I am reading Courant's Differential and Integral Calculus. Here:
Find the intermediate value $\xi $ of the mean value theorem for the following functions, and illustrate graphically:
(d) $1/(x^2+1)$
I am having a little bit of trouble to find $\xi$ in $1/(x^2+1)$. I did the following:
$$\frac{f(x_1)-f(x_2)}{x_2 - x_1}=-\frac{x_1+x_2}{\left(x_1^2+1\right) \left(x_2^2+1\right)}$$
So, there exists an $\xi\in [x_1,x_2]$ such that:
$$-\frac{x_1+x_2}{\left(x_1^2+1\right) \left(x_2^2+1\right)}=f'(\xi)$$
And then, obviously I could compute the derivative for $1/(x^2+1)$ which gives me $-\frac{2 x}{\left(x^2+1\right)^2}$ and attempt to solve:
$$-\frac{x_1+x_2}{\left(x_1^2+1\right) \left(x_2^2+1\right)}=-\frac{2 \xi}{\left(\xi^2+1\right)^2}$$
For $\xi$, but perhaps this would be "too hard" to do by hand, I suspect that there is something much simpler. I also guess that in general, any function which the derivative looks as the derivative of the function I provided, differentiating and solving for $\xi$ could be a nightmare. I have even tried on Mathematica and it gave me this:
I may be missing something truly silly here, Courant gives a very simple answer in his book.
I suggest that you just leave this problem and go on. Why? In the section of answers to the book, answers are given for (a), (b), (c) and (e), but not for (d). This indicates that it will be horrible (practically impossible). We can just speculate if that was on purpose by Courant or not.
Just for the sake of clarification, the other answers to the exercises imply that one is indeed asking for $\xi$ such that $$ f'(\xi)=\frac{f(x_2)-f(x_1)}{x_2-x_1}. $$