# How to Find the Domain of a Composite Function

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00:00
[Music] hello everybody welcome to another video hope ur ready to flex those brain muscles today we're going to talk more about composite functions but this time we're specifically talking about how to find the domain of a composite function so this is a very common issue that students have with this cuz it's a little bit tricky but basically we're often given two functions F and G we're asked to find the composite function but also find the domain so let's go and get into how to do this first how to find the domain of a composite function our first step identified domain restrictions of the inner function
00:37
remember composite functions we have an outer function and an inner function so first we look at the domain of our inner function and let's go ahead and do this example so the domain of our inner function is what we have no restrictions here this domain is all real numbers so I'm going to go and put that here okay all real numbers second find the composite function so basically I'm evaluating this I'm plugging G of X into F and I can simplify it and yeah through that stuff so I'll go ahead and plug G of X into F so I have F of well what is
01:09
G of X 2x minus 3 and we did this a lot in the last video that's why I'm gonna move a little quickly when I'm doing this in this video because we did this in the last video this video were just focusing on domain so basically I square this because I'm plugging this into my F function so 2x minus 3 is getting squared all right there's a shortcut for this if you know how to do that in fact I'm going to go ahead and do it a squared is my first term 4x squared plus 2 a B which is negative 12x
01:41
let's see plus B squared that is 9 okay so now where are we at we found the composite function we identified domain restrictions of this composite function are there any domain restrictions in this case there are not so my domain for this function is all real numbers or should I put that and that means what we include both restrictions well we don't have any restrictions either way so our final domain is all real numbers as well and the reason I started with this example is because it's the simplest of all the
02:13
cases right we have two functions that both have a domain of all real numbers so any composition function we make using these two functions will also have a domain of all real numbers okay so in general if you have polynomials quadratics linear functions anything like that and you're composing domain will be all real numbers it's other stuff where we have problems that we'll get in so there's those examples right now all right let's go ahead and try another example we're gonna find f of G of X and we're gonna find its domain as well so think about it the reason we identify the restrictions of the inner function first is because this inner
02:44
function happens first right I plug in a number it has to go through this inner function then it goes through the outer function right so if there's some number where this is not defined or where this has issues like you know negatives under square root or zeros in the denominator like this case right if I plug in zero to this function it's undefined so zero is not in the domain of this function so that means I cannot plug zero into f of G of X as well and that's why we do this because every number has to go through
03:15
the inner function first so if there's some values being excluded from the domain and the inner function we need to exclude them from the composite function as well hopefully that makes sense so what are the restrictions we've just said X cannot equal zero now let's go ahead and find this composite function so I'm plugging G of X into F let's see what does that give me so I replace everywhere I see X I replaced with 2 over X right so I have 2 over X over 2 over X plus 3 all right if I'm plugging
03:48
G of X into F so what is this equal I gotta do a little algebra let's see what algebra can I you I'll take away this equals so I can multiply top and bottom by something and there's multiple things you can do you can combine these that's one thing you can do I'm gonna multiply by x over 1 over x over 1 and the reason is because that's going to get rid of all my fractions basically look how this works x over 1 over x over 1 that gets rid of this X I'm left with two at the top x
04:20
over 1 times and again this is a parenthesis so this is going to everything x over 1 times 2 over X that just gives me to my XS cancel again plus X over 1 time times 3 that gives me 3x and this is really my final function I have 2 over 3x plus 2 so again I have what I have potentially have 0 in the denominator so I set the denominator equal to 0 right and I solved for X and those are the values that have excluding from the domain of this composite
04:51
function along with X cannot equal 0 ok so let's go ahead and set this equal to 0 2 plus 3x equals 0 subtract 2 from both sides I have 3x equals negative 2 now I can divide both sides by 3 and I have x equals negative 2 which means what in our domain of our final composite function X cannot equal 0 and X cannot equal negative 2 over 3 so if I were to write this in interval notation
05:24
it may get a little crazy here in interval notation I have domain goes from negative infinity all the way up to negative 2/3 then it starts back up at negative 2/3 is not including negative 2/3 and it goes up to what to 0 then it starts back up as 0 and it goes all the way up to infinity so this is the domain in interval notation and it may be acceptable as well just write X such that X does not equal 0 and x2
05:55
equal negative two over three let's do one more example I really encourage you to pause the video try it on your own we're gonna find f of G of X and its domain so it's going to get started first we identify domain restrictions of the inner function our inner function G of X we have a square root with X under it which means what everything under the square root we set greater than or equal to zero we don't want to have any negatives under our square root so I'm gonna do this off to the side X minus 1 greater than equal to zero so I add 1 to
06:26
both sides of the inequality and I have X is greater than equal to positive 1 so this is my restriction from the domain ok so whatever my final composite function is and whatever its domain is it has to include this restriction as well as any other additional restrictions from the domain of the composite function itself so now what do we do next find the composite function so I plug G of X into F and what does that give me well let's see I plug square root of x minus 1 into this X so
06:57
I have square root of x minus 1 squared plus 5 square root and squared those kind of cancel each other out right I'm left with X minus 1 plus 5 X minus 1 plus 5 hopefully I'll see how those cancel each other out and now I can combine these like terms and I'm left with what X plus 4 and this is the most common mistake made on these kind of problems is people ignore that first step they look at this they say oh linear function it's clear that the
07:30
domain is all over the numbers and they put all real numbers as their answer but that's not true remember we plug in an input this is a composite function so that input has to go through the inner function first this inner function is restricted to values such that X is greater than equal to 1 so that means my composite function as well has to be restricted so the domain is X such that X is greater than equal to 1 that's one way to write it my currently brackets are a little ugly sorry about that if I want to do interval notation I do a bracket right
08:01
bracket because we're including negative 1 all the way up to infinity alright last example just to get some more practice I encourage you to pause it try it on your own it's very similar to the first example so let's go ahead and do it first thing we do what we identify the domain restrictions of the inner function right and again we're finding F of G of X as well as its domain this is what we're finding as well as its domain so our inner function is G so we're gonna look at what domain restrictions we have here well we have X's in the denominator so I
08:32
can set my denominator equal to zero and that would be the value that I've excluded from the domain so when x equals 2 I get 0 in the denominator so for this domain of G of X I have X such that X does not equal positive 2 ok now I can move on to finding the composite function I'm plugging this function in to F so I have 1 over X over X minus 2 plus 3 this may get a little messy and again there are many different ways you can do this as far as
09:04
simplifying it you could combine these what I'm personally gonna do is I'm gonna multiply top and bottom by let's see X minus 2 over 1 over X minus 2 over 1 this is just another form of 1 right so I'm still allowed to do this this is legal and what this will do is get rid of that denominator so let's go ahead and do this X minus 2 times 1 that's just X minus 2 up top X minus 2 cancels with this X minus 2 I'm left with X X
09:34
minus 2 times 3 so I have plus 3 times X minus 2 make sure to be real careful with the parentheses and stuff this 3 has to go to everything so I can simplify this again maybe I'll rewrite it down here X minus 2 I'm basically putting this 3 in that's my next step is multiply this 3 out so I have 3x minus 6 so on the bottom I have X plus 3x minus 6 and on the top I still have X minus 2 and I just have one more step and that's
10:07
to combine the like terms and this is my function f of G of X I have 4x minus 6 in the bottom so what is the domain of this function let's see and you can even factor out a 2 from the bottom will that help it all that'll leave you with 2x minus 3 no that will help you it also I wouldn't even really bother the domain of this function I basically set the denominator equal to zero right same as we've been doing 4x minus 6 equals 0 I saw 4x I can add 6 to both sides and I
10:41
get 4x hopefully I can see this equals 6 now I can divide both sides by 4 and I'll rewrite this up here I get x equals 6 over 4 which is 3 over 2 so X cannot equal 2 because that's what we restricted from an inner function and from our final composite function we have a restriction of x cannot equal 3 over 2 so our final domain if I even have room to write it where can I write it I'm going to put it into this box
11:11
domain X such that X does not equal 2 and X does not equal 3 over 2 hopefully I can read that sorry it's a little got a little messy but alright that's our last example I'm hope this video helped everyday hit the like button hit subscribe leave a comment with any questions see in the next vid video keep making those brain games