I may be wrong), I meant .[/color]
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