A converse proposition to the Mean Value Theorem [duplicate]

This is just a question that popped up in my head while going through basic real analysis. The ordinary Mean Value Theorem (MVT) is given as follows.

Let $f:[a,b] \to \mathbb{R}$ be a function satisfying the following conditions: $(i)$ $f$ is continuous on $[a,b]$ and $(ii)$ $f$ is differentiable on $(a,b)$. Then $\exists c \in (a,b),$ such that $$f(b)-f(a)=f'(c) \cdot (b-a)$$

Of course, we can restrict $f$ to any interval $[x,y] \subset [a,b]$ and apply the theorem on $f|_{[c,d]}$ to state that $\exists z \in (x,y)$ such that $f(y)-f(x)=f'(z) \cdot (y-x)$

Can I formulate a converse proposition as follows?

Let $f:[a,b] \to \mathbb{R}$ be a function satisfying the following conditions: $(i)$ $f$ is continuous on $[a,b]$ and $(ii)$ $f$ is differentiable on $(a,b)$. Then $\forall c \in (a,b), \exists x,y \in [a,b]$ such that $$f(y)-f(x)=f'(c) \cdot (y-x)$$

Does it hold?

If yes, how do I prove it? (How does one find out the points $x$ and $y$, which need not be unique, that work for a given point $c?)$ Does it hold under weaker assumptions on the function?

If no, can you provide me a counter example? Can I impose stronger conditions to make it work?

Any help would be much appreciated. Thank you.

EDIT : As pointed out by @Henrik, the proposition does not hold with it's current assumptions. On the other hand, @MANMAID claims that it does hold with the additional assumption that $f$ has no point of inflection. It seems very interesting. I require a formal proof, though.

Solution 1:

I don't think so.

Consider $f(x)=x^3$ on $[-1,1]$. It's continuous and differentiable, but for $c=0\in ]-1,1[$, $f'(c)=0$ so the $x$ and $y$ you want would have to satisfy $$ y^3-x^3=0 $$ which implies $x=y$, but since you haven't required that $x\neq y$, this isn't technically a counterexample.

Solution 2:

In fact, your result is almost true. To see it, Put first $C=\{(x,y); a\leq x<y\leq b\}$. Then $C$ is the interior of a triangle (with some side accepted). Let for $(x,y)\in C$, $g(x,y)=\frac{f(y)-f(x)}{y-x}$. Then clearly $g$ is continuous from $C$ to $\mathbb{R}$. As $C$ is connected, $I=g(C)$ is a connected subset of $\mathbb{R}$, hence an interval. Now the Mean value theorem say that the image $J$ of $]a,b[$ by $f^{\prime}$ contains $I$; it is well known that a derivative has the intermediate value property, so $J$ is an interval too. Now each $f^{\prime}(x)$ is the limit as $n\to +\infty$ of $g(x,x+1/n)$; hence $J\subset \overline{I}$. As you can see, as $I\subset J \subset \overline{I}$, $I$ and $J$ differ by at most two points.