What is the domain of #f(x)= (8x)/((x-1)(x-2))#?

2 Answers

It is all the real numbers except those that nullify the denominator in our case x=1 and x=2. So the domain is #R-{1,2}#

Sep 9, 2015

The domain is all the real numbers except x can not be 1 or 2.

Explanation:

#f(x) = (8x)/[(x - 1)(x - 2)]#

The domain of a function is where that function is defined, now we can easily find the point(s) where this function is undefined and exclude them from the domain, since we can't divide by zero the roots of the denominators are the points that the function is not defined, so:

#(x - 1)(x - 2) = 0# => using the The Zero Product Property which states that if ab = 0, then either a = 0 or b = 0 (or both), we get:

#x - 1 = 0 => x = 1#
#x - 2 = 0 => x = 2#

Hence the domain is all the real numbers except 1 or 2.
in interval notation:
#(-oo , 1) uuu(1 , 2)uuu (2 , oo)#