# Homework Help: The domain of a composite function

1. Jun 21, 2004

### flashgordon

my previous calculus teacher stressed finding the domain of a composite function; he stressed more of finding which areas were not part of the function by means of three circles and saying the function didn't pass to the other circle unless it met the range of the other function. The textbook mostly just says the domain of the composite function is the union of the domains of the two functions.

Well, I'm just wandering why the domain of the composite functions is the union of the two functions domain and I think he said multiplied by the range of one of the functions(either inside or outside.

Basically, the books are not to in depth with this, and i think I need a description of the whole process with maybe a proof of why the range comes into play.(I'm thinking maybe it has to do with the way the inverse functions switch range and domain?)

2. Jun 21, 2004

### NateTG

Let's say we have two functions:
$$f: X \rightarrow Y$$
and
$$g: Y \rightarrow Z$$
then the composite function is going to be:
$$f \circ g: X \rightarrow Z$$

Now, the Domain of $$f \circ g$$ is going to be the inverse image (w.r.t $$f$$) of the domain of $$g$$.

This seems pretty obvious to me, so I may be missing something from your question.

3. Jun 21, 2004

### flashgordon

inverse image isn't part of the vocabulary as taught to me, so your using something I havn't even seen; maybe when I get higher, there will be a more proper demonstration; i'm thinking by inverse image, your meaning range. Also, what does wrt mean?

Anyways, I'm seeing a transitive property in the three lines you indicate which is perfectly intuitive, but then 'inverse image'? How is this 'inverse image'?(or range?)

I think I'm kindof seeing it; if g is from y to z, but then the composite becomes x(an f function set I guess) to z, then well, I guess that's kindof a f function version of the g function?

I'd still like to see a full proof,

4. Jun 21, 2004

### HallsofIvy

No, "image" is range, "inverse image" is domain. However, I don't understand your statement "the domain of the composite function is the union of the domains of the two functions." The composite of two functions, f and g, is f(g(x)) or g(f(x)) (the order is important).

For example, I can define a function by f(0)= 100, f(1)= 101, f(2)= 102 (so its domain or "inverse image" is the set {0, 1, 2} and its range or "image" is the set {100, 101, 102} and then define g(100)= -5, g(101)= -10 whose domain is {100,101}. The composite, g(f(x)) gives g(f(0))= g(100)= -5 and g(f(1))= g(1)= -10. g(f(2)) is not defined because g(102) is not defined: the domain of g(f(x)) is {0,1}.
The composition the other way, f(g(x)) is not defined.

Are you sure "composition" is the right word for what you are talking about?

5. Jun 23, 2004

### flashgordon

my current teacher mentioned in passing something about the domain of a composite function, but she said it wasn't something that gets talked about till a 500 level calculus class, so, hopefully, I won't have to worry about it right now; also, maybe that is why these calculus texts I can so far find do not mention it in any real depth.

6. Jun 23, 2004

### flashgordon

she also mentioned how the ranges and domains get reversed "so much."

Well, maybe I'll try to find something else to bother you all with that you'll like,

7. Feb 17, 2011

### moteach

Just to clarify. To find the domain of a composite function say f(g(x)) first find the range of g(x) this then becomes the input values or domain of the composite function. This becomes important with functions where the range is not all real numbers. And remember the domain of f(g(x)) does not always equal the domain of g(f(x)). Graphing can help determine domain and range, but it is not always accurate with composite functions.

One other note if you are performing an operation on two functions such as addition, subtraction, multiplication or division then the domain of say f(x) + g(x) is the intersection of the domain of f(x) and g(x), plus any other limitations that may occur due to the operation. (Remember you can not divide by zero, so any zero values that might pop up in a denominator are not part of the domain.)