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

$f(2)=(-3)\cdot 2^2+2\cdot 2=-12+4=-8f(2)=(-3)\backslash cdot\; 2^2+2\backslash cdot\; 2=-12+4=-8$

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

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

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

$f^{\mathrm{\prime }}(2)=(-6)\cdot 2+2=-10f\text{'}(2)=(-6)\backslash cdot\; 2+2=-10$

Setze $x_0,f(x_0),f^{\mathrm{\prime }}(x_0)x\_0,f(x\_0),f\text{'}(x\_0)$ in die Formel ein.

$g(x)=f^{\mathrm{\prime }}(x_0)(x-x_0)+f(x_0)g(x)=f\text{'}(x\_0)(x-x\_0)+f(x\_0)$ $\phantom{g(x)}=-10(x-2)-8\backslash phantom\{g(x)\}=-10(x-2)-8$

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

$f(-1)={\displaystyle \frac{2}{3}}\cdot (-1{)}^3-2=-{\displaystyle \frac{2}{3}}-2=-{\displaystyle \frac{8}{3}}f(-1)=\backslash dfrac\{2\}\{3\}\backslash cdot(-1)^3-2=-\backslash dfrac\{2\}\{3\}-2=-\backslash dfrac\{8\}\{3\}$

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

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

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

$f^{\mathrm{\prime }}(-1)=2\cdot (-1{)}^2=2f\text{'}(-1)=2\backslash cdot(-1)^2=2$

Setze $x_0,f(x_0),f^{\mathrm{\prime }}(x_0)x\_0,f(x\_0),f\text{'}(x\_0)$ in die Formel ein.

$g(x)=f^{\mathrm{\prime }}(x_0)(x-x_0)+f(x_0)g(x)=f\text{'}(x\_0)(x-x\_0)+f(x\_0)$ $\phantom{g(x)}=2(x+1)-{\displaystyle \frac{8}{3}}\backslash phantom\{g(x)\}=2(x+1)-\backslash dfrac\{8\}\{3\}$

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

$f(2)={\displaystyle \frac{1}{8}}\cdot 2^4-3\cdot 2^2+2f(2)=\backslash dfrac\{1\}\{8\}\backslash cdot\; 2^4-3\backslash cdot\; 2^2+2$ $\backslash \backslash$ $\phantom{f(x_0)}=2-12+2=-8\backslash phantom\{f(x\_0)\}=2-12+2=-8$

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

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{1}{2}}{x}^3-6x+1f\text{'}(x)=\backslash dfrac\{1\}\{2\}x^3-6x+1$

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

$f^{\mathrm{\prime }}(2)={\displaystyle \frac{1}{2}}\cdot 2^3-6\cdot 2+1f\text{'}(2)=\backslash dfrac\{1\}\{2\}\backslash cdot\; 2^3-6\backslash cdot\; 2+1$ $\backslash \backslash$ $\phantom{f^{\mathrm{\prime }}(2)}=4-12+1=-7\backslash phantom\{f\text{'}(2)\}=4-12+1=-7$

Setze $x_0,f(x_0),f^{\mathrm{\prime }}(x_0)x\_0,f(x\_0),f\text{'}(x\_0)$ in die Formel ein.

$g(x)=f^{\mathrm{\prime }}(x_0)(x-x_0)+f(x_0)g(x)=f\text{'}(x\_0)(x-x\_0)+f(x\_0)$ $\backslash \backslash$ $\phantom{g(x)}=-7(x-2)-8\backslash phantom\{g(x)\}=-7(x-2)-8$