= Solution
\b[The printed inequality has its sign reversed: $f(x)<f_2(x)$ for $x>1$.] This is present in the original scan, rather than arising in transcription. For $x>1$, <Taylor's theorem with Lagrange remainder> gives
$$
f(x)-f_2(x)=\frac{f'''(\xi)}{3!}(x-1)^3,\qquad1<\xi<x.
$$
Part (a) has $f'''(\xi)<0$, so the remainder is strictly negative. As an independent check, direct algebra gives the <quadratic Taylor majorant of a reciprocal quadratic>:
$$
\boxed{f_2(x)-f(x)=\frac{(x-1)^4}{4(1+x^2)}>0\quad(x>1).}
$$
For example, $f(2)=1/5<1/4=f_2(2)$ is already a counterexample to the requested direction. The exact identity also shows that this polynomial majorizes $f$ on the entire real line, with equality only at one.
\Image[/past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/continuous-mathematics-taylor-majorant.png]
{title=The quadratic <Taylor polynomial> lies above the reciprocal quadratic, correcting the printed inequality}
{description=The shaded gap for $x>1$ is positive and equals $(x-1)^4/(4(1+x^2))$.}
Back to article page
