A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
@KingGeorge
anonymous
 3 years ago
@KingGeorge

This Question is Open

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

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

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

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

KingGeorge
 3 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"

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

KingGeorge
 3 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 :(

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

KingGeorge
 3 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
 3 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
 3 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. \(f(c)=0\).

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

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

KingGeorge
 3 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!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.