Question 105759
{{{y=(1)/((x-3)(x+1))}}} Start with the given function



{{{(x-3)(x+1)=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.





Now set each factor equal to zero:


{{{x-3=0}}} or {{{x+1=0}}}


{{{x=3}}} or {{{x=-1}}}  Now solve for x in each case



So our solutions are {{{x=3}}} or {{{x=-1}}}




Since {{{x=-1}}} and {{{x=3}}} make the denominator equal to zero, this means we must exclude {{{x=-1}}} and {{{x=3}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\epsilon\mathbb{R} x\neq-1 and x\neq3\right}}]


which in plain English reads: x is the set of all real numbers except {{{x<>-1}}} or {{{x<>3}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, -1\right)\cup\left(-1,3 \right)\cup\left(3,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> -1 and 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, -9, 11),
blue(line(-1.5,-7,1.65,-7)),
blue(line(-1.5,-6,1.65,-6)),
blue(line(-1.5,-5,1.65,-5)),
blue(arrow(2.5,-7,10,-7)),
blue(arrow(2.5,-6.5,10,-6.5)),
blue(arrow(2.5,-6,10,-6)),
blue(arrow(2.5,-5.5,10,-5.5)),
blue(arrow(2.5,-5,10,-5)),
blue(arrow(-2.5,-7,-10,-7)),
blue(arrow(-2.5,-6.5,-10,-6.5)),
blue(arrow(-2.5,-6,-10,-6)),
blue(arrow(-2.5,-5.5,-10,-5.5)),
blue(arrow(-2.5,-5,-10,-5)),

circle(-2,-5.8,0.35),
circle(-2,-5.8,0.4),
circle(-2,-5.8,0.45),


circle(2,-5.8,0.35),
circle(2,-5.8,0.4),
circle(2,-5.8,0.45)


)}}} Graph of the domain in blue and the excluded values represented by open circles


Notice we have a continuous line until we get to the holes at {{{x=-1}}} and {{{x=3}}} (which is represented by the open circles).
This graphically represents our domain in which x can be any number except x cannot equal -1 or 3