$g\left(x\right)=\frac{1}{2}f\left(x\right)g\backslash left(x\backslash right)=\backslash frac12f\backslash left(x\backslash right)$

Lese zuerst den Scheitel von $f(x)f(x)$ ab.

$S(1;-1)S(1;-1)$

Schreibe die Funktion $f(x)f(x)$ in Scheitelform.

Bemerkung: Die Scheitelform wird noch für die nächsten Teilaufgaben benötigt. Also reicht es die an dieser Stelle zu bestimmen und in den anderen Teilaufgaben auf die hier getätigten Umformungsschritte zu verweisen.

$f\left(x\right)={(x-1)}^2-1f\backslash left(x\backslash right)=\backslash left(x-1\backslash right)^2-1$

Bestimme nun $g(x)g(x)$.

$g\left(x\right)=\frac{1}{2}\cdot f(x)=\frac{1}{2}[{(x-1)}^2-1]g\backslash left(x\backslash right)=\backslash frac12\backslash cdot\; f(x)=\backslash frac12\backslash left[\backslash left(x-1\backslash right)^2-1\backslash right]$

$\phantom{g(x)}\backslash phantom\{g(x)\}$$=\frac{1}{2}{(x-1)}^2-\frac{1}{2}=\backslash frac12\backslash left(x-1\backslash right)^2-\backslash frac12$

$g\left(x\right)=-2\cdot f\left(x\right)g\backslash left(x\backslash right)=-2\backslash cdot\; f\backslash left(x\backslash right)$

Lese zuerst den Scheitel von $f(x)f(x)$ ab.

$S(1;-1)S(1;-1)$

Schreibe die Funktion $f(x)f(x)$ in Scheitelform.

$f\left(x\right)={(x-1)}^2-1f\backslash left(x\backslash right)=\backslash left(x-1\backslash right)^2-1$

Bestimme nun $g(x)g(x)$.

$g\left(x\right)=-2[{(x-1)}^2-1]g\backslash left(x\backslash right)=-2\backslash left[\backslash left(x-1\backslash right)^2-1\backslash right]$

$\phantom{g\left(x\right)}=-2{(x-1)}^2+2\backslash phantom\{g\backslash left(x\backslash right)\}=-2\backslash left(x-1\backslash right)^2+2$

$g\left(x\right)=f\left(x\right)+\mathrm{1,5}g\backslash left(x\backslash right)=f\backslash left(x\backslash right)+1\{,\}5$

Mit $f\left(x\right)={(x-1)}^2-1f\backslash left(x\backslash right)=\backslash left(x-1\backslash right)^2-1$ gilt für $g(x)g(x)$:

$g\left(x\right)={(x-1)}^2-1+\mathrm{1,5}g\backslash left(x\backslash right)=\backslash left(x-1\backslash right)^2-1+1\{,\}5$

$\phantom{g(x)}\backslash phantom\{g(x)\}$$={(x-1)}^2+\mathrm{0,5}=\backslash left(x-1\backslash right)^2+0\{,\}5$

$g\left(x\right)={\left[f\left(x\right)\right]}^{2}g\backslash left(x\backslash right)=\backslash left[f\backslash left(x\backslash right)\backslash right]^2$

Bestimme $g(x)g(x)$ aus der Scheitelpunktsform von $f(x)=(x-1{)}^2-1f(x)=(x-1)^2-1$.

$g\left(x\right)={[{(x-1)}^2-1]}^2\phantom{g(x)}=[x^2-2x+1-1{]}^2\phantom{g(x)}=[x^2-2x{]}^2\phantom{g(x)}=x^4-4x^3+4x^{2}g\backslash left(x\backslash right)=\backslash left[\backslash left(x-1\backslash right)^2-1\backslash right]^2\backslash \backslash \backslash phantom\{g(x)\}=[x^2-2x+1-1]^2\backslash \backslash \backslash phantom\{g(x)\}=[x^2-2x]^2\backslash \backslash \backslash phantom\{g(x)\}=x^4-4x^3+4x^2$