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

1 Answer
Aug 3, 2015

Answer:

Domain: #(-oo, +oo)#
Range: #(0,1]#

Explanation:

The only possible restriction for this function's domain is when the denominator is equal to zero, but since #x^2>0, (AA)x# in #(-oo, +oo)#, #1 + x^2!# will always be positive.

This means that the domain of the function will be #RR#, or #(-oo, +oo)#.

Since the denominator will always be positive, #1/(1+x^2# will always be positive as well.

The range of the function will be #y>0# and #y<=1#, since the maximum value of #f(x)# occurs at #x=0#.

#f(0) = 1/(1 + 0) = 1 -># global maximum

That happens because you have #f(x)<1, (AA)x !=0#.

The range of the function will thus be #(0, 1]#.

graph{1/(1+x^2) [-8.89, 8.89, -4.444, 4.445]}