Prove using mean-value theorem that $x/(1+x^2)<\arctan x<x$ for $x>0$

I got the first part but how do I prove $\arctan x< x$ using the MVT?

The first part was done easily by applying MVT on $\arctan x$, should I use $\arctan x-x$ for the second part? Thanks!


closed as off-topic by 6005, Claude Leibovici, user91500, Alex M., Tom-Tom Nov 30 '15 at 13:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 6005, Claude Leibovici, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.


I think you have it. One application of the MVT gives you both parts:

Let $x>0$. Applying the Mean Value Theorem to $f(x)=\arctan x$ on the interval $[0,x]$ gives a number $c$ with $0<c<x$ such that $$ {\arctan x-\arctan 0\over x-0}={1\over 1+c^2} $$ Rearranging the above gives: $$ \arctan x={x\over 1+c^2} . $$ for some $c$ between $0$ and $x$.

Since $x>c$ and $x\gt0$, we have: $${x\over 1+x^2}\lt{x\over 1+c^2}<x;$$ whence $$ {x\over 1+x^2}\lt\arctan x\lt x . $$

  • $\begingroup$ thnx alot! perfect ans! $\endgroup$ – tirmizi Dec 31 '11 at 5:00

Not the answer you're looking for? Browse other questions tagged or ask your own question.