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

1 Answer
Nov 3, 2015

Answer:

The domain of #f(x)# is the whole of #RR#, that is #(-oo, oo)# in interval notation.

The range of #f(x)# is #(0, 1]#.

Explanation:

#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]}