finding domain of rational functiona algebraically college algebra #3
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finding domain of rational functiona algebraically college algebra #3


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
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