miniaufgabe.js ==== 21. Oktober 2019 bis 25. Oktober 2019 ==== === Dienstag 22. Oktober 2019 === Der **TR (Ti-Nspire)** ist dabei und aufgeladen. Auf einen Bruchstrich, zusammenfassen, faktorisieren, kürzen.miniAufgabe("#exopoly_bruch_add_faktor","#solpoly_bruch_add_faktor", [["$\\displaystyle \\frac{9x^2+14x+8}{3x^2-12} + \\frac{4}{3x+6}$", "$\\begin{multline*} \\frac{9x^2+14x+8}{3x^2-12} + \\frac{4}{3x+6} = \\frac{9x^2+14x+8}{3(x^2-4)} + \\frac{4}{3(x+2)} = \\frac{9x^2+14x+8}{3(x+2)(x-2)} + \\frac{4(x-2)}{3(x+2)(x-2)} = \\\\ \\frac{9x^2+14x+8}{3(x^2-4)} + \\frac{4x-8}{3(x^2-4)} = \\frac{9x^2+18x}{3(x^2-4)} = \\frac{9x(x+2)}{3(x^2-4)} = \\\\ \\frac{3\\cdot3x(x+2)}{3(x^2-4)} = \\frac{3x(x+2)}{(x+2)(x-2)} = \\frac{3x}{x-2} \\end{multline*}$"], ["$\\displaystyle \\frac{20x^2-46x-12}{5x^2-20} + \\frac{6}{5x-10}$", "$\\begin{multline*} \\frac{20x^2-46x-12}{5x^2-20} + \\frac{6}{5x-10} = \\frac{20x^2-46x-12}{5(x^2-4)} + \\frac{6}{5(x-2)} = \\frac{20x^2-46x-12}{5(x+2)(x-2)} + \\frac{6(x+2)}{5(x+2)(x-2)} = \\\\ \\frac{20x^2-46x-12}{5(x^2-4)} + \\frac{6x+12}{5(x^2-4)} = \\frac{20x^2-40x}{5(x^2-4)} = \\frac{20x(x-2)}{5(x^2-4)} = \\\\ \\frac{5\\cdot4x(x-2)}{5(x^2-4)} = \\frac{4x(x-2)}{(x+2)(x-2)} = \\frac{4x}{x+2} \\end{multline*}$"], ["$\\displaystyle \\frac{8x^2+21x+9}{2x^2-18} + \\frac{3}{2x+6}$", "$\\begin{multline*} \\frac{8x^2+21x+9}{2x^2-18} + \\frac{3}{2x+6} = \\frac{8x^2+21x+9}{2(x^2-9)} + \\frac{3}{2(x+3)} = \\frac{8x^2+21x+9}{2(x+3)(x-3)} + \\frac{3(x-3)}{2(x+3)(x-3)} = \\\\ \\frac{8x^2+21x+9}{2(x^2-9)} + \\frac{3x-9}{2(x^2-9)} = \\frac{8x^2+24x}{2(x^2-9)} = \\frac{8x(x+3)}{2(x^2-9)} = \\\\ \\frac{2\\cdot4x(x+3)}{2(x^2-9)} = \\frac{4x(x+3)}{(x+3)(x-3)} = \\frac{4x}{x-3} \\end{multline*}$"], ["$\\displaystyle \\frac{9x^2-50x-25}{3x^2-75} + \\frac{5}{3x-15}$", "$\\begin{multline*} \\frac{9x^2-50x-25}{3x^2-75} + \\frac{5}{3x-15} = \\frac{9x^2-50x-25}{3(x^2-25)} + \\frac{5}{3(x-5)} = \\frac{9x^2-50x-25}{3(x+5)(x-5)} + \\frac{5(x+5)}{3(x+5)(x-5)} = \\\\ \\frac{9x^2-50x-25}{3(x^2-25)} + \\frac{5x+25}{3(x^2-25)} = \\frac{9x^2-45x}{3(x^2-25)} = \\frac{9x(x-5)}{3(x^2-25)} = \\\\ \\frac{3\\cdot3x(x-5)}{3(x^2-25)} = \\frac{3x(x-5)}{(x+5)(x-5)} = \\frac{3x}{x+5} \\end{multline*}$"], ["$\\displaystyle \\frac{20x^2+77x+12}{4x^2-64} + \\frac{3}{4x+16}$", "$\\begin{multline*} \\frac{20x^2+77x+12}{4x^2-64} + \\frac{3}{4x+16} = \\frac{20x^2+77x+12}{4(x^2-16)} + \\frac{3}{4(x+4)} = \\frac{20x^2+77x+12}{4(x+4)(x-4)} + \\frac{3(x-4)}{4(x+4)(x-4)} = \\\\ \\frac{20x^2+77x+12}{4(x^2-16)} + \\frac{3x-12}{4(x^2-16)} = \\frac{20x^2+80x}{4(x^2-16)} = \\frac{20x(x+4)}{4(x^2-16)} = \\\\ \\frac{4\\cdot5x(x+4)}{4(x^2-16)} = \\frac{5x(x+4)}{(x+4)(x-4)} = \\frac{5x}{x-4} \\end{multline*}$"], ["$\\displaystyle \\frac{10x^2-35x-15}{2x^2-18} + \\frac{5}{2x-6}$", "$\\begin{multline*} \\frac{10x^2-35x-15}{2x^2-18} + \\frac{5}{2x-6} = \\frac{10x^2-35x-15}{2(x^2-9)} + \\frac{5}{2(x-3)} = \\frac{10x^2-35x-15}{2(x+3)(x-3)} + \\frac{5(x+3)}{2(x+3)(x-3)} = \\\\ \\frac{10x^2-35x-15}{2(x^2-9)} + \\frac{5x+15}{2(x^2-9)} = \\frac{10x^2-30x}{2(x^2-9)} = \\frac{10x(x-3)}{2(x^2-9)} = \\\\ \\frac{2\\cdot5x(x-3)}{2(x^2-9)} = \\frac{5x(x-3)}{(x+3)(x-3)} = \\frac{5x}{x+3} \\end{multline*}$"], ["$\\displaystyle \\frac{15x^2+28x+4}{5x^2-20} + \\frac{2}{5x+10}$", "$\\begin{multline*} \\frac{15x^2+28x+4}{5x^2-20} + \\frac{2}{5x+10} = \\frac{15x^2+28x+4}{5(x^2-4)} + \\frac{2}{5(x+2)} = \\frac{15x^2+28x+4}{5(x+2)(x-2)} + \\frac{2(x-2)}{5(x+2)(x-2)} = \\\\ \\frac{15x^2+28x+4}{5(x^2-4)} + \\frac{2x-4}{5(x^2-4)} = \\frac{15x^2+30x}{5(x^2-4)} = \\frac{15x(x+2)}{5(x^2-4)} = \\\\ \\frac{5\\cdot3x(x+2)}{5(x^2-4)} = \\frac{3x(x+2)}{(x+2)(x-2)} = \\frac{3x}{x-2} \\end{multline*}$"], ["$\\displaystyle \\frac{4x^2-15x-9}{2x^2-18} + \\frac{3}{2x-6}$", "$\\begin{multline*} \\frac{4x^2-15x-9}{2x^2-18} + \\frac{3}{2x-6} = \\frac{4x^2-15x-9}{2(x^2-9)} + \\frac{3}{2(x-3)} = \\frac{4x^2-15x-9}{2(x+3)(x-3)} + \\frac{3(x+3)}{2(x+3)(x-3)} = \\\\ \\frac{4x^2-15x-9}{2(x^2-9)} + \\frac{3x+9}{2(x^2-9)} = \\frac{4x^2-12x}{2(x^2-9)} = \\frac{4x(x-3)}{2(x^2-9)} = \\\\ \\frac{2\\cdot2x(x-3)}{2(x^2-9)} = \\frac{2x(x-3)}{(x+3)(x-3)} = \\frac{2x}{x+3} \\end{multline*}$"], ["$\\displaystyle \\frac{9x^2-20x-4}{3x^2-12} + \\frac{2}{3x-6}$", "$\\begin{multline*} \\frac{9x^2-20x-4}{3x^2-12} + \\frac{2}{3x-6} = \\frac{9x^2-20x-4}{3(x^2-4)} + \\frac{2}{3(x-2)} = \\frac{9x^2-20x-4}{3(x+2)(x-2)} + \\frac{2(x+2)}{3(x+2)(x-2)} = \\\\ \\frac{9x^2-20x-4}{3(x^2-4)} + \\frac{2x+4}{3(x^2-4)} = \\frac{9x^2-18x}{3(x^2-4)} = \\frac{9x(x-2)}{3(x^2-4)} = \\\\ \\frac{3\\cdot3x(x-2)}{3(x^2-4)} = \\frac{3x(x-2)}{(x+2)(x-2)} = \\frac{3x}{x+2} \\end{multline*}$"], ["$\\displaystyle \\frac{9x^2+25x+6}{3x^2-27} + \\frac{2}{3x+9}$", "$\\begin{multline*} \\frac{9x^2+25x+6}{3x^2-27} + \\frac{2}{3x+9} = \\frac{9x^2+25x+6}{3(x^2-9)} + \\frac{2}{3(x+3)} = \\frac{9x^2+25x+6}{3(x+3)(x-3)} + \\frac{2(x-3)}{3(x+3)(x-3)} = \\\\ \\frac{9x^2+25x+6}{3(x^2-9)} + \\frac{2x-6}{3(x^2-9)} = \\frac{9x^2+27x}{3(x^2-9)} = \\frac{9x(x+3)}{3(x^2-9)} = \\\\ \\frac{3\\cdot3x(x+3)}{3(x^2-9)} = \\frac{3x(x+3)}{(x+3)(x-3)} = \\frac{3x}{x-3} \\end{multline*}$"]], "
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=== Donnerstag 24. Oktober 2019 === Ihr Taschenrechner ist angeschrieben. **Mit Hilfe des Taschenrechners** erstellen Sie innerhalb von **5 Minuten** eine Tabelle mit allen Nullstellen, Kandidaten für Extremal- und Wendestellen sowie die zugehörigen Werte der folgenden Funktion, der ersten und zweiten Ableitung. Runden Sie die Resultate auf 2 Nachkommastellen. miniAufgabe("#exokurvendiskussion_funktion_tabelle","#solkurvendiskussion_funktion_tabelle", [["$f(x)=(x+1)^2\\cdot(x-1)^2$", "$$\\begin{array}{l|ccccc}\nx & -1 & -0.58 & 0 & 0.58 & 1\\\\\n\\hline f(x) & 0 & 0.44 & 1 & 0.44 & 0\\\\\nf'(x) & 0 & 1.54 & 0 & -1.54 & 0\\\\\nf''(x) & 8 & 0 & -4 & 0 & 8\\\\\n\\end{array}$$"], ["$f(x)=(x-1)\\cdot(1-x)\\cdot(x+1)^2$", "$$\\begin{array}{l|ccccc}\nx & -1 & -0.58 & 0 & 0.58 & 1\\\\\n\\hline f(x) & 0 & -0.44 & -1 & -0.44 & 0\\\\\nf'(x) & 0 & -1.54 & 0 & 1.54 & 0\\\\\nf''(x) & -8 & 0 & 4 & 0 & -8\\\\\n\\end{array}$$"], ["$f(x)=(x+2)^2\\cdot(x-2)^2$", "$$\\begin{array}{l|ccccc}\nx & -2 & -1.15 & 0 & 1.15 & 2\\\\\n\\hline f(x) & 0 & 7.11 & 16 & 7.11 & 0\\\\\nf'(x) & 0 & 12.32 & 0 & -12.32 & 0\\\\\nf''(x) & 32 & 0 & -16 & 0 & 32\\\\\n\\end{array}$$"], ["$f(x)=(x-2)\\cdot(2-x)\\cdot(x+2)^2$", "$$\\begin{array}{l|ccccc}\nx & -2 & -1.15 & 0 & 1.15 & 2\\\\\n\\hline f(x) & 0 & -7.11 & -16 & -7.11 & 0\\\\\nf'(x) & 0 & -12.32 & 0 & 12.32 & 0\\\\\nf''(x) & -32 & 0 & 16 & 0 & -32\\\\\n\\end{array}$$"], ["$f(x)=(x+1)^2\\cdot(x-2)$", "$$\\begin{array}{l|cccc}\nx & -1 & 0 & 1 & 2\\\\\n\\hline f(x) & 0 & -2 & -4 & 0\\\\\nf'(x) & 0 & -3 & 0 & 9\\\\\nf''(x) & -6 & 0 & 6 & 12\\\\\n\\end{array}$$"], ["$f(x)=(x+1)^2\\cdot(2-x)$", "$$\\begin{array}{l|cccc}\nx & -1 & 0 & 1 & 2\\\\\n\\hline f(x) & 0 & 2 & 4 & 0\\\\\nf'(x) & 0 & 3 & 0 & -9\\\\\nf''(x) & 6 & 0 & -6 & -12\\\\\n\\end{array}$$"], ["$f(x)=(x+1)\\cdot(x-2)^2$", "$$\\begin{array}{l|cccc}\nx & -1 & 0 & 1 & 2\\\\\n\\hline f(x) & 0 & 4 & 2 & 0\\\\\nf'(x) & 9 & 0 & -3 & 0\\\\\nf''(x) & -12 & -6 & 0 & 6\\\\\n\\end{array}$$"], ["$f(x)=(x+1)\\cdot(2-x)\\cdot(x-2)$", "$$\\begin{array}{l|cccc}\nx & -1 & 0 & 1 & 2\\\\\n\\hline f(x) & 0 & -4 & -2 & 0\\\\\nf'(x) & -9 & 0 & 3 & 0\\\\\nf''(x) & 12 & 6 & 0 & -6\\\\\n\\end{array}$$"], ["$f(x)=(x+1)^3\\cdot(x-1)$", "$$\\begin{array}{l|cccc}\nx & -1 & 0 & 0.5 & 1\\\\\n\\hline f(x) & 0 & -1 & -1.69 & 0\\\\\nf'(x) & 0 & -2 & 0 & 8\\\\\nf''(x) & 0 & 0 & 9 & 24\\\\\n\\end{array}$$"], ["$f(x)=(x+1)^3\\cdot(1-x)$", "$$\\begin{array}{l|cccc}\nx & -1 & 0 & 0.5 & 1\\\\\n\\hline f(x) & 0 & 1 & 1.69 & 0\\\\\nf'(x) & 0 & 2 & 0 & -8\\\\\nf''(x) & 0 & 0 & -9 & -24\\\\\n\\end{array}$$"]]);});