TITLE: A problem on the sum of the reciprocals of two derivatives
QUESTION [5 upvotes]: If $f(x)$ is continuous in the closed interval $[a,b]$ and differentiable in the open interval $a<x<b$, and if $f(a)=a$, $f(b)=b$, prove there exist points $x_1$ and $x_2$ with $a<x_1<x_2<b$ for which the following equation is true: $1/f'(x_1)+1/f'(x_2)=2$.

REPLY [3 votes]: Since $f(x)$ is continuous in the closed interval $[a,b]$ $\exists c\in (a,b)$ s.t $f(c)=\frac{f(a)+f(b)}{2}=\frac{a+b}{2}$ by intermediate value theorem.
Then by mean value theorem,
$\exists x_1\in (a,c)$s.t. $f'(x_1)=\frac{\frac{a+b}{2}-f(a)}{c-a}=\frac{\frac{a+b}{2}-a}{c-a}=\frac{b-a}{2(c-a)}$
$\exists x_2\in (c,b)$s.t. $f'(x_2)=\frac{f(b)-\frac{a+b}{2}}{b-c}=\frac{b-\frac{a+b}{2}}{b-c}=\frac{b-a}{2(b-c)}$
Hence  $1/f'(x_1)+1/f'(x_2)=2$.