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mathkillsalot
Dec31-10, 10:36 AM
1. The problem statement, all variables and given/known data

Let f be continuous at c and f(c)>0. Prove that there is an open interval (a,b) centered at c such that f(x)>0 for all x that are elements of (a,b)

2. Relevant equations

we are supposed to use the sign preserving property??

3. The attempt at a solution

I tried assigning all x as numbers greater than zero but then realized it wouldn't work since the f(x) is the one that has to be greater than 0. But the function isn't given.
Please help me....
Dick
Dec31-10, 04:02 PM
Use the epsilon/delta definition of continuity. Pick epsilon=f(c)/2.
mathkillsalot
Dec31-10, 06:49 PM
uhmmmm...can you please demonstrate??
and how did you get your epsilon?
╔(σ_σ)╝
Dec31-10, 11:18 PM
Funny someone else just made the same thread :http://www.physicsforums.com/showthread.php?t=460196


Show us that you have atleast attempted the problem.
mathkillsalot
Jan1-11, 05:49 AM
i used the definition of sign preserving property to prove that f(a) and f(b) are positive.
just assigned all x that are elements of (a,b) to be greater than 0...
My answer seemed to be correct...
mathkillsalot
Jan1-11, 05:49 AM
but it's not checked yet
HallsofIvy
Jan1-11, 05:57 AM
What is this "sign preserving property" you are talking about?
mathkillsalot
Jan1-11, 07:32 AM
here's the gist of it :))
i posted the wrong one a while ago, sorry

http://www.cut-the-knot.org/fta/brodie.shtml
╔(σ_σ)╝
Jan1-11, 09:04 AM
Do you have multiple accounts ? This is not allowed :-(.

The same question and the same mistake was posted by "another" user in the calculus and beyond section.
mathkillsalot
Jan1-11, 09:19 AM
no no i don't. I think I did post this question twice though. Didn't see the forum for homework at first.

though if you're talking about goodheavens, that person might be someone from the same school as me
╔(σ_σ)╝
Jan1-11, 10:01 AM
no no i don't. I think I did post this question twice though. Didn't see the forum for homework at first.

though if you're talking about goodheavens, that person might be someone from the same school as me

Okay,sorry about the confusion. :-)