Mean value theorem

Quote:

Originally Posted by acevipa

By the Mean value theorem show that:

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

Assume that

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

Then by differentiating both sides

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

$1 > 1 + x^2$

$x^2 < 0$ .

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

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