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

1 Answer
Feb 14, 2016

Answer:

Domain: # -oo < x < +oo#

Range: #1 >=f(x) >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<=-x^2< oo# .This is true for any value of #{x: x in RR)#

When #x#=0 then #f(x)=1#.
As #x^2# increases then #1/(1+x^2)# reduces and eventually will tend to 0