What can be said about the set of points that makes the MVT "works"? Consider a (real-valued) function $f$ that is differentiable on an open interval $I$ of $\mathbb R$. 
For $a\neq b$ in $\mathbb R$, we consider $S_{a,b}=\left\{ x\in I\; \middle|\; f'(x)=\frac{f(b)-f(a)}{b-a}\;  \right\}$. $S_{a,b}$ is non empty by the Mean Value Theorem.
Let now $\displaystyle S=\bigcup_{a\neq b \in I}S_{a,b}$
What can be said about $S$? 
Has this set been studied before? Is $S$ an interval? Is $S$ open? 
Clearly, $S$ is a subset of $I$ but the inclusion can be strict. For example, if $f(x)=x^3$, we have $0\notin S$.  This example also shows that $S$ is not always an interval.
Note: feel free to add conditions on $f$ if it allows $S$ to have nice properties.
 A: Let's look at this question more geometrically. Start with a point $x$ and calculate derivative of $f$ at $x$ and hence tangent line at $(x,f(x))$. Allow the tangent line be translated up or down. If after translation, the line cuts the graph of $f$ at two distinct points $(a,f(a)),(b,f(b))$ such that $a\lt x\lt b$, then $x\in S$ by definition.
There is a nice property of the function $f$ that guarantees the existence of such points $a,b$. If $f$ is locally convex (resp. concave) around $p$ (i.e. $f$ is convex (resp. concave) on an open interval $U$ containing $p$), then such $a,b$ must exist. Note that all points in $U$ are also in $S$ because they have the same property as $x$, thus making $x$ an interior point of $S$.
For an informal proof of the claim that local convexity (resp. concavity) is sufficient, start by picking two points $p-\varepsilon,p+\varepsilon$ for some small enough $\varepsilon\gt0$. Draw the tangent line of $f$ at $p$, and pick the segment bounded by the two vertical lines $x=p-\varepsilon$ and $x=p+\varepsilon$. Due to convexity (resp. concavity), the point $(p,f(p))$ lies below (resp. above) the line segment joining $(p-\varepsilon,f(p-\varepsilon))$ and $(p+\varepsilon,f(p+\varepsilon))$. So move the tangent segment up (resp. down) until it touches one of the two points $(p-\varepsilon,f(p-\varepsilon)),(p+\varepsilon,f(p+\varepsilon))$. Use intermediate value theorem (as $f$ is continuous) to conclude that the moved tangent segment cuts the graph of $f$ at another point $(p',f(p'))$ with $p\lt p'\le p+\varepsilon$ if the moving tangent segment touches $(p-\varepsilon,f(p-\varepsilon))$ first, and $p-\varepsilon\le p'\lt p$ if if the moving tangent segment touches $(p+\varepsilon,f(p+\varepsilon))$ first.
These arguments also shows why it is possible that $0\notin S$ for $f(x)=x^3$ and not for other points.. $0$ is an inflection point of $x^3$, where it is possible that $0\notin S$.
A: For any $y$ define
$$g_y(x) = f(x) - f'(y) (x-y).$$
Clearly $g_y'(y) = 0$, so


*

*either $y$ is a local extremum, and in this case clearly $y \in S$. 

*or $y$ is not an extremum, which is equivalent to $g_y'$ (and so also f') having an strict extremum at $y$ (i.e., as @edm showed, $f$ is strictly concave or strictly convex at that point).
This means that $\{x: f' \, \text{does not have a strict extremum at } x\} \subset S$. The other inclusion is not true: take for example $f = e^{x^2} \sin (x)$, where $S = \mathbb R$, but $f''$ vanishes in many places.
