Analysis-Kurvendiskussion-Ganzrationale Funktion



Beispiel Nr: 55
$\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)=-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5} \ $ <br/> $\bullet \text{Funktion/Ableitungen/Stammfunktion} \\ f\left(x\right)=-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5}=-5\frac{2}{5}(x+4)(x+2)(x+1)\\ f'\left(x\right)=-16\frac{1}{5}x^2-75\frac{3}{5}x-75\frac{3}{5}=-16\frac{1}{5}(x+3,22)(x+1,45)\\ f''\left(x\right)=-32\frac{2}{5}x-75\frac{3}{5}=-32\frac{2}{5}(x+2\frac{1}{3})\\ f'''\left(x\right)=-32\frac{2}{5} \\ F(x)=\int_{}^{}(-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5})dx=-1\frac{7}{20}x^4-12\frac{3}{5}x^3-37\frac{4}{5}x^2-43\frac{1}{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(-5\frac{2}{5}-\dfrac{37\frac{4}{5}}{x}-\dfrac{75\frac{3}{5}}{x^2}-\dfrac{43\frac{1}{5}}{x^3}) \\ \lim\limits_{x \rightarrow \infty}{f\left(x\right)}=[-5\frac{2}{5}\cdot \infty^3]=-\infty \\\lim\limits_{x \rightarrow -\infty}{f\left(x\right)}=[-5\frac{2}{5}\cdot (-\infty)^3]=\infty \\ \\ \bullet \text{Symmetrie zum Ursprung oder zur y-Achse } \\f\left(-x\right)=-5\frac{2}{5}\cdot (-x)^{3}-37\frac{4}{5}\cdot (-x)^{2}-75\frac{3}{5}\cdot (-x)-43\frac{1}{5} \\ \text{keine Symmetrie zur y-Achse und zum Ursprung } \\ \\ \bullet \text{Nullstellen / Schnittpunkt mit der x-Achse:} \\f(x)=-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5} = 0 \\ \\-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5}=0 \\\\ \text{Nullstelle für Polynmomdivision erraten:}-2\\ \,\small \begin{matrix} (-5\frac{2}{5}x^3&-37\frac{4}{5}x^2&-75\frac{3}{5}x&-43\frac{1}{5}&):( x +2 )=-5\frac{2}{5}x^2 -27x -21\frac{3}{5} \\ \,-(-5\frac{2}{5}x^3&-10\frac{4}{5}x^2) \\ \hline &-27x^2&-75\frac{3}{5}x&-43\frac{1}{5}&\\ &-(-27x^2&-54x) \\ \hline &&-21\frac{3}{5}x&-43\frac{1}{5}&\\ &&-(-21\frac{3}{5}x&-43\frac{1}{5}) \\ \hline &&&0\\ \end{matrix} \\ \normalsize \\ \\ -5\frac{2}{5}x^{2}-27x-21\frac{3}{5} =0 \\ x_{1/2}=\displaystyle\frac{+27 \pm\sqrt{\left(-27\right)^{2}-4\cdot \left(-5\frac{2}{5}\right) \cdot \left(-21\frac{3}{5}\right)}}{2\cdot\left(-5\frac{2}{5}\right)} \\ x_{1/2}=\displaystyle \frac{+27 \pm\sqrt{262\frac{11}{25}}}{-10\frac{4}{5}} \\ x_{1/2}=\displaystyle \frac{27 \pm16\frac{1}{5}}{-10\frac{4}{5}} \\ x_{1}=\displaystyle \frac{27 +16\frac{1}{5}}{-10\frac{4}{5}} \qquad x_{2}=\displaystyle \frac{27 -16\frac{1}{5}}{-10\frac{4}{5}} \\ x_{1}=-4 \qquad x_{2}=-1 \\ \underline{x_1=-4; \quad1\text{-fache Nullstelle}} \\\underline{x_2=-2; \quad1\text{-fache Nullstelle}} \\\underline{x_3=-1; \quad1\text{-fache Nullstelle}} \\ \\ \bullet \text{Vorzeichentabelle:} \\ \begin{array}{|c|c|c|c|c|c|c|c|c|c||} \hline & x < &-4&< x <&-2&< x <&-1&< x\\ \hline f(x)&+&0&-&0&+&0&-\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-\infty;-4[\quad \cup \quad]-2;-1[\quad f(x)>0 \quad \text{oberhalb der x-Achse}}\\ \\ \underline{\quad x \in ]-4;-2[\quad \cup \quad]-1;\infty[\quad f(x)<0 \quad \text{unterhalb der x-Achse}} \\ \\ \bullet \text{Extremwerte/Hochpunkte/Tiefpunkte:} \\f'(x)=-16\frac{1}{5}x^2-75\frac{3}{5}x-75\frac{3}{5} = 0 \\ \\ \\ -16\frac{1}{5}x^{2}-75\frac{3}{5}x-75\frac{3}{5} =0 \\ x_{1/2}=\displaystyle\frac{+75\frac{3}{5} \pm\sqrt{\left(-75\frac{3}{5}\right)^{2}-4\cdot \left(-16\frac{1}{5}\right) \cdot \left(-75\frac{3}{5}\right)}}{2\cdot\left(-16\frac{1}{5}\right)} \\ x_{1/2}=\displaystyle \frac{+75\frac{3}{5} \pm\sqrt{816\frac{12}{25}}}{-32\frac{2}{5}} \\ x_{1/2}=\displaystyle \frac{75\frac{3}{5} \pm28,6}{-32\frac{2}{5}} \\ x_{1}=\displaystyle \frac{75\frac{3}{5} +28,6}{-32\frac{2}{5}} \qquad x_{2}=\displaystyle \frac{75\frac{3}{5} -28,6}{-32\frac{2}{5}} \\ x_{1}=-3,22 \qquad x_{2}=-1,45 \\ \underline{x_4=-3,22; \quad1\text{-fache Nullstelle}} \\\underline{x_5=-1,45; \quad1\text{-fache Nullstelle}} \\f''(-3,22)=28,6>0 \Rightarrow \underline{\text{Tiefpunkt:} (-3,22/-11,4)} \\ f''(-1,45)=-28,6 \\ f''(-1,45)<0 \Rightarrow \underline{\text{Hochpunkt:} (-1,45/3,41)} \\ \\ \bullet\text{Monotonie/ streng monoton steigend (sms)/streng monoton fallend (smf) } \\ \begin{array}{|c|c|c|c|c|c|c||} \hline & x < &-3,22&< x <&-1,45&< x\\ \hline f'(x)&-&0&+&0&-\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-3,22;-1,45[\quad f'(x)>0 \quad \text{streng monoton steigend }}\\ \\ \underline{\quad x \in ]-\infty;-3,22[\quad \cup \quad]-1,45;\infty[\quad f'(x)<0 \quad \text{streng monoton fallend }} \\ \\\bullet\text{Wendepunkte:} \\f''(x)=-32\frac{2}{5}x-75\frac{3}{5} = 0 \\ \\ -32\frac{2}{5} x-75\frac{3}{5} =0 \qquad /+75\frac{3}{5} \\ -32\frac{2}{5} x= 75\frac{3}{5} \qquad /:\left(-32\frac{2}{5}\right) \\ x=\displaystyle\frac{75\frac{3}{5}}{-32\frac{2}{5}}\\ x=-2\frac{1}{3} \\ \underline{x_6=-2\frac{1}{3}; \quad1\text{-fache Nullstelle}} \\f'''(-2\frac{1}{3})=-4\\ f'''(-2\frac{1}{3}) \neq 0 \Rightarrow \\ \underline{\text{Wendepunkt:} (-2\frac{1}{3}/-4)}\\ \bullet\text{Kruemmung} \\ \begin{array}{|c|c|c|c||} \hline & x < &-2\frac{1}{3}&< x\\ \hline f''(x)&+&0&-\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-\infty;-2\frac{1}{3}[\quad f''(x)>0 \quad \text{linksgekrümmt}}\\ \\ \underline{\quad x \in ]-2\frac{1}{3};\infty[\quad f''(x)<0 \quad \text{rechtsgekrümmt}}\\ \\ \bullet\text{Eingeschlossene Fläche mit der x-Achse} \\A=\int_{-4}^{-2}\left(-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5}\right)dx=\left[-1\frac{7}{20}x^4-12\frac{3}{5}x^3-37\frac{4}{5}x^2-43\frac{1}{5}x\right]_{-4}^{-2} \\ =\left(-1\frac{7}{20}\cdot (-2)^{4}-12\frac{3}{5}\cdot (-2)^{3}-37\frac{4}{5}\cdot (-2)^{2}-43\frac{1}{5}\cdot (-2)\right)-\left(-1\frac{7}{20}\cdot (-4)^{4}-12\frac{3}{5}\cdot (-4)^{3}-37\frac{4}{5}\cdot (-4)^{2}-43\frac{1}{5}\cdot (-4)\right) \\ =\left(14\frac{2}{5}\right)-\left(28\frac{4}{5}\right)=-14\frac{2}{5} \\ A=\int_{-2}^{-1}\left(-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5}\right)dx=\left[-1\frac{7}{20}x^4-12\frac{3}{5}x^3-37\frac{4}{5}x^2-43\frac{1}{5}x\right]_{-2}^{-1} \\ =\left(-1\frac{7}{20}\cdot (-1)^{4}-12\frac{3}{5}\cdot (-1)^{3}-37\frac{4}{5}\cdot (-1)^{2}-43\frac{1}{5}\cdot (-1)\right)-\left(-1\frac{7}{20}\cdot (-2)^{4}-12\frac{3}{5}\cdot (-2)^{3}-37\frac{4}{5}\cdot (-2)^{2}-43\frac{1}{5}\cdot (-2)\right) \\ =\left(16\frac{13}{20}\right)-\left(14\frac{2}{5}\right)=2\frac{1}{4} \\ \\ $