<s>[INST] creative method to obtain range of newton function ?! I am searching for more proof that the range of $y=\frac{x}{x^2+1}$ is $ \frac{-1}{2}\leq y \leq \frac{+1}{2}$ 
  these are my tries : 
domain is $\mathbb{R}$
first : $$\quad{y=\frac{x}{x^2+1}\\yx^2+y=x \rightarrow  x^2y-x+y=0  \overset{\Delta \geq 0 }{\rightarrow} 1-4y^2 \geq 0 \rightarrow \frac{-1}{2}\leq y \leq \frac{+1}{2} }$$
  second :
$$ \quad{y=\frac{x}{x^2+1} \overset{x=tan \alpha }{\rightarrow} \space y=\frac{tan \alpha}{tan^2 \alpha+1}=\frac{tan \alpha}{\frac{1}{cos^2\alpha}}=sin \alpha cos \alpha =\\\frac{2}{2} sin \alpha cos \alpha =\frac{1}{2}sin 2\alpha\\ -1\leq sin 2\alpha \leq 1 \rightarrow -\frac{1}{2} \leq \frac{1}{2}sin 2\alpha \leq \frac{1}{2} \rightarrow   \frac{-1}{2}\leq y \leq \frac{+1}{2}}$$   3rd:
$$\quad{\frac{1}{y}=\frac{x^2+1}{x}=x+\frac{1}{x}  \overset{|x+\frac{1}{x}|\geq  2}{\rightarrow} |\frac{1}{y}| \geq 2 \rightarrow |y| \leq \frac{1}{2} \rightarrow   \frac{-1}{2}\leq y \leq \frac{+1}{2}}$$
4th :$$f(x)=\frac{x}{x^2+1} \\f'=\frac{1-x^2}{1+x^2}=0 \rightarrow x=\pm1 \\ \left\{\begin{matrix}
f(1) &=\frac{1}{2} \\ 
f(-1) & =-\frac{1}{2}\\ 
\lim_{x \rightarrow +\infty }\frac{x}{x^2+1} & =0\\ 
\lim_{x \rightarrow -\infty }\frac{x}{x^2+1} & =0
\end{matrix}\right. \overset{x \in \mathbb{R}}{\rightarrow}  \space \frac{-1}{2}\leq y \leq \frac{+1}{2}$$   
now : is there (creative ) 5th ,6th ,... method  to find range of this function  ?
 [/INST] By contradiction: assume $y > \frac{1}{2}$
$$\frac{x}{x^2 +1} > \frac{1}{2} \Rightarrow  0 > (x-1)^2$$
and $y<-\frac{1}{2}\Rightarrow 0 > (x+1)^2 $.
But we also see that $y(-1) = -\frac{1}{2}$ and $y(1) = \frac{1}{2}$.
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