aufgabe 2a

$$f(x)=\frac{x}{x+3}$$

Taylorpolynom 2. Ordnung §

Ableitungen von f(x)

$$\begin{array}{c}f(x)=\frac{x}{x+3}\\ f^{\prime }(x)=\frac{3}{(x+3{)}^2}\\ f^{\prime \prime }(x)=-\frac{6}{(x+3{)}^3}\end{array}$$

$$\begin{array}{rlrl}& T_2(f,1,x)& =& \sum _{k=0}^2\frac{f^{(k)}(1)}{k!}(x-1{)}^k\\ & & =& \frac{0}{0+3}(x-1{)}^0+\frac{3}{(1+3{)}^2}(x-1{)}^1+\\ & & & \frac{1}{2!}\cdot \frac{6\cdot 2+18}{(2+3{)}^4}(x-1{)}^2\\ & & =& 0+\frac{3}{16}\cdot (x-1)+\frac{3}{125}(x-1{)}^2\\ & & =& \frac{3}{16}x-\frac{3}{16}+\frac{3}{125}(x^2-2x+1)\\ & & =& \frac{3}{16}x-\frac{3}{16}+\frac{3}{125}{x}^2-\frac{6}{125}x+\frac{3}{125}\\ & & =& \frac{1}{2000}\left(3\cdot 125x-3\cdot 125+3\cdot 16x^2-6\cdot 16x+3\cdot 16\right)\\ & & =& \frac{1}{2000}\left(375x-375+48x^2-96x+48\right)\\ & & =& \frac{1}{2000}\left(48x^2+279x-327\right)\\ & & =& \underline{0.024x^2+0.1395x-0.1635}\end{array}$$

Falsch

$f(1)$, nicht $f(k)$.

$$\begin{array}{rl}{T}_2(f,1,x)& =\sum _{k=0}^2\frac{f^{(k)}(1)}{k!}(x-1{)}^k\\ & =\frac{1}{1+3}(x-1{)}^0+\frac{3}{(1+3{)}^2}(x-1{)}^1+\frac{1}{2!}\cdot \frac{6+18}{(1+3{)}^4}(x-1{)}^2\\ & =\frac{1}{4}+\frac{3}{16}(x-1)+\frac{24}{16\cdot 8\cdot 2}(x-1{)}^2\\ & =\frac{1}{4}+\frac{3}{16}x-\frac{3}{16}+\frac{1}{32}(x^2-2x+1)\\ & =\frac{1}{32}{x}^2+\frac{3}{16}x-\frac{1}{16}x+\frac{1}{4}-\frac{3}{16}+\frac{1}{32}\\ & =\frac{1}{32}\left(x^2+(3\cdot 2)x-(1\cdot 2)x+(1\cdot 8)-(3\cdot 2)+1\right)\\ & =\frac{1}{32}\left(x^2+4x+3\right)\\ & =\underline{\frac{1}{32}{x}^2+\frac{1}{8}x+\frac{3}{32}}\end{array}$$

Falsch

$f^{\prime \prime }(x)\ne \frac{6x+18}{(x+3{)}^4}$. $f^{\prime \prime }(x)$ ist negativ.

$$\begin{array}{rl}{T}_2(f,1,x)& =\frac{1}{4}+\frac{3}{16}x-\frac{3}{16}-\frac{1}{2}\cdot \frac{6}{4^3}\cdot (x-1{)}^2\\ & =\frac{3}{16}x+\frac{1}{16}-\frac{3}{64}\cdot \left(x^2-2x+1\right)\\ & =-\frac{3}{64}{x}^2+\frac{3}{16}x+\frac{3}{32}x+\frac{1}{16}-\frac{3}{64}\\ & =\underline{-\frac{3}{64}{x}^2+\frac{9}{32}x+\frac{1}{64}}\end{array}$$

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