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

May 18, 2018

The domain of $f$ is $\mathbb{R}$,
and the range is $\left\{f \left(x\right) \in \mathbb{R} : - \frac{1}{2} \le f \left(x\right) \le \frac{1}{2}\right\}$.

#### Explanation:

Solving for the domain of $f$, we will observe that the denominator is always positive, regardless of $x$, and indeed is least when $x = 0$. And because ${x}^{2} \ge 0$, no value of $x$ can give us ${x}^{2} = - 1$ and we can therefore rid ourself of the fear of the denominator ever equalling naught. By this reasoning, the domain of $f$ is all real numbers.

By contemplating the output of our function, we will notice that, from the right the function is decreasing until the point $x = - 1$, after which the function steadily increases. From the left, it is the opposite: the function is increasing until the point $x = 1$, after which the function steadily decreases.

From either direction, $f$ cannot ever equal $0$ except at $x = 0$ because for no number $x > 0 \mathmr{and} x < 0$ can $f \left(x\right) = 0$.

Therefore the highest point on our graph is $f \left(x\right) = \frac{1}{2}$ and the lowest point is $f \left(x\right) = - \frac{1}{2}$. $f$ can equal all numbers in between though, so the range is given by all real numbers in between $f \left(x\right) = \frac{1}{2}$ and $f \left(x\right) = - \frac{1}{2}$.

May 18, 2018

The domain is $x \in \mathbb{R}$. The range is $y \in \left[- \frac{1}{2} , \frac{1}{2}\right]$

#### Explanation:

The denominator is

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

The domain is $x \in \mathbb{R}$

To find, the range procced as follows :

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

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

$y {x}^{2} - x + y = 0$

In order for this quadratic equation to have solutions, the discriminant $\Delta \ge 0$

Therefore,

${\left(- 1\right)}^{2} - 4 \cdot y \cdot y \ge 0$

$1 - 4 {y}^{2} \ge 0$

The solution to this inequality is

$y \in \left[- \frac{1}{2} , \frac{1}{2}\right]$

The range is $y \in \left[- \frac{1}{2} , \frac{1}{2}\right]$

graph{x/(x^2+1) [-3, 3.93, -1.47, 1.992]}