«11C» 46. Hausaufgabe «PDF», «POD»

0.0.1 ↑ 46. Hausaufgabe

0.0.1.1 ↑ Selbstgestellte Aufgabe

f(x) = x^4 - 4x^2 = x^2\left(x^2 - 4\right) = x^2\left(x + 2\right)\left(x - 2\right);$f\left(x\right)=x^4-4x^2=x^2\left(x^2-4\right)=x^2\left(x+2\right)\left(x-2\right);$

Nullstellen

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

f(-x) = x^4 - 4x^2 = f(x);$f\left(-x\right)=x^4-4x^2=f\left(x\right);$ ⇒ Symmetrie zur y$y$-Achse

Extrema

f'(x) = 4x^3 - 8x = 4x\left(x^2 - 2\right) = 4x\left(x + \sqrt{2}\right)\left(x - \sqrt{2}\right) = 0; \\$f^{\prime }\left(x\right)=4x^3-8x=4x\left(x^2-2\right)=4x\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0;$ f''(x) = 12x^2 - 8;$f^{\prime \prime }\left(x\right)=12x^2-8;$ ⇒

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

f''(x_1) = f''(-\sqrt{2}) > 0; \Rightarrow P_{\mathrm{TIP}}(-\sqrt{2}, -4);$f^{\prime \prime }\left(x_1\right)=f^{\prime \prime }\left(-\sqrt{2}\right)>0;\Rightarrow P_{\text{TIP}}\left(-\sqrt{2},-4\right);$

x_2 = 0;$x_2=0;$

f''(x_2) = f''(0) < 0; \Rightarrow P_{\mathrm{HOP}}(0, 0);$f^{\prime \prime }\left(x_2\right)=f^{\prime \prime }\left(0\right)<0;\Rightarrow P_{\text{HOP}}\left(0,0\right);$

x_3 = \sqrt{2};$x_3=\sqrt{2};$

f''(x_3) = f''(\sqrt{2}) > 0; \Rightarrow P_{\mathrm{TIP}}(\sqrt{2}, -4);$f^{\prime \prime }\left(x_3\right)=f^{\prime \prime }\left(\sqrt{2}\right)>0;\Rightarrow P_{\text{TIP}}\left(\sqrt{2},-4\right);$

Monotoniebereiche

f'(x) = 4x\left(x + \sqrt{2}\right)\left(x - \sqrt{2}\right) > 0;$f^{\prime }\left(x\right)=4x\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)>0;$ ⇒

f$f$ ist in sms in \left] -\sqrt{2}, 0 \right[ \cup \left] \sqrt{2}, \infty \right[;$\left]-\sqrt{2},0\right[\cup \left]\sqrt{2},\infty \right[;$

f$f$ ist in smf in \left] -\infty, -\sqrt{2} \right[ \cup \left] 0, \sqrt{2} \right[;$\left]-\infty ,-\sqrt{2}\right[\cup \left]0,\sqrt{2}\right[;$

Graph