What is the domain and range of #f(x)=sqrt(x^2 +4)#?

1 Answer
Apr 17, 2015

#f(x)=sqrt(x^2+4)# is defined for all Real values of #x#
The Domain is #x epsilon RR#
(actually #f(x)# is valid for #x epsilon CC# but I will assume we are not interested in Complex numbers).

If we restrict #x epsilon RR#
then #f(x)# has a minimum value when #x=0# of
#sqrt(0^2+4) = 2#
and the Range of #f(x)# is #[2,+oo)#

(If we allow #x epsilon CC# the Range of #f(x)# becomes all of #CC#)