Proof of existence by I.V.T.

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  • Опубліковано 7 лют 2025
  • In this video, I showed how to use Intermediate Value Theorem to prove the existence of a number.

КОМЕНТАРІ • 90

  • @oraculum_
    @oraculum_ 11 місяців тому +58

    You're one of those few people I would actually love to click every video of theirs and immediately give a like without a further thought

    • @clionekimura9604
      @clionekimura9604 5 місяців тому +3

      I feel you, that’s exactly what I just did!

  • @demongeminix
    @demongeminix Рік тому +13

    Awesome demonstration of the use of the IVT.

  • @Necrozene
    @Necrozene 5 місяців тому +3

    I immediately answered "yes". In my mind I saw the graphs of y = x, y = x^3, and saw there were differences above and below +3, and I did not know exactly where. So, inituitively, I used the IVT. Cheers!

  • @sinanmertulucay9427
    @sinanmertulucay9427 3 місяці тому +4

    I think it could also be proved like this:
    x = x^3 + 3
    x^3 - x + 3 = 0
    let P(x) = x^3 - x + 3
    Inspecting the coefficients: 1 ∈ R, -1 ∈ R and 3 ∈ R. Hence, P(x) is a real cubic polynomial.
    Therefore it either has:
    a. 3 real solutions (x - R1)(x - R2)(x - R3)
    b. 1 real and 2 complex solutions (x - c)(x - a - bi)(x - a + bi), a ∈ R and b ∈ R
    So, there exists at least one real x such that P(x) = 0.

  • @kragiharp
    @kragiharp Рік тому +4

    Your videos are great, sir.
    I really appreciate them.
    Best wishes to you.

  • @michaelbujaki2462
    @michaelbujaki2462 4 місяці тому +4

    The number turns out to be approximately -1.67169988165716.

  • @markmajkowski9545
    @markmajkowski9545 Рік тому +8

    ??? By observation - Every cubic (odd polynomial as well) has at least one real solution. Plus there is a cubic “formula” despite its modern perceived complexity - especially when the x^2 term is 0 which was first solved before the generalized cubic.

    • @ramennoodle9918
      @ramennoodle9918 4 місяці тому

      well yeah i think this is for more general proofs of this type

  • @jan-willemreens9010
    @jan-willemreens9010 Рік тому +12

    ... A good day to you Newton despite the bad weather, Didn't I tell you ... BLACKBOARD --> SIR. NEWTON

  • @rceretta
    @rceretta 10 місяців тому

    Beautiful dialect solution!!!! I simply love it!

  • @magefreak9356
    @magefreak9356 11 місяців тому +3

    I think you could also say that:
    f(x) is continuous on all real numbers.
    As x approaches negative infinity, f(x) approaches negative infinity.
    As x approaches positive infinity, f(x) approaches positive infinity.
    I don't know if the intermediate value theorem would apply here as infinity and negative infinity isn't a number. This would also apply to all polynomials which have an odd number for the highest power of x.

    • @9adam4
      @9adam4 11 місяців тому

      I don't think you're technically allowed to do that. But what you can do is prove that f(-c) is negative and f(c) is positive, where c is any integer greater than the sum of the absolute values of the coefficients of the polynomial.

    • @magefreak9356
      @magefreak9356 11 місяців тому

      @@9adam4 why do you need the extra restriction of the coefficients?

    • @9adam4
      @9adam4 11 місяців тому

      @magefreak9356
      The coefficients will determine how far you need to go to assure you have applicable points. For instance, f(x) = x^3 - 1000x^2, you need to get past 1000 before f(-c) is negative.

    • @manpreetkhokhar5318
      @manpreetkhokhar5318 9 місяців тому +1

      As per asked question, x^3 + 3 = x; I.e x^3 - x + 3 = 0
      Let's define f(x) = x^3 - x + 3
      So, question now transforms to checking if f(x) = 0 has any real root. We can clearly examine that, f(-2) = -3 and f(-1) = 3. Since, 0 lies between -3 and 3, so, by IVT, there exists atleast one c between -2 and -1, such that f(c) = 0. This proves, that f has atleast one root.Hence, for some c between -2 and -1, c^3 - c + 3 = 0. Also, by further analyzing the limits at -inf and its increasing,decreasing nature, we can deduce that,it is the only root.

    • @Ignoranceisbliss-i2e
      @Ignoranceisbliss-i2e 5 місяців тому

      @@9adam4 yes you can because the graph is continuous and always increasing...

  • @pk2712
    @pk2712 8 місяців тому

    Great demonstration of the use of the intermediate value theorem .

  • @Faroshkas
    @Faroshkas Рік тому +2

    I love this channel

  • @jesusthroughmary
    @jesusthroughmary 7 місяців тому +7

    Every cubic function (in fact every polynomial of any odd degree) must have at least one real root because complex roots come in pairs

    • @okaro6595
      @okaro6595 5 місяців тому

      Or if one goes towards plus or minus infinity he x³ term dominates and the value also goes towards the same infinity.

    • @jesusthroughmary
      @jesusthroughmary 5 місяців тому +2

      @okaro6595 true, and therefore since it tends to both infinities it must pass through 0 on the way

    • @김민국-c9f
      @김민국-c9f 5 місяців тому

      If the coefficients of a polynomial are complex numbers, it doesn't need to have a real number root

    • @anilbera9499
      @anilbera9499 5 місяців тому

      @@김민국-c9f but they are not here..

  • @markgraham2312
    @markgraham2312 3 місяці тому

    Excellent!

  • @tcoren1
    @tcoren1 3 місяці тому +1

    You can also use the fundamental theorem of arithmatic, which is arguably more advanced but is a bit more generic and is more fundamental to the question.
    The idea is to prove every odd degree polynomial with real coefficients has a real root.
    Every n degree polynomial has exactly n roots, and can be decomposed as p(x)=a(x-x1)(x-x2)...(x-xn), with x1,x2,...xn the roots.
    Taking the complex conjugate should result in the same polynomial since it's real, so p(x)=a(x-x1*)(x-x2*)...(x-xn*)
    Equating the roots of the two versions of the polynomial shows that each coefficient is either real or comes in a pair with another coefficient such that each is the other's coefficient. Since it is impossible to arrange an odd number of numbers into pairs at least one of them must be real

    • @boguslawszostak1784
      @boguslawszostak1784 2 місяці тому

      And how will you prove this property

    • @tcoren1
      @tcoren1 2 місяці тому

      @@boguslawszostak1784which property?

    • @boguslawszostak1784
      @boguslawszostak1784 2 місяці тому

      @@tcoren1 fundamental theorem of arithmatic

    • @tcoren1
      @tcoren1 2 місяці тому

      @@boguslawszostak1784I mean it depends on what you are allowed to start with. I do not recall a proof of it over the top of my head.
      For what it's worth it's usually taught & proven before the intermediate value theorem in university, and is IIRC easier to prove, even if it makes less intuitive sense

  • @davidcawthorne7115
    @davidcawthorne7115 6 місяців тому +1

    It is real important to remember that the function has to be continuous for IVT to hold or at least continuous over the interval a b. And to always wear a cool hat when doing mathematics. 😊❤

  • @kianushmaleki
    @kianushmaleki 2 місяці тому

    Love the "natural " joke. 😅🤣

  • @abdikadirsalad1572
    @abdikadirsalad1572 Рік тому +2

    Thanks sir . Please make a video on mid term and final exam reviews calculus 1

  • @Osaode-g5j
    @Osaode-g5j 28 днів тому

    Please can you use the cubic formula to solve this question?

  • @artandata
    @artandata 9 місяців тому +2

    I kept waiting for the numerical real value of the answer!

  • @TheTrx3richie
    @TheTrx3richie 9 місяців тому

    Good demonstration!

  • @davidbrisbane7206
    @davidbrisbane7206 4 місяці тому

    x³ - x + 3 = 0 has one real root and two complex conjugate roots.
    In fact, Descartes' Rule of Signs tells us that one negative real root.

  • @davidbrisbane7206
    @davidbrisbane7206 7 місяців тому +1

    Cardano's cubic formula always find a real solution to a cubic equation, so at least one such c exists and we know f(c) = 0.

  • @tanjilsarker7678
    @tanjilsarker7678 Рік тому +2

    Thanks for the help!!

  • @JayTemple
    @JayTemple Рік тому

    I used the Sign Rule (I think that's the name) to show that there is exactly one solution that's negative. As originally stated, the number the problem asks for is a solution of x = x^3 + 3, which works out to x^3 - x + 3 = 0. If x is a solution, let y = -x. Then -y^3 + y + 3 = 0. The polynomial has exactly one sign change, so it has exactly one positive zero, but if y is positive, then x, which is -y, is negative.
    ETA: Also, I find 10 and -10 to be easier to use than smaller numbers (other than 1). My test here would be (-10)^3 - (-10) + 3 = -1000 + 100 + 3

  • @denniskisule8131
    @denniskisule8131 3 місяці тому

    what about iteration?

  • @moeberry8226
    @moeberry8226 10 місяців тому

    Good video brother but you haven’t shown that x^3-x+3 is continuous even though I know it’s a polynomial and it’s continuous on R. Taking the derivative would also be good to show. Because differentiability implies continuity.

    • @tcoren1
      @tcoren1 3 місяці тому

      That's circular logic.
      The proof for the derivative of a polynomial goes through proving it is continuous.
      It's like using L'hopital's rule to find the limit of sinx/x

  • @manitubergaming
    @manitubergaming Рік тому +1

    U soooo intelligent

  • @sjn7220
    @sjn7220 9 місяців тому +1

    Is there a solution that doesn’t involve guessing?

  • @tudorsafir2766
    @tudorsafir2766 Рік тому +2

    Isn't that also called Bolzano's Theorem?

  • @amiinmohammed8552
    @amiinmohammed8552 Місяць тому

    You are Soo good but you like to talk more please make short

  • @tcmxiyw
    @tcmxiyw 3 місяці тому

    The question is “Is x real?”. You have answered the question “Can x be real?”. In order to answer the given question, you must rule out option 2, in which two of the possibilities for x are imaginary. For a cubic equation with real coefficients, there are these possibilities: three real solutions, two real solutions (one with multiplicity two), or one real solution and two imaginary solutions. So the answer to the question could be “no”. More work to do. You did, however, answer the question in the thumbnail for the video.

  • @Ahmed-kg2gf
    @Ahmed-kg2gf 7 місяців тому

    The question is , "is x real" in order to solve that u need to get the notation for the equation and it is x³+3=x
    Reenraging the terms
    X³-x+3=0
    Any cubic polynomial with real coefficient must have at least one real solution so the answer is ye , no need to solve

  • @kennethgee2004
    @kennethgee2004 6 місяців тому

    well the question changed a little from the thumb nail to the question on the board. This one feels like a no, but it is the proof they want.

  • @9adam4
    @9adam4 11 місяців тому +3

    It's about -1.6717

    • @senseof_outrage9390
      @senseof_outrage9390 10 місяців тому

      Can you post the exact format of the solution.
      Did you solve the problem yourself... 😅

    • @neevhingrajia3822
      @neevhingrajia3822 9 місяців тому

      Thats a transcendental number right? So how can the a transcendental number be 3 more than its own cube?

    • @9adam4
      @9adam4 9 місяців тому +1

      @neevhingrajia3822
      What leads you to believe it's a transcendental number? As you correctly point out, a transcendental number can't be the solution to an equation like this.

    • @neevhingrajia3822
      @neevhingrajia3822 9 місяців тому +1

      @@9adam4 so are you saying that -1.6717 cubed would be 3 less than -1.6717?

    • @9adam4
      @9adam4 9 місяців тому

      @neevhingrajia3822
      Yes I am. Put it into the calculator yourself.

  • @yaronbracha4923
    @yaronbracha4923 7 місяців тому

    Who's Dad?

  • @williamspostoronnim9845
    @williamspostoronnim9845 Рік тому

    I like Your English very much!

  • @NathanSibali
    @NathanSibali Рік тому

    Wow

  • @physife
    @physife 3 місяці тому

    I just used -10 and 0, these are easy enough

  • @elreturner1227
    @elreturner1227 5 місяців тому

    I did x^3 -x =-3
    x(x^2 -1)=-3
    x((x+1)(x-1))=-3
    x=-3, x=-4, and x=-2 and not one is right

  • @boguslawszostak1784
    @boguslawszostak1784 2 місяці тому

    x = 1/3 (81/2 - (3 sqrt(717))/2)^(1/3) + (1/2 (27 + sqrt(717)))^(1/3)/3^(2/3)
    so exists :p

  • @Harrykesh630
    @Harrykesh630 8 місяців тому

    Using Derivatives would have been a better approach 🤔

  • @holyshit922
    @holyshit922 Рік тому +2

    It is not so difficult to calculate x
    Assume that x is sum of two unknowns,plug in into the equation
    use binomial expansion , rewrite as system of equations
    Transform this system of equations to Vieta formulas for quadratic
    Check if solution of quadratic satisfies system of equations before transformation

  • @BB-sc8ed
    @BB-sc8ed Рік тому +1

    Thank you for saving my ass

  • @Dr_piFrog
    @Dr_piFrog 3 місяці тому

    Approximately: -1.67...

  • @Noor-kq9ho
    @Noor-kq9ho Рік тому

    depressed cubics

  • @johnconrardy8486
    @johnconrardy8486 6 місяців тому

    why don't you
    why don't you create a tee shirt with your famous saying

  • @Taric25
    @Taric25 11 місяців тому +3

    Why all this proof stuff? Just solve it. x≈-1.6717 or x≈0.83585 ± 1.0469 i. That's it.

    • @xyz9250
      @xyz9250 10 місяців тому +1

      Could you post how you solved it ?

    • @deltalima6703
      @deltalima6703 14 днів тому

      Lol. Do it yourself, this isnt a classroom.
      😂

    • @Taric25
      @Taric25 13 днів тому

      ​@@xyz9250Sure, plug the coefficients of x³-x+3=0 into the cubic formula. That's it.

  • @colina64
    @colina64 Рік тому +2

    nice as usual, please try to change your blackboard to a white one or improve the light system for better visualization of your videos, best regards👍

  • @m.h.6470
    @m.h.6470 Рік тому +1

    Solution:
    is there a real solution to x = x³ + 3?
    x = x³ + 3 |-x³
    x - x³ = 3 |therefore x³ < x, which means that, as the result is an integer, x *has* to be negative!
    (x³ < x would also be possible, if 0 < x < 1, but then the result of x - x³ would not be an integer)
    Since x³ is growing very fast, x has to be quite small.
    testing left term assuming x = -1
    -1 - (-1)³ = -1 - (-1) = -1 + 1 = 0
    testing left term assuming x = -2
    -2 - (-2)³ = -2 - (-8) = -2 + 8 = 6
    Therefore there is a real solution of x between -1 and -2.

  • @sunil.shegaonkar1
    @sunil.shegaonkar1 Рік тому +1

    Nearest answer is - 1.672, it is still an approximate.
    Question remains: is there an exact solution ?
    Probably not in rational numbers, but may be an irrational one.
    Oh, That is why they are called irrational number.

  • @AbouTaim-Lille
    @AbouTaim-Lille 8 місяців тому

    X=3x³ , and excluding the trivial solution X=0. We have 3x²= 1 so x=±1/√3

  • @hridikkanjilal460
    @hridikkanjilal460 6 місяців тому

    All negative number is greater than it's cube

    • @sinexitoalmiedo
      @sinexitoalmiedo 5 місяців тому

      False because (-(1/2))^3=-1/8 but -1/8 is greater than -1/2

    • @hridikkanjilal460
      @hridikkanjilal460 5 місяців тому

      @@sinexitoalmiedo I forgot to add 'all negative ' non fractions of decimals are greater than their squres

    • @sinexitoalmiedo
      @sinexitoalmiedo 5 місяців тому

      @@hridikkanjilal460 what??

    • @hridikkanjilal460
      @hridikkanjilal460 5 місяців тому

      @@sinexitoalmiedo all negative number which are not fractions or decimals are greater than their cubes