Question 165631

{{{f(x)=(3x+1)/(x-5)}}} Start with the given function



{{{x-5=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+5}}}Add 5 to both sides



{{{x=5}}} Combine like terms on the right side






Since {{{x=5}}} makes the denominator equal to zero, this means we must exclude {{{x=5}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq5\right}}]


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


So our domain looks like this in interval notation

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


note: remember, the parenthesis <font size=4><b>excludes</b></font> 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, -5, 15),
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=5}}} (which is represented by the open circle).
This graphically represents our domain in which x can be any number except x cannot equal 5