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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!

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up vote 5 down vote accepted

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 $$ {x\over 1+x^2}\lt\arctan x\lt x . $$

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thnx alot! perfect ans! –  tirmizi Dec 31 '11 at 5:00
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No doubt on the mean value on the mean value theorem! Just take x-arctan x and you will find the dirivative smaller than 0 for all x > 0.

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The derivative of $x-\arctan(x)$ is positive when $x>0$. –  Jonas Meyer Dec 31 '11 at 8:49
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