6.4 Function Composition, Invertibility & Inverse Functions

6.4 Function Composition, Invertibility & Inverse Functions

1. The Algebra of Function Composition

Function composition is the operational process of combining two functions by using the output of one function as the input of another. Formally, given two functions $g: X \to Y$ and $f: Y \to Z$, the composite function $f \circ g$ (read "$f$ composed with $g$" or "$f$ of $g$") is defined by:

$(f \circ g)(x) = f(g(x))$

In this operational pipeline, the independent variable $x$ is processed first by the inner function $g$, generating an intermediate value $g(x)$. This intermediate output is then processed by the outer function $f$ to produce the final output $f(g(x))$.

Function composition possesses distinct algebraic properties that secondary educators must emphasize:

1. Non-Commutativity: Function composition is generally not commutative: $(f \circ g)(x) \neq (g \circ f)(x)$ For example, let $f(x) = 2x + 5$ and $g(x) = x^2$. Then: $(f \circ g)(x) = f(g(x)) = f(x^2) = 2x^2 + 5$ $(g \circ f)(x) = g(f(x)) = g(2x + 5) = (2x + 5)^2 = 4x^2 + 20x + 25$ The two resulting expressions are manifestly unequal for almost all real numbers $x$.
2. Associativity: Function composition is associative whenever the respective compositions are defined: $[f \circ (g \circ h)](x) = [(f \circ g) \circ h](x)$
3. Identity Element: The identity function $I(x) = x$ serves as the identity element under composition: $(f \circ I)(x) = (I \circ f)(x) = f(x)$.

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2. Rigorous Determination of Composite Function Domains

The domain of a composite function is one of the most frequently miscalculated concepts on the FTCE examination. A common student error is simplifying the composite algebraic expression first and determining the domain solely from the resulting formula. This practice frequently overlooks hidden domain restrictions imposed by the inner function.

Formally, the domain of $(f \circ g)(x)$ is the set of all inputs $x$ in the domain of $g$ whose corresponding outputs $g(x)$ belong to the domain of $f$:

$\operatorname{dom}(f \circ g) = \{x \in \operatorname{dom}(g) \mid g(x) \in \operatorname{dom}(f)\}$

Systematic Two-Tiered Domain Protocol

To determine $\operatorname{dom}(f \circ g)$:

• Condition 1 (Inner Domain): Find the natural domain of $g(x)$. The input $x$ must satisfy $x \in \operatorname{dom}(g)$.
• Condition 2 (Outer Compatibility): Determine what inputs are permissible for $f$. If $u \in \operatorname{dom}(f)$ requires $u \in S$, then enforce the inequality or restriction $g(x) \in S$. Solve this constraint for $x$.
• Condition 3 (Set Intersection): The final domain is the intersection of the solutions from Condition 1 and Condition 2: $\operatorname{dom}(f \circ g) = \text{Condition 1} \cap \text{Condition 2}$

Worked Composite Domain Exemplar

Let $f(x) = \frac{3}{x - 5}$ and $g(x) = \frac{2x + 1}{x - 2}$. Find the natural domain of $(f \circ g)(x)$.

Step 1: Inner function restriction. The inner function is $g(x) = \frac{2x + 1}{x - 2}$. The denominator cannot equal zero: $x - 2 \neq 0 \implies x \neq 2$ Thus, $\operatorname{dom}(g) = {x \in \mathbb{R} \mid x \neq 2}$.

Step 2: Outer function restriction on the inner output. The outer function is $f(u) = \frac{3}{u - 5}$. The outer domain requires that its input cannot equal $5$: $u \neq 5$. Therefore, the output of $g(x)$ cannot equal $5$: $g(x) \neq 5 \implies \frac{2x + 1}{x - 2} \neq 5$ Solve the boundary equation $\frac{2x + 1}{x - 2} = 5$: $2x + 1 = 5(x - 2) \implies 2x + 1 = 5x - 10 \implies 11 = 3x \implies x = \frac{11}{3}$ Thus, $x \neq \frac{11}{3}$.

Step 3: Intersect the domain conditions. Combining both conditions, $x$ must not equal $2$ and $x$ must not equal $\frac{11}{3}$: $\operatorname{dom}(f \circ g) = \left(-\infty, 2\right) \cup \left(2, \frac{11}{3}\right) \cup \left(\frac{11}{3}, \infty\right)$

[!CAUTION] If a candidate simplifies $(f \circ g)(x)$ algebraically before finding the domain: $(f \circ g)(x) = \frac{3}{\frac{2x + 1}{x - 2} - 5} = \frac{3(x - 2)}{(2x + 1) - 5(x - 2)} = \frac{3x - 6}{-3x + 11}$ Looking only at the simplified denominator $-3x + 11 \neq 0 \implies x \neq 11/3$, the candidate would completely fail to exclude $x = 2$. At $x = 2$, $g(2)$ is undefined, so the pipeline breaks at the very first step!

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3. Invertibility, Injectivity & Domain Restrictions

An inverse function reverses the action of the original function. If $f$ maps an input $x$ to an output $y$, then its inverse $f^{-1}$ maps $y$ back to $x$.

The Bijectivity Requirement & Horizontal Line Test (HLT)

A function $f: X \to Y$ possesses an inverse function $f^{-1}: Y \to X$ if and only if $f$ is a bijection:

1. Injective (One-to-One): Every distinct input produces a distinct output: $x_1 \neq x_2 \implies f(x_1) \neq f(x_2), \quad \text{or equivalently} \quad f(x_1) = f(x_2) \implies x_1 = x_2$
2. Surjective (Onto): Every element in the codomain $Y$ is realized as an output for some $x \in X$: $\operatorname{ran}(f) = Y$.

Geometrically, a function graphed on the Cartesian plane is one-to-one if and only if it passes the Horizontal Line Test (HLT): every horizontal line $y = c$ intersects the graph at most once. If any horizontal line intersects the graph at two or more points, multiple domain inputs produce the identical output value $c$, which means reversing the mapping would yield multiple outputs for a single input, violating the vertical line test for $f^{-1}$.

Restricting Domains to Create Invertible Branches

Many fundamental parent functions (such as quadratics, even polynomials, absolute value, and trigonometric functions) fail the Horizontal Line Test over their full natural domains. To define an inverse function, secondary mathematicians restrict the domain to a subset where the function is strictly monotonic (strictly increasing or strictly decreasing), thereby making it one-to-one while preserving the entire range.

• The Quadratic Parent $f(x) = x^2$: Fails HLT on $(-\infty, \infty)$ because $f(-2) = f(2) = 4$. By restricting the domain to $[0, \infty)$, the function becomes strictly increasing and one-to-one, yielding the well-known inverse $f^{-1}(x) = \sqrt{x}$ on $[0, \infty)$. Alternatively, restricting the domain to $(-\infty, 0]$ yields the negative branch inverse $f^{-1}(x) = -\sqrt{x}$.
• General Quadratic Vertex Rule: For any parabola $f(x) = a(x - h)^2 + k$, the axis of symmetry is $x = h$. Restricting the domain to $[h, \infty)$ or $(-\infty, h]$ isolates a one-to-one branch that admits an inverse.

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4. Algebraic Construction and Properties of Inverse Functions

The inverse function is denoted by $f^{-1}(x)$.

[!WARNING] Secondary teachers must explicitly warn students that the superscript $-1$ in $f^{-1}(x)$ denotes functional inverse, NOT a numerical exponent: $f^{-1}(x) \neq \frac{1}{f(x)} = [f(x)]^{-1}$ The expression $\frac{1}{f(x)}$ is the multiplicative reciprocal, whereas $f^{-1}(x)$ undoes the operation of $f$.

Fundamental Cancellation Properties

A function $f$ and its inverse $f^{-1}$ satisfy the Inverse Function Composition Identities:

$f(f^{-1}(x)) = x \quad \text{for all } x \in \operatorname{dom}(f^{-1}) = \operatorname{ran}(f)$ $f^{-1}(f(x)) = x \quad \text{for all } x \in \operatorname{dom}(f)$

To prove algebraically that two candidate functions $f$ and $g$ are inverses of each other, one must verify that both $(f \circ g)(x) = x$ and $(g \circ f)(x) = x$.

Domain-Range Duality & Geometric Reflection across $y = x$

Because the inverse function interchanges the roles of independent and dependent variables:

$\operatorname{dom}(f^{-1}) = \operatorname{ran}(f) \quad \text{and} \quad \operatorname{ran}(f^{-1}) = \operatorname{dom}(f)$

If $(a, b)$ is a point on the graph of $y = f(x)$, then $(b, a)$ is the corresponding point on the graph of $y = f^{-1}(x)$. Geometrically, the graph of $f^{-1}$ is the reflection of the graph of $f$ across the diagonal identity line $y = x$.

Systematic Algebraic Inversion Algorithm

To derive the algebraic formula for $f^{-1}(x)$:

1. Set $y = f(x)$.
2. Interchange the variable symbols $x$ and $y$ (reflecting the coordinates across $y = x$).
3. Solve the resulting equation explicitly for $y$ in terms of $x$.
4. Replace $y$ with $f^{-1}(x)$ and state any domain restrictions inherited from $\operatorname{ran}(f)$.

Worked Linear Fractional (Möbius) Inversion

Find the inverse function of $f(x) = \frac{4x - 3}{2x + 5}$. State its domain and range.

Step 1: Determine domain and range of $f$.

• Domain of $f$: $2x + 5 \neq 0 \implies x \neq -5/2$.
• Range of $f$: The horizontal asymptote is the ratio of leading coefficients: $y = \frac{4}{2} = 2$. Thus, $\operatorname{ran}(f) = {y \in \mathbb{R} \mid y \neq 2}$.

Step 2: Swap variables and solve for $y$. $x = \frac{4y - 3}{2y + 5}$ Multiply both sides by $(2y + 5)$: $x(2y + 5) = 4y - 3 \implies 2xy + 5x = 4y - 3$ Collect all terms containing $y$ on one side and terms without $y$ on the opposite side: $2xy - 4y = -5x - 3$ Factor out $y$: $y(2x - 4) = -(5x + 3)$ Divide by $(2x - 4)$: $y = \frac{-5x - 3}{2x - 4} = \frac{5x + 3}{4 - 2x}$

Step 3: State the inverse function and verify duality. $f^{-1}(x) = \frac{5x + 3}{4 - 2x}, \quad x \neq 2$ Notice that $\operatorname{dom}(f^{-1}) = {x \in \mathbb{R} \mid x \neq 2}$, which matches $\operatorname{ran}(f)$ perfectly! Furthermore, the horizontal asymptote of $f^{-1}(x)$ is $\frac{5}{-2} = -\frac{5}{2}$, meaning $\operatorname{ran}(f^{-1}) = {y \in \mathbb{R} \mid y \neq -5/2}$, which matches $\operatorname{dom}(f)$!