#1+x^2 >= 1 > 0# for all #x in RR#.
So the denominator is never #0# and #f(x)# is well defined for all #x in RR#. So the domain of #f(x)# is #RR = (-oo, oo)#
Let #y = f(x) = 1/(1+x^2)#
Multiplying both sides by #(1+x^2)# we get:
#y(1+x^2) = 1#
This has no solution if #y = 0#, so #0# is not in the range of #f(x)#.
If #y != 0# then we can divide both sides by #y# to get:
#1+x^2 = 1/y#
So: #x^2 = 1/y-1#
Since #x^2 >= 0# for all #x in RR#, we require #1/y-1 >= 0#
If #y > 1# then #0 < 1/y < 1#, so #1/y - 1 < 0#. Therefore #(1, oo)# is not part of the range.
If #y < 0# then #1/y < 0# and #1/y - 1 < 0#. So #(-oo, 0)# is not part of the range either.
Suppose #y > 0# and #y <= 1#
Since #y > 0# we can divide both sides of the second inequality by #y# to get:
#1 <= 1/y#
Then subtract #1# from both sides to get:
#1/y-1 >= 0# as required.
Hence #x = +-sqrt(1/y-1)# are two values that give #f(x) = y#.
So the range includes the whole of #(0, 1]#
graph{1/(1+x^2) [-5, 5, -2.5, 2.5]}
