«11C» 56. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 56. Hausaufgabe

0.0.1.1 ↑ Selbstgestellte Aufgabe

\mathrm{f}(x) = \frac{x}{3}\left|x^2 - 4\right|; \quad D_{\mathrm{f}} = \mathds{R};$\text{f}\left(x\right)=\frac{x}{3}\left|x^2-4\right|;D_{\text{f}}=R;$

Nullstellen

\mathrm{f}(x) = 0;$\text{f}\left(x\right)=0;$

⇒ N_1(-2, 0); \quad N_2(0, 0); \quad N_3(2, 0);$N_1\left(-2,0\right);N_2\left(0,0\right);N_3\left(2,0\right);$

Symmetrie

\mathrm{f}(-x) = \frac{-x}{3}\left|\left(-x\right)^2 - 4\right| = -\frac{x}{3}\left|x^2 - 4\right| = -\mathrm{f}(x);$\text{f}\left(-x\right)=\frac{-x}{3}\left|{\left(-x\right)}^2-4\right|=-\frac{x}{3}\left|x^2-4\right|=-\text{f}\left(x\right);$ ⇒

Punktsymmetrie zum Ursprung;

Extrema

{} \mathrm{f}(x) = \begin{cases} {} \frac{x}{3}\left(x^2 - 4\right) = \frac{1}{3}x^3 - \frac{4}{3}x {} & \text{f"ur } x \in {} \left]-\infty, -2\right] \cup \left[2, \infty\right[; \\ {} -\frac{x}{3}\left(x^2 - 4\right) = -\left(\frac{1}{3}x^3 - \frac{4}{3}x\right) {} & \text{f"ur } x \in \left]-2, 2\right[; {} \end{cases}$\text{f}\left(x\right)=\left\{\begin{array}{cc}\frac{x}{3}\left(x^2-4\right)=\frac{1}{3}{x}^3-\frac{4}{3}x\hfill & \text{f”ur }x\in \left]-\infty ,-2\right]\cup \left[2,\infty \right[;\hfill \\ -\frac{x}{3}\left(x^2-4\right)=-\left(\frac{1}{3}{x}^3-\frac{4}{3}x\right)\hfill & \text{f”ur }x\in \left]-2,2\right[;\hfill \end{array}\right.$

⇒ {} \mathrm{f}'(x) = \begin{cases} {} x^2 - \frac{4}{3} & {} \text{f"ur } x \in \left]-\infty, -2\right[ \cup \left]2, \infty\right[; \\ {} -\left(x^2 - \frac{4}{3}\right) & {} \text{f"ur } x \in \left]-2, 2\right[; {} \end{cases}${\text{f}}^{\prime }\left(x\right)=\left\{\begin{array}{cc}{x}^2-\frac{4}{3}\hfill & \text{f”ur }x\in \left]-\infty ,-2\right[\cup \left]2,\infty \right[;\hfill \\ -\left(x^2-\frac{4}{3}\right)\hfill & \text{f”ur }x\in \left]-2,2\right[;\hfill \end{array}\right.$

⇒ \pm\left(x_0^2 - \frac{4}{3}\right) = 0; \Rightarrow x_0^2 = \frac{4}{3};$\pm \left(x_0^2-\frac{4}{3}\right)=0;\Rightarrow x_0^2=\frac{4}{3};$

⇒ x_1 = -\sqrt{\frac{4}{3}} = -\frac{2}{\sqrt{3}} = -\frac{2}{3}\sqrt{3}; \quad x_2 = \frac{2}{3}\sqrt{3};$x_1=-\sqrt{\frac{4}{3}}=-\frac{2}{\sqrt{3}}=-\frac{2}{3}\sqrt{3};x_2=\frac{2}{3}\sqrt{3};$

Vorzeichenanalyse von \mathrm{f}'${\text{f}}^{\prime }$:

x_1 = -\frac{2}{3}\sqrt{3};$x_1=-\frac{2}{3}\sqrt{3};$

Vorzeichenwechsel von \mathrm{f}'${\text{f}}^{\prime }$ von -$-$ nach +$+$; ⇒ P_{\mathrm{TIP}, 1}(-\frac{2}{3}\sqrt{3}, -\frac{16}{27}\sqrt{3});$P_{\text{TIP},1}\left(-\frac{2}{3}\sqrt{3},-\frac{16}{27}\sqrt{3}\right);$

x_2 = \frac{2}{3}\sqrt{3};$x_2=\frac{2}{3}\sqrt{3};$

Vorzeichenwechsel von \mathrm{f}'${\text{f}}^{\prime }$ von +$+$ nach -$-$; ⇒ P_{\mathrm{HOP}, 1}(\frac{2}{3}\sqrt{3}, \frac{16}{27}\sqrt{3});$P_{\text{HOP},1}\left(\frac{2}{3}\sqrt{3},\frac{16}{27}\sqrt{3}\right);$

x_3 = -2;$x_3=-2;$

Vorzeichenwechsel von \mathrm{f}'${\text{f}}^{\prime }$ von +$+$ nach -$-$; ⇒ P_{\mathrm{HOP}, 2}(-2, 0);$P_{\text{HOP},2}\left(-2,0\right);$

x_4 = 2;$x_4=2;$

Vorzeichenwechsel von \mathrm{f}'${\text{f}}^{\prime }$ von -$-$ nach +$+$; ⇒ P_{\mathrm{TIP}, 2}(2, 0);$P_{\text{TIP},2}\left(2,0\right);$

Wendepunkte

{} \mathrm{f}''(x) = \begin{cases} {} 2x & {} \text{f"ur } x \in \left]-\infty, -2\right[ \cup \left]2, \infty\right[; \\ {} -2x & {} \text{f"ur } x \in \left]-2, 2\right[; {} \end{cases}${\text{f}}^{\prime \prime }\left(x\right)=\left\{\begin{array}{cc}2x\hfill & \text{f”ur }x\in \left]-\infty ,-2\right[\cup \left]2,\infty \right[;\hfill \\ -2x\hfill & \text{f”ur }x\in \left]-2,2\right[;\hfill \end{array}\right.$

\pm 2x_5 = 0; \Rightarrow x_5 = 0;$\pm 2x_5=0;\Rightarrow x_5=0;$

⇒ P_{\mathrm{WEP}, 1}(0, 0);$P_{\text{WEP},1}\left(0,0\right);$

Sowie: P_{\mathrm{WEP}, 2}(-2, 0); \quad P_{\mathrm{WEP}, 3}(2, 0);$P_{\text{WEP},2}\left(-2,0\right);P_{\text{WEP},3}\left(2,0\right);$