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

Aug 3, 2015

Domain: $\left(- \infty , + \infty\right)$
Range: $\left(0 , 1\right]$

#### Explanation:

The only possible restriction for this function's domain is when the denominator is equal to zero, but since ${x}^{2} > 0 , \left(\forall\right) x$ in $\left(- \infty , + \infty\right)$, 1 + x^2! will always be positive.

This means that the domain of the function will be $\mathbb{R}$, or $\left(- \infty , + \infty\right)$.

Since the denominator will always be positive, 1/(1+x^2 will always be positive as well.

The range of the function will be $y > 0$ and $y \le 1$, since the maximum value of $f \left(x\right)$ occurs at $x = 0$.

$f \left(0\right) = \frac{1}{1 + 0} = 1 \to$ global maximum

That happens because you have $f \left(x\right) < 1 , \left(\forall\right) x \ne 0$.

The range of the function will thus be $\left(0 , 1\right]$.

graph{1/(1+x^2) [-8.89, 8.89, -4.444, 4.445]}