$f(x)=2\cdot x^2-3\cdot x+4f(x)=2\backslash cdot\; x^2\; -\; 3\; \backslash cdot\; x\; +\; 4$

Bringe die quadratische Funktion auf die Scheitelform.

$f(x)=2{(x-\frac{3}{4})}^2+2\cdot \frac{23}{16}f(x)=2\backslash left(x-\backslash frac\{3\}\{4\}\backslash right)^2+2\backslash cdot\backslash frac\{23\}\{16\}$$\Rightarrow f(x)=2{(x-\frac{3}{4})}^2+\frac{23}{8}\backslash ;\backslash Rightarrow\backslash ;f(x)=2\backslash left(x-\backslash frac\{3\}\{4\}\backslash right)^2+\backslash frac\{23\}\{8\}$

Lies den Scheitel und die Öffnungsrichtung ab.

$S\left(\frac{3}{4}\mid \frac{23}{8}\right)S\backslash ;\backslash left(\backslash left.\backslash frac34\backslash right|\backslash ;\backslash frac\{23\}\{8\}\backslash right)$, Parabel nach oben geöffnet

Gib die Wertemenge an.

$W_f=[\frac{23}{8};\mathrm{\infty }[W\_f=\backslash left[\backslash frac\{23\}\{8\};\; \backslash ;\backslash infty\; \backslash right[$

$f(x)=-x^2+8\cdot x-2f(x)=-x^2+8\backslash cdot\; x\; -\; 2$

Bringe die quadratische Funktion auf die Scheitelform.

$f(x)=-(x-4{)}^2+14f(x)=-(x-4)^2+14$

Lese Scheitel und Öffnungsrichtung ab.

$S(4\mid 14)S\backslash ;\backslash left(4\backslash mid\; 14\; \backslash right)$, Parabel nach unten geöffnet

Gebe die Wertemenge an.

$W_f=]-\mathrm{\infty };14]W\_f=\backslash left]-\; \backslash infty\; ;\backslash ;\; 14\; \backslash right]$

Ganzrationale Funktionen dritten Grades haben keine Definitionslücken.

Außerdem gilt:

$\begin{array}{c}{\lim }_{x\to -\mathrm{\infty }}f\left(x\right)=\mathrm{\infty }\\ {\lim }_{x\to \mathrm{\infty }}f\left(x\right)=-\mathrm{\infty }\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\backslash lim\_\{x\backslash rightarrow-\backslash infty\}f\backslash left(x\backslash right)=\backslash infty\backslash \backslash \backslash lim\_\{x\backslash rightarrow\backslash infty\}f\backslash left(x\backslash right)=-\backslash infty\backslash end\{array\}$

$\Rightarrow W_f=\mathbb{R}\backslash Rightarrow\; W\_f=\backslash mathbb\{R\}$

$f(x)=16\cdot \sin (x)+3f(x)=16\backslash cdot\; \backslash sin(x)\; +\; 3$

Betrachte die Sinusfunktion. Der Sinus nimmt bekanntlich nur Werte zwischen -1 und 1 an.

$W_{\sin \left(x\right)}=[-1;1]W\_\{\backslash sin\backslash left(x\backslash right)\}=\backslash left[-1;1\backslash right]$

$\Rightarrow W_{16\cdot \sin \left(x\right)}=[-16;16]\backslash Rightarrow\; W\_\{16\backslash cdot\backslash sin\backslash left(x\backslash right)\}=\backslash left[-16;16\backslash right]$

$\Rightarrow W_f=[-13;19]\backslash Rightarrow\; W\_f=\backslash left[-13;19\backslash right]$

$f(x)=2^x-3f(x)=2^x-3$

Betrachte $2^{x}2^x$.

$2^{x}2^x$ ist eine streng monoton steigende Exponentailfunktion, die nur Werte anninmmt, die größer als 0 sind.

Betrachte die Grenzwerte.

${\displaystyle \underset{x\to -\mathrm{\infty }}{\lim }f\left(x\right)=\underset{x\to -\mathrm{\infty }}{\lim }\stackrel{\to 0}{\overbrace{2^x}}-3=-3}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow-\backslash infty\}f\backslash left(x\backslash right)=\backslash lim\_\{x\backslash rightarrow-\backslash infty\}\backslash overbrace\{2^x\}^\{\backslash rightarrow0\}-3=-3$

${\displaystyle \underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)=\underset{x\to \mathrm{\infty }}{\lim }\stackrel{\to \mathrm{\infty }}{\overbrace{2^x}}-3\to \mathrm{\infty }}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow\backslash infty\}f\backslash left(x\backslash right)=\backslash lim\_\{x\backslash rightarrow\backslash infty\}\backslash overbrace\{2^x\}^\{\backslash rightarrow\backslash infty\}-3\backslash rightarrow\backslash infty$

Gib nun den Wertebereich an.

$W_f=]-3;\mathrm{\infty }[W\_f=\backslash left]-3;\backslash ;\; \backslash infty\; \backslash right[$

$f(x)=\frac{x-3}{\ln (x-3)}f(x)=\backslash frac\{x-3\}\{\backslash ln(x-3)\}$

Bestimme den Definitionsbereich.

$D_f=]3;\mathrm{\infty }[\mathrm{\backslash }\{4\}D\_f=\backslash left]\backslash ;3;\; \backslash ;\backslash infty\; \backslash right[\backslash ;\backslash backslash\backslash ;\backslash \{4\backslash \}$

Bestimme die Grenzwerte.

${\displaystyle \underset{x\to 3}{\lim }f\left(x\right)=\underset{x\to 3}{\lim }\frac{\stackrel{\to 0}{\overbrace{x-3}}}{\underset{\to -\mathrm{\infty }}{\underbrace{\ln (x-3)}}}=0}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow3\}f\backslash left(x\backslash right)=\backslash lim\_\{x\backslash rightarrow3\}\backslash frac\{\backslash overbrace\{x-3\}^\{\backslash rightarrow0\}\}\{\backslash underbrace\{\backslash ln\backslash left(x-3\backslash right)\}\_\{\backslash rightarrow-\backslash infty\}\}=0$

Für den zweiten Grenzwert nutzt du aus, das der Logarithmus immer langsamer wächst als jedes Polynom.

${\displaystyle \underset{x\to \mathrm{\infty }}{\lim }f\left(x\right)=\underset{x\to \mathrm{\infty }}{\lim }\frac{\stackrel{\to \mathrm{\infty }}{\overbrace{x-3}}}{\underset{\to \mathrm{\infty }}{\underbrace{\ln (x-3)}}}=\mathrm{\infty }}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow\backslash infty\}f\backslash left(x\backslash right)=\backslash lim\_\{x\backslash rightarrow\backslash infty\}\backslash frac\{\backslash overbrace\{x-3\}^\{\backslash rightarrow\backslash infty\}\}\{\backslash underbrace\{\backslash ln\backslash left(x-3\backslash right)\}\_\{\backslash rightarrow\backslash infty\}\}=\backslash infty$

${\displaystyle \underset{x\to 4^{-}}{\lim }f\left(x\right)=\underset{x\to 4^{-}}{\lim }\frac{\stackrel{\to 1}{\overbrace{x-3}}}{\underset{\to 0^{-}}{\underbrace{\ln (x-3)}}}=-\mathrm{\infty }}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow4^-\}f\backslash left(x\backslash right)=\backslash lim\_\{x\backslash rightarrow4^-\}\backslash frac\{\backslash overbrace\{x-3\}^\{\backslash rightarrow1\}\}\{\backslash underbrace\{\backslash ln\backslash left(x-3\backslash right)\}\_\{\backslash rightarrow0^-\}\}=-\backslash infty$

${\displaystyle \underset{x\to 4^{+}}{\lim }f\left(x\right)=\underset{x\to 4^{+}}{\lim }\frac{\stackrel{\to 1}{\overbrace{x-3}}}{\underset{\to 0^{+}}{\underbrace{\ln (x-3)}}}=+\mathrm{\infty }}\backslash displaystyle\backslash lim\_\{x\backslash rightarrow4^+\}f\backslash left(x\backslash right)=\backslash lim\_\{x\backslash rightarrow4^+\}\backslash frac\{\backslash overbrace\{x-3\}^\{\backslash rightarrow1\}\}\{\backslash underbrace\{\backslash ln\backslash left(x-3\backslash right)\}\_\{\backslash rightarrow0^+\}\}=+\backslash infty$

Bestimme jetzt die Extrema.

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{\ln (x-3)-\frac{x-3}{x-3}}{(\ln (x-3){)}^2}}f\text{'}(x)=\backslash dfrac\{\backslash ln(x-3)-\backslash frac\{x-3\}\{x-3\}\}\{(\backslash ln(x-3))^2\}$

$f^{\mathrm{\prime }}(x)=0f\text{'}(x)=0$

$\iff \ln (x-3)-1=0\mathrm{\mid }+1\backslash Leftrightarrow\backslash ln(x-3)-1=0\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\; |+1$

$\iff \ln (x-3)=1\mathrm{\mid }e\backslash Leftrightarrow\; \backslash ln(x-3)=1\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;|e$

$\iff x-3=e\mathrm{\mid }+3\backslash Leftrightarrow\; x-3\; =\; e\; \backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;\backslash ;|+3$

$\iff x=e+3\backslash Leftrightarrow\; x=e+3$

Bestimme jetzt den y-Wert.

$f(e+3)=ef\backslash left(e+3\backslash right)=e$

Da dies das einzige Extremum ist und die beiden Grenzwerte an den jeweiligen Definitionslücken rechts und links davon (also $4^{+}4^+$und$\mathrm{\infty }\backslash ;\backslash infty$) jeweils $\mathrm{\infty }\backslash infty$ sind, ist das Extremum ein Minimum.

Gib jetzt den Wertebereich an.

$W_f=]-\mathrm{\infty };0[\cup [e;\mathrm{\infty }[W\_f=\backslash left]-\backslash ;\backslash infty;\; 0\backslash ;\backslash right[\backslash cup\backslash left[\backslash ;e;\backslash ;\backslash infty\backslash right[$