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

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

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

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

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:

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

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

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

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

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

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

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

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

\(I+II:\)

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

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

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

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

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

\(a_{3}=-1\)

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

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

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

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

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

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

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

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

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

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

\(x_{2}=-3\)

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

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

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

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

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

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

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

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

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

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

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

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

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

\(x_{1}=0\)

\(f(0)=0\)

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

\(-3x-6=0\)

\(3x=-6\)

\(x_{2}=-2\)

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

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

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

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

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

\(f'(x)=0\)

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

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

\(x_{1}=0\)

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

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

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

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

Einsetzen der Werte in f(x):

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

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

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

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

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

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

\(-2x+2=0\)

\(x=1\)

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

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