How do you find the domain of #h(x)=sqrt(4-x)+sqrt(x^2-1)#?

1 Answer
Jan 6, 2016

Answer:

Domain of #h(x)# #=(-oo,-1]uu[1,4]#

Explanation:

Domain is all possible allowable #x# value inputs.

There are generally 2 cases that are not allowed in real functions :

  1. Even square roots of negative numbers. (Not defined in #RR#).
  2. Division by zero. (Not defined at all).

In this case there are no divisions (denominators) in #h(x)# so we only consider case 1 of the square roots not allowed to be negative.
Doing so, it becomes clear that the first term is only negative if #x>4# and the second term is only negative if #x<1#.

So #

Hence the domain is #{x in RR | -oo < x<= 1 and 1 <= x <= 4}#

#=(-oo,-1]uu[1,4]#.

The graph makes it clear :

graph{sqrt(4-x)+sqrt(x^2-1) [-7.28, 10.5, -1.93, 6.95]}