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

Apr 6, 2018

The domain is $x \in \mathbb{R}$. The range is $y \in \left(0 , 1\right]$

#### Explanation:

The denominator is $= 1 + {x}^{2}$

$\forall x \in \mathbb{R}$, $1 + {x}^{2} > 0$

Therefore,

The domain of $f \left(x\right)$ is $x \in \mathbb{R}$

To determine the range, proceed as follows

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

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

$y + y {x}^{2} = 1$

$y {x}^{2} = 1 - y$

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

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

The range of $f \left(x\right)$ is the domain of $x$

$\left(\frac{1 - y}{y}\right) > 0$

$y \in {\mathbb{R}}_{+}^{\text{*}}$

$1 - y \ge 0$

$y \le 1$

Therefore,

The range is $y \in \left(0 , 1\right]$

graph{1/(1+x^2) [-11.25, 11.25, -5.625, 5.625]}