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October 11th, 2009, 02:10 PM   #1
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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
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October 11th, 2009, 02:22 PM   #2
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Re: not bounded

Looks fine. Technically it only works for c > 0, but otherwise just choose x = 1 (say).
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