Analysis-Kurvendiskussion-Ganzrationale Funktion



Beispiel Nr: 82
$\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^4-37\frac{4}{5}x^3-75\frac{3}{5}x^2-43\frac{1}{5}x \ $ <br/> $\bullet \text{Funktion/Ableitungen/Stammfunktion} \\ f\left(x\right)=-5\frac{2}{5}x^4-37\frac{4}{5}x^3-75\frac{3}{5}x^2-43\frac{1}{5}x=-5\frac{2}{5}(x+4)(x+2)(x+1)x\\ f'\left(x\right)=-21\frac{3}{5}x^3-113\frac{2}{5}x^2-151\frac{1}{5}x-43\frac{1}{5}=-21\frac{3}{5}(x+3,33)(x+1,53)(x+0,393)\\ f''\left(x\right)=-64\frac{4}{5}x^2-226\frac{4}{5}x-151\frac{1}{5}=-64\frac{4}{5}(x+2,6)(x+0,896)\\ f'''\left(x\right)=-129\frac{3}{5}x-226\frac{4}{5} \\ F(x)=\int_{}^{}(-5\frac{2}{5}x^4-37\frac{4}{5}x^3-75\frac{3}{5}x^2-43\frac{1}{5}x)dx=-1\frac{2}{25}x^5-9\frac{9}{20}x^4-25\frac{1}{5}x^3-21\frac{3}{5}x^2+c \\ \\ \bullet\text{Definitions- und Wertebereich:}\\\qquad \mathbb{D} = \mathbb{R} \qquad \mathbb{W} = ]-\infty,37,3] \\ \\ \bullet \text{Grenzwerte:} \\ f(x)=x^4(-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^4]=-\infty \\\lim\limits_{x \rightarrow -\infty}{f\left(x\right)}=[-5\frac{2}{5}\cdot (-\infty)^4]=-\infty \\ \\ \bullet \text{Symmetrie zum Ursprung oder zur y-Achse } \\f\left(-x\right)=-5\frac{2}{5}\cdot (-x)^{4}-37\frac{4}{5}\cdot (-x)^{3}-75\frac{3}{5}\cdot (-x)^{2}-43\frac{1}{5}\cdot (-x) \\ \text{keine Symmetrie zur y-Achse und zum Ursprung } \\ \\ \bullet \text{Nullstellen / Schnittpunkt mit der x-Achse:} \\f(x)=-5\frac{2}{5}x^4-37\frac{4}{5}x^3-75\frac{3}{5}x^2-43\frac{1}{5}x = 0 \\ x(-5\frac{2}{5}x^3-37\frac{4}{5}x^2-75\frac{3}{5}x-43\frac{1}{5})=0 \Rightarrow x=0 \quad \vee \quad-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}} \\\underline{x_4=0; \quad1\text{-fache Nullstelle}} \\ \\ \bullet \text{Vorzeichentabelle:} \\ \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c||} \hline & x < &-4&< x <&-2&< x <&-1&< x <&0&< x\\ \hline f(x)&-&0&+&0&-&0&+&0&-\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-4;-2[\quad \cup \quad]-1;0[\quad f(x)>0 \quad \text{oberhalb der x-Achse}}\\ \\ \underline{\quad x \in ]-\infty;-4[\quad \cup \quad]-2;-1[\quad \cup \quad]0;\infty[\quad f(x)<0 \quad \text{unterhalb der x-Achse}} \\ \\ \bullet \text{Extremwerte/Hochpunkte/Tiefpunkte:} \\f'(x)=-21\frac{3}{5}x^3-113\frac{2}{5}x^2-151\frac{1}{5}x-43\frac{1}{5} = 0 \\ \\-21\frac{3}{5}x^3-113\frac{2}{5}x^2-151\frac{1}{5}x-43\frac{1}{5}=0 \\\\ Numerische Suche: \\ \underline{x_5=-3,33; \quad1\text{-fache Nullstelle}} \\\underline{x_6=-1,53; \quad1\text{-fache Nullstelle}} \\\underline{x_7=-0,393; \quad1\text{-fache Nullstelle}} \\f''(-3,33)=-114 \\ f''(-3,33)<0 \Rightarrow \underline{\text{Hochpunkt:} (-3,33/37,3)} \\ f''(-1,53)=44,1>0 \Rightarrow \underline{\text{Tiefpunkt:} (-1,53/-5,08)} \\ f''(-0,393)=-72,1 \\ f''(-0,393)<0 \Rightarrow \underline{\text{Hochpunkt:} (-0,393/7,47)} \\ \\ \bullet\text{Monotonie/ streng monoton steigend (sms)/streng monoton fallend (smf) } \\ \begin{array}{|c|c|c|c|c|c|c|c|c|c||} \hline & x < &-3,33&< x <&-1,53&< x <&-0,393&< x\\ \hline f'(x)&+&0&-&0&+&0&-\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-\infty;-3,33[\quad \cup \quad]-1,53;-0,393[\quad f'(x)>0 \quad \text{streng monoton steigend }}\\ \\ \underline{\quad x \in ]-3,33;-1,53[\quad \cup \quad]-0,393;\infty[\quad f'(x)<0 \quad \text{streng monoton fallend }} \\ \\\bullet\text{Wendepunkte:} \\f''(x)=-64\frac{4}{5}x^2-226\frac{4}{5}x-151\frac{1}{5} = 0 \\ \\ \\ -64\frac{4}{5}x^{2}-226\frac{4}{5}x-151\frac{1}{5} =0 \\ x_{1/2}=\displaystyle\frac{+226\frac{4}{5} \pm\sqrt{\left(-226\frac{4}{5}\right)^{2}-4\cdot \left(-64\frac{4}{5}\right) \cdot \left(-151\frac{1}{5}\right)}}{2\cdot\left(-64\frac{4}{5}\right)} \\ x_{1/2}=\displaystyle \frac{+226\frac{4}{5} \pm\sqrt{12247\frac{1}{5}}}{-129\frac{3}{5}} \\ x_{1/2}=\displaystyle \frac{226\frac{4}{5} \pm111}{-129\frac{3}{5}} \\ x_{1}=\displaystyle \frac{226\frac{4}{5} +111}{-129\frac{3}{5}} \qquad x_{2}=\displaystyle \frac{226\frac{4}{5} -111}{-129\frac{3}{5}} \\ x_{1}=-2,6 \qquad x_{2}=-0,896 \\ \underline{x_8=-2,6; \quad1\text{-fache Nullstelle}} \\\underline{x_9=-0,896; \quad1\text{-fache Nullstelle}} \\f'''(-2,6)=19\\ f'''(-2,6) \neq 0 \Rightarrow \\ \underline{\text{Wendepunkt:} (-2,6/19)}\\ f'''(-0,896)=1,72\\ f'''(-0,896) \neq 0 \Rightarrow \\ \underline{\text{Wendepunkt:} (-0,896/1,72)}\\ \bullet\text{Kruemmung} \\ \begin{array}{|c|c|c|c|c|c|c||} \hline & x < &-2,6&< x <&-0,896&< x\\ \hline f''(x)&-&0&+&0&-\\ \hline \end{array}\\ \\ \underline{\quad x \in ]-2,6;-0,896[\quad f''(x)>0 \quad \text{linksgekrümmt}}\\ \\ \underline{\quad x \in ]-\infty;-2,6[\quad \cup \quad]-0,896;\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^4-37\frac{4}{5}x^3-75\frac{3}{5}x^2-43\frac{1}{5}x\right)dx=\left[-1\frac{2}{25}x^5-9\frac{9}{20}x^4-25\frac{1}{5}x^3-21\frac{3}{5}x^2\right]_{-4}^{-2} \\ =\left(-1\frac{2}{25}\cdot (-2)^{5}-9\frac{9}{20}\cdot (-2)^{4}-25\frac{1}{5}\cdot (-2)^{3}-21\frac{3}{5}\cdot (-2)^{2}\right)-\left(-1\frac{2}{25}\cdot (-4)^{5}-9\frac{9}{20}\cdot (-4)^{4}-25\frac{1}{5}\cdot (-4)^{3}-21\frac{3}{5}\cdot (-4)^{2}\right) \\ =\left(-1\frac{11}{25}\right)-\left(-46\frac{2}{25}\right)=44\frac{16}{25} \\ A=\int_{-2}^{-1}\left(-5\frac{2}{5}x^4-37\frac{4}{5}x^3-75\frac{3}{5}x^2-43\frac{1}{5}x\right)dx=\left[-1\frac{2}{25}x^5-9\frac{9}{20}x^4-25\frac{1}{5}x^3-21\frac{3}{5}x^2\right]_{-2}^{-1} \\ =\left(-1\frac{2}{25}\cdot (-1)^{5}-9\frac{9}{20}\cdot (-1)^{4}-25\frac{1}{5}\cdot (-1)^{3}-21\frac{3}{5}\cdot (-1)^{2}\right)-\left(-1\frac{2}{25}\cdot (-2)^{5}-9\frac{9}{20}\cdot (-2)^{4}-25\frac{1}{5}\cdot (-2)^{3}-21\frac{3}{5}\cdot (-2)^{2}\right) \\ =\left(-4\frac{77}{100}\right)-\left(-1\frac{11}{25}\right)=-3\frac{33}{100} \\ A=\int_{-1}^{0}\left(-5\frac{2}{5}x^4-37\frac{4}{5}x^3-75\frac{3}{5}x^2-43\frac{1}{5}x\right)dx=\left[-1\frac{2}{25}x^5-9\frac{9}{20}x^4-25\frac{1}{5}x^3-21\frac{3}{5}x^2\right]_{-1}^{0} \\ =\left(-1\frac{2}{25}\cdot 0^{5}-9\frac{9}{20}\cdot 0^{4}-25\frac{1}{5}\cdot 0^{3}-21\frac{3}{5}\cdot 0^{2}\right)-\left(-1\frac{2}{25}\cdot (-1)^{5}-9\frac{9}{20}\cdot (-1)^{4}-25\frac{1}{5}\cdot (-1)^{3}-21\frac{3}{5}\cdot (-1)^{2}\right) \\ =\left(0\right)-\left(-4\frac{77}{100}\right)=4\frac{77}{100} \\ \\ $