Question 146268
{{{f(t)=(t+3)/(t(t+2))}}} Start with the given function



{{{t(t+2)=0}}} Set the denominator equal to zero. Remember, dividing by 0 is undefined. So if we find values of t that make the denominator zero, then we must exclude them from the domain.



Set each factor equal to zero:


{{{t=0}}} or {{{t+2=0}}}


{{{t=0}}} or {{{t=-2}}} Now solve for t in each case



So our solutions are {{{t=0}}} or {{{t=-2}}}




Since {{{t=-2}}} and {{{t=0}}} make the denominator equal to zero, this means we must exclude {{{t=-2}}} and {{{t=0}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{t|t\in\mathbb{R} t\neq-2 and t\neq0\right}}]


which in plain English reads: t is the set of all real numbers except {{{t<>-2}}} or {{{t<>0}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, -2\right)\cup\left(-2,0 \right)\cup\left(0,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> -2 and 0 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, -11, 9),
blue(line(-0.5,-7,0.65,-7)),
blue(line(-0.5,-6,0.65,-6)),
blue(line(-0.5,-5,0.65,-5)),
blue(arrow(1.5,-7,10,-7)),
blue(arrow(1.5,-6.5,10,-6.5)),
blue(arrow(1.5,-6,10,-6)),
blue(arrow(1.5,-5.5,10,-5.5)),
blue(arrow(1.5,-5,10,-5)),
blue(arrow(-1.5,-7,-10,-7)),
blue(arrow(-1.5,-6.5,-10,-6.5)),
blue(arrow(-1.5,-6,-10,-6)),
blue(arrow(-1.5,-5.5,-10,-5.5)),
blue(arrow(-1.5,-5,-10,-5)),

circle(-1,-5.8,0.35),
circle(-1,-5.8,0.4),
circle(-1,-5.8,0.45),


circle(1,-5.8,0.35),
circle(1,-5.8,0.4),
circle(1,-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 {{{t=-2}}} and {{{t=0}}} (which is represented by the open circles).
This graphically represents our domain in which t can be any number except t cannot equal -2 or 0