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Hazırlayan: Kemal Duran (Matematik Öğretmeni) http://www.buders.com/kadromuz.html...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/2968351/d78706d1d3b142eecdeb98d896edc06c/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/c6RegAAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"14","reqid":"1769464672018854-15716253209542582054-balancer-l7leveler-kubr-yp-klg-329-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769464672018854-15716253209542582054-balancer-l7leveler-kubr-yp-klg-329-BAL&text=videoid:9805091187224409168","teaser":[{"list":{"type":"unordered","items":["Bu video, bir eğitmen tarafından sunulan matematik eğitim içeriğidir. Eğitmen, süreklilik konusunu uygulamalı örneklerle açıklamaktadır.","Videoda süreklilik konusu detaylı olarak ele alınmakta, fonksiyonların sürekli olduğu en geniş aralıkları bulma, süreksizlik noktalarını tespit etme ve bir noktada süreklilik olma şartlarını kontrol etme gibi konular adım adım anlatılmaktadır. Toplam dokuz farklı soru üzerinden tanımsızlıklar, parçalanma noktaları ve parçalı fonksiyonlarda süreksizlik nedenleri incelenmektedir.","Video, sınavlarda çıkabilecek soru tiplerini çözerek konuyu pekiştirmekte ve son olarak bir fonksiyonun her noktada sürekli olması durumunda a+b değerini bulma sorusunu çözmektedir. Eğitmen, bir sonraki derste süreksizlik çeşitlerinin inceleneceğini belirtmektedir."]},"endTime":1144,"title":"Süreklilik Konusu Çözümlü Örnek Sorular","beginTime":0}],"fullResult":[{"index":0,"title":"Süreklilik Kavramı ve Soru Çözümü","list":{"type":"unordered","items":["Video, süreklilik konusunun uygulamalarını içeren örnek soruları ele alıyor.","Bir fonksiyonun süreksiz olma durumları tanımsızlık noktaları ve parçalı fonksiyonlarda parçalanma noktaları olarak belirtiliyor."]},"beginTime":1,"endTime":23,"href":"/video/preview/9805091187224409168?parent-reqid=1769464672018854-15716253209542582054-balancer-l7leveler-kubr-yp-klg-329-BAL&text=Calculus+Problems+and+Solutions&t=1&ask_summarization=1"},{"index":1,"title":"Sürekli Fonksiyonlar","list":{"type":"unordered","items":["f(x) = 3x + 1 fonksiyonu herhangi bir tanımsızlık veya parçalı yapıya sahip olmadığı için tüm reel sayılarda sürekli.","f(x) = 1/(x+1) fonksiyonu x = -1 noktasında tanımsız olduğu için süreksiz, bu nedenle sürekli olduğu en geniş aralık (-∞, -1) ∪ (-1, +∞).","f(x) = 1/(x²-9) fonksiyonu x = -3 ve x = 3 noktalarında tanımsız olduğu için süreksiz."]},"beginTime":23,"endTime":229,"href":"/video/preview/9805091187224409168?parent-reqid=1769464672018854-15716253209542582054-balancer-l7leveler-kubr-yp-klg-329-BAL&text=Calculus+Problems+and+Solutions&t=23&ask_summarization=1"},{"index":2,"title":"Köklü Fonksiyonların Sürekliliği","list":{"type":"unordered","items":["f(x) = √(x³) fonksiyonu x³'in tek dereceli olduğu için tüm reel sayılarda tanımlı, sürekli olduğu en geniş aralık [0, +∞).","f(x) = √(x) + √(x+1) fonksiyonu her iki kökün de çift dereceli olduğu için tanımlı olduğu en geniş aralık [0, +∞).","f(x) = √(x²-4)/(x²-1) fonksiyonu tanımlı olduğu en geniş aralık (-∞, -2] ∪ (-1, 1) ∪ [2, +∞)."]},"beginTime":229,"endTime":455,"href":"/video/preview/9805091187224409168?parent-reqid=1769464672018854-15716253209542582054-balancer-l7leveler-kubr-yp-klg-329-BAL&text=Calculus+Problems+and+Solutions&t=229&ask_summarization=1"},{"index":3,"title":"Parçalı Fonksiyonların Sürekliliği","list":{"type":"unordered","items":["f(x) = 2, x = 5 için sürekli olmak için sağdan ve soldan limitlerin f(5) değerine eşit olması gerekir.","f(x) = {7x, x > 3; 3x-1, x ≤ 3} fonksiyonu x = 3 noktasında süreksiz çünkü sağdan ve soldan limitler eşit değil.","f(x) = {3x+1, x \u003c 2; 3, x = 2; 3x-2, x > 2} fonksiyonu x = 2 noktasında sürekli, x = 5 noktasında süreksiz.","f(x) = {2x+4, x > -2; x², x ≤ -2} fonksiyonu x = -2 noktasında süreksiz çünkü sağdan ve soldan limitler eşit değil."]},"beginTime":455,"endTime":816,"href":"/video/preview/9805091187224409168?parent-reqid=1769464672018854-15716253209542582054-balancer-l7leveler-kubr-yp-klg-329-BAL&text=Calculus+Problems+and+Solutions&t=455&ask_summarization=1"},{"index":4,"title":"Süreksiz Fonksiyonların Analizi","list":{"type":"unordered","items":["Süreksiz fonksiyonların iki durumu vardır: tanımsızlık kaynaklı ve parçalanma noktası kaynaklı süreksizlikler.","Parçalı fonksiyonlarda, tanımsız yapan değer aralığa uyması gerekir; aksi takdirde süreksizlik olarak kabul edilmez.","Verilen örnekte x=2 ve x=4 değerleri süreksizlik noktaları olarak bulunurken, x=-4 değeri aralığa uymadığı için süreksizlik olarak kabul edilmez."]},"beginTime":839,"endTime":944,"href":"/video/preview/9805091187224409168?parent-reqid=1769464672018854-15716253209542582054-balancer-l7leveler-kubr-yp-klg-329-BAL&text=Calculus+Problems+and+Solutions&t=839&ask_summarization=1"},{"index":5,"title":"Parçalanma Noktasındaki 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