Talk:Intermediate value theorem
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I'm not a mathematician, but there's a bit of this proof that seems wrong. It goes
"Suppose first that f (c) > u."
I don't see how we can suppose this since by definition
c = sup {x in [a, b] : f(x) ≤ u}.
which I understand as meaning that c is the largest number of those real numbers x such that f(x) is less or equal to u. If so, how could f(c) be greater than u??
- Wrong. sup does not mean the largest member of a set, but rather the smallest upper bound of the set, i.e., the smallest number that no member of the set exceeds. If the set has a largest member, then that is the set's sup, but the set of all numbers strictly less than 10 has no largest member, but still has a sup, which is 10. Michael Hardy 01:32, 4 Jan 2004 (UTC)
My mistakem, which I realised on Friday evening when I looked up the definition of supremum. Wondering why I didn't get this from Wikipedia, I looked at the definition there supremum, and saw a non-mathematical definition precedes it, which could cause confusion. DB.
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[edit] Calculus?
The IVT for integration is certainly a result of calculus, but the original IVT involves neither differentiation nor integration but only continuity, so shouldn't it be referred to in the first sentence as a result of analysis? --131.111.249.207 17:31, 2 Jun 2005 (UTC)
I agree, and I changed this
Grokmoo 13:29, 21 October 2005 (UTC)
Hmmm...if it only involves continuity, shouldn't it REALLY be called a result of TOPOLOGY???
[edit] Closed interval
Uncommon though the formation may be, I think the theorem can say f(x) = c for x in [a, b] not just (a, b). [article] says so too. -- Taku 01:07, 13 October 2005 (UTC)
Note that the statement with (a, b) is actually a STRONGER statement. In any case, there's no point in including a and b as endpoints, since it's impossible for them to work (plug them in and try).
It can if you like, but this case it trivial, so I wouldn't worry about it. The standard statement is for (a, b).
[edit] Proof that f(x) = x for some x
The proof right below the intermediate value theorem that there exists some x such that f(x) = x is wrong. This statement is in fact not generally true, even for f(x) continuous. There are additional requirements, for example, if the domain of f is all the reals and the function is bounded (above and below).
I reorganized this article and fixed the proof.
Grokmoo 13:29, 21 October 2005 (UTC)
Would it be useful to add an intuitively understandable example to this page? Something like: If person A is climbing a mountain from 6 to 7 am, and person B is coming down the mountain during the same time interval, then there has to be some time t in that time interval when they are both at exactly the same altitude?
The example of the "without lifting the pencil" is quite intuitive and very nice.
I agree. However, opening the page and being faced with a set of formulae isn't very nice. It would be better if someone put the following paragraph at the beginning, or in the introduction:
""This captures an intuitive property of continuous functions: given f continuous on [1, 2], if f (1) = 3 and f (2) = 5 then f must be equal to 4 somewhere between 1 and 2. It represents the idea that the graph of a continuous function can be drawn without lifting your pencil from the paper.""
Khedron 00:41, 23 December 2006 (UTC)
[edit] left and right neighbourhood
"Suppose first that f (c) > u ... whenever | x - c | < δ" I think it should be c - δ < x < c (left neighbourhood) and c < x < c + δ (right neighbourhood) for f(c) < u so we can omit absolut function. In first whenever c - δ < x < c --> f(x) - f(c) < 0 so |f(x) - f(c)| = -f(x) + f(c) —The preceding unsigned comment was added by 149.156.124.14 (talk) 15:48, 23 January 2007 (UTC).
