By the Mean value theorem show that:

$\displaystyle \arctan{x} \leq x\ for\ x \geq 0$

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- May 2nd 2010, 02:48 AMacevipaMean value theorem
By the Mean value theorem show that:

$\displaystyle \arctan{x} \leq x\ for\ x \geq 0$ - May 2nd 2010, 02:58 AMProve It
Assume that

$\displaystyle \arctan{x} > x$ for $\displaystyle x \geq 0$.

Then by differentiating both sides

$\displaystyle \frac{1}{1 + x^2} > 1$

$\displaystyle 1 > 1 + x^2$

$\displaystyle x^2 < 0$.

But $\displaystyle x^2 \geq 0$ for all $\displaystyle x \in \mathbf{R}$. So we have arrived at a contradiction. So our original statement that $\displaystyle \arctan{x} > x$ is wrong.

Therefore $\displaystyle \arctan{x} \leq x$ for all $\displaystyle x \geq 0$.