Nullstellen des Nenner

$f(x)={\displaystyle \frac{1}{2-\ln (x^2-1)}}f(x)=\backslash dfrac1\{2-\backslash ln(x^2-1)\}$

Der Nenner ist dann 0, wenn $\ln (x^2-1)=2\backslash ln\backslash left(x^2-1\backslash right)=2$ ist.

$\Rightarrow x_1=-\sqrt{e^2+1},x_2=\sqrt{e^2+1}\backslash Rightarrow\; x\_1=-\backslash sqrt\{e^2+1\},\backslash ;x\_2=\backslash sqrt\{e^2+1\}$

Definitionsbereich des Logarithmus

$f(x)={\displaystyle \frac{1}{2-\ln (x^2-1)}}f(x)=\backslash dfrac1\{2-\backslash ln(x^2-1)\}$

Der Logarithmus ist nur für positive Zahlen definiert.

$\Rightarrow \backslash Rightarrow$    $D_f=]-\mathrm{\infty };-1[\cup D\_f=\backslash rbrack-\backslash infty;\backslash ;\backslash ;-1\backslash lbrack\backslash ;\backslash cup$ $]1;+\mathrm{\infty }[\mathrm{\backslash }\{-\sqrt{e^2+1},\sqrt{e^2+1}\}\backslash ;\backslash rbrack1;\backslash ;+\backslash infty\backslash lbrack\backslash ;\backslash backslash\backslash ;\backslash left\backslash \{-\backslash sqrt\{e^2+1\},\backslash ;\backslash sqrt\{e^2+1\}\backslash right\backslash \}$

Nullstellenbestimmung

$\Rightarrow \backslash Rightarrow$   Da im Zähler der Funktion kein Element mit x vorkommt, hat die Funktion keine Nullstellen.

Ableitungen

Bilde die 1. Ableitung

$f(x)={\displaystyle \frac{1}{2-\ln (x^2-1)}}f(x)=\backslash dfrac1\{2-\backslash ln(x^2-1)\}$

Mit Hilfe der Potenzgesetze umwandeln.

$f(x)={(2-\ln (x^2-1))}^{-1}f(x)=\{\backslash textstyle\backslash left(2-\backslash ln(x^2-1)\backslash right)\}^\{-1\}$

Leite jetzt mit Hilfe der Kettenregel ab:

$f\mathrm{‘}\left(x\right)=-{(2-\ln (x^2-1))}^{-2}\cdot (-\frac{1}{x^2-1}\cdot 2x)={\displaystyle \frac{2x}{(x^2-1)\cdot {(2-\ln (x^2-1))}^2}}f`\backslash left(x\backslash right)=-\{\backslash textstyle\backslash left(2-\backslash ln(x^2-1)\backslash right)\}^\{-2\}\backslash cdot\backslash left(-\backslash frac1\{x^2-1\}\backslash cdot2x\backslash right)\; =\backslash dfrac\{2x\}\{\backslash left(x^2-1\backslash right)\backslash cdot\backslash left(2-\backslash ln(x^2-1)\backslash right)^2\}$