TITLE: $f''$ has zero on interval $(a,b)$ if graph of $f$ intersects line between $(a,f(a))$ and $(b,f(b))$
QUESTION [5 upvotes]: Let $ f$ be continuous on $[a,b] $ and assume the second derivative $f''$ exists on (a,b). The graph of $f$ and the line segment joining the points $(a,f(a))$ and $(b,f(b))$ intersect at a point $(x_{0},f(x_{0}))$ where $a<x_{0}<b$. Show that there exists a point $c \in (a,b) $ such that $f''(c) = 0$.

REPLY [6 votes]: Hint: We have 
$$ \frac{f(x_0) - f(a)}{x_0 - a} = \frac{f(b) - f(x_0)}{b-x_0} $$
as the three points $(a, f(a))$, $(x_0, f(x_0))$ and $(b, f(b))$ form a line. Now apply the mean value theorem to $[a, x_0]$ and $[x_0, b]$. Rolle's theorem applied to $f'$ will do the trick.

REPLY [3 votes]: Hint: There exist $x_1 \in (a,x_0)$, $x_2\in (x_0,b)$ with $f'(x_1)=\dfrac{f(x_0)-f(a)}{x_0-a}=\dfrac{f(b)-f(x_0)}{b-x_0}=f'(x_2).$