Twice differentiable function on the closed interval $[a,b]$

I'm stuck on a problem in calculus right now and it sounds as following:

Let $f(x)$ be a twice differentiable function on the closed interval $[a,b]$ where $a,b \in \mathbb{R}$. Suppose that the segment of the line that connects $(a,f(a))$ and $(b,f(b))$ intersects the graph $y=f(x)$ in a point $(c,f(c))$ where $c \in (a,b)$. Show that there exists at least one point $d \in (a,b)$ such that $f''(d)=0$.

I've tried multiple things, such as constructing some helping function $g(x)=f(x)-u(x)$ where $u(x)$ is the line that connects connects the points $a$ and $b$. I've also tried applying Rolle's theorem on $[a,c]$ and $[c,b]$, but realising that my calculus class haven't gotten that far, I can't use it trivially (i.e. without actually proving it, which is still a ways beyond my capabilities I'm afraid). So how would I go about solving this? Any help at all would be greatly appreciated.

Solution 1:

At some point $x_1$ between $a$ and $c$, we have $$f'(x_1) = \frac{f(c)-f(a)}{c-a} = \frac{f(b)-f(a)}{b-a}$$ by the mean value theorem. At some point $x_2$ between $c$ and $b$, we have $$ f'(x_2) = \frac{f(b)-f(c)}{b-c} = \frac{f(b)-f(a)}{b-a} $$ again by the mean value theorem.

What does Rolle's theorem say about the derivative of $f'$ (i.e. $f''$) on the interval $(x_1, x_2)$ in this case?