1:00 AM
I want to learn to graph functions like that

sure, the half-ass-time

which software can I use?

lol

what isnt @TedShifrin

I swear to you (I'd bet at least $10) that it was done with Bezier curves in a drawing program. Most of what you wrote makes no sense, @MATH. 1:00 AM but which program something like Geogebra? I took it from a book I use Illustrator. I don't know the free stuff. he was trying to teach us the intermediate value theorem heck, one could do it in paint if one had to hol up let me send the picture he drew in class OK, @MATH. So you have a continuous function on the interval$[a,b]$. If you pick a value$y$between$f(a)$and$f(b)$, then he's saying there's a$c\in (a,b)$with$f(c)=y$. What I wrote makes sense. 1:02 AM is it difficult to do it using Illustrator? Nah. You click and drag a bunch. You get good at it. Sad thing is Illustrator is expensive. I bought it when I started writing books. I see I'm sure there are free alternatives; I just don't know them. I get graphs that look alike in latex I know you can do simple shapes in LaTeX, but not weird stuff. And that isn't a piece of a circle. 1:04 AM I know im still not understanding it @TedShifrin @Danu is really good at xtik (or something like that), so you might ask him. Draw the graph of a continuous function on an interval$[a,b]$, @MATH. I can get the crack for Illustrator Draw a horizontal line between the heights$f(a)$and$f(b)$. but do I have to draw the axis and everything? 1:05 AM Then that horizontal line has to cross the graph between$a$and$b$. Yeah, @twink, they probably had a universal backdrop (maybe from some math program) that laid down the coordinate system. ok drew a line @TedShifrin @MATH, then, as I said, your line has to cross your graph. yes it crosses if its continous so no holes right, @MATH. That's what the intermediate value theorem says has to happen. no holes, no jumps ok so what about the point c thing 1:07 AM lol how do we prove that its in between the$c$is the$x$-coordinate where it crosses. all you have is the graph on$[a,b]$. This is an actual theorem, and you're not expected to prove it (I'm sure). i found the IVT to be ubsurdly states Huh? it seems ubsurd for what it actually says 1:08 AM it's intuitively obvious but requires some depth to prove. not absurd just intuitively obvious. dont get me wrong its useful I'm complaining about your word choice. lol whihc one! absurd? That's the trouble with calc I ... stuff is hard to prove but intuitively obvious. yes. hmm 1:10 AM In Calc II and III, I can derive/"prove" some things in a way that gives intuitive understanding. In Calc I I cannot do that. Show that the equation x^5 + 4x^3 - 7x + 14 = 0 has at least one real solution. Use the Intermediate Value Theorem to explain, so how would do this problem without a graphing calculator, it looks like it has no holes so should i try doing the synthetic division No, @MATH. Just show it's <0 somewhere and >0 somewhere and explain that it's continuous. how can I make this function pass through (2,1)? :S The shape is not nearly the same, though, @twink. But you can translate on both axes. Review your precalculus. (I have to teach this tomorrow morning.) I'm trying to get an approximation of the original graph not the same graph 1:12 AM OK, so review precalc. :) How do you translate vertically (easy)? How do you translate horizontally (slightly harder)? so i would still need it to use synthetic division to try and get it to be a quadratic so i can factor? @TedShifrin yeah but I also want it to pass throguh (4,4) No, no, no, @MATH. You're not going to find the root. You're just using the theorem. Read what I wrote. Oh, then you're screwed, @twink. I just need to add a constant to translate it vertically You're gonna have to do some stretching, too. 1:13 AM O.O Morning dodsy @MATHASKER If you're trying to use synthetic division or some other way of factoring, you're missing the point of the exercise entirely. late day, faust. yes but to still prove that ones less than zero and the other is greater than zero i have to find the roots and try pluggin them to find y values and see if they are greater or less than zero @TedShifrin "you're going to have to do some stretching" sounds like a willy wonka quote. 1:14 AM You are not asked to find the zero. All you have to show is it exists. No, no, @MATH, just plug in (intelligently chosen) numbers for$x$. pick x =-100 pick x = 100 one value its positive the other its negative there must exist one value between -100 and 100 where it equals 0 too many "it"s in that sentence, @Faust :) Confusing to a newbie. lol I'm sorta serious. Use a few nouns.. 1:15 AM whats a noun? smacks Faust person place or thing? right Part of the chat experience here is learning to be a good teacher 1:16 AM so you want me to replace it with function?4 Yes Im learning to be a good learner :3 ah yes i understand LOL, OK, Nate. :S 1:17 AM But you too have to do your share of helping. Ah, one day. That's how you become better at math. That's seriously one reason I did so well on GREs ... was because I'd spent so much time helping/teaching people calculus and such. he helped me drink some beer yesterday that counts right? Um, no. haha you make me sound like a country bumpkin, Faust. 1:18 AM that's a bonding experience, not a teaching one :P @TedShifrin that's true. can all Bezier curves be represented by a formula? hell no 3 @TedShifrin lol then the formula I'm looking for may not exist 1:19 AM although they often use cubic splines (gluing different cubics together) right not in the sense of$y=f(x)$when I click and drag, I don't have cubic splines @TedShifrin my prof answered me ight i chose -4 and got -1294 so than should i choose another number? yup 1:20 AM and confirmed what everyone said here. I would say there's a formula in the sense of Bezier curves being piecewise parametric @MATHASKER try a positive one but that's not a useful characterization for the present purpose Semi, I think they're piecewise-cubic, but you don't always get splines when you click and drag in Illustrator. tried positive 4 and got 1266 1:21 AM hmm, fair now what does the IVT say? that for a negative x there must be a y and for a positive x there must be a y?... no Reread what I typed up there ^^^^^ Don't go into vagueries. Be specific. 1:22 AM Damn, everyone's hanging up on poor @MATH, but it's good for him. What does the IVT state, if you have$a=-4$and$b=4$? it builds character? :P it says that there is some value c in the domain (-4,4) where f(c)= ? Weird question but does$(x-3)^{\frac{1}{3}} = \sqrt[3]{(x-3)}$do \sqrt[3]{} thanks 1:24 AM np one from my notes said that f(c) = (f(a), f(b)) that's equal though right? That's crap, @MATHASKER. or is my notation off. Remember I told you what you originally typed was garbage? 1:24 AM I'd say that's fine, @dodsy thanks semi. Yes, for any real number$a$,$a^{1/3} = \root3\of a$. Remember what$a^{m/n}$means. The only place where it could be problematic is for complex values, and well oh true, i guess i got the notes wrong let me see what you wrote gotta scorll up For complex, it's all multivalued. Note I said real. 1:25 AM complex isn't really relevant @TedShifrin why do you like to be mean? !? oh this guy again rolls eyes I don't think I've been mean. I'm direct. @MATHASKER what value would like f(c) to equal ? that would help you with the roots of the function? Don't stir up trouble, twink. I don't pamper students, but I'm not mean. 1:26 AM It's mean to tell someone what he said is garbage It made no sense (which is what I typed the first time). so i chose like -1 which is in between -4 and 4 Ted, in my opinion, it isn't worth it. but that's rude and see if there is a value for that? 1:26 AM @MATHASKER what does it mean for something to be a root of a function Well, you won't get good at math if I tell you you're wonderful when you're not. No I mean, if it was me, I wouldn't engage "Twink". that when you plug that number in you get zero as an answer I know, Nate. We're old "friends." ok good! 1:27 AM and he won't get good at math telling him that what he said is garbage Oh haha. so you would like f(c)=0 ! for some c and remember there's always people better than one @twink: Over my long teaching career, I did have some students who did not like me. But 95% liked me a lot and probably 50% took multiple courses from me. I don't need you lecturing to me. If you don't like me, then ignore me. now the IVT tells you that there is some c between -4 and 4 such that f(c)=0 1:28 AM But I'm proud that Faust is taking over and doing a great job. yes @Faust teachers learn from the students too sometimes so the IVT tells you that there is some root in there which was what your original question was asking you to show it's odd that you like to pick at me because I'm openly gay. 1:29 AM but not when you're arrogant It's not arrogant to say that something makes no sense. But I get your point. Let's drop it. LOL who's talking about sexuality? >_> You have made huge fusses in here, twink. So cut it out. if you say so.. 1:30 AM So, what's up for tomorrow, Ted? rolls 9 3/4 eyes so do like try to pick a number between those two and see which one equals to zero @Faust Teaching functions, Nate :) Teaching? Nice! Yeah, I'm teaching a class every Sunday to math geeks ;) 1:31 AM and going to the gay club later It's not rude to tell someone that what they're saying is garbage. Pretty typical in the math community. If someone takes that as insult, get over it. sighhhh. It's a critique of the argument, not of the person @MATHASKER you simply can show the calculations for f(-4) and f(4) then say by the IVT a root c exists between -4 and 4 such that f(c) =0 and your done @Dodsy: It's a small class, but so far some of them seem really smart. 1:32 AM honestly, today somebody told me something I said was nonsense and it saved me a lot of embarrassment, because I would've handed in nonsense. I tried for 2 years to volunteer tutor at-risk kids, but that's harder than getting a paid job. Probably twink thinks I'm not qualified. @TedShifrin That's great! I'm sure the students enjoy it just as much as you do. :) I find that (assofar) professors that show a great interest in what they teach seem to be more fun to listen to. @MATHASKER you were simply asked to show that a root existed not what the actual root is as a number Nate, I think you tend to learn more if the teacher wants you to. Me personally? Or in general? 1:33 AM @Faust: I think the hard part of IVT is that it's just vague existence, and students expect that a precise answer is expected. In general, Nate. I think both apply, but trying to discern your meaning. Ah, I see. I agree. oh true, @Faust ight thanks guys i guess i was misinterpeting what my teacher was teaching thats why i made it so complicated @TedShifrin you are one of the most qualified people here, but also one of the most arrogants Sure thing, @MATH. It's a new thing to have theorems/statements like that. And you're being so constructive, twink. he wrote something like c(a,b) f(c) = f(a),f(b)) and i just copied it down cause i wasnt paying attention 1:35 AM @TedShifrin well it is arguably the first time where you wave your hands around for awhile and then say a solution exists but i haven't a inclination what it is. at least it was the first time for me in mathematics I think he wrote$\in$without explaining it, @MATH. So you wrote = thats why i avoided element and domain and range Yes, Faust, you did superbly. maybe, i might go to my teacher and try to clear it up a bit more, lol i did this in pre-calc, thats why it was confusing me lol i know that confused the crap out of me in calc I 1:36 AM I tried to say "between" and stuff earlier. Aren't you supposed to be asleep, Jasper? woot woot jasper in dah house Jasps I went to see SBA's calculus class. Just don't leave out an "s" there, Nate. Oh, how's he doing? I just tried to contribute a little here and there. 1:38 AM lmfao. @MATHASKER a teachers Office Hours are more important than the actual class imho Both are important. There weren't many students, and it started a couple hours late. I am shaking my head @TedShifrin A couple of hours late!! I suppose that's ok, Nate. 1:39 AM I don't know whether the students actually understood or not, because there wasn't time for tests. But we just discussed the epsilon delta definition, that's about it. It's really tough for a youngster to be in charge of an on-line class with no teaching experience and little mathematical experience. But I applaud him for trying. I hope he'll ask us for help/advice if he needs to. I think the whole idea is crazy, personally ... Yeah. Except he isn't, Nate. no haha, I know Damn Scientologist. 1:41 AM @Dodsy why are you stalking me? just found that part of the transcript funny. I am not, sir. yes, you are, sir I think I'm going to cook dinner. Anyone who wants to drop by is welcome. Most I know don't. now I know one who does @Twink No, I overheard you talking to Ted before and was looking for that conversation, not stalking you. I'd have no benefit to stalking you, Twink. @TedShifrin enjoy your dinner! 1:43 AM sure too much paranoia in this chat yes. I don't agree believe what you like :) Hi @Twink, lol 1:44 AM the chat needs to be rid of the trolls and the dramatists @BalarkaSen I have a feeling that I would be eradicated O.O I don't care what you feel I didn't mean to offend you, Twink. just thought the transcript was funny. you didn't, you need much more to offend me What is going on? 1:46 AM hi @Jasper @Jasper not a damn clue I think Dodsy has a crisis LOL (removed) I had to remove it sounded too ridiculous. :( :) 1:50 AM I now have a suspicion that Dodsy is Twink... you are$\iff\$ mean
Wouldn't that be really weird?

Jasper now you're having hallucinations

LOL

you're getting worse and worse :(

Hey @LeakyNun

1:51 AM
you need help

But how can you two respond to each other so quickly?

idk @Jasper I didn't think it was that quick.

typing fast?

no, I am a very slow typer.

1:52 AM
Hmm, then I guess you two can be good friends. You have a lot of chemistry, lol.

:o

I prefer prostate cancer

@BalarkaSen what's that comment

thats defintly terriffing

1:53 AM
by kieran

@Twink I once had it examined. It was a weird sensation.

@Jasper be careful, you're dating yourself.

@Dodsy loool i never noticed that comment

@Twink Nope, just a weird sensation, lol.

was it a nice sensation?

1:55 AM
@Twink not sure if these are appropriate questions, Twink.

Change of topic then, lol.
So what brand of coffee do you guys like?

Tim Hortons, of course.

These days I just get those in a packet and add hot water.

Tim hortons coffee?
They have timmies where you live, Jasp?

I just drink some brandless one, lol.

1:56 AM
oh I see.

Hmm, I don't think I have seen timmies in Antarctica.

@Jasper are you still crazy?

D:
Antarctica!?
Twink, seriously bud??

@Twink Yes, I am still trying to get better. I hope I can make it soon. Pray for me.

I can't
I don't believe in god

1:58 AM
Pray means fervent hope that sends EM waves across the universe.
You don't need to believe in any god for that.

I don't believe in sending waves

OK then

Twink, how about you just believe in being nice to others and sensitive to their feelings?

but it's possible that you'll be crazy for the rest of your life

Well, I have hope.

1:59 AM
I just said it's possible

First of all, "crazy" is offensive.

@Dodsy What's your favourite brand of beer again?