How do you find the domain and range of #f(x) = 1/(1+x^2)#?

1 Answer
Apr 6, 2018

Answer:

The domain is #x in RR#. The range is #y in (0,1]#

Explanation:

The denominator is #=1+x^2#

#AA x in RR#, #1+x^2>0#

Therefore,

The domain of #f(x)# is #x in RR#

To determine the range, proceed as follows

#y=1/(1+x^2)#

#y(1+x^2)=1#

#y+yx^2=1#

#yx^2=1-y#

#x^2=(1-y)/y#

#x=sqrt((1-y)/y)#

The range of #f(x)# is the domain of #x#

#((1-y)/y)>0#

#y in RR _+^("*")#

#1-y>=0#

#y<=1#

Therefore,

The range is #y in (0,1]#

graph{1/(1+x^2) [-11.25, 11.25, -5.625, 5.625]}