Find the domain of function: h(x)=√(x-4)/(x^2-x-12)
So the domain is all of the values that x can be that will result in a solution in a way right? So right off the bat anything at would make this negative is a no no but more importantly what would make this undefined?
If i break apart the denominator i get (x-4)(x-3) if I plug in 4 or 3 i will get no solution so those cannot be included in the domain
wait is it \[\sqrt{x-4}/(x ^{2}-x-12) or \sqrt{(x-4)/(x ^{2}-x-12)}\]
its the first one?
okay perfect x has to be greater than 4!
sorry i kept too long
do you understand the entire answer?
so do i work for the denominator or? and equate it to 0
yeah you set it to zero and solve for x
only the denominator right?
yeah
oh kk.then x=-3 and 4
the reason I'm saying greater than 4 is because you need absolutely need to half the top portion of the fraction greater than 0
yeah sorry i put 3 and 4 its -3 but -3 doesnt even work because if we put that in the top becomes undefined
there is no square root of negative numbers
so what is the answer then?
x>4
that is the answer
Sorry was trying not to put all the pieces together. Trying to let you figure out some of it so you can learn from it.
oh okay.
so u plugged the numbers in the denominator or the numerator
to see if its correct
hehe you should fix you're skype I could explain this much easier
i want to know how i can check for my anawers
sec let me pause my tv to explain it
kk
okay so! the point of this problem is to see when the function is undefined because that would mean that it is not in the domain. The domain meaning what numbers can I put in for x that will give me solutions.
so the first step in this is to look okay how can I make this undefined. we look and look we see a square root- well we know square roots cannot contain negative numbers theres your first portion to you answer so we know that x-4=0 x=4 so X can be greater than or equal to 4
thats the first part.
now the denominator- we see a function lets say F(x)=root(x-4)/y y=x^2-x-12 when y=0 the function F(x) is undefined correct?
undefined, correct?
*
yes
so we need to figure out how to make this function 0 the only way possible is by factoring so we factor it out, like you did, got x+3=0 and x-4=0 giving us x=-3,4
now if we combine the conditions of the first the numerator we know that we cannot have -3 in our answer as far as the domain because it is less than and not equal to 4
does that make sense to you?
so i always need to work out for the numerator first before the denominator
haha no im sorry my point was you need to make sure both agree with each other
you wont always have a square root so sometimes the top wont even matter
so that means after fnking x and plugging x into the equation the equation shoulb be equal to zero?
finding
on the bottom yes because that will make the overall equation undefined
oh okay
if its undefined it is not part of the domain
get it now?
didnt get what u last said.i thot the domain should be undefined
if you select an x that results in the function being undefined that value for x is not included in the domain
so 4 does work because it makes the dominator zero. so it can't be part of the domain
doesnt work*
oh so any number that doesnot equate a function to zero is undefined?
hmmm okay undefined is a messy word. It's essentially you've violated a golden rule like dividing by zero
I didnt really understand what were were saying with that last phrase
oh ok .so if im getting u,,since 4 plugged into the denominator makes the function =0 which is undefined 4 cannot be a domain
hmm okay so I probably shouldnt have used function like I did without defining all the equations as functions. if I plug in 4 it makes the dominator zero making the function undefined because we're dividing by zero.
let me write it out on paper and post a picture.
oh ok
the way im going to show you is a little unorthodox but i think its the best way to understand problem like this.
lol ok
still waiting..lol
sorry making it easy to understand
okayy
ok so the answer as i see is >4
yes
\[i was thinking the anwer was gonna be(-i\[\[i was thinking the anwer was gonna be(-\infty,-3) or (-3,4) or (4,\infty)
well it can be expressed as (4,inf)
its the same thing
oh ok.
thanks so much you soo helpful
no problem!
also
your other problem that i answered
you probably arent doing derivatives yet are you?
no not yet
yeah i got that
ok will chat with you some other time.thanks very much.feeling sleepy.lol
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