Good day students welcome to mathgotserved.com

in this clip were going to be going over problem number three on how to find the domain of

rational functions algebraic we X so the instructions for example 3 are for us to find the domain

of the following rational function so you looking that Om F of X equals Y times the

quantity X minus three divided by three X square plus fourteen X minus five right so

we went over in problem off one into problems one to in order to find the domain only function

we have to exclude any value that causes the function to have an undefined outputs from

the domain so what causes a function to have an undefined outputs any value that causes

the denominator to attain the value of zero results in the function having an undefined

outputs so that number should be excluded from the domain right so the couple also Richart

you asked to find the domain of the rational function the look like this is polynomial

over polynomial is simply soul equation we set the denominator equal to zero and exclude

the solution of this equation from your domain and that will be the domain of your rational

function right so in this problem the denominator is where our focus is going to be our denominator

is three X square plus fourteen X minus five so what do we do with our denominator we said

it equal to zero indicated here this is the quadratic equation now the different ways

you can solve the quadratic the first method you always want to use is the factorization

method let’s hope that this expression is fact ripple and then we can just simply find

what the Om values to excluded from our domain our right somebody use of the X game or the

AC method to factor this quadratic equation AC goes on the top and B equals on the bottom

right in this problem a is three B is fourteen and see is negative five right there AC is

three times negative five which is negative fifteen and B is fourteen the question now

is what two numbers multiplied to give you negative fifteen and asked to give you fourteen

OSC how about fifty and one, see fifteen and if we do fifteen and Y if we one the sum to

be positive down into smaller has to negative okay so let’s write is fifteen times negative

one yields negative fifteen fifteen minus one is fourteen excellent now this quadratic

has the a value of peanuts not want so we can X take a shortcut here we actually have

to Om so after the expression by grouping okay so let’s go ahead and do that so the

have three X square now these two terms that we got from our X game right here will inserted

in place of fourteen negative X minus X actually positive fifteen X minus X minus five equals

zero right these two numbers combined to give you positive fourteen so we haven’t change

the problem this change the with looks now this down the middle and out factor by grouping

right from the first two I can extract three X of this me with X plus five and then from

these two I can extract negative one now this me with X was five also equals zero now these

two quantities in current this is must be the same in other four our factorization to

the correct okay so they are the same so let’s get so we factor them into one X plus five

and then we factor out X was five these two exercise will be left with three X minus one

and would those to India (okay three X minus one equals zero okay so now we have the fact

that form X was five times three X minus one equals zero we now have a situation that is

similar to problem number two so how do resolve this we going to use the zero product property

rights so to do that will simply set both factors equal to zero so will going to do

X plus five equals zero and three X minus one equals zero right to solve the first equation

will subtract five from both sides that gives is X equals negative five and then over here

this is a two-step algebraic equation that to solve it we just take two steps at one

to both sides in then divided by three okay so as one that gives us three X equals one

and then divide both sides by three okay so for the public site by three hundred three

five three we have X equals one third so what are these values these are the values that

causes our function to have an undefined outputs so what do we do these values must be excluded

from the domain in other for all our output values to be defined alright so let’s go ahead

and write a so what is the domain the domain of the function is susceptible X is such that

X is not equal to negative five and X is not equal to one third okay so X can of the any

of these two values in order for our function to have e find out right so that is your final

result thanks so much for taking the time to watch this presentation really appreciate

it feel free to subscribe to our channel for updates to other cool tutorials such as this

you have any comments or questions or you like to request the video for us to make for

you feel free to include that in the comments section below this video more clips can be

found on mathgotserved.com thanks again for watching and have a wonderful day

## One Comment

## Sebastian Terrel-Perez

Thank you. My teacher is clueless when it comes to teaching students on how to find the domain.