Composition of Functions - AP PreCalculus Study Guide

The Core Idea: Composition of Functions

Function composition is the process of combining two functions by applying one function to the result of another. Conceptually, it is a sequential process where the output of an "inner" function becomes the input for an "outer" function. This creates a new, composite function. The notation for the composition of a function $g$ with a function $f$ is $h(x)=g(f(x))$, where $f$ is the inner function and $g$ is the outer function.

A critical aspect of function composition is understanding the domain of the resulting function. The domain of the composite function $h(x)=g(f(x))$ is restricted by two conditions: first, any input `x$ must be valid for the inner function $f$; second, the output of the inner function, $f(x)$, must be a valid input for the outer function $g$. This concept of combining functions applies across all representations, including analytical formulas, tables of values, and graphs. The reverse process, called decomposition, involves identifying the inner and outer functions that make up a given composite function.

Key Formulas & Rules

The primary rule in this topic governs the definition and domain of a composite function.

Composition Notation

The composition of an outer function $g$ with an inner function $f$ is denoted in two main ways:

1. $h(x)=g(f(x))$

2. $h(x)=(g\circ f)(x)$

Both notations mean "g of f of x" and describe the same process: first evaluate $f(x)$, then use that result as the input for $g$.

The Rule for the Domain of a Composite Function

The domain of the composite function $h(x)=g(f(x))$ is the set of all real numbers $x$ that satisfy two conditions simultaneously:

1. $x$ must be in the domain of the inner function, $f(x)$.

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

This can be expressed more formally as:

The domain of $g\circ f$ is $x\in \text{Domain}(f)\mid f(x)\in \text{Domain}(g)$.

Understanding The Domain of Composite Functions

The most nuanced part of function composition is correctly determining the domain of the resulting function. It is not enough to simply find the formula for the composite function and then determine its domain. The domain is constrained before any simplification occurs.

Think of the process as a two-stage assembly line. The input `x$ first goes into Machine $f$.

• Condition 1: Machine $f$ must be able to accept the input `x$. If $x$ is not in the domain of $f$, the process fails immediately.

• Condition 2: Machine $f$ produces an output, $f(x)$. This output is then sent to Machine $g$. Machine $g$ must be able to accept $f(x)$ as its input. If the output $f(x)$ is not in the domain of $g$, the process fails at this second stage.

Therefore, for the entire composite machine $g(f(x))$ to work, both conditions must be met. The set of all initial inputs $x$ that successfully pass through both stages constitutes the domain of the composite function. Ignoring the first condition is a common error. For example, if $f(x)=\sqrt{x}$ and $g(x)=x^2$, the composite function is $g(f(x))=(\sqrt{x}{)}^2=x$. The domain of the final expression $y=x$ appears to be all real numbers. However, the inner function $f(x)=\sqrt{x}$ is only defined for $x\ge 0$. This initial restriction carries through, so the domain of $g(f(x))$ is $[0,\infty )$.

Core Concepts & Rules

• Definition of Composition: The composition of function $g$ with function $f$ creates a new function $h$ defined by $h(x)=g(f(x))$.

• Order of Operations: To evaluate $g(f(x))$, you must evaluate the inner function $f(x)$ first. The resulting output value is then used as the input value for the outer function $g$.

• Domain Restriction Rule: The domain of $g(f(x))$ consists of all $x$-values in the domain of $f$ for which $f(x)$ is in the domain of $g$.

• Multiple Representations: Function composition can be performed regardless of how the functions are represented (e.g., formulas, tables, or graphs).

• Decomposition: Any function $h$ can be expressed as a composition of two functions, $f$ and $g$, such that $h(x)=g(f(x))$. This process is called decomposition.

• Non-Uniqueness of Decomposition: There can be multiple, valid ways to decompose a single function. For example, $h(x)=(x+3{)}^2$ could be decomposed as $f(x)=x+3$ and $g(x)=x^2$, or as $f(x)=x$ and $g(x)=(x+3{)}^2$.

Step-by-Step Example 1: Composition and Domain from Formulas

Let $f(x)=\frac{1}{x-4}$ and $g(x)=\sqrt{x-1}$. Find the composite function $h(x)=g(f(x))$ and determine its domain.

Part A: Find the formula for $h(x)=g(f(x))$

Step 1: Write out the structure of the composition.

$h(x)=g(f(x))$

Step 2: Substitute the entire expression for the inner function $f(x)$ into the outer function $g$.

$h(x)=g\left(\frac{1}{x-4}\right)$

Step 3: Apply the rule of the outer function $g(u)=\sqrt{u-1}$ to the input u = \frac{1}{x-4}.

$h(x)=\sqrt{\left(\frac{1}{x-4}\right)-1}$

Step 4: (Optional, but good practice) Simplify the expression by finding a common denominator inside the square root.

$h(x)=\sqrt{\frac{1}{x-4}-\frac{x-4}{x-4}}$

$h(x)=\sqrt{\frac{1-(x-4)}{x-4}}$

$h(x)=\sqrt{\frac{5-x}{x-4}}$

Part B: Find the domain of $h(x)=g(f(x))$

We follow the two-condition rule for the domain.

Step 1: Determine the domain of the inner function, $f(x)=\frac{1}{x-4}$.

The denominator cannot be zero, so $x-4\ne 0$.

The domain of $f$ is $x\ne 4$. In interval notation, this is $(-\infty ,4)\cup (4,\infty )$.

Step 2: Determine the domain of the outer function, $g(x)=\sqrt{x-1}$.

The expression inside the square root must be non-negative, so $x-1\ge 0$.

The domain of $g$ is $x\ge 1$. In interval notation, this is $[1,\infty )$.

Step 3: Apply the second condition: the output of the inner function, $f(x)$, must be in the domain of the outer function $g$.

We need $f(x)\ge 1$.

$\frac{1}{x-4}\ge 1$

To solve this inequality, we subtract 1 and find a common denominator.

$\frac{1}{x-4}-1\ge 0$

$\frac{1-(x-4)}{x-4}\ge 0$

$\frac{5-x}{x-4}\ge 0$

The critical points are $x=5$ (where the numerator is zero) and $x=4$ (where the denominator is zero). We test intervals:

• If $x<4$, let $x=0$: $5/(-4)<0$. No.

• If $4<x\le 5$, let $x=4.5$: $0.5/0.5>0$. Yes.

So, this condition requires $4<x\le 5$.

Step 4: Find the intersection of the conditions from Step 1 and Step 3.

• Condition from Step 1: $x\ne 4$

• Condition from Step 3: $4<x\le 5$

The intersection of these two sets is $(4,5]$.

The domain of $h(x)=g(f(x))$ is $(4,5]$. Notice this is the same domain we would find by analyzing the final simplified form $h(x)=\sqrt{\frac{5-x}{x-4}}$.

Step-by-Step Example 2: Composition from Mixed Representations

A function $f(x)$ is defined by the table below. A function $g(x)$ is defined by the graph.

Table for $f(x)$:

Graph for $g(x)$:

(A graph showing the function $g(x)=|x|-2$)