Composite Functions - Domain

Composite Functions - Domain - How it Works - Video

Definitions

Definition:

Here we have the definition of composite functions. With composite functions, it is like double substitution. The key here is to work inside out so start in the middle and work outward.

One that often takes a back pedal is domain of a composite function. The domain of f and g is not always in your answer of (f ∘ g) or (g ∘ f). You should not take the domain after simplifying. That is where people make mistakes. Sometimes, you get lucky and it is the same, but not often. You should find the domain right after substituting the inside function into the outer function.

Composite Function Model - Domain

Composite Models - Domain:

Here we have two models to show you why the domain is not always the same when we take composite of different functions.

On the left we have (f ∘ g) = f(g[x]). The function g has a smaller domain then function f. When we first start, our domain is g. So we already have a smaller domain than f. So, when we substitute g into f, we cannot gain the numbers that work in f and not g. So our final domain will be smaller than the original function f.

On the right we have (g ∘ f) = g(f[x]). The function f has a larger domain then function g. When we first start, our domain is f. So we have a larger domain than g. But, when we substitute f into g, we lose numbers that work in f and not g. So our final domain could be smaller than the original function g.

Let's take a quick look at f(x) = x and g(x) = sqrt(x). The domain of f(x) is from (-∞, +∞) and the domain of g(x) = [0, +∞).

(f ∘ g) = f(g[x]) = sqrt(x) => The domain of g, we substitute, is all nonnegative real numbers, and we put into a function where the domain was all real numbers. The nonnegative numbers still do not work even though it worked for f. So our domain is still [0, +∞).

(g ∘ f) = g(f[x]) = sqrt(x) => The domain of f, is all real numbers, and we put into a function where the domain is all nonnegative real numbers. Since negative real numbers do not work, our domain decreases to what works for g. So our domain is now [0, +∞).

Example 1 - (f ∘ g)(x)

Example 1 (f ∘ g)(x):

Here we have two functions f(x) = sqrt(2-x) and g(x) = x² - 1.

We want to find the domain of (f ∘ g)(x) and (g ∘ f)(x). In order to do that we need to find the composition of each function. So let's start with (f ∘ g)(x).

= sqrt( 2 - [x² - 1] ) is where we find the domain. Why do we find the domain here? Well, sometimes when we simplify, operations get canceled and that affects the domain.

Now we have the two input values +sqrt(3) and -sqrt(3), which make the minimum output values for square root, which is 0.

Now we can build our domain with that information. Our domain is {x ϵ ℝ: -sqrt(3) ≤ x ≤ +sqrt(3)}. We can use the graph to help us build the domain. If the graph is not there, then we can use points before -sqrt(3) and in the middle and after +sqrt(3) to determine which outcome can happen.

Example 1 - (g ∘ f)(x)

Example 1 (g ∘ f)(x):

Here we have two functions f(x) = sqrt(2-x) and g(x) = x² - 1.

We want to find the domain of (f ∘ g)(x) and (g ∘ f)(x). In order to do that we need to find the composition of each function. We already found the domain of (f ∘ g)(x). So let's continue with (g ∘ f)(x).

= [ sqrt( 2 - x ) ]² - 1 is where we find the domain. Why do we find the domain here? Well, sometimes when we simplify, operations get canceled and that affects the domain like from step 2 to step 4.

Now we have the one input value, 2, which make the minimum output values for square root, which is 0.

Now we can build our domain with that information. Our domain is {x ϵ ℝ: x ≤ 2}. We can use the graph to help us build the domain. If the graph is not there, then we can use points before 2 and after 2 to determine which outcome can happen.

Live Worksheet

Teacher - Edpuzzle Link