Week 3 — Lecture Outline · Functions: Notation, Domain & Range

Course: College Algebra (MATH 120) · Silver Oak University (fictional sample) · Prof. Calloway
Objectives covered: Objective 3 — Use function notation to evaluate functions and describe domain and range; perform operations on functions and composition.
SLOs touched: A (apply procedures accurately) · B (connect symbolic/graphical representations; interpret in context)
Meeting pattern: 2 sessions × 75 min = 150 min. Segment minutes below total ~150; scale to your own pattern.

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Week at a Glance

The week's big question	"Given a rule, what can I plug in, what can I get out, and how do I combine two rules into one?"
By the end of the week, students can…	(1) identify a function from a mapping, table, or graph (vertical line test) and evaluate f(x) at numeric and algebraic inputs; (2) state the domain of polynomials (all reals), rational functions (exclude denominator zeros), and radical functions with even roots (radicand ≥ 0); (3) perform the four function operations and state the quotient's domain; (4) compose two functions and evaluate at a specific input.
Key vocabulary	function, input, output, relation, vertical line test, function notation, f(x), evaluate, domain, range, interval notation, restriction, operations on functions, composition, (f ∘ g)(x), inside-out
Materials	slides (Deck 3), the week's readings + video links, Desmos (or GeoGebra) and a calculator, one approved chatbot (Gemini / Claude / ChatGPT) for the AI-critique moment and the tutorial
Timing note	8 segments, ~150 min total. Session 1 = Segments 1–4 (~73 min). Session 2 = Segments 5–8 (~77 min).

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Segment 1 — Hook & the Promise (8 min) · Session 1 opens

Hook. "Think about a vending machine. You put in a code — B3 — and exactly one snack comes out. Not two snacks, not sometimes a snack, always exactly one snack. That's a function."
- "Or think about a GPS: you give it your starting address and it gives back exactly one route recommendation (its best guess). One input, one output."
- "Now think about weather: you give it today's date and… well, you might get sun or rain. Multiple outputs from one input. That's not a function — and that distinction is the whole point of this week."

The promise (write it on the board): "By the end of this week you can read f(x) = 3x − 1 and instantly evaluate f(a + 2) = 3a + 5, know exactly which values of x are 'allowed' for any function we'll ever study, and chain two functions together without mixing up the order."

Why it matters line: "Functions are the backbone of everything in this course from here on — graphs, equations, exponentials, logarithms. They're how algebra describes the world."

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Segment 2 — What Is a Function? (20 min)

Plain language first. A relation is any pairing of inputs to outputs. A function is a relation where each input produces exactly one output.
- Draw a mapping diagram: a set of inputs (domain) on the left, outputs (range) on the right, arrows connecting them.
- Function: every left-side value has exactly one arrow leaving it.
- Not a function: one left-side value has two arrows (one input, two outputs).

Key definition (put on board): A function f from a set D to a set R assigns to every input x in D exactly one output f(x) in R.

The vertical line test (graphical criterion):
- A graph represents a function if and only if every vertical line crosses the graph at most once.
- Demonstrate: graph of y = 2x + 1 (a line, passes VLT — function). Graph of a circle x² + y² = 4 (some vertical lines cross twice — NOT a function).
- Memory hook: "A vertical line is 'one x-value' — it shouldn't touch the graph twice, because that would mean one input gives two outputs."

One fully worked example (every step):

Look at four descriptions. Which is a function?
- (A) {(1,3), (2,5), (3,7), (4,3)} — input 1 → 3, input 2 → 5, etc. Each input appears once. Function (two inputs can share the same output; the rule is about inputs, not outputs).
- (B) {(1,3), (2,5), (1,7)} — input 1 appears twice (3 and 7). Not a function.
- (C) A vertical-line-test-passing parabola y = x² — every x has one y. Function.
- (D) A circle — fails VLT. Not a function.

Name the misconception + cure:
- ❌ "(A) is not a function because 3 appears twice as an output."
✅ Cure: the rule is about inputs, not outputs. Two different inputs mapping to the same output is fine. What's forbidden is one input mapping to two different outputs.

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Segment 3 — Function Notation & Evaluating (25 min)

Plain language first. When we write f(x) = 3x − 1, we mean:
- f is the name of the function (the machine).
- x is the input variable.
- 3x − 1 is the rule (what the machine does to x).
- f(x) does NOT mean f times x. It means "f evaluated at x" — the output when the input is x.

Memory hook: "f(x) is 'f of x' — it's a label for the output, not a multiplication."

Evaluating at a number — one fully worked example:

Let f(x) = 2x² − 3. Find f(−2).
1. Replace every x with (−2): f(−2) = 2(−2)² − 3
2. Exponent first: (−2)² = 4 (careful — the parentheses square the negative) → 2(4) − 3
3. Multiply: 2(4) = 8 → 8 − 3
4. Subtract: f(−2) = 5
Common distractor: 2(−2)² = 2(−4) = −8 — forgetting that (−2)² = +4, not −4.

Evaluating at an algebraic input — fully worked example:

Let f(x) = 3x − 1. Find f(a + 2).
1. Replace every x with (a + 2): f(a + 2) = 3(a + 2) − 1
2. Distribute the 3: 3a + 6 − 1
3. Combine: f(a + 2) = 3a + 5
The trap: writing 3a + 2 − 1 = 3a + 1 (not distributing the 3 to the 2). The entire expression (a + 2) replaces x; the parentheses are non-negotiable.

Name the misconception + cure (the #1 Week-3 error):
- ❌ "f(x) means f times x, so f(a + 2) = f · a + 2."
✅ Cure: f is the name of the function, not a number. f(x) = 3x − 1 is a rule. "f(a + 2)" tells us to plug (a + 2) in everywhere x lives. No multiplication between f and the parentheses.

Quick Interaction (Think-Pair-Share, ~5 min):

Put f(x) = 3x − 1 on the board. Ask: "Which is correct — f(a + 2) = 3a + 5 or f(a + 2) = 3a + 1?" Students vote, compare with neighbor, class discusses. Then ask: "What would f(−x) look like?" → f(−x) = 3(−x) − 1 = −3x − 1. One more: f(x + h) = 3(x + h) − 1 = 3x + 3h − 1.

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Segment 4 — Domain & Range (20 min) · Session 1 closes (~73 min)

Hook back to plain language. "The domain is every value of x you're allowed to plug in. The range is every output value you can actually get. Most of the time this week we focus on domain — the range often requires graphing to determine."

The three domain rules (write on board as a table):

Function type	Domain rule	Example
Polynomial	All real numbers (−∞, ∞)	f(x) = x² − 4x + 1 → domain: (−∞, ∞)
Rational (fraction)	Exclude values that make the denominator = 0	f(x) = (x+1)/(x−3) → set x−3 = 0 → x = 3; domain: x ≠ 3
Even root (√)	Radicand must be ≥ 0	f(x) = √(x−5) → set x−5 ≥ 0 → x ≥ 5; domain: [5, ∞)

Memory hook: "Three cases, three rules: polynomial = no restrictions; rational = no zero denominators; even root = no negative radicand."

Fully worked examples (every step):

Polynomial: f(x) = x² − 4x + 1. No denominators, no roots → domain: (−∞, ∞) or all real numbers.

Rational: f(x) = (x + 1)/(x − 3). Set denominator = 0: x − 3 = 0 → x = 3. Exclude x = 3.
Domain: all real x with x ≠ 3, or (−∞, 3) ∪ (3, ∞) in interval notation.

Radical: f(x) = √(x − 5). Set radicand ≥ 0: x − 5 ≥ 0 → x ≥ 5.
Domain: [5, ∞) in interval notation.

Name the misconception + cure:
- ❌ "The domain of √(x − 5) is x > 5 (strict inequality)."
✅ Cure: √0 = 0, which is fine. The radicand can equal zero — it just can't be negative. So x ≥ 5, including x = 5.
- ❌ "The domain of (x+1)/(x−3) is x ≠ 0."
✅ Cure: set the denominator (not the numerator) equal to zero. x − 3 = 0 gives x = 3.

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Segment 5 — Operations on Functions (20 min) · Session 2 opens

Hook back in: "Last session: a function is a rule. Today: what happens when you add, subtract, multiply, or divide two rules? You get a new rule — and you have to track any new domain restrictions the new rule introduces."

Plain language first. For two functions f and g with a shared domain:
- (f + g)(x) = f(x) + g(x) — add the output expressions.
- (f − g)(x) = f(x) − g(x) — subtract.
- (fg)(x) = f(x) · g(x) — multiply.
- (f/g)(x) = f(x)/g(x) — divide; domain: wherever g(x) ≠ 0 (extra restriction).

Memory hook: "The operations on functions are just operations on their formulas — except for division, where you must exclude the x-values that make g(x) = 0."

One fully worked example (every step):

Let f(x) = x + 2 and g(x) = x − 3.
- (f + g)(x) = (x + 2) + (x − 3) = 2x − 1
- (f − g)(x) = (x + 2) − (x − 3) = x + 2 − x + 3 = 5 (a constant function!)
- (fg)(x) = (x + 2)(x − 3) = x² − 3x + 2x − 6 = x² − x − 6
- (f/g)(x) = (x + 2)/(x − 3), domain: x − 3 ≠ 0 → x ≠ 3.
Can this simplify? No common factors. Result: (x + 2)/(x − 3), x ≠ 3.

The quotient that simplifies — a second example:

Let f(x) = x² − 9 and g(x) = x − 3.
(f/g)(x) = (x² − 9)/(x − 3). Factor: (x − 3)(x + 3)/(x − 3) = x + 3.
But we must keep the restriction: x ≠ 3 (g(3) = 0 even though the simplified form x + 3 looks fine at x = 3).
Domain: x ≠ 3, even after simplification.

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Segment 6 — Composition of Functions + The Famous Trap (22 min)

Plain language first. Composition is chaining two functions: the output of g becomes the input of f.
- (f ∘ g)(x) = f(g(x)) — do g first, then plug that result into f.
- This is inside-out: the inner function (g) runs first.
- "g of x goes inside f" — right-to-left evaluation, even though we write left-to-right.

Memory hook: "(f ∘ g)(x): start with g, then hand that output to f. The circle means 'g goes first.'"

One fully worked example (evaluate at a specific number):

Let f(x) = x² and g(x) = x + 1. Find (f ∘ g)(2).
1. Start with g: g(2) = 2 + 1 = 3
2. Feed that result into f: f(3) = 3² = 9
(f ∘ g)(2) = 9 ✅
Now check (g ∘ f)(2) — note the different order:
1. Start with f: f(2) = 2² = 4
2. Feed into g: g(4) = 4 + 1 = 5
(g ∘ f)(2) = 5 — a different answer! Composition is not commutative.

One fully worked example (general formula):

Let f(x) = x² and g(x) = x + 1. Find (f ∘ g)(x).
(f ∘ g)(x) = f(g(x)) = f(x + 1) = (x + 1)² = x² + 2x + 1

Compare (g ∘ f)(x) = g(f(x)) = g(x²) = x² + 1.
Two different functions — order matters.

The famous trap — name it explicitly:

❌ "(f ∘ g)(x) = g(f(x)) — swap the order, it's the same thing."
✅ Cure: f ∘ g means f AFTER g: do g first, then f. Think of the function closest to the input (g) running first. Example: (f ∘ g)(2) = f(g(2)) = f(3) = 9, but (g ∘ f)(2) = g(f(2)) = g(4) = 5. Order is everything.

Callback: "Remember last week when we subtracted from both sides in a specific order? Same idea here — sequence matters. The composition circle tells you which machine runs first, and that's g."

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Segment 7 — Technology Workflow + AI-Critique (15 min)

Technology workflow — Desmos check for composition (exact steps):
1. Open desmos.com/calculator (free, no login).
2. Define f: type f(x) = x^2 on line 1.
3. Define g: type g(x) = x + 1 on line 2.
4. Type f(g(x)) on line 3 and (x+1)^2 on line 4 — the two graphs should be identical.
5. Type g(f(x)) on line 5 and x^2 + 1 on line 6 — identical again.
6. Toggle line 3 vs. line 5 on and off — the two compositions are different curves. Desmos makes the non-commutativity visible.
7. Domain check: graph f(x) = √(x − 5) — Desmos will show a graph that starts at x = 5, confirming the domain [5, ∞).

AI-critique moment:

Paste this to an approved chatbot: "Let f(x) = 2x + 1 and g(x) = x − 3. Compute (f ∘ g)(x) and (g ∘ f)(x)."
Check its work by hand:
- (f ∘ g)(x) = f(g(x)) = f(x − 3) = 2(x − 3) + 1 = 2x − 5.
- (g ∘ f)(x) = g(f(x)) = g(2x + 1) = (2x + 1) − 3 = 2x − 2.

Chatbots sometimes reverse the order or treat (f ∘ g)(x) as f(x) · g(x) (multiplication instead of composition). Verify the key steps yourself. That habit — the tool drafts, you judge — is the whole semester in miniature.

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Segment 8 — Callback & Hand-off (10 min) · Session 2 closes (~77 min)

Callback + tease:
- Callback: "Week 1 gave us the rules for simplifying; Week 2 gave us equations to solve. Week 3 gives us the container that holds all of it: the function. Everything we simplify or solve lives inside a function now."
- Tease next week: "Next week (Week 4) we zoom in on one specific family of functions — linear functions. Their graphs are lines, their rates of change are slopes, and they're the foundation for every trend, projection, and 'cost per item' problem you'll ever see. The tool is the graph; the insight is the slope."

Hand-off (the week's graded work):
- Lecture Tutorial 3 (AI tutor, share-link submission) — function notation, domain & range, operations, composition.
- Quiz 3 (end of week, no AI) and Discussion 3 ("Is It a Function? What Goes In?" — model a real-world input→output relationship with an AI partner).
- Assignment 3 ("Mapping the Rules") — AI-coached, self-scored.

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Instructor FAQ — Common Stumbles

Student says / does	Quick cure
Writes f(x) = f · x or f(a+2) = f · (a+2).	f is the name of the function, not a variable. f(a+2) means "substitute (a+2) for x in the rule." No multiplication.
Finds f(a+2) = 3a + 1 for f(x) = 3x − 1 (distributes incompletely).	The 3 multiplies the whole quantity (a+2): 3(a+2) = 3a + 6, not 3a + 2. Show the step with the parentheses intact first.
States domain of (x+1)/(x−3) as "x ≠ 0."	Set the denominator (not numerator) equal to zero: x − 3 = 0 → x = 3. Zero in the numerator is fine; zero in the denominator is division by zero.
States domain of √(x−5) as "x > 5" (strict).	√0 = 0 is defined. The radicand must be ≥ 0, so x ≥ 5, not x > 5. Include the boundary.
Computes (f/g)(x) = (x²−9)/(x−3) = x+3, no restriction.	Even after cancellation, x ≠ 3 must be stated because g(3) = 0. The simplification is algebra; the restriction is about the original domain.
Reverses composition: (f ∘ g)(x) = g(f(x)).	(f ∘ g)(x) = f(g(x)) — g runs first. "f after g." Read the circle as "g goes first, then f." Check numerically: (f ∘ g)(2) ≠ (g ∘ f)(2) in general.
Thinks range = domain.	Domain is the set of inputs (x-values); range is the set of outputs (y-values). They can overlap, but they're different questions.

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Scope flag

This outline stays within Objective 3. The vertical line test and basic graphical identification are added context (strictly, identifying functions from sets/tables suffices); the VLT is included because it bridges to the graphing work in Week 4. Composition domain restrictions (domain of f ∘ g depends on the domain of g and the range of g landing in f's domain) are left for Objective 7 (Week 10/11); only clean compositions with no extra restrictions are used this week.