1)By "reverse" MVT I mean: eg. in d=1, for any $x\in \mathbb{R}$ there exists $a,b$ containing x s.t. for $f\in C^{1}$ we have $f'(x)=\frac{f(b)-f(a)}{b-a}$.
This fails for all functions that cross the real line eg. $f(x)=x^{3}$ fails at $x=0$. However, I think if the function is also convex and strictly positive, or convace and strictly negative, it works. Can we weaken it to locally convex?
2)How about some general formula connecting derivative with it's antiderivative i.e. for any $x\in \mathbb{R}$ there exists $a,b$ s.t. for $f\in C^{1}$ we have $f'(x)=G(f(a),f(b),r_{i})$, where $r_{i}\in \mathbb{R}$ are real numbers.
Thanks
For a locally convex function f, for any $x\in dom(f)$ there exists $(a,b)$ s.t. the tangent line at x lies below the graph of f(a,b).
Wlog $f(a)<f(b)$ and let $\theta=$slope of line connecting $f(a),f(b)$ and $\phi=$slope of the tangent line.
If $\theta>\phi$,we can lift the line vertically (preserving the slope), to a line that intersects the graphe at $\frac{f(x)+f(a)}{2}$. Because of the slope inequality, the new line will intersect the graph of f at two points: $f(c):=\frac{f(x)+f(a)}{2}$ and some other point f(d) below f(b).
Similarly, if $\theta<\phi$,we can lift the line vertically (preserving the slope), to a line that intersects at $\frac{f(x)+f(b)}{2}$.
Thus, $f'(x)$ can be expressed as $\frac{f(c)-f(d)}{c-d}$.