# What is the domain and range of f(x) = 1/(1+x^2)?

Nov 3, 2015

The domain of $f \left(x\right)$ is the whole of $\mathbb{R}$, that is $\left(- \infty , \infty\right)$ in interval notation.

The range of $f \left(x\right)$ is $\left(0 , 1\right]$.

#### Explanation:

$1 + {x}^{2} \ge 1 > 0$ for all $x \in \mathbb{R}$.

So the denominator is never $0$ and $f \left(x\right)$ is well defined for all $x \in \mathbb{R}$. So the domain of $f \left(x\right)$ is $\mathbb{R} = \left(- \infty , \infty\right)$

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

Multiplying both sides by $\left(1 + {x}^{2}\right)$ we get:

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

This has no solution if $y = 0$, so $0$ is not in the range of $f \left(x\right)$.

If $y \ne 0$ then we can divide both sides by $y$ to get:

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

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

Since ${x}^{2} \ge 0$ for all $x \in \mathbb{R}$, we require $\frac{1}{y} - 1 \ge 0$

If $y > 1$ then $0 < \frac{1}{y} < 1$, so $\frac{1}{y} - 1 < 0$. Therefore $\left(1 , \infty\right)$ is not part of the range.

If $y < 0$ then $\frac{1}{y} < 0$ and $\frac{1}{y} - 1 < 0$. So $\left(- \infty , 0\right)$ is not part of the range either.

Suppose $y > 0$ and $y \le 1$

Since $y > 0$ we can divide both sides of the second inequality by $y$ to get:

$1 \le \frac{1}{y}$

Then subtract $1$ from both sides to get:

$\frac{1}{y} - 1 \ge 0$ as required.

Hence $x = \pm \sqrt{\frac{1}{y} - 1}$ are two values that give $f \left(x\right) = y$.

So the range includes the whole of $\left(0 , 1\right]$

graph{1/(1+x^2) [-5, 5, -2.5, 2.5]}