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Re: Prove x^3 not Continuous Uniform
Hey guys,
I'm having a little trouble trying to prove a question on uniform continuity that is in one of my textbooks (which doesn't have answers, unfortunately).
The question asks:
"Is the function f(x) = x^{3}+x uniformly continuous on R? Prove your answer"
I am almost certain that the answer is no for the simple reason that as x gets increasingly large the gaps get bigger so the same epsilon value won't work, but I'm not sure as to how to go about writing this in mathematical terms.
I know that for a function to be uniformly continuous |f(x) - f(x')|<epsilon whenever |x - x'|<delta for ALL x, x' that are elements of the subset, S, which in this case is R.
I would appreciate any help although preferably a proof that I can follow so that I can make the link to what I know to how to write it, if that makes sense.
Thanks in advance,
Mark
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Re: Prove x^3 not Continuous Uniform
Thanks for this, it did help me with understanding it.
Although I am now wondering how the '+ x' at the end would adjust this proof.
I'm guessing that I would write "y^{3} + y - x^{3} - x",
Which would then simplify to (y-x)(1 + (y^{2} + yx + x^{2})). Correct?
But then when you pick values for x and y you cannot do so without there being a delta left over in your answer (ie., you can show the contradiction as there is a delta left over)
Does this mean that it is, indeed, uniformly continuous?
Or do I need to choose x and y so that y = x + d/(d+3) (d = delta)
This would then cancel the deltas out and give 1 (as it is the multiplicative inverse), which contradicts the statement that |y^{2} + yx + x^{2}| < 1.
Although my problem with this is that I'm not sure if this fits with the previous restrictions that were made on x and y.
Any help here would be great!
Thanks,
Mark