Composition of Functions: Domains & De-Compositions
Purplemath
The domain of a function is the set of values which are valid inputs. For a polynomial function, the domain will always be "all x"; for a rational function, the domain will be all x-values which do not cause division by zero; for an even-index radical function (that is, for a square root, or a fourth root, or a sixth root, etc), the domain will be all x-values which do not put a negative value inside the radical.
When composing functions, sometimes the output of one function can create issues as inputs to the other function.
As a result, you have to be careful with the domains and ranges of composite functions. And domains and ranges of compositions are an area which can generate trick questions. (Fair warning: You should expect at least one of these trick questions on the next exam.)
What is an example of finding the domain of a function composition?
- Given and g(x) = x − 2, find the domain of (f ∘ g)(x).
Since f (x) involves a square root, the inputs have to be non-negative. This means that the domain (that is, the set of x-values) for f (x) is "all x ≥ 0".
Then, in (g ∘ f )(x), where I'm plugging x first into , the domain is at least restricted to "all x ≥ 0".
Let's see what the composition looks like:
The domain for the square root is all inputs that make x − 2 non-negative; that is, all x such that x − 2 ≥ 0. Solving this for x, I get the domain of (f ∘ g)(x) as being all x ≥ 2. And this is my answer.
domain of (f ∘ g)(x): x ≥ 2
- Given and g(x) = x − 2, find the domain of (g ∘ f )(x).
This exercise has the composition working in the other order.
The domain for this is going to be all inputs that make the square root defined. Since there is only "x" inside the square root, then my answer is:
domain of (g ∘ f )(x) is x ≥ 0
Note: If your initial functions are just plain old polynomials, then their domains are "all x", and so will be the domain of the functions' composition. It's pretty much only if you're dealing with rational functions with denominators (where you can't divide by zero) or even-index radical functions like square roots (where you can't have a negative inside the radical) that the domain ever becomes an issue.
Thus far, we've taken two functions and composed them together. Sometimes, they'll want you to go the other way.
What does it mean to decompose a function?
To decompose a function is to try to find two (or more) functions which, when composed, give you the function they'd given you to started with. There can be more than one valid answer to a given decomposition exercise. It is usually pretty obvious what functions they had in mind when they wrote the exercises. But any answer you come up with — any answer that works, of course — should be an acceptable solution.
- Given h(x) = (x + 1)2 + 2(x + 1) − 3, determine two functions f (x) and g(x) which, when composed, generate h(x).
This is asking me to notice patterns and to figure out what might have been put inside something else. Because they did not multiply things out, but instead left the parentheticals, it's pretty easy to see what two functions they have in mind.
In form, this function looks similar to the quadratic x2 + 2x − 3. But, instead of squaring x, they're squaring x + 1; instead of multiplying x by 2, they're multiplying x + 1 by 2.
In other words, this appears to be a quadratic into which they've plugged the linear expresssion x + 1. So I'll make g(x) = x + 1, and then plug this function into f(x) = x2 + 2x − 3. I'll check to confirm that I get the correct result:
(f ∘ g)(x)
= f (g(x))
= f (x + 1)
= ( )2 + 2( ) − 3
= (x + 1)2 + 2(x + 1) − 3
This confirms that my choice of functions is valid; when composed, they do return the required function. So this is my answer:
f (x) = x2 + 2x − 3
g(x) = x + 1
Can there be more than one set of functions that compose to the same thing?
While it is usually pretty clear which two functions were composed to get a "Decompose this function" result, there is nothing that says that there must necessarily be only one set of functions that were composed. Any pair of functions which give the correct result will be valid mathematical answers, and should be acceptable to your grader as a correct solution — as long as you're not insultingly trivial about it.
In the previous exercise, I mentioned the fact that the author of that exercise had not multiplied things out, which made my life pretty easy. But I can find other solutions, especially if I do that multiplying and simplifying. What they had given me simplifies as:
h(x) = x2 + 4x
Is there any way to decompose this? Sure! Here's an optional set of functions:
f (x) = x + 1
g(x) = x2 + 4x − 1
Now I'll compose them, to prove that they give h(x):
(f ∘ g)(x) = f (x2 + 4x − 1)
= (x2 + 4x − 1) + 1
= (x2 + 4x) − 1 + 1
= x2 + 4x
This is the same thing they'd given me in the first place; it looks different because I just composed to get the simplified form, where they'd given me the very-much-not simplified form.
As long as you show all your steps and reasoning (in this case, by multiplying out what they'd given me, simplifying, and working with that result, instead), you should get full credit.
- Given , determine two functions f (x) and g(x) which, when composed, generate h(x).
The square root is on (or "around") the polynomial expression 4x + 1. This makes for an obvious solution: put the 4x + 1 inside the square root.
To do so, I'll use the following functions:
f (x) = √(x)
g(x) = 4x + 1
I can compose to check if these two functions will work:
This confirms that my two functions work. So my answer is:
g(x) = 4x + 1
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