 hello everyone in this video I want to go over the example of find the domain of a composite
function, in this case f of g of x where we have two distinct domains, both one
for f of x and one for
g of x so in a problem like this we always need to
find the composite function of f of g of x and we also need to
take into account again the domains of each individual function so lets try to break this down and find an
efficient method of doing this let’s first of all re-write f of g of x remember is also written as this since it’s the order is written f of g and as we can see up here in that order that translates to f of g of x written like this which means that when you evaluate a function like this you do the inside function first that means
it’s important we consider first the domain of g of x since again if we were
to evaluate this we would have to go through that function
first so it’s important that we know the domain of that function we need to know the domain or the restrictions
of it so that we can remain a function so let’s do that first we’ll find the domain of g of x so I’ll isolate my g of x which again was square root of x + 4 now if you recall, in a square root situation
or an even root we are more concerned with what is inside
the radical and specifically we want anything that’s inside
the radical has to be greater than or equal to zero so I’ll set up x + 4 is greater than or equal
to zero and I’ll solve my inequality in this case I’ll subtract 4 so x is greater than or equal to negative
4, will be my domain or if you want interval notation we’ll say negative 4 to positive infinity so now that we understand the domain of g of x
let’s look at the composite function so I’m going to put g of x inside of f of
x, which again is up here at x squared plus 1 over x – 3 so again here are my two functions so if you
recall that means that we put in for g of x we put that into the f function and then we
put that into the function f itself so in other words I take the square root of x + 4 and put it into x anywhere I see it in the
f function so in place of x I’ll have instead of x squared I’ll have the square root of x + 4 quantity squared plus 1 over the square root of x + 4 minus 3 when I evaluate I want to square my square root it’ll cancel out and I’ll have x + 4 + 1 over now the square root of x + 4 I can’t do anything
with so I’ll leave it as so so I’m left with after I simplify x + 5 over the square root of x + 4 minus 3 so this is my composite function now that I have this I’ll want to go back
to the overall domain so with the overall domain I want to look
at of course the first domain, domain of my inside function, the g and now I want to consider the domain of my
new composite function so I have the original domain now let’s look
at the domain of my new composite again which is right here so here’s my old domain of g for my new composite function, remember for
domain I’m looking for a couple things, I’m looking for a variable in the denominator and I look for radicals so I have both here, I have a variable in
the denominator if I look at the denominator as an overall whole and I have a radical, so I’m going to do both so let’s start with the denominator as a whole since it has a variable I’m going to set the
denominator equal to zero, and I solve for x in this case I add 3 so the square root of x + 4 equals 3 I’ll square both sides so x + 4 equals 9 and by subtracting 4 I find that x is equal to 5, now when we find domains when you look at domains in this way, we’re
trying to find the values that you cannot have, so in other words this means that part of this domain is that we cannot have
the value of 5 because when we have 5, we would put that
back into the function we would get a zero in the denominator, we’d
have the square root of 9, which is 3 minus 3, so we’d have zero in the denominator so recall that this means x is such that x cannot be 5 so that’s part of it the next part is we look at our individual radical so we’ve looked at the denominator as a whole,
now we look at the radical itself and that, our rules for radicals are again
that we set the radicand, the x + 4 greater than or equal to zero and when I solve this x is greater than or equal to negative 4 that is the other restriction now you’ll notice
that’s the same as the domain of g of x so that works out very nicely and last what we need to do is combine all
of these, so we want to combine all these individual parts into one overall domain so let’s list them out here, we have x is greater than or equal to negative 4 and
we have x cannot be 5 so one way I like to think of it is if I were
to look at a number line and if I were to put these two on there, we’re
looking at both of them so here is negative 4 and here’s 5 we’ll say so x can’t be 5 so I’ll put an open circle
over that and I’m saying it can be negative 4 so I put a closed circle and now, since everything is bigger than negative
4 I draw a line to the right skipping 5 of course and going on to infinity so if I were to look at this whole domain basically I’m trying to say what is the domain
shown by the number line? I would say we’d have the interval notation is we’ll say from negative 4 to 5 and then I’ll combine that union with 5 to infinity so this would be my overall domain so with my overall domain I now have the composite function, which again
was x + 5 over the square root of x + 4 minus
3 and I have the domain of said function so overall the three steps I want to take when I find
the domain of a composite function is I first of all find the domain of the inside
function then I find the domain of f of g of x while
finding f of g of x and then I combine my domains into one overall expression