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Version 6.4 von Dirk Tebbe am 2025/10/13 15:30
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1 1.
2 [[image:Kosinusfunktion.svg||width="450" style="display:block;margin-left:auto;margin-right:auto"]]
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4 2.
5 {{formula}}h(x)=cos(\frac{\pi}{4}x)+1{{/formula}}
6 {{formula}}h'(x)=\frac{\pi}{4}\cdot (-sin(\frac{\pi}{4}x))+1=-\frac{\pi}{4} sin(\frac{\pi}{4}x){{/formula}}
7 {{formula}}h'(6)=-\frac{\pi}{4}sin(\frac{\pi}{4}\cdot 6)=\frac{\pi}{4}{{/formula}}
8 {{formula}}h(6)=1{{/formula}}
9 Einsetzen von {{formula}}m=\frac{\pi}{4}{{/formula}} und {{formula}}P(6|1){{/formula}}in {{formula}}y=mx+c{{/formula}} liefert {{formula}}c=1-\frac{3}{2}\pi{{/formula}}.
10 {{formula}}t: y=\frac{\pi}{4}x+1-\frac{3}{2}\pi{{/formula}}
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12 3.
13 {{formula}}h'(x)=m{{/formula}}
14 {{formula}}-\frac{\pi}{4} sin(\frac{\pi}{4}x)=2{{/formula}}
15 {{formula}}sin(\frac{\pi}{4}x)=-\frac{8}{\pi}{{/formula}}
16 Substituiere:{{formula}}\frac{\pi}{4}x=u{{/formula}}
17 {{formula}}sin(u)=-\frac{8}{\pi}{{/formula}}
18 {{formula}}-\frac{8}{\pi}<-1{{/formula}}
19 {{formula}}-\frac{8}{\pi}{{/formula}} liegt somit ausserhalb des Wertebereichs der Sinusfunktion.
20 Deswegen hat die Gleichung keine Lösung.