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lehrkraefte:blc:miniaufgaben [2024/03/18 14:30]
Ivo Blöchliger [Aufgaben vom aktuellen Jahr] |
lehrkraefte:blc:miniaufgaben [2024/05/01 07:25] (current)
Ivo Blöchliger [29. April 2024 bis 3. Mai 2024] |
| </PRELOAD> | </PRELOAD> |
| |
| ==== 18. März 2024 bis 22. März 2024 ==== | ==== 29. April 2024 bis 3. Mai 2024 ==== |
| === Dienstag 19. März 2024 === | === Dienstag 30. April 2024 === |
| Maturaarbeitspräsentationen. Unterrichtsausfall | Die folgenden Funktionen haben genau zwei Wendestellenkandidaten. Bestimmen Sie diese.<JS>miniAufgabe("#exowende4tengrades","#solwende4tengrades", |
| === Mittwoch 20. März 2024 === | [["$f(x)=x^{4}+4x^{3}-90x^{2}+2x-2$", "$f'(x)=4x^{3}+12x^{2}-180x+2$<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"]], |
| Schreiben Sie den Term einmal als Binärbaum und einmal in Computer-Notation.<JS>miniAufgabe("#exotermanalyse1","#soltermanalyse1", | " <hr> "); |
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| " <hr> ", " <hr> "); | |
| </JS> | </JS> |
| <HTML> | <HTML> |
| <div id="exotermanalyse1"></div> | <div id="exowende4tengrades"></div> |
| |
| </HTML> | </HTML> |
| |
| <HTML> | <HTML> |
| <div id="soltermanalyse1"></div> | <div id="solwende4tengrades"></div> |
| <div style='font-size:12px;color:gray;'>ruby termanalyse.rb 1</div> | <div style='font-size:12px;color:gray;'>ruby extremalstellen-von-polynom-3ten-grades.rb 2</div> |
| </HTML> | </HTML> |
| |
| </hidden> | </hidden> |
| |
| ==== 25. März 2024 bis 29. März 2024 ==== | |
| === Dienstag 26. März 2024 === | === Mittwoch 1. Mai 2024 === |
| Leiten Sie ohne Hilfsmittel ab. Klammern Sie danach gemeinsame Faktoren. Weitere Vereinfachungen sind nicht nötig. | Eine Funktion 3. Grades hat die Form $f(x)=ax^3 + bx^2 + cx + d$ mit $a,b,c,d \in \mathbb{R}$ und $a\neq 0$. |
| <JS>miniAufgabe("#exoprod_ketten_nur_poly","#solprod_ketten_nur_poly", | |
| [["$\\left(8x+8\\right)^{36} \\cdot \\left(3x^{6}+3\\right)^{27}$", "$$\\begin{multline*}\\left(\\left(8x+8\\right)^{36} \\cdot \\left(3x^{6}+3\\right)^{27}\\right)' = \\\\\n36\\cdot \\left(8x+8\\right)^{35}\\cdot \\left(8)\\right) \\cdot \\left(3x^{6}+3\\right)^{27} + \\left(8x+8\\right)^{36} \\cdot 27 \\cdot \\left(3x^{6}+3\\right)^{26} \\cdot \\left(18x^{5})\\right) = \\\\\n9 \\cdot \\left(8x+8\\right)^{35} \\cdot \\left(3x^{6}+3\\right)^{26} \\cdot \\left[ 4 \\cdot \\left(8\\right)\\cdot\\left(3x^{6}+3\\right) + 3 \\cdot \\left(8x+8\\right) \\cdot \\left(18x^{5}\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(-8x+11\\right)^{30} \\cdot \\left(4x^{4}+4x\\right)^{25}$", "$$\\begin{multline*}\\left(\\left(-8x+11\\right)^{30} \\cdot \\left(4x^{4}+4x\\right)^{25}\\right)' = \\\\\n30\\cdot \\left(-8x+11\\right)^{29}\\cdot \\left(-8)\\right) \\cdot \\left(4x^{4}+4x\\right)^{25} + \\left(-8x+11\\right)^{30} \\cdot 25 \\cdot \\left(4x^{4}+4x\\right)^{24} \\cdot \\left(16x^{3}+4)\\right) = \\\\\n5 \\cdot \\left(-8x+11\\right)^{29} \\cdot \\left(4x^{4}+4x\\right)^{24} \\cdot \\left[ 6 \\cdot \\left(-8\\right)\\cdot\\left(4x^{4}+4x\\right) + 5 \\cdot \\left(-8x+11\\right) \\cdot \\left(16x^{3}+4\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(-2x^{4}-2\\right)^{12} \\cdot \\left(9x-13\\right)^{42}$", "$$\\begin{multline*}\\left(\\left(-2x^{4}-2\\right)^{12} \\cdot \\left(9x-13\\right)^{42}\\right)' = \\\\\n12\\cdot \\left(-2x^{4}-2\\right)^{11}\\cdot \\left(-8x^{3})\\right) \\cdot \\left(9x-13\\right)^{42} + \\left(-2x^{4}-2\\right)^{12} \\cdot 42 \\cdot \\left(9x-13\\right)^{41} \\cdot \\left(9)\\right) = \\\\\n6 \\cdot \\left(-2x^{4}-2\\right)^{11} \\cdot \\left(9x-13\\right)^{41} \\cdot \\left[ 2 \\cdot \\left(-8x^{3}\\right)\\cdot\\left(9x-13\\right) + 7 \\cdot \\left(-2x^{4}-2\\right) \\cdot \\left(9\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(-7x^{6}-7x\\right)^{18} \\cdot \\left(7x-12\\right)^{27}$", "$$\\begin{multline*}\\left(\\left(-7x^{6}-7x\\right)^{18} \\cdot \\left(7x-12\\right)^{27}\\right)' = \\\\\n18\\cdot \\left(-7x^{6}-7x\\right)^{17}\\cdot \\left(-42x^{5}-7)\\right) \\cdot \\left(7x-12\\right)^{27} + \\left(-7x^{6}-7x\\right)^{18} \\cdot 27 \\cdot \\left(7x-12\\right)^{26} \\cdot \\left(7)\\right) = \\\\\n9 \\cdot \\left(-7x^{6}-7x\\right)^{17} \\cdot \\left(7x-12\\right)^{26} \\cdot \\left[ 2 \\cdot \\left(-42x^{5}-7\\right)\\cdot\\left(7x-12\\right) + 3 \\cdot \\left(-7x^{6}-7x\\right) \\cdot \\left(7\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(6x+7\\right)^{36} \\cdot \\left(4x^{4}-3\\right)^{8}$", "$$\\begin{multline*}\\left(\\left(6x+7\\right)^{36} \\cdot \\left(4x^{4}-3\\right)^{8}\\right)' = \\\\\n36\\cdot \\left(6x+7\\right)^{35}\\cdot \\left(6)\\right) \\cdot \\left(4x^{4}-3\\right)^{8} + \\left(6x+7\\right)^{36} \\cdot 8 \\cdot \\left(4x^{4}-3\\right)^{7} \\cdot \\left(16x^{3})\\right) = \\\\\n4 \\cdot \\left(6x+7\\right)^{35} \\cdot \\left(4x^{4}-3\\right)^{7} \\cdot \\left[ 9 \\cdot \\left(6\\right)\\cdot\\left(4x^{4}-3\\right) + 2 \\cdot \\left(6x+7\\right) \\cdot \\left(16x^{3}\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(-9x-4\\right)^{33} \\cdot \\left(7x^{4}-13x\\right)^{22}$", "$$\\begin{multline*}\\left(\\left(-9x-4\\right)^{33} \\cdot \\left(7x^{4}-13x\\right)^{22}\\right)' = \\\\\n33\\cdot \\left(-9x-4\\right)^{32}\\cdot \\left(-9)\\right) \\cdot \\left(7x^{4}-13x\\right)^{22} + \\left(-9x-4\\right)^{33} \\cdot 22 \\cdot \\left(7x^{4}-13x\\right)^{21} \\cdot \\left(28x^{3}-13)\\right) = \\\\\n11 \\cdot \\left(-9x-4\\right)^{32} \\cdot \\left(7x^{4}-13x\\right)^{21} \\cdot \\left[ 3 \\cdot \\left(-9\\right)\\cdot\\left(7x^{4}-13x\\right) + 2 \\cdot \\left(-9x-4\\right) \\cdot \\left(28x^{3}-13\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(10x+11\\right)^{28} \\cdot \\left(-9x^{5}-8x\\right)^{8}$", "$$\\begin{multline*}\\left(\\left(10x+11\\right)^{28} \\cdot \\left(-9x^{5}-8x\\right)^{8}\\right)' = \\\\\n28\\cdot \\left(10x+11\\right)^{27}\\cdot \\left(10)\\right) \\cdot \\left(-9x^{5}-8x\\right)^{8} + \\left(10x+11\\right)^{28} \\cdot 8 \\cdot \\left(-9x^{5}-8x\\right)^{7} \\cdot \\left(-45x^{4}-8)\\right) = \\\\\n4 \\cdot \\left(10x+11\\right)^{27} \\cdot \\left(-9x^{5}-8x\\right)^{7} \\cdot \\left[ 7 \\cdot \\left(10\\right)\\cdot\\left(-9x^{5}-8x\\right) + 2 \\cdot \\left(10x+11\\right) \\cdot \\left(-45x^{4}-8\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(7x^{5}-9x\\right)^{10} \\cdot \\left(-7x-4\\right)^{15}$", "$$\\begin{multline*}\\left(\\left(7x^{5}-9x\\right)^{10} \\cdot \\left(-7x-4\\right)^{15}\\right)' = \\\\\n10\\cdot \\left(7x^{5}-9x\\right)^{9}\\cdot \\left(35x^{4}-9)\\right) \\cdot \\left(-7x-4\\right)^{15} + \\left(7x^{5}-9x\\right)^{10} \\cdot 15 \\cdot \\left(-7x-4\\right)^{14} \\cdot \\left(-7)\\right) = \\\\\n5 \\cdot \\left(7x^{5}-9x\\right)^{9} \\cdot \\left(-7x-4\\right)^{14} \\cdot \\left[ 2 \\cdot \\left(35x^{4}-9\\right)\\cdot\\left(-7x-4\\right) + 3 \\cdot \\left(7x^{5}-9x\\right) \\cdot \\left(-7\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(7x^{4}-5\\right)^{10} \\cdot \\left(9x-2\\right)^{35}$", "$$\\begin{multline*}\\left(\\left(7x^{4}-5\\right)^{10} \\cdot \\left(9x-2\\right)^{35}\\right)' = \\\\\n10\\cdot \\left(7x^{4}-5\\right)^{9}\\cdot \\left(28x^{3})\\right) \\cdot \\left(9x-2\\right)^{35} + \\left(7x^{4}-5\\right)^{10} \\cdot 35 \\cdot \\left(9x-2\\right)^{34} \\cdot \\left(9)\\right) = \\\\\n5 \\cdot \\left(7x^{4}-5\\right)^{9} \\cdot \\left(9x-2\\right)^{34} \\cdot \\left[ 2 \\cdot \\left(28x^{3}\\right)\\cdot\\left(9x-2\\right) + 7 \\cdot \\left(7x^{4}-5\\right) \\cdot \\left(9\\right) \\right] \\\\\n\\end{multline*}$$"], ["$\\left(-12x+8\\right)^{42} \\cdot \\left(6x^{5}-12\\right)^{18}$", "$$\\begin{multline*}\\left(\\left(-12x+8\\right)^{42} \\cdot \\left(6x^{5}-12\\right)^{18}\\right)' = \\\\\n42\\cdot \\left(-12x+8\\right)^{41}\\cdot \\left(-12)\\right) \\cdot \\left(6x^{5}-12\\right)^{18} + \\left(-12x+8\\right)^{42} \\cdot 18 \\cdot \\left(6x^{5}-12\\right)^{17} \\cdot \\left(30x^{4})\\right) = \\\\\n6 \\cdot \\left(-12x+8\\right)^{41} \\cdot \\left(6x^{5}-12\\right)^{17} \\cdot \\left[ 7 \\cdot \\left(-12\\right)\\cdot\\left(6x^{5}-12\\right) + 3 \\cdot \\left(-12x+8\\right) \\cdot \\left(30x^{4}\\right) \\right] \\\\\n\\end{multline*}$$"]], | Erklären Sie, warum eine Funktion 3. Grades |
| " <br><hr> "); | * a) mindestens eine Nullstelle haben muss. |
| | * <del>b) entweder genau 2 oder keine lokale Extrema hat.</del> |
| | * c) immer genau eine Wendestelle hat. |
| | |
| | <hidden Lösungsvorschlag> |
| | * a) Für betragsmässig genug grosse $x$ dominiert der Term $ax^3$ alle anderen Terme der Funktion. D.h. für $x \to \infty$ hat $f(x)$ das gleiche Vorzeichen wie $a$, für $x \to -\infty$ das entgegengesetzte Vorzeichen. Die Funktion ist stetig, d.h. der Funktionsgraph macht keine Sprünge und hat keine Lücken. Da der Funktionsgraph für sehr kleine $x$ und sehr grosse $x$ einmal oberhalb und einmal unterhalb der $x$-Achse verläuft, muss er die $x$-Achse dazwischen mindestens einmal schneiden, d.h. die Funktion muss eine Nullstelle haben. |
| | * b) Die Ableitung ist eine quadratische Funktion, die genau 2, eine oder keine Nullstellen hat. |
| | * Keine Nullstellen, heisst keine Extrema. |
| | * Genau eine Nullstelle heisst, die Ableitung hat die Form $f'(x)=u\cdot(x-v)^2$, mit $v$ als «doppelter» Nullstelle (mit $u\neq 0$). Damit ist die zweite Ableitung $f''(x)=2u \cdot (x-v)$ und damit ist $f''(v)=0$ und $v$ ein Wendestellenkandidat. Weiter ist $f'''(x)=2u \neq 0$, womit wir eine echte Wendestelle mit horizontaler Tangente haben, also ein Sattelpunkt und somit keine Extremalstelle. |
| | * Zwei Nullstellen heisst, die Ableitung hat als quadratische Funktion die Form $f'(x)=u\cdot(x-v)(x-w) = u(x^2-(v+w)x+vw)$ mit $v \neq w$ den Nullstellen und $u \neq 0$. Die zweite Ableitung ist $f''(x) = u \cdot (2x-(v+w))$ und damit $f''(v)=u(2v-v-w) = u(w-v) \neq 0$ und $f''(w)=u(2w-v-w)=u(v-w) \neq 0$. Damit sind $v$ und $w$ zwei «echte» Extremalstellen von $f$. |
| | * c) Die zweite Ableitung ist $f''(x)= 6ax+2b$ und hat genau eine Nullstelle, nämlich $-\frac{b}{3a}$, die immer existiert (wegen $a\neq 0$). Die dritte Ableitung ist konstant $f'''(x)=6a \neq 0$, womit eine Wendestelle vorliegt. |
| | </hidden> |
| | ==== 6. Mai 2024 bis 10. Mai 2024 ==== |
| | === Dienstag 7. Mai 2024 === |
| | Mit Hilfe des TR, berechnen Sie die Nullstellen, Extremalstellenkandidaten und Wendestellenkandidaten. |
| | Machen Sie eine Tabelle mit diesen $x$-Werten sowie die zugehörigen $y$-Werte und Steigungen. |
| | Für Extremalstellenkandidaten notieren Sie zusätzlich das Vorzeichen der zweiten Ableitung. |
| | Skizzieren Sie dann mit diesen Informationen den Graphen. |
| | <JS>miniAufgabe("#exokurvendiskussionMitTRquartic","#solkurvendiskussionMitTRquartic", |
| | [["$f(x) = -\\frac{1}{48}\\left(3x^{4}+4x^{3}-36x^{2}+0+94\\right)$", "<table style='border: 1px solid black;'><tr><td> $x$ </td><td> -3.88 </td><td> -3.00 </td><td> -1.78 </td><td> -1.65 </td><td> 0.00 </td><td> 1.11 </td><td> 2.00 </td></tr>\n<tr><td>$f(x)$ </td><td> 0.00 </td><td> 1.97 </td><td> .27 </td><td> -.00 </td><td> -1.95 </td><td> -1.23 </td><td> -.62 </td></tr>\n<tr><td>$f'(x)$ </td><td> 5.07 </td><td> 0.00 </td><td> -2.05 </td><td> -2.03 </td><td> 0.00 </td><td> 1.01 </td><td> 0.00 </td></tr>\n<tr><td>$f''(x)$ </td><td> -7.88 </td><td> -3.75 </td><td> -.00 </td><td> .27 </td><td> 1.50 </td><td> -.00 </td><td> -2.50 </td></tr></table>\n\n<br><svg height=\"197.02667\" viewBox=\"0 0 302.69333 197.02667\" width=\"302.69333\" xmlns=\"http://www.w3.org/2000/svg\"><g transform=\"matrix(.13333333 0 0 -.13333333 0 197.02667)\"><path d=\"m65.4844 74.9336v1259.7264\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m380.418 74.9336v1259.7264\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m695.348 74.9336v1259.7264\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m1010.28 74.9336v1259.7264\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m1640.14 74.9336v1259.7264\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m1955.07 74.9336v1259.7264\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m65.4844 74.9336h1889.5856\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m65.4844 389.863h1889.5856\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m65.4844 1019.73h1889.5856\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m65.4844 1334.66h1889.5856\" style=\"fill:none;stroke:#808080;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m65.4844 717.391v-25.192\" style=\"fill:none;stroke:#000;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m27.1836 619.582h76.6054v49.211h-76.6054z\" fill=\"#fff\"/><path d=\"m27.1836 619.582h76.6054v49.211h-76.6054z\" style=\"fill:none;stroke:#fff;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m380.418 717.391v-25.192\" style=\"fill:none;stroke:#000;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m342.113 619.582h76.606v49.211h-76.606z\" fill=\"#fff\"/><path d=\"m342.113 619.582h76.606v49.211h-76.606z\" style=\"fill:none;stroke:#fff;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m695.348 717.391v-25.192\" style=\"fill:none;stroke:#000;stroke-width:5;stroke-linecap:round;stroke-linejoin:round;stroke-miterlimit:10\"/><path d=\"m657.043 619.582h76.605v49.211h-76.605z\" fill=\"#fff\"/><path d=\"m657.043 619.582h76.605v49.211h-76.605z\" style=\" |