Show that $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 }$ So I'm trying to show that:
$\dfrac{ -1 }{ 2 } \le \dfrac{ x }{ 1+x^2 } \le \dfrac{ 1 }{ 2 }$
for every value of x.
I know I have to use mean value theorem so I tried to show it with cases. First I tried showing that $\dfrac{-1}{2} \le \dfrac{x}{1+x^2}$ and then $\dfrac{1}{2} \ge \dfrac{x}{1+x^2}$ using MVT. Is that correct? Can someone guide me?
Thanks!

Let $u=|x|$, then $\frac{ -1 }{ 2 } \le \frac{ x }{ 1+x^2 } \le \frac{ 1 }{ 2 } \iff  \frac{ u }{ 1+u^2 } \le \frac{ 1 }{ 2 }$.
You are using $AM-GM$ here and not Mean Value Theorem since $1+u^2\geq2u$ by $AM-GM$.
To do the calculus way, just find the derivative of $x\over {1+x^2}$.
$df\over dx$$={1+x^2-2x^2\over{(1+x^2)^2}}=0\implies x=\pm1$.
Sub in $x=1$ you get $f={1\over2}$ and sub in $x=-1$ you get $f={-{1\over2}}$.
And $lim_{x\to \infty}f=0$ shows it is a global maximum. Minimum similarly.
