Lösung Erkunden (eine Potenzfunktion)

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1. (((Randverhalten: Verhalten im Unendlichen

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1) Verhalten gegen plus Unendlich ({{formula}}+\infty{{/formula}})

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|={{formula}}x{{/formula}}| {{formula}}+1{{/formula}}| {{formula}}+10{{/formula}}| {{formula}}+100{{/formula}}| {{formula}}+1000{{/formula}}| {{formula}}+10^6{{/formula}}| {{formula}}+10^9{{/formula}}| {{formula}}+10^{12}{{/formula}}|{{formula}}+10^{+\infty}{{/formula}}

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|={{formula}}f(x){{/formula}}|1|{{formula}}\frac{1}{10}{{/formula}}|{{formula}}\frac{1}{100}{{/formula}}|{{formula}}\frac{1}{1000}{{/formula}}|{{formula}}\frac{1}{1000000}{{/formula}}|{{formula}}\frac{1}{1000000000}{{/formula}}|{{formula}}\frac{1}{1000000000000}{{/formula}}|0

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2) Verhalten gegen minus Unendlich ({{formula}}-\infty{{/formula}})

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|={{formula}}x{{/formula}}| {{formula}}-1{{/formula}}| {{formula}}-10{{/formula}}| {{formula}}-100{{/formula}}| {{formula}}-1000{{/formula}}| {{formula}}-10^6{{/formula}}| {{formula}}-10^9{{/formula}}|{{formula}}-10^{12}{{/formula}}|{{formula}}-10^{-\infty}{{/formula}}

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|={{formula}}f(x){{/formula}}|{{formula}}-1{{/formula}}|{{formula}}-\frac{1}{10}{{/formula}}|{{formula}}-\frac{1}{100}{{/formula}}|{{formula}}-\frac{1}{1000}{{/formula}}|{{formula}}-\frac{1}{1000000}{{/formula}}|{{formula}}\frac{1}{1000000000}{{/formula}}|{{formula}}\frac{1}{1000000000000}{{/formula}}|0

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)))

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1. (((Randverhalten: Verhalten nahe der Definitionslücke ({{formula}}x \approx 0{{/formula}})

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1) Verhalten links bei der Definitionslücke ({{formula}}x \approx 0{{/formula}} mit {{formula}}x<0{{/formula}})

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|={{formula}}x{{/formula}}| {{formula}}-1{{/formula}}| {{formula}}-0,1{{/formula}}| {{formula}}-0,01{{/formula}}| {{formula}}-0,001{{/formula}}| {{formula}}-10^{-6}{{/formula}}| {{formula}}-10^{-9}{{/formula}}| {{formula}}-10^{-12}{{/formula}}|0

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|={{formula}}f(x){{/formula}}|{{formula}}-1{{/formula}}|{{formula}}-10{{/formula}}|{{formula}}-100{{/formula}}|{{formula}}-1000{{/formula}}|{{formula}}-10^6{{/formula}}| {{formula}}-10^9{{/formula}}|{{formula}}-10^{-12}{{/formula}}|{{formula}}-\infty{{/formula}}

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2) Verhalten rechts bei der Definitionslücke ({{formula}}x \approx 0{{/formula}} mit {{formula}}x>0{{/formula}})

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|={{formula}}x{{/formula}}| {{formula}}+1{{/formula}}| {{formula}}+0,1{{/formula}}| {{formula}}+0,01{{/formula}}| {{formula}}+0,001{{/formula}}| {{formula}}+10^{-6}{{/formula}}| {{formula}}+10^{-9}{{/formula}}| {{formula}}+10^{-12}{{/formula}}|0

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|={{formula}}f(x){{/formula}}|{{formula}}1{{/formula}}|{{formula}}10{{/formula}}|{{formula}}100{{/formula}}|{{formula}}1000{{/formula}}|{{formula}}10^6{{/formula}}| {{formula}}10^9{{/formula}}|{{formula}}10^{-12}{{/formula}}|{{formula}}\infty{{/formula}}

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)))

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1. Erkennst du eine Symmetrie?

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Die Funktion {{formula}} f(x) = \frac{1}{x} {{/formula}} ist punktsymmetrisch zur Ursprungsgeraden, da gilt:

f(-x) = -f(x)

Das bedeutet, dass die Funktion eine Symmetrie bezüglich des Ursprungs hat.