TITLE: Is it true that $\frac{f(b)-f(a)}{b-a}-\frac{b-a}{g(b)-g(a)}=f'(c)-\frac{1}{g'(c)}$
QUESTION [5 upvotes]: Let $f$ and $g$ be continuous functions on the closed interval $[a,b]$, and differentiable on the open interval $(a,b)$, where $a<b$. If $g'(x)\neq 0$ on $(a,b)$, then does there exists $c$ on $(a,b)$ such that $\frac{f(b)-f(a)}{b-a}-\frac{b-a}{g(b)-g(a)}=f'(c)-\frac{1}{g'(c)}$?
Note that $g(b)-g(a)\neq 0$ by Rolle's theorem.
By Lagrange's mean value theorem, there exist $c$ and $d$ on $(a,b)$, such that $\frac{f(b)-f(a)}{b-a}-\frac{b-a}{g(b)-g(a)}=f'(c)-\frac{1}{g'(d)}$. I guess that $c$ and $d$ can have the same value.

REPLY [2 votes]: Consider two functions $f$, $g$ such that $f'(x)=1/g'(x)$. Then the RHS is always zero, but not necessarily the LHS.
Example: Let $g(x):=x^3$, $f'(x)=\frac{1}{3x^2}$, so $f(x)=-\frac{1}{3x}$. Take $a=1$, $b=2$, to get $$\frac{f(b)-f(a)}{b-a}-\frac{b-a}{g(b)-g(a)}=\frac{1}{42}\ne0=f'(c)-\frac{1}{g'(c)}$$ for any $c\in[1,2]$.