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Eğitmen, bu tür soruları çözmek için dört adımlı bir yöntem sunmaktadır: kapalı bölgeyi çizme, köşe koordinatlarını bulma, kritik noktaları tespit etme ve sınır fonksiyonlarının üzerinde maksimum-minimum değerlerini arama.","Video, toplam beş örnek soru üzerinden konuyu pekiştirmeyi amaçlamakta ve özellikle sınır fonksiyonlarının üzerinde maksimum-minimum değerlerini bulma aşamasında karşılaşılan zorluklar ele alınmaktadır."]},"endTime":1205,"title":"Çok Değişkenli Fonksiyonlarda Mutlak Maksimum ve Minimum Bulma","beginTime":0}],"fullResult":[{"index":0,"title":"Çok Değişkenli Fonksiyonlarda Mutlak Maksimum ve Minimum Bulma","list":{"type":"unordered","items":["Çok değişkenli fonksiyonlarda mutlak maksimum ve minimum bulma konusunda örnek soru incelenecek.","Bir fonksiyonun maksimum ve minimum değerleri ancak ve ancak kapalı bir bölge ile tanımlandıysa sorulabilir.","Çok değişkenli fonksiyonlarda maksimum-minimum demek mutlak maksimum-minimumu kast etmektedir."]},"beginTime":1,"endTime":13,"href":"/video/preview/11187584981951555564?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=1&ask_summarization=1"},{"index":1,"title":"Örnek Sorunun Analizi","list":{"type":"unordered","items":["f(x,y) = x² - y² fonksiyonu verilmiş ve köşe koordinatları (0,0), (-1,1), (1,1) olan üçgen üzerindeki f(x,y) fonksiyonunun maksimum ve minimum değerleri bulunması isteniyor.","Maksimum-minimum değerlerini bulmak için önce kapalı bölgeyi çizmek gerekir.","Herhangi bir bölge ile sınırlanılmadığında mutlak maksimum-minimum sorulamaz, ancak yerel maksimum sorulabilir."]},"beginTime":13,"endTime":213,"href":"/video/preview/11187584981951555564?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=13&ask_summarization=1"},{"index":2,"title":"Maksimum ve Minimum Değerlerin Bulunması","list":{"type":"unordered","items":["Maksimum-minimum değerlerini bulmak için üç aday nokta vardır: köşe koordinatları, kritik noktalar ve sınır noktaları.","Köşe koordinatları tartışmasız maksimum-minimum adaylarıdır ve fonksiyonda yerleştirilerek değerleri hesaplanır.","Kritik noktalar, gradient f = 0 olan noktalar olup, bu örnekte (0,0) noktası kritik nokta olarak bulunmuştur."]},"beginTime":213,"endTime":606,"href":"/video/preview/11187584981951555564?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=213&ask_summarization=1"},{"index":3,"title":"Sınır Noktalarının İncelenmesi","list":{"type":"unordered","items":["Kapalı bölgenin sınırlarında da maksimum-minimum değer olabilir ve bu sorunun en zorlu bölümüdür.","Üçgenin sınırları için denklemler bulunur: y = x, y = -x ve y = 1.","Her sınır üzerinde maksimum-minimum adayı olup olmadığını tespit etmek için, sınır denklemleri fonksiyona yerleştirilir ve tek değişkenli fonksiyon olarak ele alınır."]},"beginTime":606,"endTime":821,"href":"/video/preview/11187584981951555564?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=606&ask_summarization=1"},{"index":4,"title":"Sonuçların Değerlendirilmesi","list":{"type":"unordered","items":["(1,1) noktası en küçük değer (-1) veren aday nokta olarak bulunmuştur.","y = x ve y = -x doğruları üzerindeki noktalar, köşe koordinatlarından biriyle aynı sonucu verdiği için yeni aday nokta getirmemiştir.","Tüm aday noktaların sonuçları karşılaştırılarak maksimum ve minimum değerler belirlenir."]},"beginTime":821,"endTime":996,"href":"/video/preview/11187584981951555564?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=821&ask_summarization=1"},{"index":5,"title":"Fonksiyonların Mutlak Maksimum ve Minimum Değerleri","list":{"type":"unordered","items":["Bir fonksiyonun mutlak maksimum ve minimum değeri, kapalı bir bölge ile tanımlandığı zaman sorulabilir.","Kapalı bölge bazen üçgen, dörtgen veya yamuksal bir bölge olabilir.","Kapalı bölgeyi çizdıktan sonra, bölgenin köşeleri mutlak maksimum ve minimum adaylarıdır."]},"beginTime":1034,"endTime":1152,"href":"/video/preview/11187584981951555564?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=1034&ask_summarization=1"},{"index":6,"title":"Maksimum ve Minimum Değerlerin Bulunması","list":{"type":"unordered","items":["Fonksiyonun kritik noktaları da mutlak maksimum ve minimum adaylarıdır.","Sınır bölgelerinin fonksiyonlarını doğru denklemlerini elde edip, türevi sıfır yapan noktalar da aday olur.","Tüm adayları kıyaslayarak en büyük çıkaran mutlak maksimum, en küçük çıkaran da mutlak minimum değeri verir."]},"beginTime":1152,"endTime":1196,"href":"/video/preview/11187584981951555564?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=1152&ask_summarization=1"}],"linkTemplate":"/video/preview/11187584981951555564?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-II : Çok Değişkenli Fonksiyonlarda Mutlak Max ve Min Bulma Örnek Soru-1","related_orig_text":"Calculus Chronicles","related_porno":false,"related_less_3m_off":true,"client":"d2d","no_cnt":1,"related_src":"serp","related":"{\"porno\":false,\"vfp\":1,\"orig_text\":\"Calculus 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kanala Katılın.. Ders PDF’lerine ulaşma ve size özel diğer ayrıcalıklardan yararlanın… / @youtubeuniversitesi Mühendislik Bölümleri İçin Calculus-1 (Genel Matematik) § Eğri Altında Alan § Konu...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/2862986/ad309d04545522b667128365789a394e/564x318_1"},"target":"_self","position":"1","reqid":"1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL","summary":{"fullTextUrl":"/int_search_summary?data=http%3A%2F%2Fwww.youtube.com%2Fwatch%3Fv%3D0JN3ZkZwYtA","linkTemplate":"/video/preview/2863188196425289924?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-1 & 22. 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Hazırlayan: Kemal Duran (Matematik Öğretmeni)...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/4074897/bf9e8eecc746be298fe790bfbd3f22d7/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/g3N5NAIAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"2","reqid":"1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=videoid:4201234903955118038","teaser":[{"list":{"type":"unordered","items":["Bu video, matematik eğitimi formatında bir ders anlatımıdır. Eğitmen, bir düzlemin dışındaki bir noktaya en yakın noktasının koordinatlarını bulma yöntemini adım adım açıklamaktadır.","Video, bir örnek üzerinden başlayıp, x+y+2z=5 düzleminin (1,1,1) noktasına en yakın noktasının koordinatlarını bulma sürecini göstermektedir. Eğitmen, çözüm için üç temel adımı detaylı olarak anlatmaktadır: düzlemin normal vektörünün bulunması, düzleme dik doğrunun parametrik denkleminin yazılması ve bu doğrunun düzlemle kesişme noktasının bulunması. Video, bu temel stratejiyi tekrarlayarak sonlanmaktadır."]},"endTime":515,"title":"Bir Düzlemin Dışındaki Bir Noktaya En Yakın Noktasının Koordinatlarını Bulma","beginTime":0}],"fullResult":[{"index":0,"title":"Düzlemin Dışındaki Bir Noktaya En Yakın Nokta","list":{"type":"unordered","items":["Bu videoda bir düzlemin dışındaki bir noktaya en yakın noktasının koordinatlarını bulma yöntemi anlatılacak.","Örnek olarak x+y+2z=5 düzleminin (1,1,1) noktasına en yakın olduğu noktanın koordinatları bulunacak.","Noktanın düzlemin dışında olduğu, düzlem denklemine yerleştirildiğinde eşitliğin sağlanmamasından anlaşılır."]},"beginTime":0,"endTime":108,"href":"/video/preview/4201234903955118038?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=0&ask_summarization=1"},{"index":1,"title":"En Yakın Noktanın Bulunma Yöntemi","list":{"type":"unordered","items":["Düzlemin dışındaki bir noktaya en kısa uzaklık, bu noktadan düzleme çizilen dik uzaklıktır.","Düzlemin üzerinde bu noktaya en yakın olan nokta, dik uzunluğun düzleme değdiği noktadır.","Çözüm için dışarıdaki noktadan geçen ve düzleme dik olan doğrunun denklemini yazmak gerekir."]},"beginTime":108,"endTime":146,"href":"/video/preview/4201234903955118038?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=108&ask_summarization=1"},{"index":2,"title":"Çözüm Adımları","list":{"type":"unordered","items":["Birinci adım: Düzlemin normal vektörü bulunur, bu vektör x, y, z katsayılarıdır.","İkinci adım: Düzleme dik doğrunun parametrik denklemi yazılır: x=1+t, y=1+t, z=1+2t.","Üçüncü adım: Doğrunun parametrik denklemi düzlemin denkleminde yerine konularak t değeri 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fazlası için: http://www.khanacademy.org.tr Matematikten sanat tarihine, ekonomiden fen bilimlerine, basit toplamadan diferansiyel denklemlere, ilkokul seviyesinden üniversite seviyesine...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/1397409/7a9d210ba4b194b1cb527cb7f6a7fa21/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/lFM8WAAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"6","reqid":"1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=videoid:1158038666299291059","teaser":[{"list":{"type":"unordered","items":["Bu video, matematik eğitimi formatında bir ders anlatımıdır. Eğitmen, türevin ne olduğunu ve nasıl hesaplanacağını açıklamaktadır.","Video, türevin bir eğri boyunca herhangi bir noktadaki anlık eğimi bulma özelliğini ele alıyor. Eğitmen önce f(x) = x² fonksiyonu üzerinden x = 3 noktasındaki anlık eğimi hesaplıyor, ardından türevin genel tanımını gösteriyor ve f'(x) = 2x formülünü elde ediyor. Video, türevin fizikte, optimizasyon problemlerinde ve diğer alanlardaki uygulamalarını gelecek videolarda gösterileceğini belirterek sona eriyor."]},"endTime":420,"title":"Türev Kavramı ve Hesaplama Yöntemleri","beginTime":0}],"fullResult":[{"index":0,"title":"Türevin Tanımı ve Örnek Uygulama","list":{"type":"unordered","items":["Türev, bir eğri boyunca herhangi bir noktadaki anlık eğimi bulmayı sağlar.","f(x) = x² fonksiyonunun x = 3 noktasındaki eğimini bulmak için, x = 3 ve x = 3+h noktaları arasındaki eğim hesaplanır.","İki nokta arasındaki eğim, y koordinatındaki değişimin x koordinatındaki değişime oranıdır ve h sıfıra yaklaşırken limit alınarak anlık eğim bulunur."]},"beginTime":0,"endTime":225,"href":"/video/preview/1158038666299291059?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=0&ask_summarization=1"},{"index":1,"title":"Türevin Genel Formülü","list":{"type":"unordered","items":["Fonksiyonun herhangi bir noktadaki eğimi, limit h sıfıra giderken (f(x+h) - f(x)) / h formülüyle hesaplanır.","Bu formül, eğim formülünün x'teki değişim (h) sıfıra yaklaşırken limit alınarak elde edilmesidir.","Türev, Lagrange tarafından kullanılan f'(x) sembolüyle gösterilir."]},"beginTime":225,"endTime":293,"href":"/video/preview/1158038666299291059?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=225&ask_summarization=1"},{"index":2,"title":"f(x) = x² Fonksiyonunun Türevi","list":{"type":"unordered","items":["f(x) = x² fonksiyonunun türevi, limit h sıfıra yaklaşırken (x+h)² - x² / h formülüyle hesaplanır.","Türev hesaplandığında f'(x) = 2x olarak bulunur.","f(x) = 2x formülü, herhangi bir noktadaki anlık eğimi bulmak için kullanılabilir; örneğin (16, 256) noktasındaki eğim 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binden fazla ücretsiz ders videosu için: https://www.khanacademy.org.tr Matematikten sanat tarihine, ekonomiden fen bilimlerine, basit toplamadan diferansiyel denklemlere, ilkokul seviyesinden...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/962614/02d386b1fcb1df8fbc8842f638fa2c8a/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/zvmbnQAAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"7","reqid":"1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=videoid:4192788898677205843","teaser":[{"list":{"type":"unordered","items":["Bu video, matematik eğitimi formatında bir ders anlatımıdır. Eğitmen, bir fonksiyona polinom kullanarak yaklaşık değerler bulma yöntemini açıklamaktadır.","Video, bir fonksiyona polinom kullanarak yaklaşım yöntemlerini adım adım göstermektedir. Önce tek terimli polinom (sabit terim) ile başlayıp, ardından birinci ve ikinci türevleri de eşit olan iki terimli polinom, son olarak da tüm türevleri eşit olan Maclaurin serisi oluşturulmaktadır. Eğitmen, her bir polinomun fonksiyona yaklaşımını grafiksel olarak göstererek, Maclaurin serisinin Taylor serisinin özel bir durumu olduğunu belirtmektedir. Video, bir sonraki bölümde fonksiyon örnekleri kullanılarak konunun daha iyi anlaşılacağını söyleyerek sona ermektedir."]},"endTime":555,"title":"Maclaurin Serisi ve Polinom Yaklaşımı","beginTime":0}],"fullResult":[{"index":0,"title":"Fonksiyonlara Polinom Yakınlama","list":{"type":"unordered","items":["Bir fonksiyona polinom kullanarak yaklaşık değerler bulmak için, fonksiyonun ve türevlerinin 'daki değerlerini bilmek varsayılıyor.","Tek terimli bir polinom olarak f 'a eşit bir yatay doğru kullanılabilir, ancak bu sadece bir değerde yaklaşma sağlar.","Daha iyi yaklaşma için polinomun ve fonksiyonun 'daki türevlerinin de eşit olması istenebilir."]},"beginTime":0,"endTime":113,"href":"/video/preview/4192788898677205843?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=0&ask_summarization=1"},{"index":1,"title":"Birinci Dereceden Polinom","list":{"type":"unordered","items":["p(x) = f + f'x şeklinde bir polinom kullanıldığında, hem p(0) = f hem de p'(0) = f' olur.","Bu polinom, x=0'daki eğimlerin birbirine eşit olduğu için fonksiyon değerlerine daha yakın sonuçlar verir.","Bu polinom, x=0'da çizilen bir teğet doğruya benzer."]},"beginTime":113,"endTime":226,"href":"/video/preview/4192788898677205843?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=113&ask_summarization=1"},{"index":2,"title":"İkinci Dereceden Polinom","list":{"type":"unordered","items":["p(x) = f + f'x + (1/2)f''(0)x² şeklinde bir polinom kullanıldığında, hem p(0) = f, hem p'(0) = f', hem de p''(0) = f''(0) olur.","Bu polinom, fonksiyonun ve polinomun ikinci türevlerinin de eşit olmasını sağlar.","Her eklenen terim, polinomun 'daki n'inci türevinin fonksiyonun 'daki n'inci türevine eşit olmasını 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Süreklilik için gerekli şartlar (sağdan limit, soldan limit ve fonksiyonun o noktadaki değeri birbirine eşit olması) detaylı olarak anlatılıyor. Son olarak, süreksizliğin iki temel durumu (tanımsızlıklar ve parçalı fonksiyonların parçalanma noktaları) açıklanıyor ve bir sonraki videoda süreksizlik ile ilgili soruların inceleneceği belirtiliyor."]},"endTime":736,"title":"Matematik Dersinde Süreklilik Kavramı","beginTime":0}],"fullResult":[{"index":0,"title":"Fonksiyonların Sürekliliği","list":{"type":"unordered","items":["Fonksiyonların sürekliliği, grafiği verildiğinde kalemi kaldırmak zorunda kalınan noktalarla belirlenir.","Grafiği verilmediğinde, fonksiyonun sürekliliğini belirlemek için matematiksel bir şarta ihtiyaç vardır.","Grafiğin var olduğu durumlarda, süreksizlik içi boş noktalar, kesintili çizgiler veya sonsuza giden çizgilerle belirlenir."]},"beginTime":1,"endTime":162,"href":"/video/preview/14573587156205004783?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=1&ask_summarization=1"},{"index":1,"title":"Süreklilik ve Limit 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Eğitmen, öğrencilere adım adım bir fonksiyonun grafiğini çizme sürecini göstermektedir.","Videoda, f(x) = 3x⁴ - 4x³ + 2 fonksiyonunun grafiğini çizmek için gerekli analizler yapılmaktadır. Önce kritik noktalar (x=0 ve x=1) bulunmakta, ardından ikinci türev kullanılarak çukurluk yönü incelenmekte ve potansiyel büküm noktaları (x=0 ve x=2/3) belirlenmektedir. Bu bilgiler kullanılarak fonksiyonun grafiği çizilmektedir.","Video, türev kavramını kullanarak fonksiyonların analizi ve grafik çizimi konusunu ele almaktadır. Eğitmen, analiz becerilerini kullanarak karmaşık bir fonksiyonun grafiğini çizme sürecini detaylı bir şekilde göstermektedir."]},"endTime":979,"title":"Fonksiyonun Grafiğini Çizme: Türev ve Analiz Yöntemleri","beginTime":0}],"fullResult":[{"index":0,"title":"Fonksiyonun Grafiğini Çizme","list":{"type":"unordered","items":["Türev, çukurluk yönü, maksimum, minimum ve büküm noktaları bilgisi kullanılarak hesap makinesi olmadan bir fonksiyonun grafiği çizilebilir.","Fonksiyon f(x) = 3x⁴ - 4x³ + 2 olarak tanımlanmıştır.","Fonksiyonun grafiğini çizmek için öncelikle kritik noktalar belirlenmelidir."]},"beginTime":0,"endTime":32,"href":"/video/preview/5290259787722281402?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=0&ask_summarization=1"},{"index":1,"title":"Kritik Noktaların Bulunması","list":{"type":"unordered","items":["Kritik noktalar, türevin sıfır veya tanımsız olduğu noktalardır.","Verilen fonksiyon her yerde türevlidir, dolayısıyla kritik noktalar f'(x) = 0 denkleminin çözümleridir.","f'(x) = 12x³ - 12x² türevi bulunmuştur."]},"beginTime":32,"endTime":111,"href":"/video/preview/5290259787722281402?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=32&ask_summarization=1"},{"index":2,"title":"Kritik Noktaların Çözümü","list":{"type":"unordered","items":["f'(x) = 12x³ - 12x² = 0 denklemi çözerken 12x² ifadesi paranteze alınır.","Denklem 12x²(x - 1) = 0 şeklinde yazılır ve çözülür.","Kritik noktalar x = 0 ve x = 1 olarak bulunmuştur."]},"beginTime":111,"endTime":204,"href":"/video/preview/5290259787722281402?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=111&ask_summarization=1"},{"index":3,"title":"İkinci Türev ve Çukurluk Analizi","list":{"type":"unordered","items":["Fonksiyonun ikinci türevi f''(x) = 36x² - 24x olarak bulunmuştur.","x = 0 noktasında ikinci türev sıfır olduğundan çukurluk yönü belirsizdir, potansiyel büküm noktası olabilir.","x = 1 noktasında ikinci türev 12'dir, bu pozitif olduğu için çukurluk yukarı doğru ve bu nokta minimum noktasıdır."]},"beginTime":204,"endTime":325,"href":"/video/preview/5290259787722281402?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=204&ask_summarization=1"},{"index":4,"title":"Büküm Noktalarının Bulunması","list":{"type":"unordered","items":["İkinci türevin sıfır olduğu noktalar büküm noktaları olabilir.","f''(x) = 36x² - 24x = 0 denklemi çözülürken 12x ifadesi paranteze alınır.","Denklem 12x(3x - 2) = 0 şeklinde yazılır ve çözülür."]},"beginTime":325,"endTime":384,"href":"/video/preview/5290259787722281402?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=325&ask_summarization=1"},{"index":5,"title":"Büküm Noktalarının 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olur."]},"beginTime":566,"endTime":704,"href":"/video/preview/5290259787722281402?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=566&ask_summarization=1"},{"index":7,"title":"Fonksiyonun Kritik Noktaları","list":{"type":"unordered","items":["x = 1 noktasında eğimin olduğu ve bu noktanın minimum olduğu tespit edilmiştir.","f(1) = 1 değerini bulmak için başlangıç fonksiyonuna geri dönmüşlerdir.","x = 0 noktası büküm noktası olarak belirlenmiş ve bu noktada eğim 0'dır."]},"beginTime":704,"endTime":797,"href":"/video/preview/5290259787722281402?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=704&ask_summarization=1"},{"index":8,"title":"Büküm Noktaları ve Çizim","list":{"type":"unordered","items":["x = 2/3 noktası da büküm noktası olarak bulunmuştur.","x \u003c 2/3 için çukurluk aşağı doğru, x > 2/3 için çukurluk yukarı doğru olarak 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fazlası için: http://www.khanacademy.org.tr Matematikten sanat tarihine, ekonomiden fen bilimlerine, basit toplamadan diferansiyel denklemlere, ilkokul seviyesinden üniversite seviyesine...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/1788942/8a51a00c15fca7121e14e780a2a007b0/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/0S41IQEAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"14","reqid":"1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=videoid:15337069700319503372","teaser":[{"list":{"type":"unordered","items":["Bu video, matematik eğitimi formatında bir ders anlatımıdır. Eğitmen, minimum ve maksimum bulma konusunu optimizasyon sorularında nasıl kullanabileceğimizi göstermektedir.","Videoda, çarpımı eksi onaltı olan iki sayının karelerinin toplamı minimum olan değerlerini bulma problemi adım adım çözülmektedir. Eğitmen önce problemi matematiksel olarak ifade eder, ardından fonksiyonu tek değişkenli hale getirir, türev alır ve minimum değerini bulur. Son olarak, bulunan değerlerin gerçekten minimum olduğunu ikinci türev yardımıyla ispatlar. Video, optimizasyon problemlerinin nasıl çözüleceğini gösteren bir örnek sunmaktadır."]},"endTime":461,"title":"Optimizasyon Problemleri Çözümü Eğitim Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"Optimizasyon Problemleri ve Örnek Soru","list":{"type":"unordered","items":["Optimizasyon sorularında bazı kısıtlamalara uyan hangi sayıların maksimum veya minimum değer verdiği sorulur.","Çarpımları eksi onaltı olan ve karelerinin toplamı minimum olan iki sayı bulunması isteniyor.","Bu soruda x ve y sayıları için x×y = -16 ve karelerinin toplamı S = x² + y² şeklinde modelleniyor."]},"beginTime":0,"endTime":103,"href":"/video/preview/15337069700319503372?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=0&ask_summarization=1"},{"index":1,"title":"Fonksiyonun Türevi ve Çözüm","list":{"type":"unordered","items":["y = -16/x şeklinde ifade edilerek S = x² + (-16/x)² şeklinde tek değişkenli fonksiyona indirgeniyor.","Fonksiyonun türevi S' = 2x - 576/x³ olarak bulunuyor ve bu türev sıfır eşitlenerek x = ±4 değerleri elde ediliyor.","x = 4 için y = -4, x = -4 için y = 4 olduğu görülüyor, bu iki sayı çarpımları -16 olan ve karelerinin toplamı minimum olan sayılar."]},"beginTime":103,"endTime":309,"href":"/video/preview/15337069700319503372?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=103&ask_summarization=1"},{"index":2,"title":"Minimum Değerin İspatı","list":{"type":"unordered","items":["Fonksiyonun ikinci türevi S'' = 2 + 1536/x⁴ olarak bulunuyor ve x = 4 için S'' pozitif olduğu görüldüğü için minimum noktası olduğu ispatlanıyor.","x = 4 için karelerinin toplamı S(4) = 4² + (-4)² = 32 olarak hesaplanıyor.","Diğer çarpımları -16 olan sayı çiftleri (1 ve -16, -8 ve 2 gibi) için karelerinin toplamı 32'den büyük olduğu 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Konuşmacı, bazı üniversitelerin hocalarının \"süreksizlikleri bulun ve bunların isimlerini yazın\" şeklinde sorduğu sorulara yardımcı olmak amacıyla bu videoyu hazırladığını belirtiyor.","Video, süreksizlik çeşitlerini üç ana kategoride inceliyor: atlama süreksizliği (jump discontinuity), kaldırılabilir süreksizlik (removeble discontinuity) ve sonsuzluk kaynaklı süreksizlik (infinity discontinuity). Her bir süreksizlik türü için tanım, grafiksel gösterimi ve limit kavramı üzerinden açıklamalar yapılıyor. Ayrıca, bu süreksizlik türlerinin nasıl tespit edileceği örneklerle gösteriliyor."]},"endTime":541,"title":"Süreksizlik Çeşitleri Eğitim Videosu","beginTime":0}],"fullResult":[{"index":0,"title":"Süreksizlik Çeşitleri","list":{"type":"unordered","items":["Bu videoda süreksiz olan noktalara konulmuş isimler incelenecektir.","Bazı üniversitelerin hocaları süreksizlikleri bulup isimlerini yazmanızı isteyebilir.","Süreksizlik çeşitleri üçtür: atlama süreksizliği, kaldırılabilir süreksizlik ve sonsuzluk kaynaklı süreksizlik."]},"beginTime":1,"endTime":39,"href":"/video/preview/10144518966714258128?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=1&ask_summarization=1"},{"index":1,"title":"Atlama Süreksizliği","list":{"type":"unordered","items":["Atlama süreksizliği (İngilizce: jump discontinuity) grafiği çizmede kalemi kaldırmamıza neden olan noktadır.","Bu süreksizlik, sağdan limitle soldan limitin birbirine eşit olmamasından kaynaklanır.","Sürekliliği sağlayan şey, sağdan limit, soldan limit ve o noktadaki fonksiyon değeri birbirine eşit olmasıdır."]},"beginTime":39,"endTime":163,"href":"/video/preview/10144518966714258128?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=39&ask_summarization=1"},{"index":2,"title":"Kaldırılabilir Süreksizlik","list":{"type":"unordered","items":["Kaldırılabilir süreksizlik (İngilizce: removeble discontinuity) grafiğin içi boş bir noktadan geçmesiyle oluşur.","Bu süreksizlikte sağdan limitle soldan limit birbirine eşit olmasına rağmen, o noktadaki fonksiyon değeri onlara eşit değildir.","Bu süreksizlik, bir şeyler yapılarak kaldırılabildiği için bu isme sahiptir."]},"beginTime":163,"endTime":245,"href":"/video/preview/10144518966714258128?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=163&ask_summarization=1"},{"index":3,"title":"Sonsuzluk Kaynaklı Süreksizlik","list":{"type":"unordered","items":["Sonsuzluk kaynaklı süreksizlik (İngilizce: infinity discontinuity) sağdan veya soldan limitten herhangi birinin değerinin sonsuza gitmesinden kaynaklanır.","Bu süreksizliklerde, sağdan veya soldan limitten en az birinin artı sonsuz veya eksi sonsuzdan birine gitmesi yeterlidir.","Tanımsızlıktan kaynaklı tüm süreksizlikler, sağdan ve soldan yaklaşıldığında artı veya eksi sonsuzlara gittikleri için sonsuz kaynaklı süreksizliktir."]},"beginTime":245,"endTime":343,"href":"/video/preview/10144518966714258128?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=245&ask_summarization=1"},{"index":4,"title":"Örnekler","list":{"type":"unordered","items":["f(x) = (2x+1)/(x+3) fonksiyonunda x=-3'te tanımsızlıktan kaynaklı sonsuz süreksizlik vardır.","f(x) = (x+1)/(x-3) fonksiyonunda x=3'te sağdan ve soldan limit birbirine eşitken f(3) değeri farklı olduğu için kaldırılabilir süreksizlik vardır.","f(x) = (x²+1)/(x-4) fonksiyonunda x=4'te sağdan ve soldan limit birbirinden farklı olduğu için atlama süreksizliği vardır."]},"beginTime":343,"endTime":536,"href":"/video/preview/10144518966714258128?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=343&ask_summarization=1"}],"linkTemplate":"/video/preview/10144518966714258128?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=%%timestamp%%&ask_summarization=1"},"isAdultDoc":false,"relatedParams":{"text":"Calculus-I : Süreksizlik Çeşitleri (Type of Discontinuities)","related_orig_text":"Calculus Chronicles","related_porno":false,"related_less_3m_off":true,"client":"d2d","no_cnt":1,"related_src":"serp","related":"{\"porno\":false,\"vfp\":1,\"orig_text\":\"Calculus 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fazlası için: http://www.khanacademy.org.tr Matematikten sanat tarihine, ekonomiden fen bilimlerine, basit toplamadan diferansiyel denklemlere, ilkokul seviyesinden üniversite seviyesine...","preview":{"posterSrc":"//avatars.mds.yandex.net/get-vthumb/762067/0160f99495ec31b709a14e746a13fd21/564x318_1","videoSrc":"https://video-preview.s3.yandex.net/233ZSgEAAAA.mp4","videoType":"video/mp4"},"target":"_self","position":"19","reqid":"1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL","summary":{"isFull":true,"fullTextUrl":"/video/result?ask_summarization=1&numdoc=1&noreask=1&nomisspell=1&parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=videoid:15347343534791439065","teaser":[{"list":{"type":"unordered","items":["Bu video, bir eğitmen tarafından sunulan matematik problem çözüm dersidir. Eğitmen, izleyicilerden gelen istek üzerine kare mukavvadan en büyük hacimli üstü açık kutu problemi çözmektedir.","Videoda, kenarı 24 santimetre olan bir kare mukavvadan en büyük hacimli üstü açık kutu oluşturma problemi adım adım çözülmektedir. Eğitmen önce problemi görselleştirerek kare mukavvayı çizerek, köşelerden eşit büyüklükte kareler kesip kenarları yukarı kıvırarak kutunun nasıl oluşturulacağını açıklar. Ardından kutunun hacmini x cinsinden bir fonksiyon olarak yazarak türev alma, kritik noktaları bulma ve maksimum hacmi hesaplama sürecini detaylı şekilde gösterir. Sonuç olarak, kutunun optimum hacminin 16x16x4 santimetre olduğu bulunur."]},"endTime":506,"title":"Kare Mukavvadan En Büyük Hacimli Üstü Açık Kutu Problemi Çözümü","beginTime":0}],"fullResult":[{"index":0,"title":"Problemin Tanıtımı","list":{"type":"unordered","items":["Kenarı 24 santimetre olan bir kare mukavvadan oluşturulacak en büyük hacimli üstü açık kutunun bulunması isteniyor.","Mukavvadan köşelerden eşit büyüklükte kareler kesilip kenarları yukarı doğru kıvrılacak.","Kutunun hacminin maksimum olması için hangi boyutta kare kesilmeli sorusu çözülecek."]},"beginTime":0,"endTime":71,"href":"/video/preview/15347343534791439065?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=0&ask_summarization=1"},{"index":1,"title":"Kutunun Boyutlarının Belirlenmesi","list":{"type":"unordered","items":["Kesilen karenin kenar uzunluğuna x diyerek, kutunun tabanının kenar uzunluğu 24-2x olarak hesaplanıyor.","Kutunun yüksekliği, yukarı doğru kıvrılan kenarların uzunluğu olan x olarak belirleniyor.","Kutunun hacmi V = x × (24-2x) × (24-2x) formülüyle ifade ediliyor."]},"beginTime":71,"endTime":196,"href":"/video/preview/15347343534791439065?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=71&ask_summarization=1"},{"index":2,"title":"Hacmin Maksimum Değerinin Bulunması","list":{"type":"unordered","items":["Hacim fonksiyonu V = 4x³ - 96x² + 576x olarak hesaplanıyor.","Hacmin maksimum değerini bulmak için fonksiyonun türevi V' = 12x² - 192x + 576 olarak bulunuyor.","Türevin sıfır olduğu noktalar x = 4 ve x = 12 olarak hesaplanıyor."]},"beginTime":196,"endTime":436,"href":"/video/preview/15347343534791439065?parent-reqid=1769464923335290-13741870115270290948-balancer-l7leveler-kubr-yp-vla-177-BAL&text=Calculus+Chronicles&t=196&ask_summarization=1"},{"index":3,"title":"Maksimum Değerin İspatı","list":{"type":"unordered","items":["x = 12 değeri için 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