$\begin{array}{l} \text{Gesucht:}\\ \text{Definitions- und Wertebereich} \\ \text{Grenzwerte} \\ \text{Symmetrie} \\ \text{Nullstellen - Schnittpunkt mit der x-Achse} \\ \text{Ableitungen - Stammfunktion} \\ \text{Extremwerte - Monotonie} \\ \text{Wendepunkte - Krümmung} \\ \text{Stammfunktion} \\ \text{Eingeschlossene Fläche mit der x-Achse} \\ \text{Funktion:}f\left(x\right)= \frac{1}{10}x^3+\frac{3}{10}x^2-1\frac{3}{5}x-4\frac{4}{5} \ <br/> \bullet \text{Funktion/Ableitungen/Stammfunktion} \\ f\left(x\right)= \frac{1}{10}x^3+\frac{3}{10}x^2-1\frac{3}{5}x-4\frac{4}{5}=\frac{1}{10}(x+4)(x+3)(x-4)\\ f'\left(x\right)= \frac{3}{10}x^2+\frac{3}{5}x-1\frac{3}{5}=\frac{3}{10}(x+3,52)(x-1,52)\\ f''\left(x\right)= \frac{3}{5}x+\frac{3}{5}=\frac{3}{5}(x+1)\\ f'''\left(x\right)= \frac{3}{5} \\ F(x)=\int_{}^{}( \frac{1}{10}x^3+\frac{3}{10}x^2-1\frac{3}{5}x-4\frac{4}{5})dx= \frac{1}{40}x^4+\frac{1}{10}x^3-\frac{4}{5}x^2-4\frac{4}{5}x+c \\ \\ \bullet\text{Definitions- und Wertebereich:}\\\qquad \mathbb{D} = \mathbb{R} \qquad \mathbb{W} = \mathbb{R} \\ \\ \bullet \text{Grenzwerte:} \\ f(x)=x^3( \frac{1}{10}+\dfrac{\frac{3}{10}}{x}-\dfrac{1\frac{3}{5}}{x^2}-\dfrac{4\frac{4}{5}}{x^3}) \\ \lim\limits_{x \rightarrow \infty}{f\left(x\right)}=[\frac{1}{10}\cdot \infty^3]=\infty \\\lim\limits_{x \rightarrow -\infty}{f\left(x\right)}=[\frac{1}{10}\cdot (-\infty)^3]=-\infty \\ \\ \bullet \text{Symmetrie zum Ursprung oder zur y-Achse } \\f\left(-x\right)=\frac{1}{10}\cdot (-x)^{3}+\frac{3}{10}\cdot (-x)^{2}-1\frac{3}{5}\cdot (-x)-4\frac{4}{5} \\ \text{keine Symmetrie zur y-Achse und zum Ursprung } \\ \\ \bullet \text{Nullstellen / Schnittpunkt mit der x-Achse:} \\f(x)= \frac{1}{10}x^3+\frac{3}{10}x^2-1\frac{3}{5}x-4\frac{4}{5} = 0 \\ \\ \frac{1}{10}x^3+\frac{3}{10}x^2-1\frac{3}{5}x-4\frac{4}{5}=0 \\\\ \text{Nullstelle für Polynmomdivision erraten:}-3\\ \,\small \begin{matrix} ( \frac{1}{10}x^3&+\frac{3}{10}x^2&-1\frac{3}{5}x&-4\frac{4}{5}&):( x +3 )= \frac{1}{10}x^2 -5,55\cdot 10^{-17}x -1\frac{3}{5} \\ \,-( \frac{1}{10}x^3&+\frac{3}{10}x^2) \\ \hline &-5,55\cdot 10^{-17}x^2&-1\frac{3}{5}x&-4\frac{4}{5}&\\ &-(-5,55\cdot 10^{-17}x^2&-1,67\cdot 10^{-16}x) \\ \hline &&-1\frac{3}{5}x&-4\frac{4}{5}&\\ &&-(-1\frac{3}{5}x&-4\frac{4}{5}) \\ \hline &&&0\\ \end{matrix} \\ \normalsize \\ \\ \frac{1}{10}x^{2}-5,55\cdot 10^{-17}x-1\frac{3}{5} =0 \\ x_{1/2}=\displaystyle\frac{+5,55\cdot 10^{-17} \pm\sqrt{\left(-5,55\cdot 10^{-17}\right)^{2}-4\cdot \frac{1}{10} \cdot \left(-1\frac{3}{5}\right)}}{2\cdot\frac{1}{10}} \\ x_{1/2}=\displaystyle \frac{+5,55\cdot 10^{-17} \pm\sqrt{\frac{16}{25}}}{\frac{1}{5}} \\ x_{1/2}=\displaystyle \frac{5,55\cdot 10^{-17} \pm\frac{4}{5}}{\frac{1}{5}} \\ x_{1}=\displaystyle \frac{5,55\cdot 10^{-17} +\frac{4}{5}}{\frac{1}{5}} \qquad x_{2}=\displaystyle \frac{5,55\cdot 10^{-17} -\frac{4}{5}}{\frac{1}{5}} \\ x_{1}=4 \qquad x_{2}=-4 \\ \underline{x_1=-4; \quad1\text{-fache Nullstelle}} \\\underline{x_2=-3; \quad1\text{-fache Nullstelle}} \\\underline{x_3=4; \quad1\text{-fache Nullstelle}} \\ \\ \bullet \text{Vorzeichentabelle:} \\ \begin{array}{|c|c|c|c|c|c|c|c|c|c||} \hline & x < &-4&< x <&-3&< x <&4&< x\\ \hline f(x)&-&0&+&0&-&0&+\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-4;-3[\quad \cup \quad]4;\infty[\quad f(x)>0 \quad \text{oberhalb der x-Achse}}\\ \\ \underline{\quad x \in ]-\infty;-4[\quad \cup \quad]-3;4[\quad f(x)<0 \quad \text{unterhalb der x-Achse}} \\ \\ \bullet \text{Extremwerte/Hochpunkte/Tiefpunkte:} \\f'(x)= \frac{3}{10}x^2+\frac{3}{5}x-1\frac{3}{5} = 0 \\ \\ \\ \frac{3}{10}x^{2}+\frac{3}{5}x-1\frac{3}{5} =0 \\ x_{1/2}=\displaystyle\frac{-\frac{3}{5} \pm\sqrt{\left(\frac{3}{5}\right)^{2}-4\cdot \frac{3}{10} \cdot \left(-1\frac{3}{5}\right)}}{2\cdot\frac{3}{10}} \\ x_{1/2}=\displaystyle \frac{-\frac{3}{5} \pm\sqrt{2\frac{7}{25}}}{\frac{3}{5}} \\ x_{1/2}=\displaystyle \frac{-\frac{3}{5} \pm1,51}{\frac{3}{5}} \\ x_{1}=\displaystyle \frac{-\frac{3}{5} +1,51}{\frac{3}{5}} \qquad x_{2}=\displaystyle \frac{-\frac{3}{5} -1,51}{\frac{3}{5}} \\ x_{1}=1,52 \qquad x_{2}=-3,52 \\ \underline{x_4=-3,52; \quad1\text{-fache Nullstelle}} \\\underline{x_5=1,52; \quad1\text{-fache Nullstelle}} \\f''(-3,52)=-1,51 \\ f''(-3,52)<0 \Rightarrow \underline{\text{Hochpunkt:} (-3,52/0,188)} \\ f''(1,52)=1,51>0 \Rightarrow \underline{\text{Tiefpunkt:} (1,52/-6,19)} \\ \\ \bullet\text{Monotonie/ streng monoton steigend (sms)/streng monoton fallend (smf) } \\ \begin{array}{|c|c|c|c|c|c|c||} \hline & x < &-3,52&< x <&1,52&< x\\ \hline f'(x)&+&0&-&0&+\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-\infty;-3,52[\quad \cup \quad]1,52;\infty[\quad f'(x)>0 \quad \text{streng monoton steigend }}\\ \\ \underline{\quad x \in ]-3,52;1,52[\quad f'(x)<0 \quad \text{streng monoton fallend }} \\ \\\bullet\text{Wendepunkte:} \\f''(x)= \frac{3}{5}x+\frac{3}{5} = 0 \\ \\ \frac{3}{5} x+\frac{3}{5} =0 \qquad /-\frac{3}{5} \\ \frac{3}{5} x= -\frac{3}{5} \qquad /:\frac{3}{5} \\ x=\displaystyle\frac{-\frac{3}{5}}{\frac{3}{5}}\\ x=-1 \\ \underline{x_6=-1; \quad1\text{-fache Nullstelle}} \\f'''(-1)=-3\\ f'''(-1) \neq 0 \Rightarrow \\ \underline{\text{Wendepunkt:} (-1/-3)}\\ \bullet\text{Kruemmung} \\ \begin{array}{|c|c|c|c||} \hline & x < &-1&< x\\ \hline f''(x)&-&0&+\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-1;\infty[\quad f''(x)>0 \quad \text{linksgekrümmt}}\\ \\ \underline{\quad x \in ]-\infty;-1[\quad f''(x)<0 \quad \text{rechtsgekrümmt}}\\ \\ \bullet\text{Eingeschlossene Fläche mit der x-Achse} \\A=\int_{-4}^{-3}\left( \frac{1}{10}x^3+\frac{3}{10}x^2-1\frac{3}{5}x-4\frac{4}{5}\right)dx=\left[ \frac{1}{40}x^4+\frac{1}{10}x^3-\frac{4}{5}x^2-4\frac{4}{5}x\right]_{-4}^{-3} \\ =\left(\frac{1}{40}\cdot (-3)^{4}+\frac{1}{10}\cdot (-3)^{3}-\frac{4}{5}\cdot (-3)^{2}-4\frac{4}{5}\cdot (-3)\right)-\left(\frac{1}{40}\cdot (-4)^{4}+\frac{1}{10}\cdot (-4)^{3}-\frac{4}{5}\cdot (-4)^{2}-4\frac{4}{5}\cdot (-4)\right) \\ =\left(6\frac{21}{40}\right)-\left(6\frac{2}{5}\right)=\frac{1}{8} \\ A=\int_{-3}^{4}\left( \frac{1}{10}x^3+\frac{3}{10}x^2-1\frac{3}{5}x-4\frac{4}{5}\right)dx=\left[ \frac{1}{40}x^4+\frac{1}{10}x^3-\frac{4}{5}x^2-4\frac{4}{5}x\right]_{-3}^{4} \\ =\left(\frac{1}{40}\cdot 4^{4}+\frac{1}{10}\cdot 4^{3}-\frac{4}{5}\cdot 4^{2}-4\frac{4}{5}\cdot 4\right)-\left(\frac{1}{40}\cdot (-3)^{4}+\frac{1}{10}\cdot (-3)^{3}-\frac{4}{5}\cdot (-3)^{2}-4\frac{4}{5}\cdot (-3)\right) \\ =\left(-19\frac{1}{5}\right)-\left(6\frac{21}{40}\right)=-25\frac{29}{40} \\ \\ \end{array}$