Question 97435
1.


{{{K(v)=(4v^2 - 9)/(v^2 - v - 72)}}} Start with the given function 





{{{v^2-v-72=0=0}}} Set the denominator equal to zero


{{{(v-9)(v+8)=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:


{{{v-9=0}}} or {{{v+8=0}}}


{{{v=9}}} or {{{v=-8}}}  Now solve for v in each case


Since {{{v=9}}} or {{{v=-8}}} make the denominator equal to zero, that means we must exclude these values from the domain.


So our domain is: v is the set of all real numbers except {{{v<>-8}}} or {{{v<>9}}}


Which looks like this in interval notation:

*[Tex \Large \left(-\infty, -8\right)\cup\left(-8, 9\right)\cup\left(9,\infty \right)]





----------------------------------------------------------------------------------------


2. 


{{{f(x)=1/(x-11)}}}Start with the given function



{{{x-11=0}}} Set the denominator equal to zero



{{{x=0+11}}}Add 11 to both sides



{{{x=11}}} Combine like terms on the right side



So when {{{x=11}}}, our denominator is zero. So we must exclude this value.



So our domain is: x is the set of all real numbers except {{{x<>11}}}



Which looks like this in interval notation:


*[Tex \Large \left(-\infty, 11\right)\cup\left(11,\infty \right)]