# How do you find the domain and range of x/(x^2+1)?

The domain is $R$ and the range is $\left[- \frac{1}{2} , \frac{1}{2}\right]$

#### Explanation:

The domain is $R$ the set of all reals because $1 + {x}^{2} > 0$.

To find the range we have that

$y = \frac{x}{{x}^{2} + 1} \implies y \left({x}^{2} + 1\right) = x \implies y {x}^{2} + x + y = 0 =$

this is trinomial with respect to x hence

${1}^{2} - 4 y \cdot y \ge 0 \implies 1 - 4 {y}^{2} \ge 0 \implies - \frac{1}{2} \le y \le \frac{1}{2}$