Question 166656
Remember, the domain is simply the set of all x values that produce a y value. 


# 1




{{{h(t)=(t-5)/(t^2-25)}}} Start with the given function



{{{t^2-25=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.





{{{(t-5)(t+5)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)





Now set each factor equal to zero:


{{{t-5=0}}} or {{{t+5=0}}}


{{{t=5}}} or {{{t=-5}}}  Now solve for t in each case



So our solutions are {{{t=5}}} or {{{t=-5}}}




Since {{{t=-5}}} and {{{t=5}}} make the denominator equal to zero, this means we must exclude {{{t=-5}}} and {{{t=5}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{t|t\in\mathbb{R} t\neq-5 and t\neq5\right}}]


which in plain English reads: t is the set of all real numbers except t CANNOT equal {{{-5}}} or t CANNOT equal {{{5}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, -5\right)\cup\left(-5,5 \right)\cup\left(5,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> -5 and 5 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, -10, 10),
blue(line(-4.5,-7,4.65,-7)),
blue(line(-4.5,-6,4.65,-6)),
blue(line(-4.5,-5,4.65,-5)),
blue(arrow(5.5,-7,10,-7)),
blue(arrow(5.5,-6.5,10,-6.5)),
blue(arrow(5.5,-6,10,-6)),
blue(arrow(5.5,-5.5,10,-5.5)),
blue(arrow(5.5,-5,10,-5)),
blue(arrow(-5.5,-7,-10,-7)),
blue(arrow(-5.5,-6.5,-10,-6.5)),
blue(arrow(-5.5,-6,-10,-6)),
blue(arrow(-5.5,-5.5,-10,-5.5)),
blue(arrow(-5.5,-5,-10,-5)),

circle(-5,-5.8,0.35),
circle(-5,-5.8,0.4),
circle(-5,-5.8,0.45),


circle(5,-5.8,0.35),
circle(5,-5.8,0.4),
circle(5,-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=-5}}} and {{{t=5}}} (which is represented by the open circles).
This graphically represents our domain in which t can be any number except t cannot equal -5 or 5





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# 2






{{{g(x)=(x^3-27)/(4x)}}} Start with the given function



{{{4x=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)/(4)}}} Divide both sides by 4 to isolate x




{{{x=0}}} Divide






Since {{{x=0}}} makes the denominator equal to zero, this means we must exclude {{{x=0}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq0\right}}]


which in plain English reads: x is the set of all real numbers except x CANNOT equal {{{0}}}


So our domain looks like this in interval notation

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


note: remember, the parenthesis <font size=4><b>excludes</b></font> 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, -10, 10),
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=0}}} (which is represented by the open circle).
This graphically represents our domain in which x can be any number except x cannot equal 0