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

Answer 1

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

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 be aware of is that you're working with a square root, which can only be taken of positive numbers for real numbers.

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

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

For, this requirement is met.

#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)#

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

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Answer 2

Eli Assouline

Exclude values that make the expression undefined. Domain: (x \in (-\infty, 0) \cup (2, \infty)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.