I'm going to try to solve the problem presented below. But as you will realize, I'll run into a problem. I hope you can give me somet tips.
Let $f(x)$ be a function which is contiuous on $[a, b]$ and such that $f''(x)$ exists at every point in $(a, b)$. Suppose that the line segment joining $(a, f(a))$ and $(b, f(b))$ intersects the graph of $f(x)$ at a point $(c, f(c))$ where $a<c<b$. Prove that there is at least one point $d\in (a, b)$ at which $f''(d) = 0$.
Proof: To begin with, realize that $f(x)$ is twice differentiable, and that this implicates that $f'(x)$ is continous.
Furthermore, we determine the slope $k_S$ of the segment joining the two points given above. We get:
$$ k_S= \frac{f(b)-f(a)}{b-a}$$
The information provided in the excercise, shows us that all critieras, in order to use the mean value theorem, are fulfilled. Hence, there $\exists \xi \in (a,b) | \frac{f(b)-f(a)}{b-a} = f'(\xi)$.
We also know that the secant intersects the graph of $f(x)$ on the point $(c,f(c))$, therefore, we can determine the function of the secant, as:
$ f_S(x) = f'(\xi)(1-c)x+f(c)$
In order for there to exist a $d | f''(d) = 0$, we have to be able to use Rolle's theorem, which can be used if we can show that $f'(a) = f'(b)$, is my guess? But since we aren't given that $f'(x)$ isn't defined for the closed interval $[a,b]$, I don't really see how I can continue from here on.
Thank you!