Chapter 1 — Front Matter
Purpose and Overview
Mathematics is, at its core, the study of relationships. Before you can understand how things change — before you can find the rate at which a satellite moves across the sky, or predict how a drug concentration fades in the bloodstream — you need a precise language for describing relationships between quantities. That language is the language of functions.
Chapter 1 is the opening gateway to precalculus. It reviews and deepens the function concept you first encountered in Algebra I and Algebra II, and it builds the foundation you will need for every chapter that follows. You will study functions as relationships between inputs and outputs, as graphs that reveal behavior visually, as equations that describe rules algebraically, as tables that organize values numerically, and as models that connect mathematics to real situations.
This chapter has a clear purpose: to make sure that when you meet a trigonometric function, a rational function, an exponential function, or a limit in later chapters, you already have a firm understanding of what a function is, what its graph tells you, how to find its domain and range, how to read its intercepts and intervals of increase and decrease, and how to compute and interpret its rate of change. Precalculus is largely the study of functions before limits and derivatives. This chapter makes that clear from the beginning.
Opening Question
A hospital pharmacy records the concentration of a medication in a patient's bloodstream every hour after an injection. At hour 0, the concentration is 12 milligrams per liter. At hour 1, it is 9. At hour 2, it is 6.75. At hour 3, it is about 5.06. The concentration keeps dropping in a consistent pattern.
Can you describe the rule connecting time and concentration? How would you represent it as an equation, a table, and a graph? How would you use that representation to predict the concentration at hour 6, or to find when the concentration falls below 2 milligrams per liter? These are function questions. By the end of this chapter, you will have the tools to answer them precisely.
14Sections
14Objectives
14Key Terms
12+Worked Examples
Learning Objectives
- 1Describe what precalculus studies and why functions are central to it.
- 2Define a function as a relationship that assigns exactly one output to each input.
- 3Use function notation to represent, evaluate, and interpret functions.
- 4Determine the domain and range of a function from its equation, graph, or context.
- 5Read and interpret graphs of functions, including intercepts, zeros, and increasing and decreasing intervals.
- 6Compute and interpret average rate of change over an interval.
- 7Write, graph, and evaluate piecewise functions.
- 8Construct function models for real situations and interpret their outputs.
- 9Compare different representations of the same function: equation, graph, table, and verbal description.
- 10Use graphing technology responsibly to explore function behavior.
- 11Recognize and correct common function mistakes.
- 12Prepare to study transformations and parent functions in Chapter 2.
Sections 1.1 – 1.2
What Precalculus Studies and What a Function Is
The Function Concept · Representations · The Vertical Line Test
Section 1.1
What Precalculus Studies
Precalculus is not simply a list of algebra topics with some trigonometry added on. It is a carefully organized preparation for the study of calculus and for advanced applications in mathematics, science, engineering, and technology. You have already studied arithmetic, algebra, and geometry. Precalculus asks a deeper question: What are all the different kinds of relationships between quantities, and how do we study their behavior systematically? That question is the question of functions.
In Part I, you will build a complete understanding of functions: what they are, how to represent them, how to transform them, how to combine them, and how to reverse them. In Part II, you will study trigonometric functions — the mathematical description of rotation, periodic motion, waves, and circular behavior. In Part III, you will deepen your study of polynomial, rational, exponential, and logarithmic functions. In Part IV, you will study conic sections, parametric equations, polar coordinates, and vectors. In Part V, you will meet limits, continuity, and rates of change — the ideas that calculus will develop fully.
| Part | Topics | Why It Matters |
|---|
| Part I | Functions, graphs, transformations, inverses | Foundation for everything that follows |
| Part II | Angles, unit circle, trig graphs, identities, triangles | Models rotation, waves, oscillation |
| Part III | Polynomial, rational, exponential, logarithmic, sequences | Models growth, decay, long-run behavior |
| Part IV | Conics, parametric equations, polar, vectors | Models motion, position, shape |
| Part V | Limits, continuity, rates of change, modeling | Bridge to Calculus I |
Why Functions?
You cannot take a derivative without understanding how functions behave. You cannot interpret an integral without understanding what accumulation means. Precalculus is not a detour — it is the essential foundation. Every chapter in this book is preparation for Calculus I and for the applied mathematics of physics, engineering, computing, and finance.
Section 1.2
Functions as Relationships
Definition — Function
A function is a relationship between two sets of quantities in which each input is paired with exactly one output. The word "exactly one" is the key. A function may produce different outputs for different inputs, or the same output for multiple inputs — but it may never produce two different outputs for the same input.
Example 1.2.1
Deciding whether a relationship is a function
- ProblemDetermine whether each relationship is a function. (a) Each student in a class assigned their GPA. (b) Each GPA assigned to all students who hold it. (c) Each real number x assigned the value x². (d) Each positive y assigned x such that x² = y.
- ThinkAsk: can one input value lead to two different outputs? If yes, it is not a function.
- Solve
(a) Each student has one GPA at a given time. → Function.
(b) One GPA may be held by many students. → Not a function from GPA to individual.
(c) Every x has exactly one square x². → Function. (x=3 and x=−3 both give 9; that is allowed.)
(d) If y = 9, then x could be 3 or −3. → Not a function (two outputs for one input).
- CheckReview each against the "exactly one output" rule. ✓
Answer: (a) Function. (b) Not a function. (c) Function. (d) Not a function as stated.
Four Representations of a Function
| Representation | Description | Example for f(x) = x² |
|---|
| Verbal | A description in words | "Assign to each input its square." |
| Algebraic | An equation or formula | f(x) = x² |
| Numerical | A table of input-output pairs | x: 1, 2, 3, 4 → f(x): 1, 4, 9, 16 |
| Graphical | A plot of all ordered pairs (x, f(x)) | The parabola opening upward with vertex at origin |
Vertical Line Test
A graph represents a function if and only if every vertical line intersects the graph at most once. If a vertical line crosses a graph at two points, there exists an input with two different outputs — violating the definition.
Graph Instruction: Draw the parabola y = x² and the circle x² + y² = 4. For the parabola, a vertical line at x = 2 touches only (2, 4): one output. For the circle, a vertical line at x = 1 touches (1, √3) and (1, −√3): two outputs. The parabola is a function; the circle is not.
- In your own words, explain the difference between arithmetic and algebra. Then explain why the word "function" takes that idea further.
- A vending machine gives exactly one item per valid selection. Is this a function? What is the input? What is the output?
- Determine whether each table could represent a function. Explain your reasoning.
Table A: x → y: (1,3), (2,5), (3,5), (4,8), (5,3)
Table B: x → y: (1,4), (2,7), (1,9), (3,2)
- Give an example from daily life of a relationship that is a function and one that is not. Explain what makes each one a function or not.
Sections 1.3 – 1.4
Function Notation, Domain, and Range
Reading f(x) · Evaluating at Expressions · Domain Restrictions · Range from Graphs
Section 1.3
Function Notation
When mathematicians write f(x), they are not multiplying f by x. They are naming a function f and indicating the output when the input is x. Read f(x) as "f of x." The notation f(x) = 2x + 3 defines a function named f with the rule: take an input, multiply it by 2, add 3.
| Expression | Meaning | Value when f(x) = 2x + 3 |
|---|
f(5) | Output when input is 5 | 2(5) + 3 = 13 |
f(−1) | Output when input is −1 | 2(−1) + 3 = 1 |
f(a) | Output when input is a | 2a + 3 |
f(x + h) | Output when input is x + h | 2(x + h) + 3 = 2x + 2h + 3 |
Example 1.3.1
Evaluating functions using notation
- ProblemLet g(x) = x² − 3x + 1. Find g(0), g(4), g(−2), and g(x + 1).
- ThinkReplace every instance of x in the formula with the given input. Use parentheses carefully, especially for expressions.
- Solve
g(0) = (0)² − 3(0) + 1 = 1
g(4) = (4)² − 3(4) + 1 = 16 − 12 + 1 = 5
g(−2) = (−2)² − 3(−2) + 1 = 4 + 6 + 1 = 11
g(x+1) = (x+1)² − 3(x+1) + 1
= x² + 2x + 1 − 3x − 3 + 1
= x² − x − 1
- CheckVerify g(4): 16 − 12 + 1 = 5. ✓
Answer: g(0) = 1, g(4) = 5, g(−2) = 11, g(x+1) = x² − x − 1.
Student Warning — f(x + h) ≠ f(x) + f(h)
A very common error is writing f(x + h) = f(x) + f(h). This is almost never true. For f(x) = x²: f(x + h) = (x + h)² = x² + 2xh + h², which is not the same as f(x) + f(h) = x² + h². Always substitute the entire expression x + h for x in the formula before simplifying.
Section 1.4
Domain and Range
The domain of a function is the set of all input values for which the function is defined. Two common algebraic sources of domain restrictions are division by zero (exclude values that make a denominator zero) and even roots of negative numbers (require the radicand to be non-negative). The range is the set of all output values the function can produce.
Example 1.4.1
Finding the domain of a function
- ProblemFind the domain of: (a) f(x) = 3x² − 7, (b) g(x) = 1/(x − 5), (c) h(x) = √(2x − 4), (d) k(x) = √(x + 3)/(x − 1).
- ThinkFor each function, identify what values of x cause division by zero or a square root of a negative number.
- Solve
(a) f(x) = 3x² − 7: no restrictions.
Domain: (−∞, ∞).
(b) g(x) = 1/(x − 5): x − 5 ≠ 0 → x ≠ 5.
Domain: (−∞, 5) ∪ (5, ∞).
(c) h(x) = √(2x − 4): 2x − 4 ≥ 0 → x ≥ 2.
Domain: [2, ∞).
(d) k(x) = √(x+3)/(x−1): x+3 ≥ 0 → x ≥ −3,
and x − 1 ≠ 0 → x ≠ 1.
Domain: [−3, 1) ∪ (1, ∞).
- CheckFor (d), test x = 0: k(0) = √3/(−1) = −√3. Defined. ✓ Test x = 1: denominator = 0. Not defined. ✓
Answer: (a) All reals. (b) x ≠ 5. (c) [2, ∞). (d) [−3, 1) ∪ (1, ∞).
Reading Domain and Range from a Graph
On a graph, the domain is the set of all x-values where the graph has a point (read left to right). The range is the set of all y-values where the graph has a point (read bottom to top). A filled dot means the endpoint is included; an open circle means it is excluded.
f(x) = 5x − 2
Find f(0), f(3), f(−4), f(a + 1)g(x) = x² + x − 6
Find g(0), g(2), g(−3), g(x + h)h(x) = (x+1)/(x−3)
Find h(0), h(5). State the domain.p(x) = √(3x + 9)
Find p(0), p(4). State the domain.- Domain of f(x) = 7x − 4
- Domain of g(x) = 1/(x² − 4)
- Domain of h(x) = √(5 − 2x)
- Domain of f(x) = √(x + 1)/(x − 4)
For domain problems, write answers in interval notation.
- For f(x) = 4x² − 1, find and simplify the difference quotient: [f(x + h) − f(x)] / h.
- For g(x) = 3x², find [g(x + h) − g(x)] / h and simplify. What does the result approach as h → 0?
Sections 1.5 – 1.8
Graphs, Intercepts, Intervals, and Rate of Change
Reading Graphs · Zeros · Increasing/Decreasing · Average Rate of Change · Difference Quotient
Section 1.5
Graphs of Functions
A graph is a visual representation of a function. Rather than seeing one input-output pair at a time, you see the entire behavior at once: where it is large or small, where it rises or falls, where it crosses the axes, and where it is undefined. Every point on the graph has coordinates (x, f(x)) — the horizontal coordinate is the input; the vertical coordinate is the output.
When you look at a graph, ask these questions in order: What is the domain? What is the range? Where does the graph cross the axes? Where does it rise and fall? Where does it reach a high or low point? Are there any gaps or breaks?
Example 1.5.1
Reading a graph completely
- GraphA smooth curve that starts at (−3, 0) with a filled dot, rises to a local maximum at (−1, 4), falls to a local minimum at (2, −2), then rises again, ending at (4, 3) with an open circle. No breaks.
- ThinkRead left to right for domain; read vertically for range; note filled vs. open circles at endpoints.
- Solve
Domain: [−3, 4) (x = −3 included; x = 4 excluded)
Range: [−2, 4] (lowest y = −2; highest y = 4)
x-intercept: (−3, 0)
y-intercept: read the graph near x = 0 — between the peak and valley
Increasing on: (−3, −1) and (2, 4)
Decreasing on: (−1, 2)
Local max: (−1, 4) Local min: (2, −2)
Answer: Domain [−3, 4); Range [−2, 4]; Increasing on (−3,−1) and (2,4); Max at (−1, 4).
Section 1.6
Intercepts and Zeros
The x-intercepts of a function are points where the graph crosses the horizontal axis. At those points y = 0, so the input values are called zeros of the function. To find zeros algebraically, set f(x) = 0 and solve. The y-intercept is where the graph crosses the vertical axis; find it by evaluating f(0). A function has at most one y-intercept but may have zero, one, or many x-intercepts.
Example 1.6.1
Finding intercepts and zeros
- ProblemFind the x-intercepts, zeros, and y-intercept of f(x) = x² − 5x + 6.
- ThinkFor the y-intercept, evaluate f(0). For zeros, solve f(x) = 0 by factoring.
- Solve
y-intercept: f(0) = 0 − 0 + 6 = 6 → point (0, 6)
Zeros: x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
x = 2 or x = 3
- Checkf(2) = 4 − 10 + 6 = 0 ✓. f(3) = 9 − 15 + 6 = 0 ✓.
Answer: Zeros at x = 2 and x = 3; x-intercepts (2, 0) and (3, 0); y-intercept (0, 6).
Section 1.7
Increasing, Decreasing, and Constant Intervals
Formal Definitions
Increasing on I: For all x₁ < x₂ in I, f(x₁) < f(x₂). The graph rises left to right.
Decreasing on I: For all x₁ < x₂ in I, f(x₁) > f(x₂). The graph falls left to right.
Constant on I: For all x₁, x₂ in I, f(x₁) = f(x₂). The graph is horizontal.
Always state these intervals as open intervals of x-values, not y-values. Use open parentheses at the endpoints where the function transitions between increasing and decreasing.
Student Warning — Use x-values, Not y-values
Write "increasing on (−2, 1)" using x-values. Never write "increasing on y ∈ [3, 7]." The interval describes where along the input axis the function is rising — not what the output values are. This is a very common error.
Example 1.7.1
Identifying intervals of increase and decrease
- GraphA function decreasing from the left to a local minimum at (−2, −5), then increasing to a local maximum at (1, 3), then decreasing toward +∞.
- Solve
Decreasing on: (−∞, −2) and (1, ∞)
Increasing on: (−2, 1)
Local min: (−2, −5) Local max: (1, 3)
Answer: Increasing on (−2, 1); decreasing on (−∞, −2) and (1, ∞).
Section 1.8
Average Rate of Change
The average rate of change of f from x = a to x = b measures how much the output changes per unit of input over that interval. It equals the slope of the line through (a, f(a)) and (b, f(b)) — called a secant line.
Average Rate of Change Formula
Average rate of change = [f(b) − f(a)] / (b − a)
Units: (output units) per (input unit). When time is the input and distance is the output, this is average speed. When input is time and output is population, it is average growth rate.
Example 1.8.1
Computing average rate of change
- ProblemLet f(x) = x² + 2x. Find the average rate of change from x = 1 to x = 4.
- Solve
f(1) = 1 + 2 = 3
f(4) = 16 + 8 = 24
Average rate of change = (24 − 3) / (4 − 1) = 21 / 3 = 7
- CheckThe function is increasing and accelerating on this interval, so a rate of 7 is plausible. ✓
Answer: The average rate of change from x = 1 to x = 4 is 7.
Example 1.8.2
The difference quotient — average rate of change in general form
- ProblemLet f(x) = 3x². Find the average rate of change from x to x + h, where h ≠ 0.
- ThinkThis is the difference quotient — the key expression in calculus. Substitute x and x + h into the formula, then simplify and factor out h.
- Solve
f(x + h) = 3(x + h)² = 3(x² + 2xh + h²) = 3x² + 6xh + 3h²
f(x) = 3x²
[f(x+h) − f(x)] / h = (3x² + 6xh + 3h² − 3x²) / h
= (6xh + 3h²) / h
= h(6x + 3h) / h
= 6x + 3h
- CheckAs h approaches 0, the expression approaches 6x — the derivative of 3x² in Calculus I. ✓
Answer: The average rate of change from x to x + h is 6x + 3h.
Bridge to Calculus I
The difference quotient [f(x+h) − f(x)] / h is the central expression that leads to the derivative. As h shrinks toward zero, the average rate of change over a shorter and shorter interval approaches the instantaneous rate of change — the slope of the tangent line. You will study this formally in Calculus I.
- Find zeros and y-intercept of f(x) = x² − x − 6
- Find zeros and y-intercept of g(x) = 2x² + x − 3
- State increasing/decreasing intervals for a function with local min at (−1, −3) and local max at (2, 5).
- Average rate of change of f(x) = x³ from x = −1 to x = 2
- Average rate of change of g(x) = 2x + 1 from x = 0 to x = 4
- Average rate of change of h(x) = x² from x = 3 to x = 5
- Simplify: [f(x+h) − f(x)] / h for f(x) = 2x² − x
- Simplify: [f(x+h) − f(x)] / h for f(x) = x² + 3x + 1
Sections 1.9 – 1.12
Piecewise Functions, Models, Representations, and Technology
Piecewise Functions · Absolute Value · Modeling · Comparing Representations · Calculator Use
Section 1.9
Piecewise Functions
Some functions use different formulas on different parts of their domain. These are called piecewise functions. They arise naturally wherever a rule changes at a threshold: tax brackets, shipping rates, subscription tiers, or the absolute value function itself.
Example 1.9.1
Evaluating a piecewise function
- ProblemLet f(x) = { 2x + 1 if x < 0, and x² − 1 if x ≥ 0 }. Find f(−3), f(0), and f(4).
- ThinkDetermine which rule applies by checking the condition for each input.
- Solve
f(−3): since −3 < 0, use 2x + 1. f(−3) = 2(−3) + 1 = −5.
f(0): since 0 ≥ 0, use x² − 1. f(0) = 0 − 1 = −1.
f(4): since 4 ≥ 0, use x² − 1. f(4) = 16 − 1 = 15.
- CheckEach input was matched to the correct piece. ✓
Answer: f(−3) = −5, f(0) = −1, f(4) = 15.
Example 1.9.2
Graphing a piecewise function with three pieces
- ProblemGraph f(x) = { −x if x < −1 ; 2 if −1 ≤ x < 2 ; x − 1 if x ≥ 2 }.
- ThinkGraph each piece on its domain interval. Pay careful attention to open and closed endpoints.
- Graph
Piece 1 (x < −1): Ray y = −x, ending at (−1, 1) with an open circle.
Piece 2 (−1 ≤ x < 2): Horizontal segment y = 2 from (−1, 2) [closed] to (2, 2) [open].
Piece 3 (x ≥ 2): Ray y = x − 1, starting at (2, 1) [closed] and rising.
- InterpretThere are jump discontinuities at x = −1 (graph jumps from y → 1 to y = 2) and at x = 2 (graph drops from y → 2 to y = 1). You will study discontinuities formally in Chapter 13.
Answer: Three-piece graph with jumps at x = −1 and x = 2. Domain: all real numbers. Range: (1, ∞) together with y = 2 on [−1, 2).
Absolute Value as a Piecewise Function
The absolute value |x| = { x if x ≥ 0, and −x if x < 0 } is the most familiar piecewise function. Its graph is a V-shape with vertex at the origin, slope 1 on the right and slope −1 on the left. Every transformation you study in Chapter 2 applies to it.
Section 1.10
Function Models in Real Situations
Problem-Solving Method
Read → Represent → Analyze → Solve → Check → Explain
Always define variables with descriptions and units. Write the equation. Use it to answer the question. Interpret the result in context — a model that ends with a number but no explanation of what it means has not finished the job.
Example 1.10.1
Building and using a piecewise linear model
- SituationA cell phone data plan costs $30/month for the first 5 GB, plus $10 for each additional GB above 5. Write C(d) for monthly cost in dollars as a function of data used d in GB, for d ≥ 0.
- Represent
C(d) = 30 if 0 ≤ d ≤ 5
C(d) = 30 + 10(d−5) if d > 5
= 10d − 20 if d > 5
- SolveHow much for 8 GB? Since 8 > 5: C(8) = 10(8) − 20 = 60.
- Check8 GB = 5 base GB + 3 extra GB at $10 each. $30 + $30 = $60. ✓
Answer: A customer using 8 GB pays $60 that month.
Example 1.10.2
Building a quadratic model — projectile height
- SituationA ball thrown upward from 4 feet has height h(t) = −16t² + 48t + 4 feet, where t is seconds after the throw. Find: (a) initial height, (b) maximum height, (c) when it lands.
- ThinkThis is a downward-opening parabola. Maximum occurs at the vertex. The ball lands when h(t) = 0.
- Solve
(a) h(0) = 4 feet. ✓ (Confirms initial height.)
(b) Vertex at t = −b/(2a) = −48/(2·(−16)) = 48/32 = 1.5 sec
h(1.5) = −16(2.25) + 48(1.5) + 4 = −36 + 72 + 4 = 40 feet.
(c) Set h(t) = 0: −16t² + 48t + 4 = 0
Divide by −4: 4t² − 12t − 1 = 0
t = [12 ± √(144 + 16)] / 8 = [12 ± √160] / 8
= [12 ± 4√10] / 8 = [3 ± √10] / 2
Positive solution: t ≈ (3 + 3.162)/2 ≈ 3.08 seconds.
- CheckBall starts at 4 ft, peaks at 40 ft, lands at ≈ 3.08 sec. Physically reasonable. ✓
Answer: Starts at 4 feet. Maximum height 40 feet at t = 1.5 sec. Lands at approximately t ≈ 3.08 seconds.
Section 1.11
Comparing Function Representations
Every function can be described verbally, algebraically, numerically, and graphically. Each representation reveals different information most easily. A skilled precalculus student moves fluidly among all four. Consider the function: "The output is three less than twice the input."
| Representation | Form | What It Shows Best |
|---|
| Verbal | "Three less than twice the input" | Communicating meaning to others |
| Algebraic | f(x) = 2x − 3 | Exact computation, simplification, analysis |
| Numerical | x: −2, 0, 1, 3, 5 → f(x): −7, −3, −1, 3, 7 | Comparing specific values; revealing patterns |
| Graphical | Line with slope 2 and y-intercept (0, −3) | Overall shape, intercepts, direction, behavior |
Example 1.11.1
Representing a real model in all four ways
- SituationA parking garage charges $5 to enter and $2 per hour. Represent cost C as a function of time t (hours) in all four ways.
- VerbalThe total cost starts at $5 and increases by $2 for every hour parked.
- Algebraic
C(t) = 2t + 5, for t ≥ 0.
- Numerical
t = 0 hr → C = $5
t = 1 hr → C = $7
t = 2 hr → C = $9
t = 3 hr → C = $11
t = 5 hr → C = $15
- GraphicalRay starting at (0, 5) with slope 2, extending right. Label axes: "Hours" (horizontal), "Cost in dollars" (vertical). The y-intercept (0, 5) represents the $5 entry fee.
Answer: Four complete representations shown. All describe the same relationship: C(t) = 2t + 5.
Section 1.12
Calculator and Graphing Technology Awareness
Graphing calculators and computer tools like Desmos are powerful for exploring function behavior, checking algebraic work, and building models. But they are tools, not substitutes for reasoning. Technology cannot tell you why a function has a certain domain, or what a zero means in a real situation, or how to set up a model. You need to understand the mathematics first.
| Technology Does Well | Technology Does Not Do for You |
|---|
| Graphs functions quickly | Explain why a function has a domain restriction |
| Estimates zeros, maxima, minima numerically | Tell you what a zero means in a real situation |
| Builds tables of values | Choose the right model for a data set |
| Fits regression curves to data | Simplify or manipulate algebraic expressions |
| Checks algebraic work with a numerical test | Recognize whether a window setting is appropriate |
Window Settings Matter
A graphing tool only shows the portion of the graph inside its current window. If the window is too small, you miss behavior; too large, and the graph appears flat. Example: Graphing f(x) = x³ − 10x in a window of [−2, 2] by [−5, 5] may not show all three zeros and both turning points. Try [−5, 5] by [−15, 15] for a complete picture.
- Evaluate: f(x) = { x + 4 if x < 0 ; x² if x ≥ 0 }. Find f(−3), f(0), f(5).
- Graph: g(x) = { 3 if x < 1 ; 2x − 1 if x ≥ 1 }. Is there a jump at x = 1?
- A gym membership costs $50/month. Write C(m) for total cost after m months. Find C(12). What is the average rate of change over the year?
- A car rental charges $40/day plus $15 fee. Write C(d). Find C(1), C(3), C(7). How many days can you rent for $175?
- h(t) = −16t² + 64t + 6 models the height (feet) of a projectile. Find: (a) initial height, (b) max height and when it occurs, (c) when it lands.
- Write a function modeling a shipping cost: $8 for first pound, $3 each additional pound, for w ≥ 1.
Sections 1.13 – 1.14
Common Mistakes and Preparing for Chapter 2
Seven Recurring Errors · Parent Functions Preview · Readiness Check
Section 1.13
Common Function Mistakes
Mistake 1 — Confusing f(x) with multiplication
f(x) = f · x, so f(3) = f · 3
f(x) is function notation, not multiplication. It means "the output of function f at input x." There is no multiplication symbol here.
Mistake 2 — Writing f(x + h) = f(x) + f(h)
f(x + h) = f(x) + h or f(x) + f(h)
Substitute x + h everywhere x appears in the formula and expand. For f(x) = x²: f(x + h) = (x + h)² = x² + 2xh + h². This is not x² + h².
Mistake 3 — Including values that cause division by zero
"The domain of 1/(x − 2) includes x = 2."
x = 2 makes the denominator zero, which is undefined. Domain: (−∞, 2) ∪ (2, ∞).
Mistake 4 — Wrong radicand condition for domain
"Domain of √(x² − 4) is x ≥ 0."
Require x² − 4 ≥ 0, so x² ≥ 4, giving x ≤ −2 or x ≥ 2. Domain: (−∞, −2] ∪ [2, ∞).
Mistake 5 — Stating intervals in terms of y-values
"The function is increasing on y ∈ [3, 7]."
Increasing and decreasing intervals are stated using x-values. "The function is increasing on x ∈ (1, 4)."
Mistake 6 — Confusing local and global maxima
"The highest point on the graph is the local maximum at (2, 5)."
A local maximum is a peak relative to nearby values only. The function could reach higher values elsewhere. Clarify "local" vs. "absolute" (global).
Mistake 7 — Using a closed bracket at an excluded endpoint
"Domain of 1/(x − 3) is (−∞, 3]."
Since x = 3 is excluded, use an open bracket: (−∞, 3) ∪ (3, ∞). A closed bracket [ means the endpoint is included in the domain.
Section 1.14
Preparing for Transformations and Parent Functions
Chapter 2 will build on everything you have studied here. Instead of studying individual functions, you will study families of functions and how they are related through transformations. A parent function is the simplest, most basic member of a function family.
| Parent Function | Equation | Graph Shape |
|---|
| Linear | f(x) = x | Line through origin with slope 1 |
| Quadratic | f(x) = x² | Parabola opening upward, vertex at origin |
| Cubic | f(x) = x³ | S-shaped curve through origin |
| Square Root | f(x) = √x | Starts at origin, increases to the right |
| Absolute Value | f(x) = |x| | V-shape with vertex at origin |
| Reciprocal | f(x) = 1/x | Two branches; never touches axes |
From each parent function, you build an entire family by shifting, stretching, compressing, and reflecting the graph. For example, f(x) = (x − 3)² + 2 is the parent quadratic shifted right 3 units and up 2 units, with vertex at (3, 2) instead of (0, 0). Understanding transformations will allow you to study trigonometric, rational, and exponential functions far more efficiently.
- A student writes "f(x + 2) = f(x) + f(2)." For f(x) = x², show by example that this is wrong.
- A student finds the domain of √(x² − 9) as "x ≥ 0." Find the correct domain.
- A student says "g is decreasing on y ∈ [−2, 4]." Rewrite correctly using x-values if the function decreases on the interval from x = 1 to x = 5.
- Evaluate f(−3) and f(0) and f(3) for the parent functions f(x) = x², f(x) = x³, f(x) = |x|, and f(x) = √x (where defined). Organize your results in a table.
- For f(x) = x², predict what happens to the graph if you replace f(x) with f(x − 3). Check by evaluating both at x = 3 and x = 5.
Chapter 1 — Back Matter
Chapter Summary
- Sec 1.1Precalculus is the systematic study of functions and their behavior as preparation for calculus, physics, engineering, and applied mathematics across all five parts of this book.
- Sec 1.2A function assigns exactly one output to each input. Functions can be represented verbally, algebraically, numerically, and graphically. The Vertical Line Test confirms whether a graph represents a function.
- Sec 1.3f(x) is function notation, not multiplication. Evaluating f(x + h) requires substituting the entire expression x + h before simplifying. f(x + h) ≠ f(x) + f(h).
- Sec 1.4Domain is the set of allowed inputs; range is the set of possible outputs. Domain restrictions arise from division by zero and even roots of negatives. Read domain and range from graphs using endpoint circles.
- Sec 1.5A graph shows domain, range, intercepts, increasing and decreasing intervals, and local maxima and minima all at once. Systematic graph reading is an essential precalculus skill.
- Sec 1.6Zeros of f are inputs where f(x) = 0; they correspond to x-intercepts. The y-intercept is found from f(0). A function has at most one y-intercept but may have many x-intercepts.
- Sec 1.7Increasing and decreasing intervals are stated in x-value open intervals. A local maximum is a peak relative to nearby values; a local minimum is a valley. Neither is necessarily the global extreme.
- Sec 1.8Average rate of change = [f(b) − f(a)] / (b − a) = slope of the secant line. The difference quotient [f(x+h) − f(x)] / h generalizes this and prepares directly for the derivative.
- Sec 1.9Piecewise functions use different rules on different domain intervals. Evaluate by matching each input to its correct rule. Graphs may have jumps at the boundaries between pieces.
- Sec 1.10Mathematical models represent real situations with defined variables, equations, and contextual interpretation. Always use Read → Represent → Analyze → Solve → Check → Explain.
- Sec 1.11Verbal, algebraic, numerical, and graphical representations each reveal different aspects of a function. Moving among all four is an essential precalculus competency.
- Sec 1.12Technology supports exploration and checking but does not replace reasoning. Use appropriate windows. Always understand the mathematics before using a graph or table from technology.
- Sec 1.13Seven common errors: misreading f(x), splitting f(x+h), including excluded domain values, incorrect radicand conditions, using y-values for intervals, confusing local and global extrema, wrong bracket types.
- Sec 1.14Chapter 2 will study parent functions and transformations — shifts, reflections, stretches, and compressions — organizing all function families into a unified, predictable system.
Key Terms Review
function
A rule that assigns exactly one output value to each input value.
domain
The set of all allowed input values for a function.
range
The set of all possible output values of a function.
function notation
f(x) notation indicating the output at input x. Not multiplication.
Vertical Line Test
A graph represents a function if no vertical line intersects it more than once.
zero
An input value where f(x) = 0. Corresponds to an x-intercept.
x-intercept
A point where the graph crosses the x-axis; a zero of the function.
y-intercept
The point where the graph crosses the y-axis, found by evaluating f(0).
increasing interval
An x-interval where outputs rise as inputs increase. Stated as open x-intervals.
decreasing interval
An x-interval where outputs fall as inputs increase.
average rate of change
[f(b) − f(a)] / (b − a); the slope of the secant line through (a, f(a)) and (b, f(b)).
difference quotient
[f(x+h) − f(x)] / h; generalized average rate of change; bridges to the derivative.
piecewise function
A function defined by different rules on different parts of its domain.
mathematical model
A function, equation, or graph representing a real situation with defined variables.
Skill Practice
f(x) = 5x − 2
Find f(0), f(3), f(−4), f(a+1)g(x) = x² + x − 6
Find g(0), g(2), g(−3), g(x+h)h(x) = (x+1)/(x−3)
Find h(0), h(5), h(−1). State domain.p(x) = √(3x + 9)
Find p(0), p(4). State domain.q(x) = |2x − 5|
Find q(0), q(2.5), q(5), q(−1)- Simplify [f(x+h)−f(x)]/h for f(x) = 4x² − 1
Find the domain of each function. Write answers in interval notation.
f(x) = 7x − 4g(x) = x² − 9h(x) = (x+2)/(x−6)f(x) = √(x − 7)g(x) = 1/(x² − 4)h(x) = √(5 − 2x)f(x) = √(x+1)/(x−4)g(x) = 3x/(x² − x − 12)
- Zeros and y-int of x² − x − 6
- Zeros and y-int of 2x² + x − 3
- Average rate of change of x³ from x = −1 to x = 2
- Average rate of change of x² from x = 3 to x = 5
- Simplify [f(x+h)−f(x)]/h for f(x) = x² + 3x + 1
- Simplify [f(x+h)−f(x)]/h for f(x) = 2x² − x
Word Problems and Modeling
Define your variable(s), write the function model, and answer the specific question. Interpret results in context.
- Temperature Model. The temperature (°F) in a city on a summer day is modeled by T(t) = −0.5t² + 8t + 65, where t is hours after midnight, 0 ≤ t ≤ 24. (a) Find the temperature at midnight. (b) Find the temperature at noon (t = 12). (c) Find when the temperature is highest and the maximum temperature. (d) Evaluate T(20) and interpret.
- Shipping Model. A shipping company charges $8 for the first pound and $3 for each additional pound. (a) Write C(w) for w ≥ 1. (b) Find C(1), C(5), C(10). (c) For how many pounds can you ship for exactly $23?
- Drug Concentration. A medication concentration is modeled by C(t) = 12t / (t² + 4), where C is in mcg/mL and t is hours after taking the medication, t ≥ 0. (a) Find C(0), C(2), C(4). (b) Use technology to estimate when concentration is greatest. (c) What appears to happen to C as t grows very large?
- Distance and Time. A car travels at 65 mph. (a) Write d(t) for distance in t hours. (b) What is the average rate of change over any 2-hour interval? (c) What does the slope represent?
- Profit Model. A store sells candles for $18 each. Production cost: C(n) = 200 + 6n for n candles. (a) Write revenue R(n). (b) Write profit P(n) = R(n) − C(n). (c) Find the break-even quantity. (d) Find profit for 50 candles sold.
- Candle Height. A candle is 12 inches tall and burns at 3/4 inch per hour. (a) Write h(t) for height after t hours. (b) State the domain in context. (c) When is it burned out? (d) Find average rate of change from t = 2 to t = 6 and interpret.
Error Analysis
Each problem shows a student's work. Identify the error and write the correct solution.
Problem 1
For f(x) = 3x + 1: f(x + 2) = f(x) + f(2) = (3x + 1) + 7 = 3x + 8
What is the error? What is f(x + 2) correctly?
Problem 2
Domain of √(x² − 9) is x ≥ 0, because you need the radicand positive.
What condition is correct? Find the correct domain in interval notation.
Problem 3
f(x) = (x+4)/(x−4) is undefined at x = −4 because the numerator becomes zero.
Identify the error. At what value is f actually undefined?
Problem 4
Average rate of change of f(x) = x² from x = 1 to x = 3 is f(3)/f(1) = 9/1 = 9.
What is the correct formula for average rate of change? Compute the correct value.
Problem 5
"The function is increasing on y ∈ [2, 8] because the y-values go up from 2 to 8."
Rewrite this statement correctly, assuming the function increases from x = 1 to x = 4.
Problem 6
For f(x) = { 2x if x ≤ 3 ; x² − 1 if x > 3 }: f(3) = 3² − 1 = 8 because 3 is close to the second piece.
Which piece applies at x = 3? What is the correct value of f(3)?
Chapter 1 Checkpoint
Chapter 1 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 1
Part A — Function Identification
- Does the equation x² + y² = 25 define y as a function of x? Explain using the Vertical Line Test.
- Data pairs: (2, 5), (4, 7), (6, 5), (8, 3). Does this represent a function? Explain.
Part B — Notation and Evaluation
- Let
f(x) = 2x² − 3x + 1. Find f(−2), f(a), and simplify [f(x+h) − f(x)] / h.
Part C — Domain
- Find the domain in interval notation:
f(x) = (x − 1) / (x² − 9)
- Find the domain in interval notation:
g(x) = √(8 − 2x)
Part D — Graph Analysis
- A function has y-intercept (0, 3), decreases on (−∞, 2), has a minimum at (2, −1), and increases on (2, ∞). (a) What is the range? (b) Sketch a possible graph. (c) Is f positive or negative near x = 0?
Part E — Average Rate of Change
- Let
g(x) = x³ − x. (a) Find average rate of change from x = 0 to x = 2. (b) Find average rate of change from x = −1 to x = 1. (c) Interpret part (b) in a sentence.
Part F — Piecewise Functions
- Let
f(x) = { x + 4 if x < −1 ; −2 if −1 ≤ x ≤ 2 ; x² − 3 if x > 2 }. (a) Find f(−4), f(−1), f(0), f(2), f(3). (b) What is the domain? (c) Is there a jump at x = −1?
Part G — Modeling
- A candle is 12 inches tall and burns at 3/4 inch per hour. (a) Write h(t) for height. (b) State the domain in context. (c) When does it burn out? (d) Find average rate of change from t = 2 to t = 6.
Part H — Error Identification
- A student says: "The range of f(x) = x² + 3 is all real numbers, because x can be any real number." Identify and correct the error.
- A student evaluates
f(x) = −x² + 4 at x = −2 and gets −8. Find and explain the error. What is correct?
Looking Ahead
Chapter 2: Transformations, Inverses, and Function Operations
Chapter 1 has established the function concept as the central organizing idea of precalculus. You can define, evaluate, graph, and model functions. You can identify domain, range, intercepts, intervals, and rate of change.
Chapter 2 will use this foundation to study how functions are related to one another through transformations. You will learn that f(x) = (x − 3)² + 4 is the parabola y = x² shifted right 3 and up 4. You will learn that f(x) = −|x| is the reflection of y = |x| across the x-axis. Instead of studying every function from scratch, you will learn a small set of parent functions and a set of transformation rules — one of the most powerful organizing ideas in precalculus. Chapter 2 will then turn to inverse functions, which reverse input-output relationships, and to operations on functions including composition.
MGU Library Connection
- Chapter 2: Transformations, Inverses, and Function Operations — continues directly from this chapter
- Appendix A: Algebra II Readiness Review — linear, quadratic, exponential functions from Algebra II
- Appendix C: Function Parent Graphs Reference — visual reference for all parent functions
- Appendix D: Transformation Rules Reference — shift, reflection, stretch, compression rules
- Appendix E: Domain and Range Reference — procedures and examples by function type
- MGU Dictionary entries: function, domain, range, average rate of change, difference quotient, piecewise function