My Math Forum not bounded

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 October 11th, 2009, 01:10 PM #1 Senior Member   Joined: Sep 2009 Posts: 115 Thanks: 0 not bounded how could i prove that x to the third is not bounded: we know for a function to be bounded it has to be f(x) < or equal to a boundary say c so for it to be not bounded there exists an x such that f(x) > c so for x to the third > c x> sqare root third of c so x = square root third of c +1 so plug (square root third of c +1) into the function but then i get: c + (3*square root third of c^2)+ (3*square root third of c)+1 is that good enough or is this too complicated
 October 11th, 2009, 01:22 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: not bounded Looks fine. Technically it only works for c > 0, but otherwise just choose x = 1 (say).

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