\(\begin{eqnarray} f(x)=0\\ \frac{1}{8}x^{4}+\frac{1}{2}x^{3}=0\\ \frac{1}{2}\cdot x^{3}\cdot(\frac{x}{4}+1)=0 \end{eqnarray} \)

\(\begin{eqnarray} x_{1}\qquad^{3}=0 \hspace{20}\to \hspace{20} N_{1}^{(3)}(0|0)\\ \frac{x}{4}+1=0 \hspace{140}\\ x_{2}=-4 \hspace{20}\to \hspace{20} N_{2}(-4|0) \end{eqnarray}\)

\(\begin{eqnarray} f(x)=\frac{1}{8}x^{4}+\frac{1}{2}x^{3}\\ \vspace{10}\\ f'(x)=\frac{1}{2}x^{3}+\frac{3}{2}x^{2}\\ \vspace{10}\\ f''(x)=\frac{3}{2}x^{2}+3x\\ \end{eqnarray}\)

\(\begin{eqnarray} f'(x)=0\\ \vspace{3}\\ \frac{1}{2}x^{3}+\frac{3}{2}x^{2}=0\\ \vspace{3}\\ \frac{1}{2}\cdot x^{2}\cdot (x+3)=0\\ \end{eqnarray}\)

\(\begin{eqnarray} x_{1}\,^{2}=0 \Rightarrow f(x_{1})=0\\ \vspace{5}\\ x+3=0 \Rightarrow x_{2}=-3 \Rightarrow f(x_{2})=-\frac{27}{8}=-3,38\\ \end{eqnarray}\)

\(\begin{eqnarray} f''(x_{1})=f''(0) = 0 \Rightarrow keine\,Entscheidung!\\ \vspace{5}\\ f''(x_{2})= f''(-3)= \frac{27}{2}-9>0 \Rightarrow Tiefpunkt\,T(-3/-3,38) \end{eqnarray}\)

\(\begin{eqnarray} f(x)=\frac{1}{8}x^{4}+\frac{1}{2}x^{3}\\ \vspace{10}\\ f'(x)=\frac{1}{2}x^{3}+\frac{3}{2}x^{2}\\ \vspace{10}\\ f''(x)=\frac{3}{2}x^{2}+3x\\ \end{eqnarray}\)

\(\begin{eqnarray} f''(x)=0\\ \frac{3}{2}x^{2}+3x=0 \qquad \Rightarrow \qquad 3x\cdot(\frac{x}{2}+1)=0 \end{eqnarray}\)

\(\begin{eqnarray} x_{1}=0 \qquad \Rightarrow \qquad f(x_{1})=0\\ \vspace{5}\\ \frac{x}{2}+1=0\qquad \Rightarrow\qquad x_{2}=-2\qquad \Rightarrow\qquad f(x_{2})=-2 \end{eqnarray}\)

\(W_{1}(0/0); \qquad W_{2}(-2/-2)\)

\(\begin{eqnarray} f(x)=x^{3}+2x^{2}-3x-4\\ f'(x)= 3x^{2}+4x-3\\ f''(x)=6x+4\\ f'(x)=0 \Rightarrow \end{eqnarray}\)

\(\begin{eqnarray} \vspace{5}\\ \Rightarrow 3x^{2}+4x-3=0\\ \vspace{5}\\ x^{2}+\frac{4}{3}\cdot x-1=0\\ x_{1,2}=-\frac{2}{3}\pm\sqrt{\frac{4}{9}+1} \end{eqnarray}\)

\(\begin{eqnarray} x_{1,2}=-\frac{2}{3}\pm\frac{\sqrt{13}}{3}\\ \Rightarrow \qquad x_{1}=\frac{-2+\sqrt{13}}{3}=0,535\\ \vspace{5}\\ \Rightarrow x_{2}=\frac{-2-\sqrt{13}}{3}=-1,869 \end{eqnarray} \)

\(\begin{eqnarray} f(x_{1})=-4,879 \hspace{20} f(x_{2})=2,065\\ f''(x_{1})=7,2>0 \hspace{20} f''(x_{2})=-7,2<0\\ \Rightarrow T(0,535/-4,879) \hspace{20}\Rightarrow H(-1,869/2,065) \end{eqnarray}\)

\(\begin{eqnarray} f(x)=x^{3}+2x^{2}-3x-4\\ f'(x)= 3x^{2}+4x-3\\ f''(x)=6x+4\\ f''(x)=0\Rightarrow \end{eqnarray} \)

\(\begin{eqnarray} \Rightarrow6x+4=0\\ \vspace{5}\\ x=-\frac{2}{3}=-0,67\\ \vspace{5}\\ f(-\frac{2}{3})=-\frac{38}{27}=-1,41 \end{eqnarray}\)

\(\Rightarrow \hspace{20} W(-\frac{2}{3}/-\frac{38}{27})\)

\(\begin{eqnarray} f(x)=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\\ \vspace{5}\\ f'(x)=3a_{3}x^{2}+2a_{2}x+a_{1}\\ \vspace{5}\\ f''(x)=6a_{3}x+2a_{2} \end{eqnarray}\)

Die Angabe legt vier Gleichungen fest:

\(\begin{eqnarray} f(-3)=4 \Rightarrow -27a_{3}+9a_{2}-3a_{1}+a_{0}=4\\ f'(-3)=0 \Rightarrow 27a_{3}-6a_{2}+a_{1}=0\\ \end{eqnarray}\)

\(\begin{eqnarray} f''(-2)=0 \Rightarrow -12a_{3}+2a_{2}=0\\ f(-1)=0 \Rightarrow -a_{3}+a_{2}-a_{1}+a_{0}=0\\ \end{eqnarray}\)

Die Lösung des Gleichungssystems liefert:

\(\begin{eqnarray} a_{0}=4\\ a_{1}=9\\ a_{2}=6\\ a_{3}=1 \end{eqnarray}\)

Ergebnis:

\(f(x)=x^{3}+6x^{2}+9x+4\)

\(\begin{eqnarray} f(x)=x^{3}+6x^{2}+9x+4\\ f'(x)=3x^{2}+12x+9\\ f''(x)=6x+12 \end{eqnarray}\)

\(\begin{eqnarray} f'(x)=0\\ 3x^{2}+12x+9=0\\ x^{2}+4x+3=0\\ \end{eqnarray}\)

\(\begin{eqnarray} x_{1,2}=-2\pm\sqrt{4-3}=-2\pm1\\ x_{1}=-3 \Rightarrow f(x_{1})=4\\ x_{2}=-1 \Rightarrow f(x_{2})=0 \end{eqnarray}\)

Klassifikation der Extremwerte:

\(\begin{eqnarray} f''(x_{1})=f''(-3)=-6 \qquad <0 \Rightarrow Hochpunkt \qquad H(-3/4)\\ \vspace{5}\\ f''(x_{2})=f''(-1)=6 \qquad >0 \Rightarrow Tiefpunkt \qquad T(-1/0) \end{eqnarray} \)

\(\begin{eqnarray} f(x)=x^{3}+6x^{2}+9x+4\\ f'(x)=3x^{2}+12x+9\\ f''(x)=6x+12\\ f'''(x)=6 \end{eqnarray} \)

\(\begin{eqnarray} f''(x)=0\\ \vspace{5}\\ 6x+12=0\\ \vspace{10}\\ x=-2 \qquad \Rightarrow \qquad f(x)=2 \end{eqnarray}\)

\(Wendepunkt \qquad W(-2/2)\)

\(\begin{eqnarray} f(x)=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\\ f'(x)=3a_{3}x^{2}+2a_{2}x+a_{1}\\ f''(x)=6a_{3}x+2a_{2}\\ \end{eqnarray}\)

\(\begin{eqnarray} f(0)=2 \Rightarrow a_{0}=2\\ f(2)=6 \Rightarrow 8a_{3}+4a_{2}+2a_{1}+a_{0}=6\\ f'(2)=0 \Rightarrow 12a_{3}+4a_{2}+a_{1}=0\\ f''(2)=0 \Rightarrow 12a_{3}+2a_{2}=0 \end{eqnarray}\)

\(\begin{eqnarray} 8a_{3}+4a_{2}+2a_{1}=4\\ 12a_{3}+4a_{2}+a_{1}=0\\ 12a_{3}+2a_{2}=0\\ \rule[-10mm]{150cm}{1mm} \end{eqnarray}\)

Vereinfacht:

\(\begin{eqnarray} 4a_{3}+2a_{2}+a_{1}=2\\ 12a_{3}+4a_{2}+a_{1}=0\\ 6a_{3}+a_{2}=0\\ \Rightarrow a_{2}=-6a_{3}\\ \rule[-10mm]{150cm}{1mm} \end{eqnarray}\)

 \(\begin{eqnarray} 4a_{3}-12a_{3}+a_{1}=2\\ 12a_{3}-24a_{3}+a_{1}=0\\ \rule[-10mm]{150cm}{1mm} \end{eqnarray}\)

\(\begin{eqnarray} -8a_{3}+a_{1}=2\\ -12a_{3}+a_{1}=0\\ \rule[-10mm]{150cm}{1mm} \end{eqnarray}\)

\(\begin{eqnarray} 4a_{3}=2\\ \vspace{10}\\ \to a_{3}=\frac{1}{2}\\ \vspace{5}\\ \to a_{1}=6\\ \vspace{5}\\ \to a_{2}=-3 \end{eqnarray}\)

Ergebnis:

\(f(x)=\frac{1}{2}x^{3}-3x^{2}+6x+2\)

\(\begin{eqnarray} f(x)=\frac{1}{2}x^{3}-3x^{2}+6x+2\\ f'(x)=\frac{3}{2}x^{2}-6x+6 \end{eqnarray}\)

\(\begin{eqnarray} f'(x)=0\\ \frac{3}{2}x^{2}-6x+6=0\\ x^{2}-4x+4=0\\ \vspace{10}\\ x_{1,2}=2\pm\sqrt{4-4}=2 \end{eqnarray}\)

Außer der Stelle x=2 besitzt diese Funktion keine weitere Stelle mit horizontaler Tangente.

Da die zweite Ableitung an dieser Stelle aber 0 ist, besitzt die Funktion daher keinen Extremwert.

\(\begin{eqnarray} f(x)=x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\\ f'(x)=3x^{2}+2a_{2}x+a_{1}\\ f''(x)=6x+2a_{2} \end{eqnarray}\)

Aus der Tatsache, dass S(1/4) ein Sattelpunkt ist, lassen sich drei Gleichungen ableiten:

\(\begin{eqnarray} f(1)=4 \qquad \Rightarrow \qquad 1+a_{2}+a_{1}+a_{0}=4\\ f'(1)=0 \qquad \Rightarrow \qquad 3+2a_{2}+a_{1}=0\\ f''(1)=0 \qquad \Rightarrow \qquad 6+2a_{2}=0\\ \end{eqnarray}\)

Ergebnis:

\(\begin{eqnarray} a_{2}=-3\\ a_{1}=3\\ a_{0}=3 \end{eqnarray}\)

\(f(x)=x^{3}-3x^{2}+3x+3\)

\(\begin{eqnarray} f(x)=x^{3}-3x^{2}+3x+3\\ f'(x)=3x^{2}-6x+3\\ f''(x)=6x-6 \end{eqnarray}\)

\(\begin{eqnarray} f'(x)=0\\ \vspace{5}\\ 3x^{2}-6x+3=0\\ \vspace{5}\\ x^{2}-2x+1=0\\ \vspace{5}\\ (x-1)^{2}=0\\ \vspace{5}\\ x_{1}=x_{2}=1 \end{eqnarray}\)

Es liegt eine Doppelnullstelle der 1. Ableitungsfunktion vor.

Das deutet auf einen Wendepunkt hin. (Wie man zb mit der zweiten Ableitung nachweisen kann, die an dieser Stelle 0 ist.)

Es müssen fünf Gleichungen für die folgenden Unbekannten gefunden werden: \(a_{0},\qquad a_{1},\qquad a_{2},\qquad a_{3},\qquad a_{4}\)

Zuerst:

\(P_{4}(x)=a_{4}\cdot x^{4}+a_{3}\cdot x^{3}+a_{2}\cdot x^{2}+a_{1}\cdot x+a_{0}\)

\(P_{4}'(x)=4a_{4}\cdot x^{3}+3a_{3}\cdot x^{2}+2a_{2}\cdot x+a_{1}\)

\(P_{4}''(x)=12a_{4}\cdot x^{2}+6a_{3}\cdot x+2a_{2}\)

Nun werden die Informationen aus der Angabe ausgewertet:

\(P_{4}(0)=0\qquad \to \qquad a_{0}=0 \)

\(P_{4}'(0)=0\qquad \to \qquad a_{1}=0\)

\(P_{4}''(0)=0\qquad \to \qquad a_{2}=0\)

\(P_{4}(-1)=\frac{3}{4}\qquad \to \qquad a_{4}-a_{3}+a_{2}-a_{1}+a_{0}=\frac{3}{4}\)

\(P_{4}'(-1)=-2\qquad \to \qquad -4a_{4}+3a_{3}-2a_{2}+a_{1}=-2\)

Dieses 5x5 Gleichungssystem reduziert sich sofort auf ein 2x2 Gleichungssystem:

\(a_{4}-a_{3}=\frac{3}{4}\)

\(-4a_{4}+3a_{3}=-2\)

\(\rule[-2]{150}{1}\)

\(4a_{4}-4a_{3}=3\)

\(-4a_{4}+3a_{3}=-2\)

\(\rule[-2]{150}{1}\)

\(a_{3}=-1\)

\(a_{4}=-\frac{1}{4}\)

Ergebnis:

\(P_{4}(x)=-\frac{1}{4}x^{4}-x^{3}\)

\(P_{4}(x)=-\frac{1}{4}x^{4}-x^{3}\)

\(P_{4}'(x)=-x^{3}-3x^{2}\)

\(P''_{4}(x)=-3x^{2}-6x\)

\(P_{4}'(x)=0\)

\(-x^{3}-3x^{2}=0\)

\(-x^{2}\cdot(x+3)=0\)

\(x_{1}^{(2)}=0\)

\(P_{4}(x_{1})=0\)

\(P_{4}''(x_{1})=0\)

\(x_{2}=-3\)

\(P_{4}(x_{2})=6,75\)

\(P_{4}''(x_{2})=-9<0\)

\(\to\qquad H(-3/6,75)\)

\(P_{4}(x)=-\frac{1}{4}x^{4}-x^{3}\)

\(P_{4}'(x)=-x^{3}-3x^{2}\)

\(P''_{4}(x)=-3x^{2}-6x\)

\(P_{4}''(x)=0\)

\(-3x^{2}-6x=0\)

\(-3x\cdot(x+2)=0\)

\(x_{1}=0\)

\(P_{4}(x_{1})=0\)

\(W_{1}(0/0)\)

\(x_{2}=-2 \)

\(P_{4}(x_{2})=4\)

\(W_{2}(-2/4)\)

\(f(x)=a_{3}x^{3}+a_{2}x^{2}\)

\(f'(x)=3a_{3}x^{2}+2a_{2}x\)

\(f(4)=4 \to\qquad 64a_{3}+16a_{2}=4\)

\(f'(4)=0 \to\qquad 48a_{3}+8a_{2}=0\)

\(\rule[-10]{150}{1}\)

\(64a_{3}+16a_{2}=4\qquad /:\qquad4\)

\(48a_{3}+8a_{2}=0 \qquad /:\qquad 8\)

\(\rule[-10]{150}{1}\)

\(16a_{3}+4a_{2}=1\)

\(6a_{3}+a_{2}=0\)

\(\rule[-10]{150}{1}\)

\(a_{2}=-6a_{3}\)

\(16a_{3}+4\cdot(-6a_{3})=1\)

\(16a_{3}-24a_{3}=1\)

\(a_{3}=-\frac{1}{8}\)

\(a_{2}=\frac{3}{4}\)

\(\to f(x)= -\frac{1}{8}\cdot x^{3}+\frac{3}{4}\cdot x^{2}\)

\(f(x)= -\frac{1}{8}x^{3}+\frac{3}{4}x^{2}\)

\(f(x)=0\)

\(-\frac{1}{8}x^{3}+\frac{3}{4}x^{2}=0\)

\(x^{2}(-\frac{1}{8}x+\frac{3}{4})=0\)

\(x_{1,2}=0\)

\(-\frac{1}{8}x+\frac{3}{4}=0\)

\(\frac{1}{8}x=\frac{3}{4}\)

\(x_{3}=6\)

\(\to N_{1}\,^{(2)}(0/0)\)

\(\to N_{2}(6/0)\)

\(f(x)= -\frac{1}{8}x^{3}+\frac{3}{4}x^{2}\)

\(f'(x)= -\frac{3}{8}x^{2}+\frac{3}{2}x\)

\(f''(x)=-\frac{3}{4}x+\frac{3}{2}\)

\(f'(x)=0\)

\(-\frac{3}{8}x^{2}+\frac{3}{2}x=0\)

\(x(-\frac{3}{8}x+\frac{3}{2})=0\)

\(x_{1}=0\)

\(-\frac{3}{8}x+\frac{3}{2}=0\)

\(\frac{3}{8}x=\frac{3}{2}\qquad \to \qquad x_{2}=4\)

\(f''(0)=\frac{3}{2}\qquad \to >\qquad 0\qquad \to T\)

\(f''(4)=-\frac{3}{2}\qquad \to < \qquad 0\qquad \to H\)

Einsetzen der Werte in f(x):

\(\to T(0/0)\)

\(\to H(4/4)\)

\(f(x)= -\frac{1}{8}x^{3}+\frac{3}{4}x^{2}\)

\(f'(x)= -\frac{3}{8}x^{2}+\frac{3}{2}x\)

\(f''(x)=-\frac{3}{4}x+\frac{3}{2}\)

\(f''(x)=0\)

\(-\frac{3}{4}x+\frac{3}{2}=0\)

\(\frac{3}{4}x=\frac{3}{2}\)

\(x=2\)

Einsetzen des Wertes in f(x):

\(\to \qquad W(2/2)\)

\(P_{4}(x)=a_{4}x^{4}+a_{3}x^{3}-x^{2}\)

\( P'_{4}(x)=4a_{4}x^{3}+3a_{3}x^{2}-2x\)

\( P''_{4}(x)=12a_{4}x^{2}+6a_{3}x-2\)

\(I:\qquad \qquad \qquad P_{4}(2)=-4 \)

\(II:\qquad P''_{4}(2)=0\)

\(\rule[10]{150}{1}\)

\(I:\qquad \qquad \qquad 16a_{4}+8a_{3}-4=-4\)

\(II:\qquad 48a_{4}+12a_{3}-2=0\)

\(\rule[10]{150}{1}\)

\(I:\qquad \qquad \qquad 2a_{4}+a_{3}=0\)

\(II:\qquad 24a_{4}+6a_{3}=1\)

\(\rule[10]{150}{1}\)

\(I: \qquad a_{3}=-2a_{4}\)

\(I\qquad in \qquad II:\)

\(24a_{4}+6(-2a_{4})=1\)

\(12a_{4}=1 \to \qquad a_{4}=\frac{1}{12}\)

\(a_{3}=-2a_{4} = -\frac{1}{6}\)

\(\to \qquad f(x)=P_{4}(x)=\frac{1}{12}x^{4}-\frac{1}{6}x^{3}-x^{2}\)

\(f(x)=\frac{1}{12}x^{4}-\frac{1}{6}x^{3}-x^{2}\)

\(f(x)=0\)

\(\frac{1}{12}x^{4}-\frac{1}{6}x^{3}-x^{2}=0\)

\(x^{2}\cdot (x^{2}-2x-12)=0\)

\(1.\)

\(x^{2}=0\)

\(x_{1,2}=0\)

\(\to \qquad N_{1}\,^{(2)}(0/0)\)

\(2.\)

\(x^{2}-2x-12=0\)

Kleine Lösungsformel:

\(x_{3,4}=1\pm\sqrt{1+12}\)

\(x_{3}=1+\sqrt{13}=4,61\)

\(x_{4}=1-\sqrt{13}=-2,61\)

\(\to \qquad N_{2}(4,61/0)\)

\(\to \qquad N_{3}(-2,61/0)\)

\(f(x)=\frac{1}{12}x^{4}-\frac{1}{6}x^{3}-x^{2}\)

\(f'(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x\)

\(f''(x)=x^{2}-x-2\)

\(f'''(x)=2x-1\)

\(f'(x)=0\)

\(\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x=0\)

\(x\cdot (\frac{1}{3}\cdot x^{2}-\frac{1}{2}\cdot x-2)=0\)

\(1.\)

\(x_{1}=0\)

\(f(0)=0\)

\(f''(0)=-2 \qquad \to \qquad <\qquad 0 \qquad \to H(0/0)\)

\(2.\)

\(\frac{1}{3}\cdot x^{2}-\frac{1}{2}\cdot x-2=0\)

große Lösungsformel:

\(x_{2}=\frac{3+\sqrt{105}}{4}=3,31 \qquad \to \qquad f(3,31)=-7,00\)

\(x_{3}=\frac{3-\sqrt{105}}{4}=-1,81 \qquad \to \qquad f(-1,81)=-1,39\)

\(f''(3,31)=5,66 \qquad \to \qquad > \qquad 0 \qquad \to \qquad T(3,31/-7,00)\)

\(f''(-1,81)=3,09 \qquad \to \qquad > \qquad 0 \qquad \to \qquad T(-1,81/-1,39)\)

\(f(x)=\frac{1}{12}x^{4}-\frac{1}{6}x^{3}-x^{2}\)

\(f'(x)=\frac{1}{3}x^{3}-\frac{1}{2}x^{2}-2x\)

\(f''(x)=x^{2}-x-2\)

\(f'''(x)=2x-1\)

\(f''(x)=0\)

\(x^{2}-x-2=0\)

Kleine Lösungsformel:

\(x_{1,2}=\frac{1}{2}\pm \sqrt{\frac{1}{4}+{2}}\)

\(x_{1}=-1 \qquad \to \qquad f(-1)=-\frac{3}{4}\)

\(x_{2}=2 \qquad \to \qquad f(2)=-4\)

\(W_{1}(-1/-0,75)\)

\(W_{2}(2/-4)\)

\(f(x)=-\frac{1}{8}x^{3}+a_{2}x^{2}+2\)

\(f'(x)=-\frac{3}{8}x^{2}+2a_{2}x\)

\(f''(x)=-\frac{3}{4}x+2a_{2}\)

\(f''(2)=0\)

\(-\frac{3}{4}\cdot 2+2a_{2}=0\)

\(a_{2}=\frac{3}{4}\)

\(f(x)=-\frac{1}{8}x^{3}+\frac{3}{4}x^{2}+2\)

\(f(x)=-\frac{1}{8}x^{3}+\frac{3}{4}x^{2}+2\)

\(f'(x)=-\frac{3}{8}x^{2}+\frac{3}{2}x\)

\(f''(x)=-\frac{3}{4}x+\frac{3}{2}\)

\(f'(x)=0\)

\(-\frac{3}{8}x^{2}+\frac{3}{2}x=0\)

\(x(-\frac{3}{8}x+\frac{3}{2})=0\)

\(1.\)

\(x_{1}=0\)

\(f(0)=2\)

\(2.\)

\(-\frac{3}{8}x+\frac{3}{2}=0\)

\(\frac{3}{8}x=\frac{3}{2}\)

\(x_{2}=4\)

\(f(4)=6\)

\(f''(0)=\frac{3}{2} \qquad \to \qquad >\, 0 \qquad \to \qquad T(0/2)\)

\(f''(4)=-\frac{3}{2} \qquad \to \qquad <\, 0 \qquad \to \qquad H(4/6)\)

\(f(x)=-\frac{1}{8}x^{3}+\frac{3}{4}x^{2}+2\)

\(f'(x)=-\frac{3}{8}x^{2}+\frac{3}{2}x\)

\(f''(x)=-\frac{3}{4}x+\frac{3}{2}\)

\(f''(x)=0\)

\(-\frac{3}{4}x+\frac{3}{2}=0\)

\(\frac{3}{4}x=\frac{3}{2}\)

\(x=2\)

\(f(2)=4\)

\(\to \qquad W(2/4)\)

\(f(x)=x^{3}+a_{1}x+a_{0}\)

\(f'(x)=3x^{2}+a_{1}\)

\(f''(x)=6x\)

 Wendepunkt:

 \(6x=0 \\ x_w=0\) 

Tangente am Wendepunkt: \(f(x_w)+f'(x_w)(x-x_w)= a_0+a_1x= -\frac 32 x +2\)

\(a_1= -\frac 32 \\ a_2= 2\)

\(f(x)=a_{3}x^{3}+a_{2}x^{2}-\frac{9}{2}x+a_{0}\)

\(f'(x)=3a_{3}x^{2}+2a_{2}x-\frac{9}{2}\)

\(f''(x)=6a_{3}x+2a_{2}\)

\(I:\qquad f(4)=6\)

\(II:\qquad f'(4)=\frac{3}{2}\)

\(III: \qquad f''(4)=0\)

\(\rule[10]{150}{1}\)

\(I: \qquad \qquad 64a_{3}+16a_{2}+a_{0}=24\)

\(II: \qquad \qquad 48a_{3}+8a_{2}=6\)

\(III:\qquad 24a_{3}+2a_{2}=0\)

\(II-2\cdot III:\)

\(48a_{3}+8a_{2}=6\)

\(48a_{3}+4a_{2}=0\)

\(\rule[10]{150}{1}\)

\(a_{2}=\frac{3}{2}\)

\(\rule[10]{150}{1}\)

\(24a_{3}+2a_{2}=0\)

\(24a_{3}+2\cdot \frac{3}{2}=0\)

\(a_{3}=-\frac{1}{8}\)

\(64a_{3}+16a_{2}+a_{0}=24\)

\(64\cdot (-\frac{1}{8})+16\cdot (\frac{3}{2}) +a_{0}=24\)

\(a_{0}=8\)

\(\to \qquad f(x)=-\frac{1}{8}x^{3}+\frac{3}{2}x^{2}-\frac{9}{2}x+8\)

\(f(x)=-\frac{1}{8}x^{3}+\frac{3}{2}x^{2}-\frac{9}{2}x+8\)

\(f'(x)=-\frac{3}{8}x^{2}+3x-\frac{9}{2}\)

\(f''(x)=-\frac{3}{4}x+3\)

\(f'(x)=0\)

\(-\frac{3}{8}x^{2}+3x-\frac{9}{2}=0 \qquad /\qquad \cdot\qquad -\frac{8}{3}\)

\(x^{2}-8x+12=0\)

Kleine Lösungsformel:

\(x_{1,2}=4\pm\sqrt{16-12}\)

\(x_{1}=6\)

\(x_{2}=2\)

\(f(6)=8\)

\(f''(6)=-1,5 \qquad \to \qquad < \qquad 0 \qquad \to \qquad H(6/8)\)

\(f(2)=4\)

\(f''(2)=1,5 \qquad \to \qquad > \qquad 0 \qquad \to \qquad T(2/4)\)

\(f(x)=-\frac{1}{8}x^{3}+\frac{3}{2}x^{2}-\frac{9}{2}x+8\)

\(f'(x)=-\frac{3}{8}x^{2}+3x-\frac{9}{2}\)

\(f''(x)=-\frac{3}{4}x+3\)

\(f''(x)=0\)

\(-\frac{3}{4}x+3=0\)

\(\frac{3}{4}x=3\)

\(x=4\)

\(f(4)=6\)

\(\to \qquad W(4/6)\)

\(f(x)=-\frac{1}{8}x^{3}+\frac{3}{2}x^{2}-\frac{9}{2}x+8\)

\(f'(x)=-\frac{3}{8}x^{2}+3x-\frac{9}{2}\)

\(f'(4)=-\frac{3}{8}\cdot 16+3\cdot 4-\frac{9}{2}\)

\(f'(4)=\frac{3}{2}=k\)

\(y=kx+d\)

\(y=y_{w}=6\)

\(x=x_{w}=4\)

\(6=\frac{3}{2}\cdot 4+d\)

\(d=0\)

\(t_{w}\,:\qquad y=\frac{3}{2}x\)

\(f(x)=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\)

\(f'(x)=3a_{3}x^{2}+2a_{2}x+a_{1}\)

\(f''(x)=6a_{3}x+2a_{2}\)

Aufstellen von 4 Gleichungen:

\(f(2)=3\)

\(f(0)=1\)

\(f''(0)=0\)

\(f'(0)=-3\)

\(I: \qquad 8a_{3}+4a_{2}+2a_{1}+a_{0}=3\)

\(II: \qquad a_{0}=1\)

\(III: \qquad a_{2}=0\)

\(IV: \qquad a_{1}=-3\)

\(8a_{3}+4a_{2}+2a_{1}+a_{0}=3\)

\(8a_{3}+4\cdot 0+2\cdot(-3)+1=3\)

\(a_{3}=1\)

\(\to \qquad f(x)=x^{3}-3x+1\)

\(f(x)=x^{3}-3x+1\)

\(f'(x)=3x^{2}-3\)

\(f''(x)=6x\)

\(f'(x)=0\)

\(3x^{2}-3=0\)

\(x^{2}=1 \)

\(x_{1,2}=\pm1\)

\(f(1)=-1\)

\(f''(1)=6 \qquad \to \qquad >\, 0 \qquad \to \qquad T(1/-1)\)

\(f(-1)=3\)

\(f''(-1)=-6 \qquad \to \qquad <\, 0 \qquad \to \qquad H(-1/3)\)

\(f(x)=x^{3}-3x+1\)

\(f'(x)=3x^{2}-3\)

\(f''(x)=6x\)

\(f''(x)=0\)

\(6x=0\)

\(x=0\)

\(f(0)=1\)

\(\to \qquad W(0/1)\)

\(W(0/1)\)

\(f(x)=x^{3}-3x+1\)

\(f'(x)=3x^{2}-3\)

\(f'(x_{w})=k\)

\(f'(0)=-3\)

\(k=-3\)

\(y=kx+d\)

\(y=y_{w}=1\)

\(x=x_{w}=0\)

\(1=-3\cdot 0+d\)

\(d=1\)

\(\to \qquad t_{w}\,: \qquad y=-3x+1\)

\(f(x)=a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}\)

\(f'(x)=3a_{3}x^{2}+2a_{2}x+a_{1}\)

\(f''(x)=6a_{3}x+2a_{2}\)

Verwertung der Informationen aus der Angabe:

\(I: \qquad \qquad f(0)=0\)

\(II: \qquad f'(0)=0\)

\(III: \, f''(1)=0\)

\(IV: \, f(1)=\frac{2}{3}\)

\(I: \qquad \qquad a_{0}=0\)

\(II: \qquad a_{1}=0\)

\(III: \qquad 6a_{3}+2a_{2}=0 \to \qquad 3a_{3}+a_{2}=0\)

\(IV: \qquad a_{3}+a_{2}+a_{1}+a_{0}=\frac{2}{3} \to \qquad a_{3}+a_{2}=\frac{2}{3}\)

\(\rule[10]{150}{1}\)

\(III: \qquad a_{2}=-3a_{3}\)

\(III\, in\, IV: \qquad a_{3}-3a_{3}=\frac{2}{3}\)

\(-2a_{3}=\frac{2}{3}\)

\(a_{3}=-\frac{1}{3}\)

\(a_{2}=-3\cdot (-\frac{1}{3})\)

\(a_{2}=1\)

\(\to \qquad f(x)=-\frac{1}{3}x^{3}+x^{2}\)

\(f(x)=-\frac{1}{3}x^{3}+x^{2}\)

\(f(x)=0\)

\(-\frac{1}{3}x^{3}+x^{2}=0\)

\(x^{2}(-\frac{1}{3}x+1)=0\)

\(1.\)

\(x_{1}\,^{2}=0\)

\(N_{1}\,^{(2)}(0/0) \to \qquad "Zweifachnullstelle"\)

\(2.\)

\(-\frac{1}{3}x+1=0\)

\(x_{2}=3\)

\(N_{2}(3/0)\)

\(f(x)=-\frac{1}{3}x^{3}+x^{2}\)

\(f'(x)=-x^{2}+2x\)

\(f''(x)=-2x+2\)

\(f'(x)=0\)

\(-x^{2}+2x=0\)

\(x(-x+2)=0\)

\(x_{1}=0\)

\(-x+2=0\)

\(x_{2}=2\)

\(f(0)=0\)

\(f''(0)=2\qquad \to \qquad >\, 0 \qquad \to \qquad T(0/0)\)

\(f(2)=1,33\)

\(f''(2)=-2\qquad \to \qquad <\, 0 \qquad \to \qquad H(2/1,33)\)

\(f(x)=-\frac{1}{3}x^{3}+x^{2}\)

\(f'(x)=-x^{2}+2x\)

\(f''(x)=-2x+2\)

\(f''(x)=0\)

\(-2x+2=0\)

\(x=1\)

\(f(1)=-\frac{1}{3}+1=\frac{2}{3}\)

\(\to \qquad W(1/\frac{2}{3})\)

\(f(x)=-\frac{1}{3}x^{3}+x^{2}\)

\(f'(x)=-x^{2}+2x\)

\(W(1/\frac{2}{3})\)

\(f'(1)=k\)

\(f'(1)=-1+2=1\)

\(k=1\)

\(y=kx+d\)

\(\frac{2}{3}=1\cdot1+d\)

\(d=-\frac{1}{3}\)

\(t_{w}\,:\qquad y=x-\frac{1}{3}\)

\(f(x)=a_{4}x^{4}+a_{3}x^{3}\)

\(f'(x)=4a_{4}x^{3}+3a_{3}x^{2}\)

\(f''(x)=12a_{4}x^{2}+6a_{3}x\)

\(I: \qquad f(-3)=\frac{27}{4}\)

\(II: \qquad f'(-3)=0\)

\(I: \qquad 81a_{4}-27a_{3}=\frac{27}{4}\)

\(II: \qquad -108a_{4}+27a_{3}=0\)

\(I+II:\)

\(-27a_{4}=\frac{27}{4}\)

\(a_{4}=-\frac{1}{4}\)

\(a_{4}\qquad in \qquad II:\)

\(-108\cdot(-\frac{1}{4})+27a_{3}=0\)

\(27+27a_{3}=0\)

\(a_{3}=-1\)

\(f(x)=-\frac{1}{4}x^{4}-x^{3}\)

\(f(x)=-\frac{1}{4}x^{4}-x^{3}\)

\(f(x)=0\)

\(-\frac{1}{4}x^{4}-x^{3}=0\)

\(x^{3}(-\frac{1}{4}x-1)=0\)

\(x_{1}=0\)

\(\to \qquad N_{1}\,^{(3)}(0/0)\)

\(-\frac{1}{4}x-1=0\)

\(\frac{1}{4}x=-1\)

\(x_{2}=-4\)

\(\to \qquad N_{2}(-4/0)\)

\(f(x)=-\frac{1}{4}x^{4}-x^{3}\)

\(f'(x)=-x^{3}-3x^{2}\)

\(f''(x)=-3x^{2}-6x\)

\(f'(x)=0\)

\(-x^{3}-3x^{2}=0\)

\(x^{2}(-x-3)=0\)

\(x_{1}=0\)

\(f(0)=0\)

\(f''(0)=0\)

\(\to \qquad kein\qquad globales \qquad Extremum\)

\(-x-3=0\)

\(x_{2}=-3\)

\(f(-3)=\frac{27}{4}\)

\(f''(-3)=-9\qquad \to \qquad <\, 0 \qquad \to \qquad H(-3/\frac{27}{4})\)

\(f(x)=-\frac{1}{4}x^{4}-x^{3}\)

\(f'(x)=-x^{3}-3x^{2}\)

\(f''(x)=-3x^{2}-6x\)

\(f''(x)=0\)

\(-3x^{2}-6x=0\)

\(x(-3x-6)=0\)

\(x_{1}=0\)

\(f(0)=0\)

\(\to \qquad W_{1}(0/0)\)

\(-3x-6=0\)

\(3x=-6\)

\(x_{2}=-2\)

\(f(-2)=-\frac{1}{4}\cdot16+8=4\)

\(\to \qquad W_{2}(-2/4)\)

\(W(-2/4)\)

\(f(x)=-\frac{1}{4}x^{4}-x^{3}\)

\(f'(x)=-x^{3}-3x^{2}\)

\(f'(-2)=k\)

\(f'(-2)=8-12=-4\)

\(k=-4\)

\(y=kx+d\)

\(4=-4\cdot(-2)+d\)

\(d=-4\)

\(\to \qquad t_{w}\,:\qquad y=-4x-4\)