Wiki-Quellcode von Lösung Tangente in einem Kurvenpunkt II
Zuletzt geändert von Martin Stern am 2025/10/13 14:28
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author | version | line-number | content |
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6.3 | 1 | 1. |
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6.2 | 2 | [[image:Exponentialfunktion.svg||width="450" style="display:block;margin-left:auto;margin-right:auto"]] |
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7.1 | 3 | |
4 | |||
5 | 2. | ||
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1.1 | 6 | {{formula}}f\left(x\right)=0{{/formula}} |
7 | {{formula}}4-\frac{1}{2} e^x=0{{/formula}} | ||
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1.2 | 8 | {{formula}}4=\frac{1}{2} e^x{{/formula}} |
9 | {{formula}}8=e^x{{/formula}} | ||
10 | {{formula}}ln(8)=x{{/formula}} | ||
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1.1 | 11 | {{formula}}f´\left(x\right)=-\frac{1}{2} e^x{{/formula}} |
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1.2 | 12 | {{formula}}f´\left(ln(8)\right)=-\frac{1}{2} e^{ln(8)}=-\frac{1}{2}\cdot 8=-4{{/formula}} |
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1.3 | 13 | Einsetzen von {{formula}}m=-4{{/formula}} in {{formula}}y=mx+c{{/formula}}: |
14 | {{formula}}y= -4x+c{{/formula}} | ||
15 | und {{formula}}N(ln(8)|0){{/formula}} | ||
16 | {{formula}} 0= -4 \cdot ln(8)+c{{/formula}} | ||
17 | {{formula}} c = 4 \cdot ln(8){{/formula}} | ||
18 | {{formula}}y=-4\cdot x+ 4 \cdot ln(8){{/formula}} | ||
19 | |||
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7.1 | 20 | |
21 | 3. | ||
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1.4 | 22 | {{formula}}4-\frac{1}{2} e^x=4{{/formula}} |
23 | {{formula}}-\frac{1}{2} e^x=0{{/formula}} | ||
24 | {{formula}} e^x=0{{/formula}} | ||
25 | Diese Gleichung hat keine Lösung, da {{formula}} e^x\neq 0{{/formula}} | ||
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1.3 | 26 | |
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7.1 | 27 | |
28 | 4. | ||
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1.4 | 29 | {{formula}}f´\left(x\right)=-\frac{1}{2} e^x< 0{{/formula}} für alle x. |