4.3 Extrema-Wendepunkte

Beispiel 1: $\begin{aligned} f (x) & = \frac{1}{3} x^{3} - 2 x^{2} + 3 x + 1 \\ f^{'} (x) & = x^{2} - 4 x + 3 \end{aligned}$

\begin{aligned} 0 & = x^{2} - 4 x + 3 \\ x_{e_{1 / 2}} & = 2 \pm \sqrt{4 - 3} \\ x_{e 1 / 2} & = 2 \pm 1 \\ x_{e 1} & = 3 \\ x_{e 2} & = 1 \end{aligned}

f^{''} | x | = 2 x - 4

\begin{array}{lll} f^{''} (3) = 6 - 4 > 0 & (Links) & Minimalstelle \\ f^{''} (1) = 2 - 4 < 0 & (Rechts) & Maximalstelle \end{array}

\begin{array}{ll} f (3) = 1 & ⟹ T (3 ∣ 1) \\ f (1) = \frac{\underset{―}{7}}{3} & ⟹ T (3 ∣ \frac{7}{3}) \end{array}

\begin{aligned} 0 = 2 x - 4 \\ x_{w} = 2 \end{aligned}

f^{'''} (x) = 2 > 0 R L - Wendepunkt

f (2) = \frac{5}{3} ⟹ W (2 ∣ \frac{5}{3})