I want to prove the following proposition:
Proposition. There exist real polynomials $f$ and $g$ such that $g(0)\neq g(1)$ and $$ \frac{f(1)-f(0)}{g(1)-g(0)}\neq \frac{f^{'}(\xi)}{ g^{'}(\xi)}$$ for all $\xi\in (0,1)$.
It seems directly related to the generalized mean value theorem (MVT) but am not sure how to prove it. The negation of the statement is:
For all real polynomials $f$ and $g$ such that $g(0)\neq g(1)$ there exist $\xi\in (0,1)$ such that $$ \frac{f(1)-f(0)}{g(1)-g(0)}= \frac{f^{'}(\xi)}{ g^{'}(\xi)}$$
which is true if $g^{'}(\xi)\neq 0$ by the generalized MVT. Any help is greatly appreciated.
Best Answer
What you can prove is that (in particular) for every real polynomial functions $f$,$g$ such that $g(0) \neq g(1)$ you can find a $c \in (0,1)$ such that $$ \left[ {f(1) - f(0)} \right]g'(c) = \left[ {g(1) - g(0)} \right]f'(c) $$ thus either $g'(c)=0$ so that $f'(c)=0$ also, or $$ \frac{{f(1) - f(0)}} {{g(1) - g(0)}} = \frac{{f'(c)}} {{g'(c)}} $$ Hence you can build an example of $f$ and $g$ where there exists a $c \in (0,1)$ such that $f'(c)=g'(c)=0$ and such that for every other $\xi \in (0,1)$ with $\xi \neq c$ it is $$ \frac{{f(1) - f(0)}} {{g(1) - g(0)}} \ne \frac{{f'(\xi )}} {{g'(\xi )}} $$ For instance you can choose $$ f(x) = \left( {x - \frac{1} {2}} \right)^3 + \left( {x - \frac{1} {2}} \right)^2 $$ $$ g(x) = \left( {x - \frac{1} {2}} \right)^3 - \left( {x - \frac{1} {2}} \right)^2 $$ You have that $$ \frac{{f(1) - f(0)}} {{g(1) - g(0)}} = 1 $$ $f'(1/2)=g'(1/2)=0$ and for every $\xi \neq 1/2$ it is
$$ \frac{{f'(\xi )}} {{g'(\xi )}} = \frac{{3\left( {\xi - \frac{1} {2}} \right) + 2}} {{3\left( {\xi - \frac{1} {2}} \right) - 2}} $$ which, of course , can never be $1$.