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lehrkraefte:blc:miniaufgaben [2021/09/26 20:23]
Ivo Blöchliger [27. September 2021 bis 1. Oktober 2021] |
lehrkraefte:blc:miniaufgaben [2024/04/23 09:46]
Ivo Blöchliger |
| | ~~NOTOC~~ |
| ===== Miniaufgaben ===== | ===== Miniaufgaben ===== |
| * Auf jede Lektion (ausser Prüfungslektionen) ist eine Miniaufgabe vorzubereiten. Am Anfang der Lektion wird eine Münze geworfen. Damit der Münzwurf gültig ist, muss sich die Münze mindestens 10 mal in der Luft drehen. Zeigt die Münze **Zahl**, wird eine Aufgabe in Form eines Kurztests geprüft. | * Auf jede Lektion (ausser Prüfungslektionen) ist eine Miniaufgabe vorzubereiten. Am Anfang der Lektion wird ein Würfel geworfen. Zeigt der Würfel eine Vier, Fünf oder Sechs, wird eine Aufgabe in Form eines Kurztests geprüft. |
| * Jeder Schüler hat 3 Joker für das 1. Semester. Bei Meldung per e-mail oder Threema (HX3WS583) bis spätestens 12 h vor Lektionsbeginn wird der Schüler vom eventuellen Kurztest ersatzlos dispensiert. Zeigt die Münze Kopf, ist der Joker aber auch aufgebraucht! | * Jeder Schüler hat 5 Joker für das ganze Jahr. Diese werden über die [[lehrkraefte:blc:informatik:glf22:crypto:joker-chain|JokerChain]] verwaltet und können bis 23:59 am Vortag eingelöst werden. Bei Einsatz eines Jokers wird der Schüler vom eventuellen Kurztest ersatzlos dispensiert. Zeigt der Würfel 1-3, ist der Joker aber auch aufgebraucht! |
| | //Beachten Sie, dass via andere Kanäle keine Joker mehr eingelöst werden können. Bei Problemen werde ich Sie aber nach Möglichkeit unterstützen (mit genügend zeitlichem Vorlauf).// |
| * Der Minikurztest ist auf mitgebrachtem **A4-Papier im Hochformat** zu lösen. Ausgefranste Ränder, zerknittertes Papier, abgerissene Ecken und Übergrössen führen zu **Abzug**. | * Der Minikurztest ist auf mitgebrachtem **A4-Papier im Hochformat** zu lösen. Ausgefranste Ränder, zerknittertes Papier, abgerissene Ecken und Übergrössen führen zu **Abzug**. |
| * Der Name ist **oben rechts** zu notieren. | * Der Name ist **oben rechts** zu notieren. |
| * Die Prüfungsblätter können mehrmals verwendet werden, die Aufgaben sind aber sauber abzugrenzen. | * Die Prüfungsblätter können mehrmals verwendet werden, die Aufgaben sind aber sauber abzugrenzen. |
| * Der Durchschnitt aller Miniaufgaben zählt als eine volle 4. Prüfungsnote. | * Schreiben Sie nicht mit Rot oder einer schlecht lesbaren Farbe, wie z.B. gelb. (Ja, ja, jede Regel hat eine Geschichte). |
| | * Der Durchschnitt aller Miniaufgaben zählt als eine volle 6. Prüfungsnote. |
| |
| <PRELOAD> | <PRELOAD> |
| </PRELOAD> | </PRELOAD> |
| |
| ==== 20. September 2021 bis 24. September 2021 ==== | ==== 22. April 2024 bis 26. April 2024 ==== |
| === Donnerstag 23. September 2021 === | === Dienstag 23. April 2024 === |
| Ausrechnen, Resultat als gekürzter Bruch:<JS>miniAufgabe("#exonumbercrunch2","#solnumbercrunch2", | Leiten Sie von Hand und ohne Unterlagen ab: |
| [["$\\displaystyle \\frac{\\frac{3}{\\frac{9}{5}}+\\frac{-\\frac{7}{2}}{-7}}{\\frac{-2}{\\frac{24}{7}}+\\frac{-\\frac{13}{6}}{2}}$", "$\\displaystyle \\frac{\\frac{3}{\\frac{9}{5}}+\\frac{-\\frac{7}{2}}{-7}}{\\frac{-2}{\\frac{24}{7}}+\\frac{-\\frac{13}{6}}{2}} = \\frac{3\\cdot\\frac{5}{9}+-\\frac{7}{2}\\cdot-\\frac{1}{7}}{-2\\cdot\\frac{7}{24}+-\\frac{13}{6}\\cdot\\frac{1}{2}} = \\frac{\\frac{5}{3}+\\frac{1}{2}}{-\\frac{7}{12}-\\frac{13}{12}} = \\frac{\\frac{10}{6}+\\frac{3}{6}}{-\\frac{7}{12}-\\frac{13}{12}} = \\frac{\\frac{13}{6}}{-\\frac{5}{3}} = \\frac{13}{6} \\cdot -\\frac{3}{5} = -\\frac{13}{10}$"], ["$\\displaystyle \\frac{\\frac{-2}{\\frac{9}{5}}+\\frac{\\frac{16}{9}}{4}}{\\frac{3}{-\\frac{13}{2}}+\\frac{\\frac{12}{13}}{-3}}$", "$\\displaystyle \\frac{\\frac{-2}{\\frac{9}{5}}+\\frac{\\frac{16}{9}}{4}}{\\frac{3}{-\\frac{13}{2}}+\\frac{\\frac{12}{13}}{-3}} = \\frac{-2\\cdot\\frac{5}{9}+\\frac{16}{9}\\cdot\\frac{1}{4}}{3\\cdot-\\frac{2}{13}+\\frac{12}{13}\\cdot-\\frac{1}{3}} = \\frac{-\\frac{10}{9}+\\frac{4}{9}}{-\\frac{6}{13}-\\frac{4}{13}} = \\frac{-\\frac{10}{9}+\\frac{4}{9}}{-\\frac{6}{13}-\\frac{4}{13}} = \\frac{-\\frac{2}{3}}{-\\frac{10}{13}} = -\\frac{2}{3} \\cdot -\\frac{13}{10} = \\frac{13}{15}$"], ["$\\displaystyle \\frac{\\frac{2}{\\frac{24}{17}}+\\frac{-\\frac{20}{3}}{4}}{\\frac{2}{-\\frac{12}{5}}+\\frac{\\frac{5}{3}}{-5}}$", "$\\displaystyle \\frac{\\frac{2}{\\frac{24}{17}}+\\frac{-\\frac{20}{3}}{4}}{\\frac{2}{-\\frac{12}{5}}+\\frac{\\frac{5}{3}}{-5}} = \\frac{2\\cdot\\frac{17}{24}+-\\frac{20}{3}\\cdot\\frac{1}{4}}{2\\cdot-\\frac{5}{12}+\\frac{5}{3}\\cdot-\\frac{1}{5}} = \\frac{\\frac{17}{12}-\\frac{5}{3}}{-\\frac{5}{6}-\\frac{1}{3}} = \\frac{\\frac{17}{12}-\\frac{20}{12}}{-\\frac{5}{6}-\\frac{2}{6}} = \\frac{-\\frac{1}{4}}{-\\frac{7}{6}} = -\\frac{1}{4} \\cdot -\\frac{6}{7} = \\frac{3}{14}$"], ["$\\displaystyle \\frac{\\frac{6}{-\\frac{20}{3}}+\\frac{\\frac{5}{2}}{3}}{\\frac{-5}{\\frac{15}{4}}+\\frac{-\\frac{10}{3}}{-3}}$", "$\\displaystyle \\frac{\\frac{6}{-\\frac{20}{3}}+\\frac{\\frac{5}{2}}{3}}{\\frac{-5}{\\frac{15}{4}}+\\frac{-\\frac{10}{3}}{-3}} = \\frac{6\\cdot-\\frac{3}{20}+\\frac{5}{2}\\cdot\\frac{1}{3}}{-5\\cdot\\frac{4}{15}+-\\frac{10}{3}\\cdot-\\frac{1}{3}} = \\frac{-\\frac{9}{10}+\\frac{5}{6}}{-\\frac{4}{3}+\\frac{10}{9}} = \\frac{-\\frac{27}{30}+\\frac{25}{30}}{-\\frac{12}{9}+\\frac{10}{9}} = \\frac{-\\frac{1}{15}}{-\\frac{2}{9}} = -\\frac{1}{15} \\cdot -\\frac{9}{2} = \\frac{3}{10}$"], ["$\\displaystyle \\frac{\\frac{2}{\\frac{20}{13}}+\\frac{-\\frac{18}{5}}{6}}{\\frac{2}{-\\frac{16}{5}}+\\frac{-\\frac{5}{2}}{10}}$", "$\\displaystyle \\frac{\\frac{2}{\\frac{20}{13}}+\\frac{-\\frac{18}{5}}{6}}{\\frac{2}{-\\frac{16}{5}}+\\frac{-\\frac{5}{2}}{10}} = \\frac{2\\cdot\\frac{13}{20}+-\\frac{18}{5}\\cdot\\frac{1}{6}}{2\\cdot-\\frac{5}{16}+-\\frac{5}{2}\\cdot\\frac{1}{10}} = \\frac{\\frac{13}{10}-\\frac{3}{5}}{-\\frac{5}{8}-\\frac{1}{4}} = \\frac{\\frac{13}{10}-\\frac{6}{10}}{-\\frac{5}{8}-\\frac{2}{8}} = \\frac{\\frac{7}{10}}{-\\frac{7}{8}} = \\frac{7}{10} \\cdot -\\frac{8}{7} = -\\frac{4}{5}$"], ["$\\displaystyle \\frac{\\frac{2}{\\frac{20}{7}}+\\frac{\\frac{9}{2}}{-3}}{\\frac{-8}{-\\frac{14}{3}}+\\frac{-\\frac{16}{7}}{2}}$", "$\\displaystyle \\frac{\\frac{2}{\\frac{20}{7}}+\\frac{\\frac{9}{2}}{-3}}{\\frac{-8}{-\\frac{14}{3}}+\\frac{-\\frac{16}{7}}{2}} = \\frac{2\\cdot\\frac{7}{20}+\\frac{9}{2}\\cdot-\\frac{1}{3}}{-8\\cdot-\\frac{3}{14}+-\\frac{16}{7}\\cdot\\frac{1}{2}} = \\frac{\\frac{7}{10}-\\frac{3}{2}}{\\frac{12}{7}-\\frac{8}{7}} = \\frac{\\frac{7}{10}-\\frac{15}{10}}{\\frac{12}{7}-\\frac{8}{7}} = \\frac{-\\frac{4}{5}}{\\frac{4}{7}} = -\\frac{4}{5} \\cdot \\frac{7}{4} = -\\frac{7}{5}$"], ["$\\displaystyle \\frac{\\frac{-2}{-\\frac{8}{5}}+\\frac{-\\frac{17}{6}}{2}}{\\frac{-2}{\\frac{24}{7}}+\\frac{\\frac{8}{3}}{-8}}$", "$\\displaystyle \\frac{\\frac{-2}{-\\frac{8}{5}}+\\frac{-\\frac{17}{6}}{2}}{\\frac{-2}{\\frac{24}{7}}+\\frac{\\frac{8}{3}}{-8}} = \\frac{-2\\cdot-\\frac{5}{8}+-\\frac{17}{6}\\cdot\\frac{1}{2}}{-2\\cdot\\frac{7}{24}+\\frac{8}{3}\\cdot-\\frac{1}{8}} = \\frac{\\frac{5}{4}-\\frac{17}{12}}{-\\frac{7}{12}-\\frac{1}{3}} = \\frac{\\frac{15}{12}-\\frac{17}{12}}{-\\frac{7}{12}-\\frac{4}{12}} = \\frac{-\\frac{1}{6}}{-\\frac{11}{12}} = -\\frac{1}{6} \\cdot -\\frac{12}{11} = \\frac{2}{11}$"], ["$\\displaystyle \\frac{\\frac{2}{-\\frac{13}{7}}+\\frac{-\\frac{14}{13}}{-2}}{\\frac{-6}{-\\frac{13}{2}}+\\frac{-\\frac{18}{13}}{9}}$", "$\\displaystyle \\frac{\\frac{2}{-\\frac{13}{7}}+\\frac{-\\frac{14}{13}}{-2}}{\\frac{-6}{-\\frac{13}{2}}+\\frac{-\\frac{18}{13}}{9}} = \\frac{2\\cdot-\\frac{7}{13}+-\\frac{14}{13}\\cdot-\\frac{1}{2}}{-6\\cdot-\\frac{2}{13}+-\\frac{18}{13}\\cdot\\frac{1}{9}} = \\frac{-\\frac{14}{13}+\\frac{7}{13}}{\\frac{12}{13}-\\frac{2}{13}} = \\frac{-\\frac{14}{13}+\\frac{7}{13}}{\\frac{12}{13}-\\frac{2}{13}} = \\frac{-\\frac{7}{13}}{\\frac{10}{13}} = -\\frac{7}{13} \\cdot \\frac{13}{10} = -\\frac{7}{10}$"], ["$\\displaystyle \\frac{\\frac{-2}{-\\frac{11}{2}}+\\frac{\\frac{16}{11}}{2}}{\\frac{2}{\\frac{22}{5}}+\\frac{\\frac{30}{11}}{-2}}$", "$\\displaystyle \\frac{\\frac{-2}{-\\frac{11}{2}}+\\frac{\\frac{16}{11}}{2}}{\\frac{2}{\\frac{22}{5}}+\\frac{\\frac{30}{11}}{-2}} = \\frac{-2\\cdot-\\frac{2}{11}+\\frac{16}{11}\\cdot\\frac{1}{2}}{2\\cdot\\frac{5}{22}+\\frac{30}{11}\\cdot-\\frac{1}{2}} = \\frac{\\frac{4}{11}+\\frac{8}{11}}{\\frac{5}{11}-\\frac{15}{11}} = \\frac{\\frac{4}{11}+\\frac{8}{11}}{\\frac{5}{11}-\\frac{15}{11}} = \\frac{\\frac{12}{11}}{-\\frac{10}{11}} = \\frac{12}{11} \\cdot -\\frac{11}{10} = -\\frac{6}{5}$"], ["$\\displaystyle \\frac{\\frac{4}{-\\frac{11}{2}}+\\frac{\\frac{12}{11}}{-2}}{\\frac{2}{-\\frac{16}{19}}+\\frac{-\\frac{15}{8}}{-3}}$", "$\\displaystyle \\frac{\\frac{4}{-\\frac{11}{2}}+\\frac{\\frac{12}{11}}{-2}}{\\frac{2}{-\\frac{16}{19}}+\\frac{-\\frac{15}{8}}{-3}} = \\frac{4\\cdot-\\frac{2}{11}+\\frac{12}{11}\\cdot-\\frac{1}{2}}{2\\cdot-\\frac{19}{16}+-\\frac{15}{8}\\cdot-\\frac{1}{3}} = \\frac{-\\frac{8}{11}-\\frac{6}{11}}{-\\frac{19}{8}+\\frac{5}{8}} = \\frac{-\\frac{8}{11}-\\frac{6}{11}}{-\\frac{19}{8}+\\frac{5}{8}} = \\frac{-\\frac{14}{11}}{-\\frac{7}{4}} = -\\frac{14}{11} \\cdot -\\frac{4}{7} = \\frac{8}{11}$"]], | <JS>miniAufgabe("#exonurpolynome","#solnurpolynome", |
| " ", " <hr> "); | [["a) $f(x)=-\\frac{4}{3}x^{11}-\\frac{1}{9}x^{9}+\\frac{2}{3}x^{3}\\quad$ b) $f(x)=-\\frac{1}{2}x^{12}+\\frac{1}{5}x^{7}+\\frac{1}{2}x^{4}\\quad$ ", "a) $f'(x)=-\\frac{44}{3}x^{10}-x^{8}+2x^{2}\\quad$ b) $f'(x)=-6x^{11}+\\frac{7}{5}x^{6}+2x^{3}\\quad$ "], ["a) $f(x)=-\\frac{4}{3}x^{11}-\\frac{4}{9}x^{6}+\\frac{2}{9}x^{2}\\quad$ b) $f(x)=\\frac{1}{5}x^{12}-\\frac{1}{2}x^{7}+\\frac{4}{3}x^{5}\\quad$ ", "a) $f'(x)=-\\frac{44}{3}x^{10}-\\frac{8}{3}x^{5}+\\frac{4}{9}x\\quad$ b) $f'(x)=\\frac{12}{5}x^{11}-\\frac{7}{2}x^{6}+\\frac{20}{3}x^{4}\\quad$ "], ["a) $f(x)=-\\frac{1}{3}x^{11}-\\frac{1}{2}x^{10}+\\frac{1}{6}x^{5}\\quad$ b) $f(x)=-\\frac{2}{3}x^{11}-\\frac{2}{7}x^{9}+\\frac{2}{9}x^{8}\\quad$ ", "a) $f'(x)=-\\frac{11}{3}x^{10}-5x^{9}+\\frac{5}{6}x^{4}\\quad$ b) $f'(x)=-\\frac{22}{3}x^{10}-\\frac{18}{7}x^{8}+\\frac{16}{9}x^{7}\\quad$ "], ["a) $f(x)=-\\frac{1}{2}x^{9}+\\frac{1}{3}x^{5}-\\frac{3}{8}x^{2}\\quad$ b) $f(x)=-\\frac{4}{3}x^{12}-\\frac{3}{5}x^{11}-\\frac{4}{9}x^{4}\\quad$ ", "a) $f'(x)=-\\frac{9}{2}x^{8}+\\frac{5}{3}x^{4}-\\frac{3}{4}x\\quad$ b) $f'(x)=-16x^{11}-\\frac{33}{5}x^{10}-\\frac{16}{9}x^{3}\\quad$ "], ["a) $f(x)=\\frac{3}{5}x^{12}-\\frac{2}{3}x^{5}+\\frac{1}{4}x^{4}\\quad$ b) $f(x)=\\frac{3}{4}x^{9}-\\frac{3}{7}x^{5}-\\frac{2}{5}x^{2}\\quad$ ", "a) $f'(x)=\\frac{36}{5}x^{11}-\\frac{10}{3}x^{4}+x^{3}\\quad$ b) $f'(x)=\\frac{27}{4}x^{8}-\\frac{15}{7}x^{4}-\\frac{4}{5}x\\quad$ "], ["a) $f(x)=\\frac{2}{3}x^{12}+\\frac{3}{8}x^{8}+\\frac{4}{3}x^{7}\\quad$ b) $f(x)=-\\frac{1}{5}x^{12}-\\frac{1}{3}x^{6}-\\frac{1}{2}x^{3}\\quad$ ", "a) $f'(x)=8x^{11}+3x^{7}+\\frac{28}{3}x^{6}\\quad$ b) $f'(x)=-\\frac{12}{5}x^{11}-2x^{5}-\\frac{3}{2}x^{2}\\quad$ "], ["a) $f(x)=\\frac{1}{4}x^{12}-\\frac{1}{2}x^{10}-\\frac{2}{5}x^{7}\\quad$ b) $f(x)=-\\frac{1}{8}x^{12}-\\frac{1}{5}x^{5}-\\frac{3}{4}x^{4}\\quad$ ", "a) $f'(x)=3x^{11}-5x^{9}-\\frac{14}{5}x^{6}\\quad$ b) $f'(x)=-\\frac{3}{2}x^{11}-x^{4}-3x^{3}\\quad$ "], ["a) $f(x)=\\frac{3}{2}x^{12}-\\frac{2}{3}x^{10}+\\frac{1}{8}x^{3}\\quad$ b) $f(x)=\\frac{3}{4}x^{12}+\\frac{1}{3}x^{9}-\\frac{2}{5}x^{4}\\quad$ ", "a) $f'(x)=18x^{11}-\\frac{20}{3}x^{9}+\\frac{3}{8}x^{2}\\quad$ b) $f'(x)=9x^{11}+3x^{8}-\\frac{8}{5}x^{3}\\quad$ "], ["a) $f(x)=\\frac{4}{9}x^{12}+\\frac{4}{9}x^{6}-\\frac{3}{8}x^{3}\\quad$ b) $f(x)=\\frac{1}{3}x^{6}+\\frac{1}{4}x^{5}+\\frac{2}{3}x^{2}\\quad$ ", "a) $f'(x)=\\frac{16}{3}x^{11}+\\frac{8}{3}x^{5}-\\frac{9}{8}x^{2}\\quad$ b) $f'(x)=2x^{5}+\\frac{5}{4}x^{4}+\\frac{4}{3}x\\quad$ "], ["a) $f(x)=\\frac{1}{8}x^{10}-\\frac{2}{7}x^{8}-\\frac{4}{7}x^{3}\\quad$ b) $f(x)=\\frac{1}{3}x^{12}-\\frac{1}{5}x^{6}-\\frac{1}{2}x^{2}\\quad$ ", "a) $f'(x)=\\frac{5}{4}x^{9}-\\frac{16}{7}x^{7}-\\frac{12}{7}x^{2}\\quad$ b) $f'(x)=4x^{11}-\\frac{6}{5}x^{5}-x\\quad$ "]], |
| | " <br> "); |
| </JS> | </JS> |
| <HTML> | <HTML> |
| <div id="exonumbercrunch2"></div> | <div id="exonurpolynome"></div> |
| |
| </HTML> | </HTML> |
| <hidden Lösungen> | <hidden Lösungen> |
| | |
| <HTML> | <HTML> |
| <div id="solnumbercrunch2"></div> | <div id="solnurpolynome"></div> |
| | <div style='font-size:12px;color:gray;'>ruby ableiten-von-hand.rb 4</div> |
| </HTML> | </HTML> |
| | |
| </hidden> | </hidden> |
| |
| | === Mittwoch 24. April 2024 === |
| === Freitag 24. September 2021 === | Die folgenden Funktionen haben genau zwei Extremalpunkte. Bestimmen Sie diese.<JS>miniAufgabe("#exoextrema3tengrades","#solextrema3tengrades", |
| Primfaktorzerlegung: Schreiben Sie als Produkt von Potenzen von Primzahlen, nach aufsteigender Basis geordnet.<JS>miniAufgabe("#exoprimfaktorzerlegung","#solprimfaktorzerlegung", | [["$f(x)=\\frac{1}{3}x^{3}-\\frac{3}{2}x^{2}-10x+2$", "Ableitung: $f'(x)=x^{2}-3x-10$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -3, Produkt -10): $\\left(x+2\\right)\\left(x-5\\right)=0$. Daraus liest man ab: $x_1=-2$, $x_2=5$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{3\\pm\\sqrt{9-4\\cdot-10}}{2}$ und daraus $x_1=-2$, $x_2=5$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{40}{3}$ und $y_2=f(x_2)=-\\frac{263}{6}$.<br>\n$E_1 = \\left(-2, \\frac{40}{3}\\right)$ $E_2 = \\left(5, -\\frac{263}{6}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}+\\frac{1}{2}x^{2}-12x+4$", "Ableitung: $f'(x)=x^{2}+x-12$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe 1, Produkt -12): $\\left(x+4\\right)\\left(x-3\\right)=0$. Daraus liest man ab: $x_1=-4$, $x_2=3$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{-1\\pm\\sqrt{1-4\\cdot-12}}{2}$ und daraus $x_1=-4$, $x_2=3$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{116}{3}$ und $y_2=f(x_2)=-\\frac{37}{2}$.<br>\n$E_1 = \\left(-4, \\frac{116}{3}\\right)$ $E_2 = \\left(3, -\\frac{37}{2}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}-\\frac{1}{2}x^{2}-20x+2$", "Ableitung: $f'(x)=x^{2}-x-20$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -1, Produkt -20): $\\left(x+4\\right)\\left(x-5\\right)=0$. Daraus liest man ab: $x_1=-4$, $x_2=5$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{1\\pm\\sqrt{1-4\\cdot-20}}{2}$ und daraus $x_1=-4$, $x_2=5$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{158}{3}$ und $y_2=f(x_2)=-\\frac{413}{6}$.<br>\n$E_1 = \\left(-4, \\frac{158}{3}\\right)$ $E_2 = \\left(5, -\\frac{413}{6}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}-\\frac{1}{2}x^{2}-20x+4$", "Ableitung: $f'(x)=x^{2}-x-20$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -1, Produkt -20): $\\left(x+4\\right)\\left(x-5\\right)=0$. Daraus liest man ab: $x_1=-4$, $x_2=5$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{1\\pm\\sqrt{1-4\\cdot-20}}{2}$ und daraus $x_1=-4$, $x_2=5$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{164}{3}$ und $y_2=f(x_2)=-\\frac{401}{6}$.<br>\n$E_1 = \\left(-4, \\frac{164}{3}\\right)$ $E_2 = \\left(5, -\\frac{401}{6}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}+\\frac{3}{2}x^{2}-10x+4$", "Ableitung: $f'(x)=x^{2}+3x-10$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe 3, Produkt -10): $\\left(x+5\\right)\\left(x-2\\right)=0$. Daraus liest man ab: $x_1=-5$, $x_2=2$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{-3\\pm\\sqrt{9-4\\cdot-10}}{2}$ und daraus $x_1=-5$, $x_2=2$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{299}{6}$ und $y_2=f(x_2)=-\\frac{22}{3}$.<br>\n$E_1 = \\left(-5, \\frac{299}{6}\\right)$ $E_2 = \\left(2, -\\frac{22}{3}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}-x^{2}-8x+4$", "Ableitung: $f'(x)=x^{2}-2x-8$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -2, Produkt -8): $\\left(x+2\\right)\\left(x-4\\right)=0$. Daraus liest man ab: $x_1=-2$, $x_2=4$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{2\\pm\\sqrt{4-4\\cdot-8}}{2}$ und daraus $x_1=-2$, $x_2=4$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{40}{3}$ und $y_2=f(x_2)=-\\frac{68}{3}$.<br>\n$E_1 = \\left(-2, \\frac{40}{3}\\right)$ $E_2 = \\left(4, -\\frac{68}{3}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}-x^{2}-8x-5$", "Ableitung: $f'(x)=x^{2}-2x-8$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -2, Produkt -8): $\\left(x+2\\right)\\left(x-4\\right)=0$. Daraus liest man ab: $x_1=-2$, $x_2=4$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{2\\pm\\sqrt{4-4\\cdot-8}}{2}$ und daraus $x_1=-2$, $x_2=4$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{13}{3}$ und $y_2=f(x_2)=-\\frac{95}{3}$.<br>\n$E_1 = \\left(-2, \\frac{13}{3}\\right)$ $E_2 = \\left(4, -\\frac{95}{3}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}-x^{2}-8x+4$", "Ableitung: $f'(x)=x^{2}-2x-8$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -2, Produkt -8): $\\left(x+2\\right)\\left(x-4\\right)=0$. Daraus liest man ab: $x_1=-2$, $x_2=4$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{2\\pm\\sqrt{4-4\\cdot-8}}{2}$ und daraus $x_1=-2$, $x_2=4$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{40}{3}$ und $y_2=f(x_2)=-\\frac{68}{3}$.<br>\n$E_1 = \\left(-2, \\frac{40}{3}\\right)$ $E_2 = \\left(4, -\\frac{68}{3}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}-\\frac{3}{2}x^{2}-10x-3$", "Ableitung: $f'(x)=x^{2}-3x-10$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -3, Produkt -10): $\\left(x+2\\right)\\left(x-5\\right)=0$. Daraus liest man ab: $x_1=-2$, $x_2=5$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{3\\pm\\sqrt{9-4\\cdot-10}}{2}$ und daraus $x_1=-2$, $x_2=5$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{25}{3}$ und $y_2=f(x_2)=-\\frac{293}{6}$.<br>\n$E_1 = \\left(-2, \\frac{25}{3}\\right)$ $E_2 = \\left(5, -\\frac{293}{6}\\right)$ "], ["$f(x)=\\frac{1}{3}x^{3}+\\frac{1}{2}x^{2}-6x-2$", "Ableitung: $f'(x)=x^{2}+x-6$.<br>\nNullstellen von $f'(x)$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe 1, Produkt -6): $\\left(x+3\\right)\\left(x-2\\right)=0$. Daraus liest man ab: $x_1=-3$, $x_2=2$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{-1\\pm\\sqrt{1-4\\cdot-6}}{2}$ und daraus $x_1=-3$, $x_2=2$.<br>\nDamit haben wir zwei Kandidaten für die Extremalstellen. (Dass es wirklich welche sind, könnte man mit der zweiten Ableitung überprüfen).<br>\nDie Extremalpunkte erhalten wir durch Einsetzen: $y_1=f(x_1) = \\frac{23}{2}$ und $y_2=f(x_2)=-\\frac{28}{3}$.<br>\n$E_1 = \\left(-3, \\frac{23}{2}\\right)$ $E_2 = \\left(2, -\\frac{28}{3}\\right)$ "]], |
| [["$84 \\cdot 2310 \\cdot 1100$", "$84 \\cdot 2310 \\cdot 1100 \\quad = \\quad 2^{2} \\cdot 3 \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 7 \\cdot 10 \\cdot 11 \\,\\, \\cdot \\,\\, 10^{2} \\cdot 11 \\quad = \\quad 2^{2} \\cdot 3 \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 7 \\cdot (2 \\cdot 5) \\cdot 11 \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{2} \\cdot 11 \\quad = \\quad 2^{2} \\cdot 3 \\cdot 7 \\,\\, \\cdot \\,\\, 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11 \\,\\, \\cdot \\,\\, 2^{2} \\cdot 5^{2} \\cdot 11 \\quad = \\quad 2^{5} \\cdot 3^{2} \\cdot 7^{2} \\cdot 11^{2}$"], ["$735 \\cdot 231000 \\cdot 2900$", "$735 \\cdot 231000 \\cdot 2900 \\quad = \\quad 3 \\cdot 5 \\cdot 7^{2} \\,\\, \\cdot \\,\\, 3 \\cdot 7 \\cdot 10^{3} \\cdot 11 \\,\\, \\cdot \\,\\, 10^{2} \\cdot 29 \\quad = \\quad 3 \\cdot 5 \\cdot 7^{2} \\,\\, \\cdot \\,\\, 3 \\cdot 7 \\cdot (2 \\cdot 5)^{3} \\cdot 11 \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{2} \\cdot 29 \\quad = \\quad 3 \\cdot 5 \\cdot 7^{2} \\,\\, \\cdot \\,\\, 2^{3} \\cdot 3 \\cdot 5^{3} \\cdot 7 \\cdot 11 \\,\\, \\cdot \\,\\, 2^{2} \\cdot 5^{2} \\cdot 29 \\quad = \\quad 2^{5} \\cdot 3^{2} \\cdot 5^{6} \\cdot 7^{3} \\cdot 11 \\cdot 29$"], ["$315 \\cdot 1650 \\cdot 290000$", "$315 \\cdot 1650 \\cdot 290000 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 5 \\cdot 10 \\cdot 11 \\,\\, \\cdot \\,\\, 10^{4} \\cdot 29 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 5 \\cdot (2 \\cdot 5) \\cdot 11 \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{4} \\cdot 29 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 2 \\cdot 3 \\cdot 5^{2} \\cdot 11 \\,\\, \\cdot \\,\\, 2^{4} \\cdot 5^{4} \\cdot 29 \\quad = \\quad 2^{5} \\cdot 3^{3} \\cdot 5^{7} \\cdot 7 \\cdot 11 \\cdot 29$"], ["$36 \\cdot 105000 \\cdot 130000$", "$36 \\cdot 105000 \\cdot 130000 \\quad = \\quad 2^{2} \\cdot 3^{2} \\,\\, \\cdot \\,\\, 3 \\cdot 5 \\cdot 7 \\cdot 10^{3} \\,\\, \\cdot \\,\\, 10^{4} \\cdot 13 \\quad = \\quad 2^{2} \\cdot 3^{2} \\,\\, \\cdot \\,\\, 3 \\cdot 5 \\cdot 7 \\cdot (2 \\cdot 5)^{3} \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{4} \\cdot 13 \\quad = \\quad 2^{2} \\cdot 3^{2} \\,\\, \\cdot \\,\\, 2^{3} \\cdot 3 \\cdot 5^{4} \\cdot 7 \\,\\, \\cdot \\,\\, 2^{4} \\cdot 5^{4} \\cdot 13 \\quad = \\quad 2^{9} \\cdot 3^{3} \\cdot 5^{8} \\cdot 7 \\cdot 13$"], ["$1225 \\cdot 4200 \\cdot 23000$", "$1225 \\cdot 4200 \\cdot 23000 \\quad = \\quad 5^{2} \\cdot 7^{2} \\,\\, \\cdot \\,\\, 2 \\cdot 3 \\cdot 7 \\cdot 10^{2} \\,\\, \\cdot \\,\\, 10^{3} \\cdot 23 \\quad = \\quad 5^{2} \\cdot 7^{2} \\,\\, \\cdot \\,\\, 2 \\cdot 3 \\cdot 7 \\cdot (2 \\cdot 5)^{2} \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{3} \\cdot 23 \\quad = \\quad 5^{2} \\cdot 7^{2} \\,\\, \\cdot \\,\\, 2^{3} \\cdot 3 \\cdot 5^{2} \\cdot 7 \\,\\, \\cdot \\,\\, 2^{3} \\cdot 5^{3} \\cdot 23 \\quad = \\quad 2^{6} \\cdot 3 \\cdot 5^{7} \\cdot 7^{3} \\cdot 23$"], ["$315 \\cdot 6300 \\cdot 1700$", "$315 \\cdot 6300 \\cdot 1700 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 3^{2} \\cdot 7 \\cdot 10^{2} \\,\\, \\cdot \\,\\, 10^{2} \\cdot 17 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 3^{2} \\cdot 7 \\cdot (2 \\cdot 5)^{2} \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{2} \\cdot 17 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 2^{2} \\cdot 3^{2} \\cdot 5^{2} \\cdot 7 \\,\\, \\cdot \\,\\, 2^{2} \\cdot 5^{2} \\cdot 17 \\quad = \\quad 2^{4} \\cdot 3^{4} \\cdot 5^{5} \\cdot 7^{2} \\cdot 17$"], ["$315 \\cdot 23100 \\cdot 290000$", "$315 \\cdot 23100 \\cdot 290000 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 7 \\cdot 10^{2} \\cdot 11 \\,\\, \\cdot \\,\\, 10^{4} \\cdot 29 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 7 \\cdot (2 \\cdot 5)^{2} \\cdot 11 \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{4} \\cdot 29 \\quad = \\quad 3^{2} \\cdot 5 \\cdot 7 \\,\\, \\cdot \\,\\, 2^{2} \\cdot 3 \\cdot 5^{2} \\cdot 7 \\cdot 11 \\,\\, \\cdot \\,\\, 2^{4} \\cdot 5^{4} \\cdot 29 \\quad = \\quad 2^{6} \\cdot 3^{3} \\cdot 5^{7} \\cdot 7^{2} \\cdot 11 \\cdot 29$"], ["$84 \\cdot 16500 \\cdot 1100$", "$84 \\cdot 16500 \\cdot 1100 \\quad = \\quad 2^{2} \\cdot 3 \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 5 \\cdot 10^{2} \\cdot 11 \\,\\, \\cdot \\,\\, 10^{2} \\cdot 11 \\quad = \\quad 2^{2} \\cdot 3 \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 5 \\cdot (2 \\cdot 5)^{2} \\cdot 11 \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{2} \\cdot 11 \\quad = \\quad 2^{2} \\cdot 3 \\cdot 7 \\,\\, \\cdot \\,\\, 2^{2} \\cdot 3 \\cdot 5^{3} \\cdot 11 \\,\\, \\cdot \\,\\, 2^{2} \\cdot 5^{2} \\cdot 11 \\quad = \\quad 2^{6} \\cdot 3^{2} \\cdot 5^{5} \\cdot 7 \\cdot 11^{2}$"], ["$126 \\cdot 10500 \\cdot 7000$", "$126 \\cdot 10500 \\cdot 7000 \\quad = \\quad 2 \\cdot 3^{2} \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 5 \\cdot 7 \\cdot 10^{2} \\,\\, \\cdot \\,\\, 7 \\cdot 10^{3} \\quad = \\quad 2 \\cdot 3^{2} \\cdot 7 \\,\\, \\cdot \\,\\, 3 \\cdot 5 \\cdot 7 \\cdot (2 \\cdot 5)^{2} \\,\\, \\cdot \\,\\, 7 \\cdot (2 \\cdot 5)^{3} \\quad = \\quad 2 \\cdot 3^{2} \\cdot 7 \\,\\, \\cdot \\,\\, 2^{2} \\cdot 3 \\cdot 5^{3} \\cdot 7 \\,\\, \\cdot \\,\\, 2^{3} \\cdot 5^{3} \\cdot 7 \\quad = \\quad 2^{6} \\cdot 3^{3} \\cdot 5^{6} \\cdot 7^{3}$"], ["$735 \\cdot 4200 \\cdot 13000$", "$735 \\cdot 4200 \\cdot 13000 \\quad = \\quad 3 \\cdot 5 \\cdot 7^{2} \\,\\, \\cdot \\,\\, 2 \\cdot 3 \\cdot 7 \\cdot 10^{2} \\,\\, \\cdot \\,\\, 10^{3} \\cdot 13 \\quad = \\quad 3 \\cdot 5 \\cdot 7^{2} \\,\\, \\cdot \\,\\, 2 \\cdot 3 \\cdot 7 \\cdot (2 \\cdot 5)^{2} \\,\\, \\cdot \\,\\, (2 \\cdot 5)^{3} \\cdot 13 \\quad = \\quad 3 \\cdot 5 \\cdot 7^{2} \\,\\, \\cdot \\,\\, 2^{3} \\cdot 3 \\cdot 5^{2} \\cdot 7 \\,\\, \\cdot \\,\\, 2^{3} \\cdot 5^{3} \\cdot 13 \\quad = \\quad 2^{6} \\cdot 3^{2} \\cdot 5^{6} \\cdot 7^{3} \\cdot 13$"]], | " <hr> "); |
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| <div id="exoprimfaktorzerlegung"></div> | <div id="exoextrema3tengrades"></div> |
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| <div id="solprimfaktorzerlegung"></div> | <div id="solextrema3tengrades"></div> |
| | <div style='font-size:12px;color:gray;'>ruby extremalstellen-von-polynom-3ten-grades.rb 1</div> |
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| | ==== 29. April 2024 bis 3. Mai 2024 ==== |
| ==== 27. September 2021 bis 1. Oktober 2021 ==== | === Dienstag 30. April 2024 === |
| === Donnerstag 30. September 2021 === | Die folgenden Funktionen haben genau zwei Wendestellenkandidaten. Bestimmen Sie diese.<JS>miniAufgabe("#exowende4tengrades","#solwende4tengrades", |
| Prüfung, keine Miniaufgabe | [["$f(x)=x^{4}+4x^{3}-90x^{2}+2x-2$", "$f'(x)=4x^{3}+12x^{2}-180x+2$<br>\n$f''(x) = 12x^{2}+24x-180 = 12\\left(x^{2}+2x-15\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}+2x-15=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe 2, Produkt -15): $\\left(x+5\\right)\\left(x-3\\right)=0$. Daraus liest man ab: $x_1=-5$, $x_2=3$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{-2\\pm\\sqrt{4-4\\cdot-15}}{2}$ und daraus $x_1=-5$, $x_2=3$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}-2x^{3}-120x^{2}+2x-3$", "$f'(x)=4x^{3}-6x^{2}-240x+2$<br>\n$f''(x) = 12x^{2}-12x-240 = 12\\left(x^{2}-x-20\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}-x-20=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -1, Produkt -20): $\\left(x+4\\right)\\left(x-5\\right)=0$. Daraus liest man ab: $x_1=-4$, $x_2=5$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{1\\pm\\sqrt{1-4\\cdot-20}}{2}$ und daraus $x_1=-4$, $x_2=5$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}-2x^{3}-120x^{2}+2x-3$", "$f'(x)=4x^{3}-6x^{2}-240x+2$<br>\n$f''(x) = 12x^{2}-12x-240 = 12\\left(x^{2}-x-20\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}-x-20=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -1, Produkt -20): $\\left(x+4\\right)\\left(x-5\\right)=0$. Daraus liest man ab: $x_1=-4$, $x_2=5$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{1\\pm\\sqrt{1-4\\cdot-20}}{2}$ und daraus $x_1=-4$, $x_2=5$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}-4x^{3}-48x^{2}+3x+3$", "$f'(x)=4x^{3}-12x^{2}-96x+3$<br>\n$f''(x) = 12x^{2}-24x-96 = 12\\left(x^{2}-2x-8\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}-2x-8=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -2, Produkt -8): $\\left(x+2\\right)\\left(x-4\\right)=0$. Daraus liest man ab: $x_1=-2$, $x_2=4$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{2\\pm\\sqrt{4-4\\cdot-8}}{2}$ und daraus $x_1=-2$, $x_2=4$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}-2x^{3}-72x^{2}-2x-2$", "$f'(x)=4x^{3}-6x^{2}-144x-2$<br>\n$f''(x) = 12x^{2}-12x-144 = 12\\left(x^{2}-x-12\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}-x-12=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -1, Produkt -12): $\\left(x+3\\right)\\left(x-4\\right)=0$. Daraus liest man ab: $x_1=-3$, $x_2=4$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{1\\pm\\sqrt{1-4\\cdot-12}}{2}$ und daraus $x_1=-3$, $x_2=4$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}+6x^{3}-60x^{2}+3x-5$", "$f'(x)=4x^{3}+18x^{2}-120x+3$<br>\n$f''(x) = 12x^{2}+36x-120 = 12\\left(x^{2}+3x-10\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}+3x-10=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe 3, Produkt -10): $\\left(x+5\\right)\\left(x-2\\right)=0$. Daraus liest man ab: $x_1=-5$, $x_2=2$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{-3\\pm\\sqrt{9-4\\cdot-10}}{2}$ und daraus $x_1=-5$, $x_2=2$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}-2x^{3}-36x^{2}-5x-3$", "$f'(x)=4x^{3}-6x^{2}-72x-5$<br>\n$f''(x) = 12x^{2}-12x-72 = 12\\left(x^{2}-x-6\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}-x-6=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -1, Produkt -6): $\\left(x+2\\right)\\left(x-3\\right)=0$. Daraus liest man ab: $x_1=-2$, $x_2=3$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{1\\pm\\sqrt{1-4\\cdot-6}}{2}$ und daraus $x_1=-2$, $x_2=3$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}-2x^{3}-120x^{2}-3x-2$", "$f'(x)=4x^{3}-6x^{2}-240x-3$<br>\n$f''(x) = 12x^{2}-12x-240 = 12\\left(x^{2}-x-20\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}-x-20=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -1, Produkt -20): $\\left(x+4\\right)\\left(x-5\\right)=0$. Daraus liest man ab: $x_1=-4$, $x_2=5$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{1\\pm\\sqrt{1-4\\cdot-20}}{2}$ und daraus $x_1=-4$, $x_2=5$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}+4x^{3}-48x^{2}+5x+5$", "$f'(x)=4x^{3}+12x^{2}-96x+5$<br>\n$f''(x) = 12x^{2}+24x-96 = 12\\left(x^{2}+2x-8\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}+2x-8=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe 2, Produkt -8): $\\left(x+4\\right)\\left(x-2\\right)=0$. Daraus liest man ab: $x_1=-4$, $x_2=2$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{-2\\pm\\sqrt{4-4\\cdot-8}}{2}$ und daraus $x_1=-4$, $x_2=2$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"], ["$f(x)=x^{4}-6x^{3}-60x^{2}-5x-3$", "$f'(x)=4x^{3}-18x^{2}-120x-5$<br>\n$f''(x) = 12x^{2}-36x-120 = 12\\left(x^{2}-3x-10\\right)$.<br>\nNullstellen: $f''(x)=0 \\Leftrightarrow x^{2}-3x-10=0$: Entweder faktorisieren oder Mitternachtsformel:<br>\nFaktorisiert (Summe -3, Produkt -10): $\\left(x+2\\right)\\left(x-5\\right)=0$. Daraus liest man ab: $x_1=-2$, $x_2=5$.<br>\nOder Mitternachtsformel: $\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a} = \\frac{3\\pm\\sqrt{9-4\\cdot-10}}{2}$ und daraus $x_1=-2$, $x_2=5$.<br>\nDamit haben wir zwei Kandidaten für die Wendestellen. (Dass es wirklich welche sind, könnte man mit der dritten Ableitung überprüfen).<br>\n"]], |
| === Freitag 1. Oktober 2021 === | " <hr> "); |
| | |
| Schreiben Sie den Term einmal als Binärbaum und einmal in Computer-Notation.<JS>miniAufgabe("#exotermanalyse1","#soltermanalyse1", | |
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stroke-width=\".3985\" transform=\"matrix(1 0 0 -1 239.924 134.599)\"/></g></svg> <tt>u*(k*y)+(u/(u/s)-y^s)</tt>"]], | |
| " <hr> ", " <hr> "); | |
| </JS> | </JS> |
| <HTML> | <HTML> |
| <div id="exotermanalyse1"></div> | <div id="exowende4tengrades"></div> |
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| </HTML> | </HTML> |
| <hidden Lösungen> | <hidden Lösungen> |
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| <HTML> | <HTML> |
| <div id="soltermanalyse1"></div> | <div id="solwende4tengrades"></div> |
| | <div style='font-size:12px;color:gray;'>ruby extremalstellen-von-polynom-3ten-grades.rb 2</div> |
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| |
| | === Mittwoch 1. Mai 2024 === |
| ==== Aufgaben vom aktuellen Jahr ==== | ==== Aufgaben vom aktuellen Jahr ==== |
| * [[lehrkraefte:blc:miniaufgaben:kw33-2021|KW33, 16. August 2021: Bruchrechnen]] | * [[lehrkraefte:blc:miniaufgaben:kw18-2024|KW18, 29. April 2024: ]] |
| * [[lehrkraefte:blc:miniaufgaben:kw34-2021|KW34, 23. August 2021: Lineare Gleichungen mit Bruchkoeffizienten, Multiplikationen von Termen.]] | * [[lehrkraefte:blc:miniaufgaben:kw17-2024|KW17, 22. April 2024: ]] |
| * [[lehrkraefte:blc:miniaufgaben:kw35-2021|KW35, 30. August 2021: Ausklammern]] | * [[lehrkraefte:blc:miniaufgaben:kw13-2024|KW13, 25. März 2024: Produkt- und Kettenregel auf Polynomterme anweden. Quotienten- und Kettenregel auf Polynomterme anweden.]] |
| * [[lehrkraefte:blc:miniaufgaben:kw37-2021|KW37, 13. September 2021: Potenzen, Produkte und Quotienten von ganzen Zahlen. Lineare Gleichungen mit Brüchen]] | * [[lehrkraefte:blc:miniaufgaben:kw12-2024|KW12, 18. März 2024: Terme als Baum und Computernotation notieren]] |
| * [[lehrkraefte:blc:miniaufgaben:kw38-2021|KW38, 20. September 2021: ]] | * [[lehrkraefte:blc:miniaufgaben:kw10-2024|KW10, 4. März 2024: Ableiten mit Ketten- und Produktregel, Ableiten mit Ketten- und Quotientenregel]] |
| * [[lehrkraefte:blc:miniaufgaben:kw39-2021|KW39, 27. September 2021: ]] | * [[lehrkraefte:blc:miniaufgaben:kw09-2024|KW9, 26. Februar 2024: Ableiten mit Kettenregel, Ableiten mit Produktregel]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw08-2024|KW8, 19. Februar 2024: $f'(x)=f(x)\cdot f'(0)$ für $f(x)=a^x$ zeigen, Funktionen als Verknüpfung zweier Funktionen schreiben.]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw07-2024|KW7, 12. Februar 2024: $x^2$ und $x^3$ mit Grenzwert ableiten, Polynome mit Regeln ableiten.]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw06-2024|KW6, 5. Februar 2024: Grafisch ableiten.]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw03-2024|KW3, 15. Januar 2024: Logarithmusgleichungen mit nötigem Basiswechsel]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw02-2024|KW2, 8. Januar 2024: Logarithmengesetze anwenden, Logarthmusgleichung lösen, die auf eine quadratische Gleichung führt]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw51-2023|KW51, 18. Dezember 2023: Logarithmusfunktionen ablesen, Exponentialgleichungen durch Logarthmieren lösen]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw50-2023|KW50, 11. Dezember 2023: Einfache Exponentialgleichungen von Hand ohne Logarithmen, Einfache Logarithmen von Hand]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw49-2023|KW49, 4. Dezember 2023: Exponentialfunktionen ablesen, Exponentialfunktion aus Text]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw48-2023|KW48, 27. November 2023: Wertetabellen von Potenzfunktionen mit rationalen Basen, Funktionsgraphen transformieren einfach]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw47-2023|KW47, 20. November 2023: Potenzgesetze in $\mathbb{N}$ beweisen, Potenzgesetze in Vereinfachungen anwenden.]] |
| | * KW46, 13. November 2023: Keine Miniaufgaben |
| | * [[lehrkraefte:blc:miniaufgaben:kw45-2023|KW45, 6. November 2023: Arithmetische Reihe berechnen, $a_0$, $a_1$ als quadratische Polynome gegben, berechne $a_2$.]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw44-2023|KW44, 30. Oktober 2023: Summenzeichen ausschreiben, Implizite Teilsummen von AF und AG mit Summenzeichen schreiben.]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw43-2023|KW43, 23. Oktober 2023: GF oder AF aus drei Gliedern bestimmen (mit Bruchzahlen)]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw39-2023|KW39, 25. September 2023: Parameter von AF aus zwei Gliedern, Parameter von GF aus zwei Gliedern]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw38-2023|KW38, 18. September 2023: Ganzzahlige Potenzen auswendig lernen, AF/GF implizit zu explizit]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw37-2023|KW37, 11. September 2023: Strecke zu gleichseitigem Dreieck ergänzen.]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw36-2023|KW36, 4. September 2023: Vektoren auf gewünschte Länge skalieren (mit Brüchen), Strecke zum Quadrat ergänzen.]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw35-2023|KW35, 28. August 2023: Länge von Vektoren in Normalform]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw34-2023|KW34, 21. August 2023: POV-Ray Code für Rotationen und Translation eines orientierten Torus produzieren, Gleichmässige Bewegung beschreiben, in mathematischer Notation und POV-Ray Code]] |
| | * [[lehrkraefte:blc:miniaufgaben:kw33-2023|KW33, 14. August 2023: Kugeln, Zylinder und Kegel in POV-Ray Syntax beschreiben]] |
| |
| |
| === Ältere Aufgaben === | === Ältere Aufgaben === |
| | * [[lehrkraefte:blc:miniaufgaben:zweite-klasse22-23|Aufgaben vom 2. Jahr 22/23]] |
| | * [[lehrkraefte:blc:miniaufgaben:erste-klasse21-22|Aufgaben vom 1. Jahr 21/22]] |
| * [[lehrkraefte:blc:miniaufgaben:vierte-klasse19-20|Aufgaben vom 4. Jahr 19/20]] | * [[lehrkraefte:blc:miniaufgaben:vierte-klasse19-20|Aufgaben vom 4. Jahr 19/20]] |
| * [[lehrkraefte:blc:miniaufgaben:vierte-klasse18-19|Aufgaben vom 4. Jahr 18/19]] | * [[lehrkraefte:blc:miniaufgaben:vierte-klasse18-19|Aufgaben vom 4. Jahr 18/19]] |