Composite Functions: Definitions, Evaluation, and Domain (Section 6.1 Study Notes)

Composite Functions

Introduction to Composite Functions

Composite functions are a fundamental concept in precalculus, allowing us to combine two functions to form a new function. Understanding how to evaluate and determine the domain of composite functions is essential for further study in mathematics.

• Composite Function: A function created by applying one function to the results of another.

• Notation: The composite of functions f and g is written as f ˆ g (read as "f composed with g").

• Definition: $(f \circ g)(x) = f(g(x))$

Visualizing Composite Functions

The process of forming a composite function can be visualized as passing the input through g first, then through f. The domain of the composite function consists of all values of x in the domain of g such that g(x) is in the domain of f.

• Domain of $f \circ g$: All $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$.

Evaluating Composite Functions

To evaluate a composite function at a specific value, substitute the value into the inner function, then use the result as the input for the outer function.

• Step 1: Evaluate the inner function $g(x)$ at the given value.

• Step 2: Use the result as the input for the outer function $f$.

Example: Let $f(x) = 5x - 1$ and $g(x) = 3x^2$. Find:

• $(f \circ g)(1) = f(g(1)) = f(3) = 5 \times 3 - 1 = 14$

• $(g \circ f)(1) = g(f(1)) = g(4) = 3 \times 4^2 = 48$

• $(f \circ f)(-2) = f(f(-2)) = f(-11) = 5 \times (-11) - 1 = -56$

• $(g \circ g)(-2) = g(g(-2)) = g(12) = 3 \times 12^2 = 432$

Finding the Domain of a Composite Function

Determining the domain of a composite function requires careful consideration of both the inner and outer functions.

• Rule 1: Exclude any $x$ not in the domain of $g$.

• Rule 2: Exclude any $x$ for which $g(x)$ is not in the domain of $f$.

Example: Let $f(x) = x^2 + 2x - 5$ and $g(x) = 3x + 1$. Find $(f \circ g)(x)$ and $(g \circ f)(x)$ and their domains.

• $(f \circ g)(x) = f(g(x)) = f(3x + 1) = (3x + 1)^2 + 2(3x + 1) - 5$

• Domain: Both $f$ and $g$ are defined for all real numbers, so the domain of each composite is all real numbers.

• $(g \circ f)(x) = g(f(x)) = 3(x^2 + 2x - 5) + 1 = 3x^2 + 6x - 15 + 1 = 3x^2 + 6x - 14$

• Domain: All real numbers.

Examples with Restricted Domains

When functions involve denominators or square roots, additional restrictions may apply.

• Example: $f(x) = \frac{1}{x+4}$, $g(x) = \frac{1}{x-3}$

• Domain of $g$: $x \neq 3$

• Domain of $f$: $x \neq -4$

• To find the domain of $f \circ g$, solve $g(x) = -4$ for $x$ to find additional exclusions.

• Domain of $f \circ g$: $x \neq 2$, $x \neq 3$

Summary Table: Domain Restrictions in Composite Functions

Step	Restriction	Reason
1	$x$ not in domain of $g$	Cannot evaluate $g(x)$
2	$g(x)$ not in domain of $f$	Cannot evaluate $f(g(x))$

Showing Equality of Composite Functions

Sometimes, two composite functions may be equal for all $x$ in their domain. To show this, compute both $(f \circ g)(x)$ and $(g \circ f)(x)$ and compare the results.

• Example: $f(x) = x + 5$, $g(x) = 2x - 5$

• $(f \circ g)(x) = f(g(x)) = (2x - 5) + 5 = 2x$

• $(g \circ f)(x) = g(f(x)) = 2(x + 5) - 5 = 2x + 10 - 5 = 2x + 5$

• In this case, $(f \circ g)(x) \neq (g \circ f)(x)$ unless $5 = 0$.

• Note: In some cases, as in the original example, both composites simplify to $x$ for all $x$ in the domain.

Decomposing a Function into Composites

Given a function $H(x)$, it is sometimes useful to express it as a composition of two simpler functions $f$ and $g$ such that $H(x) = (f \circ g)(x)$.

• Example: $H(x) = (3x^2 - 5)^4$

• Let $g(x) = 3x^2 - 5$, $f(x) = x^4$

• Then $H(x) = f(g(x))$

Additional Examples of Decomposition

• Example: $H(x) = \frac{1}{x + 3}$

• Let $g(x) = x + 3$, $f(x) = \frac{1}{x}$

• Then $H(x) = f(g(x))$

Key Takeaways

• Composite functions combine two functions into one by applying one after the other.

• The domain of a composite function is determined by the domains of both functions and any additional restrictions from the composition.

• Evaluating and decomposing composite functions are essential skills in precalculus.