# How do you find the range of f(x)= 1/(x^2 + 9)?

Sep 16, 2015

Range { y in RR, 0< y<= 1/9}#

#### Explanation:

Solve $y = \frac{1}{{x}^{2} + 9}$ for x

${x}^{2} = \frac{1 - 9 y}{y}$,

$x = \sqrt{\frac{1 - 9 y}{y}}$

This implies that $y \ne 0$ and $1 - 9 y \ge 0$, that is $y \le \frac{1}{9}$ Hence range of f(x) would be

$\left\{y \in \mathbb{R} , 0 < y \le \frac{1}{9}\right\}$