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swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Its my weekly question :)
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
If you dont like this one I can give you another one
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
It's been a little while since I've done actual analysis :P
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Ya so let me get you smth easier
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Well, the first part is pretty easy. It's basically just the definition of "not uniformly continuous"
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
BTW Congratulations. Where is ur offer from?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
U of Rochester. I'm still waiting to hear back from most schools I applied to. Unfortunately, I also just recently got a rejection from Princeton :(
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
awwwwww :( That is sad They are stupiidddddd
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Ah. This new one I can do. It's just the IVP (intermediate value theorem). To be honest, I wasn't expecting them to accept me.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Let's see how much I remember. Last time proved this, we started with the assumption that \(f(a)f(b)\le 0\). In this particular case, if \(f(a)=0\) or \(f(b)=0\), we're done. If \(f(a)f(b)<0\), then one of \(f(a),f(b)\) is negative, and the other positive. At this point, we had another theorem that proved there was then some \(c\in[a,b]\) s.t. \(f(c)=0\). Then define \(g(x)=f(x)+y\). Then, from what we just proved, it immediately follows that \(g(c)=y\). If we started with \(g(b)\le y\le g(a)\), take \(f(x)=g(x)y\), so \(f(b)\le 0 \le g(a)\). From this we get that \(f(b)f(a)\le 0\) and we're done.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
However, I cannot immediately remember how we proved that if \(f(a)f(b)\le0\), then there was some \(c\in[a,b]\) s.t. \(f(c)=0\).
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
hmmm gonna check my notes
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
hmmm I found a proof in my book using the supremum
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Well, I'm sorry, but I've got to go now. I need to get up early tomorrow, so I need to head to bed. Have a good night, and good luck with this!
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Thanks @KG :)
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Good night
 one year ago
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