TITLE: Cauchy's Mean Value Theorem. What can we say about $c$ with more information.
QUESTION [6 upvotes]: My questions is about Cauchy's Mean Value theorem which states:

If functions f and g are both continuous on the closed interval [a,b],
  and differentiable on the open interval (a, b), then there exists some
  c ∈ (a,b), such that
  \begin{align*}
(f(b)-f(a))g'(c)=(g(b)-g(a))f'(c)
\end{align*}

From my understanding $c$ is not that trivial to find and it might not even be unique. 
Here are is my questions? For $f \neq g$
1) Under what conditions on $g$ and $ f$ is $c$ unique? For example, is monotonicity or concavity enough? 
2) If we give more information on $g$ and $f$. 
Can we say more about the location of $c$ in the range $(a,b)$? For example, say that $g$ and $f$ are non-negative and $g'',f''> 0$, what can we say then? ** 
For example, is it true if $f$ and $g$ are concave then
$ c \in( \frac{b+a}{2},b)$
I skimmed through several reference but wasn't able to find anything concrete about c based on properties of $f$ and $g$? 
Thank you for any help in advance. Also, any references would be greatly appreciated. 
Edit: As pointed out by @LeonAragones monotonicity of $f,g$ is not enuogh to guarantee uniquens of $c$. 
** Edit:** Please the answer to the first part of the question by @san.

REPLY [3 votes]: The points $c$ depend on the function $h(x):=(g(b)-g(a))f(x)-(f(b)-f(a))g(x)$. 
They are the points where $h'(c)=0$. So concavity should be required for $h$, not for $f,g$. For example, take $a=0$, $b=6\pi$, $f(x)=x^2+1$, $g(x)=x^2+1-sin(x)$. Then $f,g,f'',g''$ are positive on the interval, however, since $h(x)=(6\pi)^2 sin(x)$, we have $c_0,\dots,c_6$ with $c_k:=(2k+1)\frac{\pi}{2}$ such that $h'(c_k)=0$.  
So, if $h''(x)>0$ for all $x$, or $h''(x)<0$ for all $x$, then the solution is unique, or more general, if $h$ is convex or concave, the solution is unique.
With respect to the second question, the answer is no, we can't say anything about the location of $c$.  "is it true if f and g are concave then $c\in (\frac{b+a}{2},b)$?" No, for example, take $f=-x^2$, $g=-1/x-x$, $a=\frac 1{10}$ and $b=10$. Then $f,g$ are concave (even $h$ is concave), but the unique solution $c=1\notin (\frac{b+a}{2},b)=(5\frac{1}{20},10)$.