How do you find the domain and range algebraically of #f(x)=1/sqrt(x-4)#?

1 Answer
Feb 7, 2015

Your function has a square root but also this square root is in the denominator of a fraction.

So, what you wan is to avoid values of #x# that:

1) make the argument of the square root NEGATIVE (you cannot find a real number as solution of a negative square root);
2) make the denominator equal to ZERO (you cannot evaluate a division by zero).

To express the above points mathematically you write that #x-4# MUST be #>0#.

This means that you can accept only values that satisfy:
#x-4>0#
#x>4#

#4# is not acceptable because if you use it you get #sqrt(4-4)# that can be calculated and gives #0# but it would be in the denominator and this is not acceptable.
When you get near #4# your function gets very big tending to #+oo# while going towards #x# values very big the function tends to zero (you can try by yourself substituting values in your function and calculating it).
So your domain will be all the positive #x# bigger than #4# up to #+oo#.
With a range for #y=f(x)# of #+oo# to #0#.
Or graphically:

graph{1/sqrt(x-4) [-10, 10, -5, 5]}