Let $f : [a, b] → \mathbb R$ be continuous on $[a, b]$ and twice differentiable on $(a, b)$. Suppose also that the line segment joining the points $(a, f(a))$ and $(b, f(b))$ meets the graph of $f$ at a point $c$, where $a < c < b$. Prove that there exists $d \in (a, b)$ such that $f''(d) = 0$.
Any help would be appreciated (Preferably using Rolle's theorem and/or MVT). Thanks!
Sketch:
Between $a$ and $c$, there is a point $d_1$ so that $f'(d_1)=\frac{f(c)-f(a)}{c-a}$.
Between $c$ and $b$, there is a point $d_2$ so that $f'(d_2)=\frac{f(b)-f(c)}{b-c}$.
Since $(a,f(a))$, $(b,f(b))$, and $(c,f(c))$ are on a line, can you show that $f'(d_1)=f'(d_2)$.
Then there is a point $d_3$ between $d_1$ and $d_2$ so that $f''(d_3)=\frac{f'(d_2)-f'(d_1)}{d_2-d_1}=0$.
Let $g(x) = \frac{f(b)-f(a)}{b-a}(x-a) + f(a)$. Thus the graph of $g$ is the line passing through $(a,f(a))$ and $(b,f(b))$.
Let $h(x) = f(x) - g(x)$. Then $h$ satisfies the conditions to apply Rolle's theorem twice. Once on $(a,c)$ and once on $(c,b)$. So we get that there exists $x_1\in (a,c)$ such that $h'(x_1) = 0$ and $x_2\in (c,b)$ such that $h'(x_2) = 0$. By applying again Rolle to $h'$, there exists $d\in (x_1,x_2)$ such that $h''(d) = 0$. Conclude by noticing that $h'' = f''$.