How do you find the domain of #sqrt((x/(x-2)))#?

1 Answer
Oct 13, 2015

Answer:

#x in (-oo, 0] uu (2, + oo)#

Explanation:

The first thing to look out for is any value of #x# that will maek the denominator of the fraction equal to zero. That happens when

#x -2 = 0 implies x = 2#

so this value of #x# will be excluded from the domain of the function.

The second thing to look out for is the fact that you're dealing with a square root, which, for real numbers, can only be taken of positive numbers.

This means that the fraction #x/(x-2)# must be greater than or equal to zero.

#x/(x-2) >= 0 #

This condition is satisfied for

#x <= 0 implies {(x <=0), (x-2 < 0) :} implies x/(x-2) >=0#

and

#x >2 implies {( x>2), (x - 2 > 0) :} implies x/(x-2) >= 0#

Therefore, the domain of the function will include any value of #x# that is smaller than or equal to #0# or greter than #2#. In interval notation, this is equivalent to

#x in (-oo, 0] uu (2, + oo)#

graph{sqrt(x/(x-2)) [-10, 10, -5, 5]}