According to the Mean Value Theorem, for a function which is continuous on the closed interval $[a,b]$, and differentiable on the open interval $(a,b)$, there exists a point $c$ in $(a,b)$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}$$
Geometrically, there exists an internal point $c$ at which, the tangent at $c$ is parallel to the chord/secant through the endpoints.
From this comment for my answer to my previous question, I realised that the point where the tangent becomes parallel to the chord/secant, is also the point which is at maximum distance from the chord/secant. I haven't learnt about that kind of property so far.
On analysing further, I came to the following conclusion:
The internal point(s) at which the tangent becomes parallel to the chord/secant joining the endpoints, are stationary points if we rotate the graph such that the chord/secant becomes horizontal. (Note: This is not the stationary point of the original function.) Further, if concavity(with respect to the new rotated position of the graph) remains the same, these points could be at maximum or minimum distance compared to the neighbouring points, from the chord/secant.
I wish to know whether my conclusion is correct or not. And, are there any other things related to Graphical Interpretation of Mean Value Theorem I must know (apart from this)?