8.2 Kurvendiskussion

Beispiel: $f (x) = \frac{x^{2}}{x - 1}$

$x_{p} = 1$ ist Polstelle, da $Z (1) = 1 \neq 0$ gilt
$Z (x) > 0$ für alle $x$
- $N (x) < 0$ fir alle $x < 1$
- $N (x) > 0$ fur alle $x > 1$
  $⟹$ Polstelle mit Vorzeichenwechsel

\begin{aligned} x^{2} : (x - 1) = x + 1 + \frac{1}{x - 1} \\ \frac{- (x^{2} - x)}{x + 0} \\ \frac{- (x - 1)}{1} \\ \begin{matrix} ⟹ f (x) = \underset{Asymptote}{\underset{⏟}{x + 1}} \underset{Restterm}{\underset{⏟}{+ \frac{11}{x - 2}}} \end{matrix} \end{aligned}

\begin{aligned} ⟹ 0 & = \frac{x^{2}}{x - 1} \\ 0 & = x^{2} \\ x_{0} & = 0 \end{aligned}

$f^{'} (x) f (x) = \frac{x^{2}}{x - 1}$ Quotientenregel

\begin{aligned} u = x^{2} u^{'} = 2 x f (x) = \frac{u}{v} \\ v = x - 1 v^{'} = 1 f^{'} (x) = \frac{u^{'} \cdot v - v^{'} \cdot n}{v^{2}} \\ ⟹ f^{'} (x) = \frac{2 x (x - 1) - 1 \cdot x^{2}}{(x - 1)^{2}} \\ f^{'} (x) = \frac{x^{2} - 2 x}{(x - 1)^{2}} \end{aligned}

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

\begin{aligned} 0 = \frac{x^{2} - 2 x}{(x - 1)^{2}} \\ ⟹ x^{2} - 2 x = 2 \\ x_{e 1 / 2} = 1 \pm \sqrt{1 - 0} \\ x_{e 1} = 1 - 1 = 0 \\ x_{e 2} = 1 + 1 = 2 \end{aligned}

\begin{aligned} f^{''} (0) = \frac{2}{(0 - 1)^{3}} = - 2 < 0 \\ ⟹ Maximum \\ f^{''} (2) = \frac{2}{(2 - 1)^{2}} = 2 > 0 \\ ⟹ Minimum \end{aligned}

\begin{aligned} f (0) = 0 \\ f (2) = \frac{4}{2 - 1} = 4 \end{aligned}

$⟹ 0 = \frac{2}{(x - 1)^{3}} ⟹$ Keine Wendepunkte