Composing Functions and Finding Their Domains

1. The Concept of Function Composition

Imagine a two-stage process. An input goes into the first machine, which modifies it and produces an output. That output is then immediately sent into a second machine, which performs its own operation and yields a final result. This is the essence of function composition.

We have two functions, say $f$ and $g$. We start with an input $x$, apply the "inner" function $g$ to it to get $g(x)$, and then apply the "outer" function $f$ to that result to get $f(g(x))$.

The notation for this is $(f\circ g)(x)$, which is defined as:
$(f\circ g)(x)=f(g(x))$

It's crucial to read this from the inside out. The function on the right, $g$, is applied first.

Let's look at a concrete mapping diagram.

As the diagram illustrates for $x=2$:

1. Input $x=2$ goes into $g(x)$, resulting in $g(2)=2^2=4$.
2. The output $4$ becomes the input for $f(x)$, resulting in $f(4)=4+3=7$.
3. Therefore, $(f\circ g)(2)=7$.

The following resource formally defines this concept and highlights an important property: function composition is generally not commutative, meaning the order matters and $(f\circ g)(x)$ is usually different from $(g\circ f)(x)$.

This next image provides a more general, abstract view of the process.

Notice the notation gof in the diagram. This is equivalent to $(g\circ f)(x)$, where $f$ is the inner function applied first. The order in this notation can feel backwards, which is why thinking of it as $g(f(x))$ is often clearer.

2. Constructing Composite Functions Algebraically

To find the formula for a composite function, we use substitution. To find $f(g(x))$, we take the entire expression for $g(x)$ and substitute it into every instance of the variable in the function $f(x)$.

Let's take $f(x)=x^2+4$ and $g(x)=x-3$.

• To find $(f\circ g)(x)=f(g(x))$:
  We start with $f(x)=x^2+4$. We replace the $x$ with the entire expression for $g(x)$.

  $$f(g(x))=(g(x){)}^2+4=(x-3{)}^2+4$$

  Expanding this gives:

  $$(x^2-6x+9)+4=x^2-6x+13$$

• To find $(g\circ f)(x)=g(f(x))$:
  We start with $g(x)=x-3$. We replace the $x$ with the entire expression for $f(x)$.

  $$g(f(x))=f(x)-3=(x^2+4)-3=x^2+1$$

As you can see, $x^2-6x+13\ne x^2+1$, confirming that the order of composition matters.

3. The Domain of a Composite Function

Just as with arithmetic combinations, determining the domain is a critical step. The rule for the domain of a composite function $(f\circ g)(x)$ is a two-part check:

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

A common mistake is to find the simplified formula for the composite function and determine its domain from that formula alone. This is incorrect because it can miss restrictions inherited from the inner function.

Professor Leonard provides an excellent explanation of this, memorably describing the inner function's domain restrictions as "baggage" or a "skeleton in the closet" that you can never get rid of, even if it seems to disappear after simplification.

Let's work through an example to solidify this systematic process.

Worked Example

Find the composite function $(f\circ g)(x)$ and its domain, given $f(x)=\frac{3}{x-1}$ and $g(x)=\sqrt{x-2}$.

Step 1: Find the domain of the inner function, $g(x)$.
The function $g(x)=\sqrt{x-2}$ involves a square root. The expression inside the root must be non-negative.
$x-2\ge 0⟹x\ge 2$.
So, the domain of $g$ is $[2,\infty )$.
Any valid $x$ for the composite function must satisfy this condition.

Step 2: Construct the composite function $(f\circ g)(x)$.
Substitute $g(x)$ into $f(x)$:

Step 3: Find the domain restrictions from the new composite form.
The simplified form has a denominator that cannot be zero.
$\sqrt{x-2}-1\ne 0$
$\sqrt{x-2}\ne 1$
Squaring both sides gives:
$x-2\ne 1⟹x\ne 3$.
This is the second condition.

Step 4: Combine all restrictions to find the final domain.
We must satisfy both conditions simultaneously:

1. From Step 1: $x\ge 2$
2. From Step 3: $x\ne 3$

Combining these, the domain is all numbers greater than or equal to 2, except for 3.
In interval notation, the domain of $(f\circ g)(x)$ is $[2,3)\cup (3,\infty )$.

Note how we first established the "inherited" restriction from $g(x)$ and then found the new restriction created by the composition itself.

For more practice with a variety of functions, including composing a function with itself (e.g., $f(f(x))$), you can review the video "Composite functions and their domains" by Krista King. It provides several clear, step-by-step examples.

Conclusion

In this lesson, we have added function composition to our toolkit for creating new functions. This method is conceptually different from the arithmetic combinations we studied previously but follows a similar, rigorous process for determining the domain.

Key Takeaways:

• Function composition, $(f\circ g)(x)=f(g(x))$, is the process of applying one function to the result of another.
• The procedure works from the inside out: evaluate the inner function $g(x)$ first, then use its output as the input for the outer function $f$.
• The order of composition matters; in general, $(f\circ g)(x)\ne (g\circ f)(x)$.
• The domain of $(f\circ g)(x)$ is determined by two constraints:
  1. The input x must be in the domain of the inner function g.
  2. The output g(x) must be in the domain of the outer function f.
• You must always account for domain restrictions from the inner function, as they are inherited by the composite function even if they aren't obvious in its final simplified form.

We have now completed our survey of the main ways to construct new functions from existing ones. In the next module, we will begin our journey into the core of calculus by investigating the concept of a limit. We will explore what it means for a function's value to get "arbitrarily close" to some number as its input approaches a certain point. This powerful idea forms the very foundation upon which derivatives and integrals are built.