Composing Functions and Determining Domain

The "How-To" of Function Composition

Function composition is the process of plugging one function into another. If we have two functions, $f(x)$ and $g(x)$, the composition "f of g of x" is written as $f(g(x))$ or using the composition operator, $(f\circ g)(x)$.

The key is to always work from the inside out. To find $f(g(x))$, you first evaluate the inner function, $g(x)$, and then use that entire result as the input for the outer function, $f(x)$.

Let's start by reading an introduction to this concept. This resource defines the notation and provides some clear real-world examples to build your intuition. Pay close attention to the examples that show that the order of composition matters—in general, $f(g(x))$ is not the same as $g(f(x))$.

Now that you understand the concept and notation, let's see how to evaluate a composite function for a specific numerical value.

Suppose we have $f(t)=t^2-t$ and $h(x)=3x+2$, and we want to evaluate $f(h(1))$.

1. Evaluate the inner function: First, find the value of $h(1)$. $$h(1)=3(1)+2=5$$
2. Use the output as the new input: Now, substitute this result into the outer function, $f(t)$. $$f(h(1))=f(5)=5^2-5=25-5=20$$

So, $f(h(1))=20$. This "inside-out" process is the fundamental mechanic of function composition.

Finding the Domain of a Composite Function

This is the most critical part of our lesson. Finding the domain of a composite function is more nuanced than finding the domain of a simple function because there are two potential sources of restrictions.

A common mistake is to find the final simplified form of the composite function and then find its domain. This is incomplete and often leads to the wrong answer. Why? Because before a number can be processed by the outer function, it must first successfully pass through the inner function.

This short video provides an excellent, intuitive explanation of why the inner function's domain is so important.

As the video explained, any input $x$ must be valid for the inner function first. This leads us to a two-step process for finding the domain of $f(g(x))$:

1. Restriction 1: The input $x$ must be in the domain of the inner function, $g(x)$.
2. Restriction 2: The output of the inner function, $g(x)$, must be in the domain of the outer function, $f(x)$.

The final domain of the composite function is the set of all $x$ values that satisfy both of these conditions.

Let's put this into practice with a few examples.

Example 1: The Square Root "Trap"

Consider the functions $f(x)=x^2+5$ and $g(x)=\sqrt{x-1}$. Let's find the domain of $(f\circ g)(x)=f(g(x))$.

1. Inner function's domain: The inner function is $g(x)=\sqrt{x-1}$. For the square root to be defined, the expression inside must be non-negative.
   $x-1\ge 0⟹x\ge 1$.
   So, our first restriction is that $x$ must be greater than or equal to 1.

2. Composite function's domain: Now, let's find the composite function and its restrictions.

   $$f(g(x))=f(\sqrt{x-1})=(\sqrt{x-1}{)}^2+5=(x-1)+5=x+4$$

   Looking at the simplified function $x+4$, it appears to have a domain of all real numbers. This is the "trap." We cannot ignore our first restriction.

3. Combine the restrictions: The input $x$ must satisfy $x\ge 1$ (from the inner function) AND it must be valid for $x+4$ (which poses no new restrictions). Therefore, the domain of $f(g(x))$ is $[1,\infty )$.

The following video segment works through a very similar example. Watching it will help solidify this crucial point.

Example 2: Rational Functions

Let's find the domain of $(f\circ g)(x)$ where $f(x)=\frac{5}{x-1}$ and $g(x)=\frac{4}{3x-2}$.

1. Inner function's domain: The inner function is $g(x)=\frac{4}{3x-2}$. The denominator cannot be zero.
   $3x-2\ne 0⟹3x\ne 2⟹x\ne \frac{2}{3}$.

2. Composite function's domain: The output of $g(x)$ becomes the input for $f(x)$. The domain of $f(x)$ requires its input to not be 1. So, we must find which values of $x$ cause $g(x)$ to be equal to 1, and exclude them.

   $$g(x)=1⟹\frac{4}{3x-2}=1$$ $$4=3x-2⟹6=3x⟹x=2$$

   So, we must also exclude $x=2$.

3. Combine the restrictions: We must exclude $x=\frac{2}{3}$ from the inner function's domain, and we must exclude $x=2$ because it produces an invalid input for the outer function.

The final domain is all real numbers except $\frac{2}{3}$ and $2$. In interval notation, this is $(-\infty ,\frac{2}{3})\cup (\frac{2}{3},2)\cup (2,\infty )$.

For a step-by-step reading of these types of problems, the LibreTexts article we saw earlier is very helpful.

Example 3: Practice with Two Rational Functions

Let's try one more example that combines these ideas. Find the domain of $f(g(x))$ for $f(x)=\frac{3}{x-2}$ and $g(x)=\frac{4}{x+1}$.

1. Inner function $g(x)$: The denominator cannot be zero, so $x+1\ne 0⟹x\ne -1$.
2. Outer function $f(x)$: Its input cannot be 2. So we find the value of $x$ that makes $g(x)=2$. $$\frac{4}{x+1}=2⟹4=2(x+1)⟹4=2x+2⟹2=2x⟹x=1$$ So, we must also exclude $x=1$.
3. Combined Domain: The domain includes all real numbers except -1 and 1. In interval notation: $(-\infty ,-1)\cup (-1,1)\cup (1,\infty )$.

The following video works through this exact problem, which can serve as a good check for your understanding.

Conclusion

Today we've added function composition to our toolkit, a method that allows us to build complex functions from simpler ones. More importantly, we've established a rigorous method for determining the domain of these new functions.

Here are the key takeaways:

• Function composition, denoted $(f\circ g)(x)=f(g(x))$, involves substituting an inner function into an outer function.
• To find the domain of a composite function $f(g(x))$, you must identify and combine two sets of restrictions:
  1. Any values of $x$ that are not in the domain of the inner function $g(x)$.
  2. Any values of $x$ that cause the output of the inner function, $g(x)$, to be outside the domain of the outer function $f(x)$.
• Simply finding the domain of the final, simplified function is not enough and can lead to incorrect answers.

In our next lesson, we will investigate a special property of functions called being "one-to-one." This property is the key that will allow us to "undo" a function by finding its inverse, a concept that is deeply connected to function composition.