Intermediate value theorem | Existence theorems | AP Calculus AB | Khan Academy

[@mxvortex2341]

It's amazing how textbooks can somehow make the most intuitive thing confusing

[@pauselab5569]

well continuity has a more rigorous definition than just "don't lift your pen" and it's pretty interesting to try to prove this theorem using the f(x)=f(x+h) for small h.

[@sheepsong5681]

I had no clue what my teacher was doing. Now I see it’s common sense like you say. Thank you. Now to try doing the homework…

[@bondwithbonds]

r+aised: $22,65Lb- And, that's how that was did, like a bib in the sheets of piggy bank heaven. Thanks G. Khan

[@enthused_panda]

boyss, if there's one thing you know about sal, it is that he can rap. overlay lofi hip hop on this video and bam... rapping sal.

[@tinadorr-kapczynski7480]

I TRIED THIS AND I'M CRACKING UP

[@nooriman5944]

boys and girls, increase the playback speed to top that

[@Thegamemakur]

Can you do a video proving 1+1=2?

[@grafficconcepts9029]

That would be impossible as the answer to 1+1 is not 2 it's a window.

[@218Flows]

If you factor out a 1 you get: 1(1+1) = 1(2) and then divide 2 you get: 1(1+1)/2 =1 and then: 2/2 = 1

[@grafficconcepts9029]

@Joe L good attempt but we are going to need a carpenter to prove this one I think champ

[@vuufke4327]

that's a math axiom, it can not be proven yet it's universally accepted (frl no joke)

[@bhoxzivanlangnamanpfhoe7869]

what is the possibility that if you have 2 eyes and if you count it you suddenly have 3

[@seamusswan3043]

wow the quality is so much nicer now

[@NoelleLajoie-n4d]

Sal, thanks for making a BSAE/CS minor possible for a little girl like me. You really are the best tutor our field could have asked for.

[@justinswag3403]

literally god bless your soul my teacher cannot explain for ####

[@balendravirsinghchauhan6006]

Oh my god.!! I had a smile on my face at 7:27 these videos are just mind blowing.!! extremely awesome and extremely knowledgeable.!!

[@Zavs___]

WYSI

[@zelsuki]

WYSI

[@bourjac]

2:40 whooop

[@abhirajarora7631]

I laughed too too hard at that.

[@oghenekomeegbevurie9205]

Your voice is so soothing to listen to🥴

[@lukegraham2881]

7:10 The suspense was beautiful... (well that's all for today folks...)

[@dakshkumar8982]

Thank you sir

[@theewanger3786]

you got some clean hand writing

[@ziintle]

Have you *seen* PatrickJMT's though 😩

[@firat.c]

7:44 my pretext skipping subjects before midterm exam

[@AbhiEducationFactory]

How you know everything? You are great. It is very helpful

[@sayidabyan5828]

mmh, i love you. i guess i'll have you teach me for the next few years

[@HotstarJio]

but this does not work for f(X)=(x-1)(x-2)when interval is [1,2]

[@broschetta4417]

good explanation thank you

[@palaksharma9926]

Hehe bold of you to assume that I will have a 'Mathematical career'...

[@fornogg]

Would it be possible to know what application you are using to make these videos? I'm a teacher and was just given a tablet PC and this is an excellent use for it, but I can't find anything as clean as this. Thank you!

[@jonsampson3030]

Khash Afshar In case you haven't found it, he uses a program called Smoothdraw3.

[@wajahatkhan9455]

Thanks...

[@SakthiPandiSocialMedia]

this video isn't a L

[@reflex7510]

that is the best graph for a continuous function XD

[@anaskaddoura4309]

Should it be differntiable or is it enough to be continuous ?

[@theyoutubetube....akaamit7786]

Continuous function is the need of an hour.

[@NewWesternFront]

@@theyoutubetube....akaamit7786 what

[@wemiphiri3466]

​@@NewWesternFront think they meant need of the hour

[@ABobroff]

at 2:43, you say the function, which I assume includes a vertical asymptote, is not continuous, however the definition of continuity only states that WITHIN THE DOMAIN, limx-->a = f(a), and the vertical asymptote is excluded from the domain, therefore the function still satisfies the definition of continuity.

[@mochimochi6357]

Limit x+ is not equal to Limit x- therefore it’s not a continuous function

[@katpizo]

love u forever

[@prabinshrestha7163]

I would really like someone to answer this question If there is a function y = 7-(x-5)^2 and it has a gap discontinuity at x = 4 Then is the function continuous according to the intermediate value theorem between the interval 1 to 7??

[@ubongabasigeorge5727]

No it's a piecewise function

[@lolygagger5991]

thx

[@Kudravets-Diana]

How to proof it with the help of cantor lemma ?

[@faithelias2798]

God loves you and he wants to save everyone, but in order for him to do that, you need to repent and be baptized. Also share his gospel with you come into contact with and keep his commandments 🙏🏾😘

[@marcushendriksen8415]

Only about 3 minutes in, but could not this theorem be generalised to include functions that are discontinuous at some c within some interval? Assuming that the limit near or at c exists, of course.

[@yayza7881]

no

[@petarspengler_ai]

is this otherwise known as cauchy-bolzano theorem?

[@W__Edits]

you are khan?

[@aaro1268]

So... stating the obvious property of continuous functions (and/or function intervals); they contain all values within their domain. Aside from its use in mathematical proofs, I don't see the use of such a tautological statement. If someone were to say "The sky is blue.", it would not help me determine why clouds produce rain.

[@doggybuns8573]

Aaron King It's supposed to be a fundamental and intuitive idea, not a theorem that solves everything. Math builds upon itself. No need to be pretentious, Einstein.

[@TheGamersTVChannel]

Aaron King get destroyed

[@aaro1268]

Anthony Acosta I'm still around. The self-evident nature of the theorems makes it seem silly that they are given so much more attention than operations, but there really isn't a point to arguing when I agree that they're fundamental.

[@Peugot905]

The intermediate value theorem is actually derived from Balzano's theorem. You don't use it much unless you are taking a class in Analysis.

[@zurviver_3747]

Aaron, you are being mislead, the sky actuality reflects about 65% purple light and only 35% blue light. Our eyes are simply not as keen in the purple-violet range.

[@piyushkoranne558]

what if we take the L as point of Global minima in your second graph ?

It must be in [a,b] and point of minima lies there.. Minima can too be that L

[@benkalu2508]

What a useless theorem

[@avg.guy.1012]

😂

[@sorindyy8622]

Strongly agrees

[@rohitsd7732]

Lol

[@SathickJaffer-hs5sj]

I too agree

[@UmerZaiem]

Haha

[@TylerMatthewHarris]

WHOOP . not continuous

[@prabinshrestha7163]

I would really like someone to answer this question If there is a function y = 7-(x-5)^2 and it has a gap discontinuity at x = 4 Then is the function continuous according to the intermediate value theorem between the interval 1 to 7?? or For function y = (((x-5)^2)*((x-4)/(x-4))) Is the function continuous according to the intermediate value theorem between the interval 1 to 7??

[@zurviver_3747]

the first function is discountinous at x=4 because f(4) does not exist. you could fix this by adding x=4, y= limit, and then use the theorem, or you could limit the therorum to [1,4) U (4,7]

[@john66rocky]

The Intermediate Value Theorem does not conclude that ANY function is continuous. Rather, the theorem only applies if two conditions are met, and one of those is that the function is continuous on a given interval. The function you provided is not continuous on the interval 1 to 7 and so the IVT does not apply to this function on this particular interval. One thing that my students struggle with is that, even if the theorem does not apply on a given interval, the conclusion can still be true. In your example, the input 1 produces an output of -9 and the input 7 produces an output of 3. If we choose an output between -9 and 3, like, say y = 0, then we can find an input (about x = 2.534) which we might call "c" so that y(c) = 0. The existence of that number is not, however, something that we can conclude by applying the IVT because, as stated previously, the IVT does not apply to this function on the interval from x = 1 to x = 7 because of the discontinuity at x = 4. The question, "Is this function continuous according to the intermediate value theorem?" may reflect a misunderstanding of what the theorem does. All the IVT does is guarantee the existence of an input value (meeting certain requirements) on a given interval on which the continuity of a particular function is already known. I hope you found this helpful.

[@GanstaCatCT]

legendary

[@JebusCross]

where can i find it on khan academy?

[@adityashankar5723]

If I get into IIT, i'll make sure 10% of my salary goes to khan academy.

[@Roan.bot.]

What is L

[@SanjivKumar-gx6zd]

Laura

[@SanjivKumar-gx6zd]

Lund

[@Optimus_Prime290]

Who came up with this theorem

[@justdvl]

7:27 so funny so dramatic :D

[@johnsridaoduan2368]

Hahaha! XD

[@dirtysocks_blackcoffee9928]

i lol'd. When you're right, you're right khan

[@xtl8506]

lol your voice is just like Ross in Friends....

[@Mailisa98]

HAHAHA .... pivot

[@tedchirvasiu]

ah lol

[@marialimon6318]

thanks for not teaching me anything in this video. I literally need help proving it.

[@dakshkumar8982]

Thank you sir

[@dakshkumar8982]

Thank you sir