Wiki-Quellcode von Lösung Tangente in einem Kurvenpunkt
Zuletzt geändert von Dirk Tebbe am 2025/10/13 12:28
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| author | version | line-number | content |
|---|---|---|---|
| 1 | 1. | ||
| 2 | {{formula}}f(x)=\frac{1}{5}x^3-\frac{16}{5}x{{/formula}} | ||
| 3 | {{formula}}f'(x)=\frac{3}{5}x^2-\frac{16}{5}{{/formula}} | ||
| 4 | {{formula}}f'(3)=\frac{11}{5}{{/formula}} | ||
| 5 | {{formula}}f(3)=-\frac{21}{5}{{/formula}} | ||
| 6 | Einsetzen von {{formula}}m=\frac{11}{5}{{/formula}} und {{formula}}P(3|-\frac{21}{5}){{/formula}} in {{formula}}y=mx+c{{/formula}}: | ||
| 7 | {{formula}}-\frac{21}{5}=\frac{11}{5}\cdot 3+c{{/formula}} | ||
| 8 | {{formula}}t: y= \frac{11}{5}x-\frac{54}{5}{{/formula}} | ||
| 9 | |||
| 10 | |||
| 11 | 2. | ||
| 12 | Variante 1: {{formula}}f'(x)=m_t{{/formula}} | ||
| 13 | {{formula}}\frac{3}{5}x^2-\frac{16}{5}=\frac{11}{5}{{/formula}} | ||
| 14 | {{formula}}3x^2=27{{/formula}} | ||
| 15 | {{formula}}x^2=9{{/formula}} | ||
| 16 | {{formula}}x_1=3{{/formula}} und {{formula}}x_2=-3{{/formula}} | ||
| 17 | {{formula}}f(-3)=\frac{21}{5}=g(-3){{/formula}} | ||
| 18 | |||
| 19 | Variante 2: Argumentation mit Punktsymmetrie |