Hi people,
I must show that: $\displaystyle \forall x \in \mathbb{R}: Arctanx>\frac{x}{1+x^2}$
I don't know how to do it???? can you help me please???
It's only true for $\displaystyle x>0 $.
To prove this, show $\displaystyle \tan^{-1}(0)=\frac{0}{0^2+1} $.
Then look at the derivatives.
$\displaystyle \frac{d}{dx}\left(\tan^{-1}(x)\right) = \frac{1}{x^2+1} $
$\displaystyle \frac{d}{dx}\left(\frac{x}{x^2+1}\right) = \frac{1}{x^2+1}-\frac{4x^2}{(x^2+1)^2} $
What does this tell you?