6.1 Function Notation, Domain, Range, and Composition

6.1 Function Notation, Domain, Range, and Composition

Quick Answer: A function is a relation where each input $x$ in the domain maps to exactly one output $y$ in the range, verified graphically by the Vertical Line Test (VLT). The domain is the set of all allowable real inputs, constrained in Algebra II by two primary rules: denominators cannot equal zero and radicands under even roots must be non-negative. In function composition, $(f \circ g)(x) = f(g(x))$, the inner function $g(x)$ evaluates first, and its domain requires that $x$ is in the domain of $g$ and the output $g(x)$ is in the domain of $f$.

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Formal Definition of a Function & Graphical Validation

[!NOTE] Standards alignment for this section. The Next Generation Algebra II blueprint assumes the definition of a function and function notation from earlier courses; within F-IF, only AII-F.IF.3 (sequences as functions), AII-F.IF.4, AII-F.IF.6, AII-F.IF.7(c, e), AII-F.IF.8b, and AII-F.IF.9 are assessed. Domain determination is tested indirectly through AII-A.REI.2 (extraneous roots) and AII-F.BF.4a (inverses and their domain restrictions). Function composition is (+)F-BF.1c, a plus standard - it is included here because verifying an inverse by composition is the standard justification for AII-F.BF.4a.

In mathematics, a relation is simply any set of ordered pairs $(x, y)$. A function is a specialized relation with a strict uniqueness requirement: for every element in the input set (the domain), there corresponds exactly one element in the output set (the range). An input may not produce multiple distinct outputs.

The Vertical Line Test (VLT)

When a relation is graphed on the Cartesian coordinate plane, the definition of a function can be tested visually:

• Vertical Line Test Theorem: A curve in the coordinate plane represents the graph of a function of $x$ if and only if no vertical line intersects the curve at more than one point.
• Mathematical Justification: A vertical line represents the equation $x = c$ for some constant real number $c$. If a vertical line intersects a graph at two distinct points $(c, y_1)$ and $(c, y_2)$ where $y_1 \neq y_2$, then the single input $x = c$ produces two distinct outputs $y_1$ and $y_2$. This violates the definition of a function.

Mapping Diagrams: One-to-One vs. Many-to-One

A mapping diagram illustrates relations by drawing directed arrows from elements in domain set $X$ to elements in range set $Y$:

• Valid Function (One-to-One or Many-to-One): Each element in the domain has exactly one arrow departing from it. For example, $f(x) = x^2$ maps both $x = 2$ and $x = -2$ to the single output $y = 4$. This "many-to-one" behavior is entirely legal—different inputs can share the same output.
• Not a Function (One-to-Many): If a single element in the domain has two or more arrows departing from it (e.g., input $x = 9$ mapping to both $y = 3$ and $y = -3$, as in the relation $y^2 = x$), the relation fails the definition of a function.

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Domain and Range Notations: Interval, Set-Builder, and Inequalities

The domain comprises all possible independent variable values ($x$) for which the function is defined. The range comprises all possible dependent variable values ($y$) produced by the function. On the New York Regents Examination, domain and range are communicated through three standard notations:

[!IMPORTANT] Parentheses vs. Brackets Rule: Always use square brackets $[;]$ for numbers that are included in the set (corresponding to $\le$ or $\ge$ and solid closed dots on a number line). Always use round parentheses $(;)$ for excluded values (corresponding to $<$, $>$, open circles, holes, and vertical asymptotes). Infinity symbols ($-\infty$ and $\infty$) always take round parentheses because infinity is an unbounded conceptual direction, not an attainable real number.

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Algebraic Domain Restrictions in Algebra II

Unless a contextual scenario or explicit prompt restricts the domain, the domain of any algebraic function is assumed to be the set of all real numbers, denoted as $\mathbb{R}$ or $(-\infty, \infty)$. In Algebra II, restrictions arise exclusively when an operation is mathematically undefined within the real number system.

The Two Master Domain Rules

1. Division by Zero is Undefined (Rational Expressions): For any function $f(x) = \frac{P(x)}{Q(x)}$, the denominator cannot equal zero: $Q(x) \neq 0$ Set the denominator equal to zero, solve for $x$, and exclude those roots from the domain.

2. Even-Index Roots of Negative Numbers are Non-Real (Radical Expressions): For any even root $f(x) = \sqrt[2n]{R(x)}$ where $n \in {1, 2, 3, \dots}$, the radicand must be non-negative: $R(x) \ge 0$ Solve this inequality to establish the allowable domain interval.

[!NOTE] Odd Roots Have No Radicand Restriction: For odd roots such as $f(x) = \sqrt[3]{x - 8}$, negative numbers possess real odd roots (for example, $\sqrt[3]{-8} = -2$). Consequently, odd-index radicals impose no domain restriction; their radicands can be any real number.

Summary of Domain Restriction Scenarios

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Function Evaluation & The Difference Quotient

Evaluating a function involves substituting an input quantity into every instance of the independent variable $x$. When the input is an algebraic binomial such as $(x + h)$, you must substitute the entire grouping in parentheses and expand algebraically.

The Difference Quotient

A central algebraic skill connecting Algebra II to calculus is simplifying the difference quotient, which represents the average rate of change between $x$ and $x + h$:

$\frac{f(x + h) - f(x)}{h}, \quad h \neq 0$

To compute the difference quotient:

1. Construct $f(x + h)$ by replacing every $x$ with $(x + h)$ and expanding all powers.
2. Subtract the original function $f(x)$, enclosing $f(x)$ in parentheses to ensure negative distribution.
3. Simplify the numerator; all terms lacking a factor of $h$ must cancel to zero.
4. Factor out $h$ from the remaining numerator terms and divide by the $h$ in the denominator.

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Function Composition & Composite Domain Restrictions (Enrichment: (+)F-BF.1c)

The composite function $(f \circ g)(x)$, read as "$f$ composed with $g$ of $x$" or "$f$ of $g$ of $x$", is defined by:

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

In this operation, the output of the inner function $g(x)$ becomes the direct input for the outer function $f$. Function composition is generally non-commutative: $f(g(x)) \neq g(f(x))$.

   Input x  ───> [ Inner Machine: g ]  ───> g(x)  ───> [ Outer Machine: f ]  ───> f(g(x))

The Two-Condition Protocol for the Domain of $(f \circ g)(x)$

A common Regents error is looking only at the simplified final formula of $f(g(x))$ to determine its domain. You must enforce two conditions simultaneously:

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

$\text{Domain of } (f \circ g) = \{x \in \mathbb{R} \mid x \in \text{Domain}(g) \text{ and } g(x) \in \text{Domain}(f)\}$

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Worked Examples

Worked Problem 1: Determining Restricted Domain with Radicals and Rational Expressions

Problem: Algebraically determine the domain of the function $f(x) = \frac{\sqrt{2x + 10}}{x^2 - 4x - 12}$. Express your answer in interval notation.

• Step 1: Enforce the non-negative radicand restriction for the even root. $2x + 10 \ge 0 \implies 2x \ge -10 \implies x \ge -5$ In interval notation, this condition is $[-5, \infty)$.

• Step 2: Enforce the non-zero denominator restriction. $x^2 - 4x - 12 \neq 0$ Factor the quadratic trinomial: $(x - 6)(x + 2) \neq 0 \implies x \neq 6 \quad \text{and} \quad x \neq -2$

• Step 3: Intersect the restrictions. The input $x$ must satisfy $x \ge -5$ while excluding $x = -2$ and $x = 6$. Both excluded points lie within the interval $[-5, \infty)$.

• Step 4: State the domain in interval notation. $[-5, -2) \cup (-2, 6) \cup (6, \infty)$

Worked Problem 2: Difference Quotient Simplification for a Quadratic Function

Problem: Given $f(x) = 2x^2 - 3x + 5$, determine and simplify the difference quotient $\frac{f(x + h) - f(x)}{h}$ where $h \neq 0$.

• Step 1: Evaluate $f(x + h)$ by algebraic substitution. $f(x + h) = 2(x + h)^2 - 3(x + h) + 5$ Expand $(x + h)^2 = x^2 + 2xh + h^2$: $f(x + h) = 2(x^2 + 2xh + h^2) - 3x - 3h + 5 = 2x^2 + 4xh + 2h^2 - 3x - 3h + 5$

• Step 2: Set up the numerator $f(x + h) - f(x)$. $[2x^2 + 4xh + 2h^2 - 3x - 3h + 5] - [2x^2 - 3x + 5]$ Distribute the negative sign: $= 2x^2 + 4xh + 2h^2 - 3x - 3h + 5 - 2x^2 + 3x - 5$

• Step 3: Combine like terms. $(2x^2 - 2x^2) + (-3x + 3x) + (5 - 5) + 4xh + 2h^2 - 3h = 4xh + 2h^2 - 3h$

• Step 4: Divide by $h$. $\frac{4xh + 2h^2 - 3h}{h} = \frac{h(4x + 2h - 3)}{h} = 4x + 2h - 3$

Worked Problem 3: Composite Function Formula and Domain Determination

Problem: Let $f(x) = \frac{2}{x - 4}$ and $g(x) = \sqrt{x + 5}$. State the equation for $(f \circ g)(x)$ and determine its complete domain in interval notation.

• Step 1: Find the composite formula $(f \circ g)(x) = f(g(x))$. Substitute $g(x) = \sqrt{x + 5}$ into $f$: $(f \circ g)(x) = \frac{2}{\sqrt{x + 5} - 4}$

• Step 2: Determine the domain restriction of the inner function $g(x)$. The radicand must be non-negative: $x + 5 \ge 0 \implies x \ge -5$, which is $[-5, \infty)$.

• Step 3: Determine when the inner function output violates the outer function's domain. The outer function $f(u) = \frac{2}{u - 4}$ is undefined when its input equals $4$. Therefore, $g(x) \neq 4$: $\sqrt{x + 5} \neq 4$ Square both sides: $x + 5 \neq 16 \implies x \neq 11$

• Step 4: Combine all domain restrictions. Starting with $x \ge -5$ and removing $x = 11$ gives: $[-5, 11) \cup (11, \infty)$

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Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Overlooking the inner function's domain in compositions. If $g(x) = \sqrt{x}$ and $f(x) = x^2$, $(f \circ g)(x) = (\sqrt{x})^2 = x$. Students frequently state the domain is all real numbers because the simplified formula is just $x$. This is incorrect! The domain is $[0, \infty)$ because negative numbers cannot enter the initial inner function $g(x)$.
• Pitfall 2: Illegally distributing function notation. $f(x + h) \neq f(x) + f(h)$. Function notation is not multiplication. You must substitute $(x + h)$ into the function's variable slots.
• Pitfall 3: Binomial expansion errors in difference quotients. Expanding $(x + h)^2$ as $x^2 + h^2$ omits the middle term $+2xh$. Always write $(x + h)(x + h) = x^2 + 2xh + h^2$.