Science:Math Exam Resources/Courses/MATH110/April 2016/Question 08 (d)/Solution 1

The second degree Taylor polynomial of $\frac{1}{x}$ at 4 is given by

$f(4)+f^{\prime }(4)(x-4)+\frac{f^{″}(4)}{2}(x-4{)}^2$

We already computed $f(4)=\frac{1}{4}$ and $f^{\prime }(4)=\frac{-1}{16}$. We have that $f^{″}(x)=\frac{2}{x^3}$, so $f^{″}(4)=\frac{2}{64}=\frac{1}{32}$.

So the Taylor polynomial is

$\frac{1}{4}-\frac{1}{16}(x-4)+\frac{1}{64}(x-4{)}^2$.

Answer: $\frac{1}{4}-\frac{1}{16}(x-4)+\frac{1}{64}(x-4{)}^2$