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 July 28th, 2012, 08:42 PM #1 Newbie   Joined: Feb 2012 Posts: 19 Thanks: 0 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.
 July 29th, 2012, 01:23 PM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Can anyone prove this? [color=#000000]No, take $f(x)=x(x-3)^3$ as a counter example.[/color]
 July 29th, 2012, 01:40 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Can anyone prove this? Surely, $(x- 3)^3$ is not "strictly decreasing"?
July 30th, 2012, 12:04 AM   #4
Math Team

Joined: Nov 2010
From: Greece, Thessaloniki

Posts: 1,989
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Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus
Re: Can anyone prove this?

Quote:
 Originally Posted by HallsofIvy Surely, $(x- 3)^3$ 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 $f(x)=x(x-3)^2$.[/color]

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