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

\(f(x)=0\)

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

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

\(1.\)

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

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

\(2.\)

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

\(x_{2}=3\)

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

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

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

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

\(f'(x)=0\)

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

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

Kleine Lösungsformel:

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

\(x_{1}=6\)

\(x_{2}=2\)

\(f(6)=8\)

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

\(f(2)=4\)

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

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

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

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

\(f'(x)=0\)

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

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

\(x_{1}=0\)

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

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

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

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

Einsetzen der Werte in f(x):

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

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

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:

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

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

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

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

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

\(f'(x)=0\)

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

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

\(1.\)

\(x_{1}=0\)

\(f(0)=0\)

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

\(2.\)

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

große Lösungsformel:

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

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

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

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

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

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

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

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

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

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

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

\(-2x+2=0\)

\(x=1\)

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

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

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

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

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

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