Real Analysis 32 | Intermediate Value Theorem

[@lucaug10]

One of my favorite theorems from Analysis, together with MVT :)

[@epicmorphism2240]

no way bruh

[@sharonnuri]

Very cool theorem. Interesting to note that the opposite is not true. Let x in [a,b] and f be continious. Then f(x) is not necessarlly in [f(a),f(b)]. The proof pretty much uses the bissection method to find the root of f tilde.

[@frederickgriffth4431]

Helps me a lot. Great proof process!

[@brightsideofmaths]

Glad it helped! And thanks for your support :)

[@kingarth0r]

Funny that the "Intermediate value theorem" is the halfway point of the playlist

[@brightsideofmaths]

Haha :D

[@tychovanderouderaa6154]

You define \tilde{f} to be -g if g(a) > 0. Then it is very clear that always \tilde{f}(a) ≤ 0, but to me, it is not directly obvious why also \tilde{f}(b)≥0. Let's say instead of the drawn function f, f is a parabola that is flipped at the minimum to get g. Wouldn't then both g(a) and g(b) be less or equal than zero? Thanks in advance! I'm probably missing something obvious.

[@brightsideofmaths]

Thanks for the question! Don't forget that we first translate the function by y. And because y is intermediate point, it is not possible to have both end points on the same side of the x-axis. Just look again how we have to choose y and how we define g. Best wishes! :)

[@Lous_-st6hi]

Thanks for sharing such an insightful proof.

[@brightsideofmaths]

Glad it was helpful!

[@TranquilSeaOfMath]

Nice illustration and explanation.

[@brightsideofmaths]

Thank you! Cheers!

[@Independent_Man3]

when you define f tilde in 4:29, shouldn't it be f tilde = -g when g(a) > g(b) and f tilde = g otherwise ? Because at 3:58 you said that you wanted the value on the right to be larger than the value on the left.

[@brightsideofmaths]

Yeah, but remember we shifted the function to the bottom. Therefore g(a) > 0 has the same meaning as g(a) > g(b).

[@meshachistifanus8336]

Proof, suggest a textbook to read along with.

[@Hold_it]

Interesting Theorem. Keep them coming! :D

[@someperson9052]

With this construction, what if \tilde{f} has multiple 0's? Couldn't you then miss the \tilde{y} you're looking for, and end up converging to a different 0? Well actually now after writing this I suppose it's enough that there exist sequences which converge to \tilde{y}...

[@brightsideofmaths]

Yes, we only have existence here :)

[@aidynubingazhibov8933]

Are {an} and {bn} Cauchy sequences because these are monotonic and bounded (and thus convergent)? Just like in Bolzano-Weierstrass theorem? Also, what books are you using for the videos if any?

[@brightsideofmaths]

Yeah, you could argue like this or simply look at the Cauchy property | a_n - a_m |. This distance can be immediately calculated and one sees that we have Cauchy sequences.

[@duckymomo7935]

A discontinuous function can have intermediate value property It was a long debate whether IVT defined continuity and unfortunately we discovered that does doesn’t and in fact we get various levels of continuity (Lipschitz and absolutely continuity)

[@chair547]

The function f(x) = x (x= 0.5) on [0,1] is discontinuous but IVT holds, no?

[@chainetravail2439]

Why unfortunately?

[@duckymomo7935]

@@chainetravail2439 because just plain old vanilla continuity isn't enough for a lot of things e.g. continuity is not enough to ensure that the set of all functions is complete (see uniform limit theorem)

[@fullfungo]

@@chair547 what?? a=0.25; b=1 By IVT we should have f(c)=0.1, but it’s not in the interval. You seem to be wrong

[@williamwarren5234]

I'm happy to say I came up with such a function myself: f(x) = sin(1/x) for x>0, f(0)=0 Discontinuous at 0, but will attain every value on [0,b] for b>0

[@fabiomendez-cordoba3902]

Hi, is there problem with the website? It keeps telling me that it can't be found :c

[@brightsideofmaths]

Thanks! It should work again. I changes some links without changing them in the description.

[@johnstroughair2816]

We have shown [f(a), f(b)] is an interval but is this enough to show f[[a,b]] is an interval?

[@brightsideofmaths]

No, that is not enough :)

[@johnstroughair2816]

Ah, so we have to apply IVT to [xmin,xmax]

[@julioschneiders8844]

@antoniocleibo