Finding absolute extrema on a closed interval | AP Calculus AB | Khan Academy

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  • Опубликовано: 28 окт 2024

Комментарии • 27

  • @nombla1111
    @nombla1111 10 лет назад +14

    I feel like you're following my Calc 1 class syllabus. Good stuff!

  • @OrangeCat371
    @OrangeCat371 20 дней назад

    Sal be saving my calc grade rn

  • @johannjames8941
    @johannjames8941 5 лет назад

    Your videos are just amazing...it helps me understand the concepts easier.

  • @prateek0536
    @prateek0536 5 лет назад +1

    Love from india❤❤❤❤❤❤

  • @DBlakleyosu62
    @DBlakleyosu62 10 лет назад

    Excellent!

  • @5gallonsofwater495
    @5gallonsofwater495 Год назад

    according to the graph of this function, i guess theres no absolute minimum. but is there a relative minimum?

  • @juju141
    @juju141 3 года назад +2

    So basically you get the absolute max when you set f’(x) equal to 0?

    • @juju141
      @juju141 3 года назад +1

      And you test the endpoints with it too

  • @murtez22
    @murtez22 6 лет назад

    همزين فاهمها من ايام السادس 😂هذا موضوع النقط الحرحة و النهايات المحلية 😆

  • @juju141
    @juju141 3 года назад +1

    How do you find the absolute minimum?

    • @viditjain2508
      @viditjain2508 3 года назад

      The abs. Max and abs. Min occur either at critical points i.e when f’(x) equal to 0 or at the endpoints of a closed interval [a,b]. Now you test all your critical points and endpoints in f(x). The max y value is abs. Max and the min y value is abs. Min

  • @ayahabdeldayem9377
    @ayahabdeldayem9377 2 года назад

    so how do i get min value

  • @mdmohiuddin9375
    @mdmohiuddin9375 10 лет назад +1

    nice

  • @headcutter09
    @headcutter09 10 лет назад

    Wowzer

  • @shashindevsare518
    @shashindevsare518 10 лет назад

    Hi mate to
    be able with. Gqr

  • @legolad444
    @legolad444 8 лет назад +2

    Isn't 8ln1-1^2=positive 1 due to the fact (-1^2)?

    • @mrustham1970
      @mrustham1970 7 лет назад +1

      here 8ln1-(1^2)

    • @TrapKingz.
      @TrapKingz. 7 лет назад

      Yeah man he messed up but either way the absolute max was still the one he pointed out!

    • @ernestbeqiri2788
      @ernestbeqiri2788 6 лет назад

      -1^2= -1; (-1)^2=1

    • @authorttaelias4483
      @authorttaelias4483 10 месяцев назад

      @@TrapKingz.nah it’s -x^2 not (-x^ 2)

  • @sergey_zatsepin
    @sergey_zatsepin 7 лет назад

    Really, Sal? You used calculator? Why no intevals like we did before? I mean you can take the derivative and then on interval place points "1","2","4" and so look where derivative is positive and where is negative, that's how you easily can define that "2" is the maximum point without any calculator.

    • @Rocky-me5cw
      @Rocky-me5cw 6 лет назад +1

      He didn't want to use the idea of second derivative.Many people don't know that.

  • @mayasooklall4014
    @mayasooklall4014 6 лет назад +2

    I’m only nine years old I’m to young for this is this for collage cause I think it is

    • @diegohawk100
      @diegohawk100 6 лет назад +3

      wait 9 more years

    • @zombieslayer2225
      @zombieslayer2225 2 года назад

      You will encounter this in your future classes and definitely in college.

    • @OrangeCat371
      @OrangeCat371 20 дней назад

      Four years left, it's only a matter of time now...