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

1 Answer
May 31, 2016

Answer:

Domain is #RR#+, Range is #RR^+#

Explanation:

Domain is given by #x^2 +1# >0. That means all real values of x, that is, it would be #RR#

For range, exchange x and y in #y= ln (x^2+1)# and find the domain. Accordingly, #x= ln(y^2 +1)#

#y^2= e^x-1#. The domain of this function is all #x>= 0# that means all real numbers #>=#=0

Hence the range of given function would be all Real numbers #>=0#