Question 99371


{{{K(x)=(x-2)/(x-3)}}} Start with the given function



{{{x-3=0}}} Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of x that make the denominator zero, then we must exclude them from the domain.




{{{x=0+3}}}Add 3 to both sides



{{{x=3}}} Combine like terms on the right side


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Answer:

So our answer is {{{x=3}}} 




Since {{{x=3}}} makes the denominator equal to zero, this means we must exclude {{{x=3}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\epsilon\mathbb{R} x\neq3\right}}]


which in plain English reads: x is the set of all real numbers except {{{x<>3}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, 3\right)\cup\left(3,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> 3 from the domain


If we wanted to graph the domain on a number line, we would get:


{{{drawing(500,50,-10,10,-10,10,
number_line( 500, -7, 13),
blue(arrow(0.2,-7,10,-7)),
blue(arrow(0.2,-6.5,10,-6.5)),
blue(arrow(0.2,-6,10,-6)),
blue(arrow(0.2,-5.5,10,-5.5)),
blue(arrow(0.2,-5,10,-5)),
blue(arrow(-0.2,-7,-10,-7)),
blue(arrow(-0.2,-6.5,-10,-6.5)),
blue(arrow(-0.2,-6,-10,-6)),
blue(arrow(-0.2,-5.5,-10,-5.5)),
blue(arrow(-0.2,-5,-10,-5)),

circle(0,-5.8,0.35),
circle(0,-5.8,0.4),
circle(0,-5.8,0.45),
circle(0,-5.8,0.4),
circle(0,-5.8,0.45)
)}}} Graph of the domain in blue and the excluded value represented by open circle


Notice we have a continuous line until we get to the hole at {{{x=3}}} (which is represented by the open circle).
This graphically represents our domain in which x can be any number except x cannot equal 3