Question
Find the range of the following function : $f(x)=\frac{x}{9+x^2}$

Answer

$f(x)=\frac{x}{9+x^2}=y$ (say)
$\therefore \mathrm{x}^2 \mathrm{y}-\mathrm{x}+9 \mathrm{y}=0$
For real $x$, Discriminant $>0$
$ \therefore 1-4(y)(9 y) \geq 0$
$\therefore y^2 \leq \frac{1}{36}$
$\therefore \frac{-1}{6} \leq y \leq \frac{1}{6}$
$\therefore \text { Range of } f=\left[\frac{-1}{6}, \frac{1}{6}\right] $

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