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

Feb 14, 2016

Domain: $- \infty < x < + \infty$

Range: $1 \ge f \left(x\right) > 0$

Explanation:

The basic 'rule' is that you are not 'allowed' to divide by 0. The proper term for this is that it is not defined.

${x}^{2}$ can only be such that $0 \le - {x}^{2} < \infty$ .This is true for any value of $\left\{x : x \in \mathbb{R}\right)$

When $x$=0 then $f \left(x\right) = 1$.
As ${x}^{2}$ increases then $\frac{1}{1 + {x}^{2}}$ reduces and eventually will tend to 0