If $f$ is differentiable on $[x_1,x_2]$, $x_1x_2>0$, prove that $\frac{x_1f(x_2)-x_2f(x_1)}{x_1-x_2}=f(c)-cf'(c)$, where $c\in(x_1,x_2)$ So I thought of making up a function $\phi(x)$ (just as how the mean value theorem was proven) and by using rolle's theorem($\phi(x_1)=\phi(x_2)=0$), I could get the answer-- $f(c)-cf'(c)$ reminds me of....something related to $e^{-x}f(x)$? but is that how its done? At least, I couldn't devise the function; couldn't link the dots together... any help would be appreciated! Thanks!

You can write the LHS as $$ \frac{\displaystyle \frac{f(x_2)}{x_2} - \frac{f(x_1)}{x_1}}{\displaystyle \frac{1}{x_2} - \frac{1}{x_2}} $$
Which reminds us of cauchy mean value theorom which just proves the result as $x_1x_2 > 0$ so both $x_1 , x_2$ are at the same side of zero so both the functions are derivable.
