Kurvendiskussion mit Parameter

Nullstellen bestimmen: Zähler gleich 0 setzen

$x^2+(k+1)x+k=0x^2+\backslash left(k+1\backslash right)x+k=0$

Diskriminante bestimmen und Mitternachtsformel anwenden.

$\begin{array}{c}D={(k+1)}^2-4k={(k-1)}^2\\ D\ge 0\mathrm{\forall }k\\ D=0\iff k=1\\ x_{1;2}=\frac{-k-1\pm \sqrt{{(k-1)}^2}}{2}\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}D=\backslash left(k+1\backslash right)^2-4k=\backslash left(k-1\backslash right)^2\backslash \backslash D\backslash geq\; 0\; \backslash \; \backslash forall\; k\backslash \backslash D=0\backslash Leftrightarrow\; k=1\backslash \backslash x\_\{1;2\}=\backslash frac\{-k-1\backslash pm\backslash sqrt\{\backslash left(k-1\backslash right)^2\}\}2\backslash end\{array\}$

Fall 1:

$x_1=\frac{-k-1+(1-k)}{2}=-kx\_1=\backslash frac\{-k-1+(1-k)\}\{2\}=-k$ $\backslash \backslash$ $\text{}{x}_2=\frac{-k-1-(1-k)}{2}=-1x\_2=\backslash frac\{-k-1-(1-k)\}\{2\}=-1$

Fall 2: $k>1k>1$

$x_1=\frac{-k-1+k-1}{2}=-1x\_1=\backslash frac\{-k-1+k-1\}\{2\}=-1$ $\backslash \backslash$ $x_2=\frac{-k-1-k+1}{2}=-kx\_2=\backslash frac\{-k-1-k+1\}\{2\}=-k$

Fall 1: Grenzwert gegen $kk$ für alle $k\in \mathbb{R}\setminus \{-1;0\}k\backslash in\backslash mathbb\{R\}\backslash setminus\backslash left\backslash \{-1;\backslash ;0\backslash right\backslash \}$ $\backslash \backslash$ Nicht hebbare Definitionslücken.

${\displaystyle \underset{x\to k\pm 0}{\lim }{f}_k\left(x\right)=\underset{x\to k\pm 0}{\lim }\frac{x^2+(k+1)x+k}{\underset{\to \pm 0}{\underbrace{x-k}}}=\pm \mathrm{\infty }}\backslash displaystyle\backslash lim\backslash limits\_\{x\backslash rightarrow\; k\backslash pm0\}f\_k\backslash left(x\backslash right)=\backslash lim\backslash limits\_\{x\backslash rightarrow\; k\backslash pm0\}\backslash frac\{x^2+\backslash left(k+1\backslash right)x+k\}\{\backslash underbrace\{x-k\}\_\{\backslash rightarrow\; \backslash pm0\}\}=\backslash pm\backslash infty$

Fall 2: $k=-1k=-1$ $\backslash \backslash$ Hebbare Definitionslücke.

${\displaystyle \begin{array}{c}\underset{x\to -1}{\lim }{f}_{-1}\left(x\right)\\ =\underset{x\to -1}{\lim }\frac{x^2+(-1+1)x+(-1)}{x+1}\\ =\underset{x\to -1}{\lim }\frac{x^2-1}{x+1}\\ =\underset{x\to -1}{\lim }x-1=-2\end{array}}\backslash def\backslash arraystretch\{1.25\}\; \backslash displaystyle\backslash begin\{array\}\{l\}\backslash lim\backslash limits\_\{x\backslash rightarrow-1\}f\_\{-1\}\backslash left(x\backslash right)\backslash \backslash =\backslash lim\backslash limits\_\{x\backslash rightarrow-1\}\backslash frac\{x^2+\backslash left(-1+1\backslash right)x+(-1)\}\{x+1\}\backslash \backslash =\backslash lim\backslash limits\_\{x\backslash rightarrow-1\}\backslash frac\{x^2-1\}\{x+1\}\backslash \backslash =\backslash lim\backslash limits\_\{x\backslash rightarrow-1\}x-1=-2\backslash end\{array\}$

Leite ${\displaystyle f_k\left(x\right)=\frac{x^2+(k+1)x+k}{x-k}}\backslash displaystyle\; f\_k\backslash left(x\backslash right)=\backslash frac\{x^2+\backslash left(k+1\backslash right)x+k\}\{x-k\}$ ab.

$f_k^{\mathrm{\prime }}(x)=\frac{(x-k)(2x+k+1)-(x^2+(k+1)x+k)}{(x-k{)}^2}=\frac{x^2-2kx-k^2-2k}{(x-k{)}^2}f\_k\text{'}(x)=\backslash frac\; \{(x-k)(2x+k+1)-(x^2+\; (k+1)x+k)\}\{(x-k)^2\}=\backslash frac\{x^2-2kx-k^2-2k\}\{(x-k)^2\}$

Setze die Ableitung gleich 0.

Bestimme die Diskriminante.

$\text{}D=4k^2+4(k^2+2k)D=4k^2+4(k^2+2k)$ $\backslash \backslash$ $D=0\iff 4k^2+4(k^2+2k)=0\iff 8k(k+1)=0\iff k=0\text{ oder }k=-1D=0\backslash \backslash \backslash Leftrightarrow\; 4k^2+4(k^2+2k)=0\backslash \backslash \backslash Leftrightarrow\; 8k(k+1)=0\backslash \backslash \backslash Leftrightarrow\; k=0\; \backslash text\{\; oder\; \}\; k=-1$

Löse mit der Mitternachtsformel. $\backslash \backslash$ Fall 1: $k=-1k=-1$ oder $k=0k=0$

${\displaystyle x=\frac{2k\pm \sqrt{4k^2+4(k^2+2k)}}{2}=k}\backslash displaystyle\; x=\backslash frac\{2k\backslash pm\backslash sqrt\{4k^2+4\backslash left(k^2+2k\backslash right)\}\}2=\backslash textstyle\; k$

Beachte: In diesem Fall gibt es keine Extrema, da $k\in ̸D_{f_k}k\backslash not\backslash in\; D\_\{f\_k\}$.

Fall 2: oder $k>0k>0$

$x_{1;2}=\frac{2k\pm \sqrt{4k^2+4(k^2+2k)}}{2}=k\pm \sqrt{8k(k+1)}x\_\{1;2\}=\backslash frac\{2k\backslash pm\backslash sqrt\{4k^2+4\backslash left(k^2+2k\backslash right)\}\}\{2\}=k\backslash pm\backslash sqrt\{8k\backslash left(k+1\backslash right)\}$