October 11th, 2009, 02:10 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, 02:22 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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