How do you find the domain and range of #f(x) = (x - 3)^(1/2)#?

1 Answer
Jul 27, 2015

Answer:

#(x-3)^(1/2)# is the same as #sqrt(x-3)#

Explanation:

For square roots the argument must be non-negative, or:
#x-3>=0->x>=3#. There is no upper limit.
So the domain is #3<=x< oo#

When #x=3->f(x)=0#, in other words #f(x)>=0#
Also here there is no upper limit as #x# gets larger.
So the range is #0<=f(x)< oo#
graph{sqrt(x-3) [-6.56, 25.48, -4.6, 11.43]}