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swissgirl
 2 years ago
@KingGeorge
swissgirl
 2 years ago
@KingGeorge

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swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Its my weekly question :)

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0If you dont like this one I can give you another one

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1It's been a little while since I've done actual analysis :P

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0Ya so let me get you smth easier

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

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0BTW Congratulations. Where is ur offer from?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1U 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 :(

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0awwwwww :( That is sad They are stupiidddddd

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1Ah. 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.

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1Let'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.

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

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0hmmm gonna check my notes

swissgirl
 2 years ago
Best ResponseYou've already chosen the best response.0hmmm I found a proof in my book using the supremum

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.1Well, 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!
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