Taylorpolynom/Quadratwurzel aus kubischem Polynom/Aufgabe/Lösung

Wir schreiben

${}f(x)={\sqrt {x^{3}-x+2}}={\left(x^{3}-x+2\right)}^{\frac {1}{2}}\,,$

es ist ${}f(2)={\sqrt {8}}=2{\sqrt {2}}$. Es ist

${}{\begin{aligned}f'(x)&={\frac {1}{2}}{\left(x^{3}-x+2\right)}^{-{\frac {1}{2}}}\cdot {\left(3x^{2}-1\right)}\\&={\frac {3}{2}}x^{2}{\left(x^{3}-x+2\right)}^{-{\frac {1}{2}}}-{\frac {1}{2}}{\left(x^{3}-x+2\right)}^{-{\frac {1}{2}}}\end{aligned}}$

und insbesondere

${}f'(2)={\frac {1}{2}}8^{-{\frac {1}{2}}}\cdot 11={\frac {11}{2\cdot 8^{\frac {1}{2}}}}={\frac {11}{4{\sqrt {2}}}}={\frac {11{\sqrt {2}}}{8}}\,.$

Es ist

${}{\begin{aligned}f^{\prime \prime }(x)&=3x{\left(x^{3}-x+2\right)}^{-{\frac {1}{2}}}-{\frac {3}{4}}x^{2}{\left(x^{3}-x+2\right)}^{-{\frac {3}{2}}}{\left(3x^{2}-1\right)}+{\frac {1}{4}}{\left(x^{3}-x+2\right)}^{-{\frac {3}{2}}}{\left(3x^{2}-1\right)}\\&=3x{\left(x^{3}-x+2\right)}^{-{\frac {1}{2}}}+{\frac {-3x^{2}+1}{4}}{\left(x^{3}-x+2\right)}^{-{\frac {3}{2}}}{\left(3x^{2}-1\right)}\end{aligned}}$

und insbesondere

${}{\begin{aligned}f^{\prime \prime }(2)&=6\cdot 8^{-{\frac {1}{2}}}-{\frac {11}{4}}\cdot 8^{-{\frac {3}{2}}}\cdot 11\\&={\frac {3}{\sqrt {2}}}-{\frac {121}{4\cdot 8^{\frac {3}{2}}}}\\&={\frac {3{\sqrt {2}}}{2}}-{\frac {121}{64\cdot {\sqrt {2}}}}\\&={\frac {192{\sqrt {2}}}{128}}-{\frac {121{\sqrt {2}}}{128}}\\&={\frac {71{\sqrt {2}}}{128}}.\end{aligned}}$

Somit ist das Taylorpolynom vom Grad ${}2$ im Entwicklungspunkt ${}2$ gleich

$2{\sqrt {2}}+{\frac {11{\sqrt {2}}}{8}}(x-2)+{\frac {71{\sqrt {2}}}{256}}(x-2)^{2}.$

Zur gelösten Aufgabe