Continuity over an interval | Limits and continuity | AP Calculus AB | Khan Academy

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  • Опубліковано 28 чер 2017
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    A function Ä is continuous over the open interval (a,b) iff it's continuous on every point in (a,b). Ä is continuous over the closed interval [a,b] iff it's continuous on (a,b), the right-sided limit of Ä at x=a is Ä(a) and the left-sided limit of Ä at x=b is Ä(b).
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КОМЕНТАРІ • 39

  • @Mimi127
    @Mimi127 6 років тому +76

    How would you figure this out algebraically?

  • @wajahatkhan9455
    @wajahatkhan9455 5 років тому +2

    Thanks Sir...

  • @susmitabanerjee216
    @susmitabanerjee216 7 років тому +10

    Now I'm very clear....... thanks!!!!

  • @Nino-ne3kl
    @Nino-ne3kl 4 роки тому +2

    Thank you so much

  • @apothe6
    @apothe6 20 днів тому

    good explanation

  • @kehalitadesse5165
    @kehalitadesse5165 3 роки тому

    Thanks that really helped

  • @luka1696
    @luka1696 3 роки тому

    Very cool

  • @user-ll7ui5ym5b
    @user-ll7ui5ym5b 2 місяці тому

    Can anyone help me if
    (negative infinity, 2] is continuous or not?

  • @johnferentoulaloum
    @johnferentoulaloum 6 років тому +1

    I need help in one question. There is an exercise that asks to prove this. Suppose f satisfies the conclusion of the middle value theorem (that in an interval [a,b] f takes every value between f(a) and f(b))and f takes on each value only once (suppose it means a 1-1 function) prove that the function is continuous at [a,b]. Ok it had an ε,δ contradiction proof which made no sense at all. Because it used the middle value theorem WITHOUT knowing that the function is continuous.Circular. Fact is the exercise looks totally wrong, because I can find any function of this kind which is NOT continuous and qualifies. You only have to find a function with no limit at certain x es. (Different - and + limit a x1,x2 etc). And it STILL is 1-1 and takes ALL VALUES between f(a) and f(b).It is a very interesting calculus book that I am studying but either I am missing something or there is a huge mistake. Any help?

  • @loona8398
    @loona8398 5 років тому +1

    Why last one is not continues?

    • @Trendy_Frogg
      @Trendy_Frogg Місяць тому

      Becaude the limit of f(x) as x approaches -2 from the right needs to equal the value of f(x) when x is -2, in this case the limit from the right is -3 and the value of the function at x equals -2 is equal to 0 and since they arent equal, It is NOT continues over that closed interval (hope you understand)

  • @parameshwarballu1278
    @parameshwarballu1278 6 років тому

    Clear and To the point!

  • @lucasargandona4658
    @lucasargandona4658 4 роки тому +1

    Would (3,4) be continuous?

    • @echodiasenemb4611
      @echodiasenemb4611 4 роки тому

      Yes... i think

    • @uwish5356
      @uwish5356 3 роки тому +2

      @sxzm isn't an infinite discontinuity considered discontinuous?

    • @uwish5356
      @uwish5356 3 роки тому

      @sxzm wait my answer in my exer just change bc everyone is like telling me that it's still continuous bc of the open interval. since it's not really going to 4 just the points between 3 and 4.

    • @uwish5356
      @uwish5356 3 роки тому

      @sxzm the answer sheets is still not released. i'll update if it's already released and anyone is still interested with this conversation

    • @lestersolaina6734
      @lestersolaina6734 3 роки тому

      @@uwish5356 what is the answer?

  • @emmanuelhimongon5026
    @emmanuelhimongon5026 4 роки тому

    What graph is this

  • @MushuIsontherun
    @MushuIsontherun 3 роки тому +1

    indi tkon ka intindi angud😤

    • @polaris445
      @polaris445 2 роки тому

      HAHAHAHAH parehas ta dai 😭

  • @fawul.7589
    @fawul.7589 7 років тому

    1st

  • @youtubeuser5822
    @youtubeuser5822 4 роки тому +1

    You should have solved some problems.

  • @mohamedelassar3982
    @mohamedelassar3982 7 років тому +1

    First!