How do you find the domain of # f(x) = 1/(sqrt(5 - x) - 1)#?

1 Answer

Answer:

The domain, in this case. is all real numbers except for the values of #x# that would result in a non-real value.

Explanation:

If #x# causes the denominator to be #0#, then the function would be undefined at that point.

#0 = sqrt(5 - x) - 1#

#1 = sqrt(5 - x)#

#1^2 = 5 - x#

#1 = 5 - x#

#-4 = -x#

#x = 4#

The function would also be imaginary at #x# if there was a square root of a negative value. Therefore...

#sqrt(5 - x) >= 0#

#5 - x >= 0^2#

#5 >= x#

The domain of #x# is all real values less than or equal to #5# and not including #4#.