Berechne $f(x_0)f(x\_0)$

$f(-2)=-2\cdot (-2{)}^3-2\cdot (-2)=16+4=20f(-2)=-2\backslash cdot\; (-2)^3-2\backslash cdot\; (-2)=16+4=20$

Bestimme $f^{\mathrm{\prime }}(x)f\text{'}(x)$

$f^{\mathrm{\prime }}(x)=-6x^2-2f\text{'}(x)=-6x^2-2$

Berechne $f^{\mathrm{\prime }}(x_0)f\text{'}(x\_0)$

$f^{\mathrm{\prime }}(-2)=-6\cdot (-2{)}^2-2=-24-2=-26f\text{'}(-2)=-6\backslash cdot\; (-2)^2-2=-24-2=-26$

Setze in die Normalenformel ein.

$g(x)=-{\displaystyle \frac{1}{f^{\mathrm{\prime }}(x_0)}}(x-x_0)+f(x_0)g(x)=-\backslash dfrac\{1\}\{f\text{'}(x\_0)\}(x-x\_0)+f(x\_0)$ $\backslash \backslash$ $\phantom{g(x)}={\displaystyle \frac{1}{26}}(x+2)+20\backslash phantom\{g(x)\}=\backslash dfrac\{1\}\{26\}(x+2)+20$

Berechne $f(x_0)f(x\_0)$

$f(1)=-{\displaystyle \frac{5}{4}}\cdot 1^2+3={\displaystyle \frac{7}{4}}f(1)=-\backslash dfrac\{5\}\{4\}\backslash cdot\; 1^2+3=\backslash dfrac\{7\}\{4\}$

Bestimme $f^{\mathrm{\prime }}(x)f\text{'}(x)$

$f^{\mathrm{\prime }}(x)=-{\displaystyle \frac{5}{2}}xf\text{'}(x)=-\backslash dfrac\{5\}\{2\}x$

Berechne $f^{\mathrm{\prime }}(x_0)f\text{'}(x\_0)$

$f^{\mathrm{\prime }}(1)=-{\displaystyle \frac{5}{2}}\cdot 1=-{\displaystyle \frac{5}{2}}f\text{'}(1)=-\backslash dfrac\{5\}\{2\}\backslash cdot\; 1=-\backslash dfrac\{5\}\{2\}$

Setze in die Normalenformel ein.

$g(x)=-{\displaystyle \frac{1}{f^{\mathrm{\prime }}(x_0)}}(x-x_0)+f(x_0)g(x)=-\backslash dfrac\{1\}\{f\text{'}(x\_0)\}(x-x\_0)+f(x\_0)$ $\backslash \backslash$ $\phantom{g(x)}={\displaystyle \frac{2}{5}}(x-1)+{\displaystyle \frac{7}{4}}\backslash phantom\{g(x)\}=\backslash dfrac\{2\}\{5\}(x-1)+\backslash dfrac\{7\}\{4\}$

Berechne $f(x_0)f(x\_0)$

$f(-1)=-e^{-4}=-{\displaystyle \frac{1}{e^4}}f(-1)=-e^\{-4\}=-\backslash dfrac\{1\}\{e^4\}$

Bestimme $f^{\mathrm{\prime }}(x)f\text{'}(x)$

$f^{\mathrm{\prime }}(x)=-4e^{4x}f\text{'}(x)=-4e^\{4x\}$

Berechne $f^{\mathrm{\prime }}(x_0)f\text{'}(x\_0)$

$f^{\mathrm{\prime }}(-1)=-4e^{-4}=-{\displaystyle \frac{4}{e^4}}f\text{'}(-1)=-4e^\{-4\}=-\backslash dfrac\{4\}\{e^4\}$

Setze in die Normalenformel ein.

$g(x)=-{\displaystyle \frac{1}{f^{\mathrm{\prime }}(x_0)}}(x-x_0)+f(x_0)g(x)=-\backslash dfrac\{1\}\{f\text{'}(x\_0)\}(x-x\_0)+f(x\_0)$ $\backslash \backslash$ $\phantom{g(x)}={\displaystyle \frac{e^4}{4}}(x+1)-{\displaystyle \frac{1}{e^4}}\backslash phantom\{g(x)\}=\backslash dfrac\{e^4\}\{4\}(x+1)-\backslash dfrac\{1\}\{e^4\}$