Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (b)/Solution 1

As in problem 11(a), we again set $h(x)=f(x)-{\frac {x+1}{2}}$, then $f''(x)=h''(x)$, so we only need to prove there are infinitely many $c$ such that $h''(c)=0$.

From 11(a), there are infinitely many $c$ such that $h(c)=0$, so we can choose three of them denoted as $c_{1}<c_{2}<c_{3}$, such that $h(c_{1})=h(c_{2})=h(c_{3})=0$. Then according to Rolle's theorem, there exists $d_{1}\in (c_{1},c_{2})$, $d_{2}\in (c_{2},c_{3})$, such that $h'(d_{1})=0$, $h'(d_{2})=0$. We use Rolle's theorem again, then there exists $e\in (d_{1},d_{2})$ such that $h''(e)=0$. There are infinitely triple groups of $c_{1},c_{2}$ and $c_{3}$ with no overlapping region of $(c_{1},c_{3})$, so there are infintely many $e\in (c_{1},c_{3})$ such that $h''(e)=0$.