Question 132287
{{{f(x)=(7-x)/(5x^2-18x-8)}}} Start with the given function



{{{5x^2-18x-8=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-4)(5x+2)=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:


{{{x-4=0}}} or {{{5x+2=0}}}


{{{x=4}}} or {{{x=-2/5}}}  Now solve for x in each case



So our solutions are {{{x=4}}} or {{{x=-2/5}}}




Since {{{x=-2/5}}} and {{{x=4}}} make the denominator equal to zero, this means we must exclude {{{x=-2/5}}} and {{{x=4}}} from our domain


So our domain is:  *[Tex \LARGE \textrm{\left{x|x\in\mathbb{R} x\neq-\frac{2}{5} or x\neq4\right}}]


which in plain English reads: x is the set of all real numbers except {{{x<>-2/5}}} or {{{x<>4}}}


So our domain looks like this in interval notation

*[Tex \Large \left(-\infty, -\frac{2}{5}\right)\cup\left(-\frac{2}{5},4 \right)\cup\left(4,\infty \right)]


note: remember, the parenthesis <font size=4><b>excludes</b></font> {{{-2/5}}} and 4 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, -8.2, 11.8),
blue(line(-1.7,-7,1.85,-7)),
blue(line(-1.7,-6,1.85,-6)),
blue(line(-1.7,-5,1.85,-5)),
blue(arrow(2.7,-7,10,-7)),
blue(arrow(2.7,-6.5,10,-6.5)),
blue(arrow(2.7,-6,10,-6)),
blue(arrow(2.7,-5.5,10,-5.5)),
blue(arrow(2.7,-5,10,-5)),
blue(arrow(-2.7,-7,-10,-7)),
blue(arrow(-2.7,-6.5,-10,-6.5)),
blue(arrow(-2.7,-6,-10,-6)),
blue(arrow(-2.7,-5.5,-10,-5.5)),
blue(arrow(-2.7,-5,-10,-5)),

circle(-2.2,-5.8,0.35),
circle(-2.2,-5.8,0.4),
circle(-2.2,-5.8,0.45),


circle(2.2,-5.8,0.35),
circle(2.2,-5.8,0.4),
circle(2.2,-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=-2/5}}} and {{{x=4}}} (which is represented by the open circles).
This graphically represents our domain in which x can be any number except x cannot equal {{{-2/5}}} or 4