Every later week of this course speaks the language set up here. A limit is a statement about a function's values; a derivative is defined as a limit of a function built by combining and transforming other functions; and half of the standard examples in every chapter are trigonometric. Time spent here is not review for its own sake — it is the vocabulary the rest of the course assumes you already have.
1.1 Why Functions First
Calculus answers two questions: how fast is something changing right now (the derivative), and how much of something accumulates over an interval (the integral). Both questions are meaningless until you can say precisely what "something" is — a rule that assigns one output to each input. That rule is a function, and everything in this course is a statement about functions: their limits, their derivatives, their integrals, their series expansions.
This week has two halves. The first (§1.2–1.9 below) is about functions in general: what they are, how to build new ones from old ones, and how their graphs move when you shift or scale them. The second (§1.10–1.13) specialises to the trigonometric functions, which reappear in almost every worked example for the rest of the semester. A short, non-examinable detour into mathematical induction closes the week, because it is useful for reading proofs later even though you will not be asked to write one on an exam.
1.2 What Is a Function?
A function f from a set D to a set Y is a rule that assigns a single element f(x) ∈ Y to each element x ∈ D. The set D is the domain; the set of all output values actually produced is the range. The variable x is called the independent variable and y = f(x) the dependent variable.
Unless a domain is stated explicitly, the natural domain of a function given by a formula is the largest set of real x for which the formula produces a real number. Two restrictions account for almost every natural-domain calculation in this course:
- You cannot divide by zero, so any x that makes a denominator vanish is excluded.
- You cannot take a real square root (or any even root) of a negative number, so the expression under an even root must be ≥ 0.
Example 1.1 — Finding a natural domain
Find the domain of f(x) = √(x + 2)x − 3.
Step 1 — The square root needs a non-negative radicand
x + 2 ≥ 0 ⇒ x ≥ −2
Step 2 — The denominator cannot be zero
x − 3 ≠ 0 ⇒ x ≠ 3
Step 3 — Combine the two conditions
Domain: [−2, 3) ∪ (3, ∞)
A curve in the xy-plane is the graph of some function of x exactly when every vertical line meets it at most once — the vertical line test. This is just a restatement of the definition: if a vertical line x = a met the curve twice, the rule would have to assign two different outputs to the single input a, which a function is not allowed to do. A circle therefore fails the test and is not the graph of a function of x, even though its upper and lower halves separately are.
1.3 Piecewise Functions and the Absolute Value Function
Some functions are defined by different formulas on different parts of the domain. These piecewise-defined functions are not exotic — the absolute value, the single most useful piecewise function in the course, is one of them:
Geometrically |x| is the distance from x to 0 on the number line, and more generally |a − b| is the distance between a and b. This distance reading is exactly what you will need in Week 2 to make sense of the precise definition of a limit, where a statement like |x − a| < δ says "x is within δ of a."
Example 1.2 — Graphing a piecewise function
Sketch f(x) = x2 for x < 1, and f(x) = 2x − 1 for x ≥ 1.
Step 1 — Graph each piece only on its own interval
Draw y = x2 but keep only the part with x < 1; draw y = 2x − 1 but keep only the part with x ≥ 1.
Step 2 — Check what happens at the join, x = 1
The parabola piece approaches (1, 1) as x → 1− but does not include it (open circle). The line piece gives f(1) = 2(1) − 1 = 1 (solid dot).
The two pieces meet exactly at (1, 1), so the graph has no visible break even though it is defined by two different formulas — a preview of continuity, Week 3.
1.4 Increasing, Decreasing, Even and Odd
A function f is increasing on an interval I if x1 < x2 in I always gives f(x1) < f(x2), and decreasing if it always gives f(x1) > f(x2). These are the everyday words "the graph goes up" and "the graph goes down," made precise enough to prove things with — and precise enough that Week 6 can characterise them exactly using the sign of the derivative.
A function is even if f(−x) = f(x) for every x in the domain, and odd if f(−x) = −f(x). Symmetry is worth checking before you graph anything, because it halves the work:
- An even function's graph is symmetric about the y-axis (fold the paper along the y-axis and the two halves match).
- An odd function's graph is symmetric about the origin (rotate the graph 180° about the origin and it lands back on itself).
Example 1.3 — Testing for symmetry
Classify g(x) = x3 − x as even, odd, or neither.
Step 1 — Compute g(−x)
g(−x) = (−x)3 − (−x) = −x3 + x = −(x3 − x)
Step 2 — Compare with ±g(x)
g(−x) = −g(x), so the odd condition holds for every x.
g is odd; its graph is symmetric about the origin.
1.5 A Catalog of Essential Functions
Thomas' Calculus groups the functions you will meet all semester into a short list of families. You do not need to memorise this table, but you should recognise a function's family on sight, because each family behaves differently near infinity, near zero, and under differentiation.
| Family | General form | Example |
|---|---|---|
| Linear | f(x) = mx + b | y = 2x − 1 |
| Power | f(x) = xa | y = √x, y = x−1 |
| Polynomial | f(x) = anxn + … + a0 | y = x3 − x |
| Rational | ratio of two polynomials | y = (x+2)/(x2−1) |
| Algebraic | built from roots, powers, and the four operations | y = √(x2−1) |
| Trigonometric | — | y = sin x, y = tan x |
| Exponential | f(x) = ax | y = 2x |
| Logarithmic | f(x) = logax | y = ln x |
Exponential and logarithmic functions are introduced properly in Week 11 (§7.2–7.3), once derivatives are available to define them rigorously; for now, recognising them by shape is enough.
1.6 Combining Functions Algebraically
If f and g are functions, new functions can be built pointwise:
Each combination is defined only where both f(x) and g(x) are defined; the quotient additionally requires g(x) ≠ 0. In symbols, the domain of f + g is Df ∩ Dg — the intersection, not the union. This trips people up more than any other rule in this section: adding two functions can only ever shrink the domain, never grow it.
Example 1.4 — Domain of a sum
Let f(x) = √x and g(x) = √(2−x). Find the domain of f + g.
Step 1 — Domain of each piece
Df = [0, ∞); Dg = (−∞, 2]
Step 2 — Intersect
Domain of f + g: [0, 2]
1.7 Composite Functions
If f and g are functions, the composite f ∘ g ("f of g") is defined by
Read this from the inside out: first apply g to x, then apply f to whatever g produced. The domain of f ∘ g consists of every x in the domain of g for which g(x) also lies in the domain of f — two conditions, not one, and both must hold.
f ∘ g and g ∘ f are generally different functions with different domains. If f(x) = √x and g(x) = x − 1, then (f ∘ g)(x) = √(x−1) has domain [1, ∞), while (g ∘ f)(x) = √x − 1 has domain [0, ∞). Always compute the domain of a composite directly from the definition; never assume it matches either piece's domain alone.
Example 1.5 — Domain of a composite
Let f(x) = √x and g(x) = 4 − x2. Find the domain of f ∘ g.
Step 1 — Write down what the composite needs
(f ∘ g)(x) = √(4−x2) is only real when the radicand is non-negative.
Step 2 — Solve the inequality
4 − x2 ≥ 0 ⇒ x2 ≤ 4 ⇒ −2 ≤ x ≤ 2
Domain of f ∘ g: [−2, 2]
1.8 Shifting a Graph
Adding a constant inside or outside a function's formula moves its graph without changing its shape:
| To obtain | From y = f(x), do this |
|---|---|
| y = f(x) + k | Shift the graph up k units (k > 0) |
| y = f(x) − k | Shift the graph down k units (k > 0) |
| y = f(x + h) | Shift the graph left h units (h > 0) |
| y = f(x − h) | Shift the graph right h units (h > 0) |
The horizontal rule feels backwards the first time you meet it: f(x − 2) shifts right, even though the sign is negative. The reason is that the new graph reaches the old height f(0) when x − 2 = 0, i.e. x = 2 — two units to the right of where the old graph reached that height. Substitute a specific number if you ever doubt the direction.
1.9 Scaling and Reflecting a Graph
Multiplying by a constant stretches, compresses, or flips a graph:
| Formula | Effect on y = f(x), c > 1 |
|---|---|
| y = c·f(x) | Stretches the graph vertically by a factor of c |
| y = (1/c)·f(x) | Compresses the graph vertically by a factor of c |
| y = f(cx) | Compresses the graph horizontally by a factor of c |
| y = f(x/c) | Stretches the graph horizontally by a factor of c |
| y = −f(x) | Reflects the graph across the x-axis |
| y = f(−x) | Reflects the graph across the y-axis |
As with shifting, the horizontal rules feel inverted compared to the vertical ones: multiplying x by c > 1 compresses the graph, because the same output now appears at a smaller input. Whenever a problem combines several transformations, Thomas' Calculus recommends applying them in the order: horizontal shift, horizontal scale/reflect, vertical scale/reflect, vertical shift — matching how the formula is built up algebraically from the inside out.
1.10 Radian Measure and the Six Trigonometric Functions
An angle in radian measure is defined by the arc it subtends on a circle of radius 1 (the unit circle): an angle of θ radians corresponds to an arc of length θ. Since the full circle has circumference 2π, one full revolution is 2π radians, giving the conversion
Radians are not an arbitrary alternative to degrees — they are the units in which every calculus formula involving trigonometric functions comes out simplest. From Week 4 onward, every angle is assumed to be in radians unless stated otherwise; a derivative formula like "the derivative of sin x is cos x" is only true when x is measured in radians.
Place the vertex of angle θ at the origin with its initial side along the positive x-axis, measuring θ counterclockwise as positive. If P = (x, y) is the point where the terminal side meets the unit circle, the six trigonometric functions are defined by
Because the unit circle has radius 1, every point on it satisfies x2 + y2 = 1, which immediately gives the most important trigonometric identity, restated in §1.12.
Common angles
| Degrees | Radians | sin | cos | tan |
|---|---|---|---|---|
| 0° | 0 | 0 | 1 | 0 |
| 30° | π/6 | 1/2 | √3/2 | √3/3 |
| 45° | π/4 | √2/2 | √2/2 | 1 |
| 60° | π/3 | √3/2 | 1/2 | √3 |
| 90° | π/2 | 1 | 0 | undefined |
| 180° | π | 0 | −1 | 0 |
| 270° | 3π/2 | −1 | 0 | undefined |
The 30–60–90 and 45–45–90 reference triangles reproduce every entry in the table above from ordinary geometry, so you never need to memorise the table itself — only those two triangles and the unit-circle definition.
1.11 Graphs of the Trigonometric Functions
As θ increases past 2π, the point P simply goes around the unit circle again, so sin and cos repeat every 2π radians: they are periodic with period 2π. Tangent repeats twice as often, with period π, because tan θ depends only on the line through the origin and P, and that line returns to itself after half a revolution.
sin and cos have domain (−∞, ∞) and range [−1, 1]. tan and cot are undefined wherever the corresponding cos or sin is zero (vertical asymptotes there), and csc, sec are undefined wherever sin, cos respectively are zero. Sketching any trigonometric graph is really an exercise in §1.8–1.9: start from the parent graph in Figure 1.3 and apply shifts and scalings.
1.12 Trigonometric Identities
These identities are used constantly, both in this course and in the ones that follow it. The Pythagorean identity comes directly from x2 + y2 = 1 on the unit circle:
Dividing Eq. 1.8 through by cos2θ or sin2θ gives its two companions:
Because cos is even and sin is odd (compare §1.4 with Figure 1.2: reflecting θ to −θ flips the sign of y but not x):
The addition formulas let you expand sin or cos of a sum or difference of angles — you will use these in Week 4 to differentiate sin x and cos x from the definition of the derivative:
Example 1.6 — Solving a trigonometric equation
Solve 2 sin2x − 1 = 0 for x ∈ [0, 2π).
Step 1 — Isolate sin2x
sin2x = 1/2 ⇒ sin x = ±√2/2
Step 2 — Find all solutions in one period using the reference angle π/4
sin x = √2/2 at x = π/4 and x = 3π/4 (Quadrants I, II).
sin x = −√2/2 at x = 5π/4 and x = 7π/4 (Quadrants III, IV).
x = π/4, 3π/4, 5π/4, 7π/4
1.13 Transformations of Trigonometric Graphs
Combining §1.8–1.9 with the sine graph of Figure 1.3 gives the general sinusoid that appears in almost every application of trigonometry to real data (sound waves, tides, alternating current, seasonal temperature):
| Symbol | Name | Effect |
|---|---|---|
| |A| | Amplitude | Vertical stretch: the graph's height above and below its midline (§1.9) |
| 2π/|B| | Period | Horizontal stretch/compression: length of one full cycle (§1.9) |
| C | Phase shift | Horizontal shift: moves the starting point of the cycle (§1.8) |
| D | Vertical shift | Moves the midline of the oscillation (§1.8) |
Example 1.7 — Modelling with a sinusoid
The depth of water in a harbour varies between 2 m and 8 m, completing one full tidal cycle every 12 hours, with high tide (maximum depth) at t = 3 h. Model the depth d(t) as a sinusoid.
Step 1 — Amplitude and vertical shift from the range
Midline D = (8+2)/2 = 5 m; amplitude A = (8−2)/2 = 3 m.
Step 2 — Period gives B
2π/B = 12 ⇒ B = π/6.
Step 3 — Use the high-tide time to fix the phase shift
A plain sine reaches its maximum a quarter-period after C, so use cosine instead, which peaks exactly at its phase shift: d(t) = A cos(B(t−C)) + D with C = 3.
d(t) = 3 cos((π/6)(t − 3)) + 5
1.14 Mathematical Induction (Not Examinable)
Per the course schedule, §A.2 is covered only briefly and is not examinable this term. It is included here because a handful of later results (for example, some series formulas in Weeks 14–15) are conventionally proved by induction, and recognising the pattern of such a proof is useful even if you will never be asked to produce one yourself.
Mathematical induction is a method for proving that a statement P(n) holds for every integer n ≥ n0. It has two parts:
- Base case. Verify P(n0) directly.
- Inductive step. Assume P(k) is true for some arbitrary k ≥ n0 (the inductive hypothesis), and use that assumption to prove P(k+1).
Together, the two parts show the statement is true for n0 (base case), hence for n0+1 (apply the inductive step once), hence for n0+2 (apply it again), and so on for every integer above n0 — like an infinite row of dominoes where the base case knocks over the first one and the inductive step guarantees each domino knocks over the next.
Example 1.8 — A sum formula by induction
Prove that 1 + 2 + 3 + … + n = n(n+1)/2 for every integer n ≥ 1.
Step 1 — Base case, n = 1
Left side: 1. Right side: 1(2)/2 = 1. They match, so P(1) holds.
Step 2 — Inductive hypothesis
Assume 1 + 2 + … + k = k(k+1)/2 for some k ≥ 1.
Step 3 — Inductive step, show P(k+1) follows
1 + 2 + … + k + (k+1) = k(k+1)2 + (k+1) = (k+1)k2 + (k+1) = (k+1)·k+22
= (k+1)((k+1)+1)/2, which is exactly the formula with n = k+1. By induction, the formula holds for every n ≥ 1.
1.15 Worked Capstone Example
Example 1.9 — Domain of a composite involving a trigonometric function
Let f(x) = √x and g(x) = 2 cos x − 1. Find the domain of (f ∘ g)(x) for x ∈ [0, 2π), and evaluate it at x = 0.
Step 1 — Write the composite and its requirement
(f ∘ g)(x) = √(2 cos x − 1), which needs 2 cos x − 1 ≥ 0, i.e. cos x ≥ 1/2 (§1.7, §1.2).
Step 2 — Solve the trigonometric inequality (§1.10–1.11)
From the common-angle table, cos x = 1/2 at x = π/3 and x = 5π/3. Since cosine is largest at x = 0 and decreases toward x = π, cos x ≥ 1/2 exactly on [0, π/3] ∪ [5π/3, 2π).
Domain of f ∘ g on [0, 2π): [0, π/3] ∪ [5π/3, 2π)
Step 3 — Evaluate at x = 0 (§1.2, §1.10)
g(0) = 2 cos(0) − 1 = 2(1) − 1 = 1
(f ∘ g)(0) = √1 = 1
1.16 Nomenclature
| Symbol | Meaning |
|---|---|
| Df | Domain of f |
| Rf | Range of f |
| f ∘ g | Composite function, (f ∘ g)(x) = f(g(x)) |
| θ | Angle, conventionally in radians in this course |
| A, B, C, D | Amplitude, angular frequency, phase shift, vertical shift of a sinusoid |
| P(n) | A statement depending on the integer n, used in induction |
| n0 | Starting integer (base case) of an induction argument |
1.17 Problems
- Domain. Find the natural domain of f(x) = (x−1)/√(9−x2).
- Vertical line test. Explain why x2 + y2 = 4 is not the graph of a function of x, and give the two functions whose graphs together make up the circle.
- Piecewise graph. Sketch f(x) = −x for x ≤ 0 and f(x) = x2 for x > 0. Is the graph broken at x = 0?
- Symmetry. Classify each as even, odd, or neither: (a) h(x) = x4 − 3x2; (b) h(x) = x + sin x; (c) h(x) = x2 + x.
- Sum and quotient domains. Let f(x) = √(x+3) and g(x) = √(5−x). Find the domains of f + g and f/g.
- Composite domains, both orders. Let f(x) = 1/x and g(x) = x + 2. Find (f ∘ g)(x) and (g ∘ f)(x) with their domains, and confirm the two composites are different functions.
- Shift and scale. Starting from y = √x, describe the sequence of transformations that produces y = 2√(x+3) − 1, and state the domain of the result.
- Reading a transformed graph. The graph of y = f(x) passes through (2, 5). Give one point that must lie on the graph of y = f(x−3)+1, and one point that must lie on y = −f(x).
- Radian conversion. Convert 150° to radians and 7π/4 to degrees.
- Exact values. Using the reference triangles, find sin(5π/6), cos(5π/6) and tan(5π/6) without a calculator.
- Solving a trig equation. Solve 2 cos x + √3 = 0 for x ∈ [0, 2π).
- Identity practice. Use Eq. 1.11–1.12 to show that sin(x + π/2) = cos x, confirming the phase relationship shown under Figure 1.3.
- Sinusoidal model. The temperature in a city varies between 18°C and 32°C over a year, reaching its maximum at day 200 of the year. Modelling day 0 as 1 January, write a sinusoid T(t) for the temperature on day t, assuming a 365-day period.
- Induction, not examinable. Prove by induction that 1 + 3 + 5 + … + (2n−1) = n2 for every integer n ≥ 1.
- Induction, not examinable. Prove by induction that 2n > n for every integer n ≥ 1.
- Combined. Let f(x) = √(1−2 sin x). Find the domain of f for x ∈ [0, 2π), and evaluate f(π).
1.18 References
- Thomas, G. B., Weir, M. D. and Hass, J. Thomas' Calculus, SI Units, 15th ed. Pearson, 2024. — §A.2 (mathematical induction, briefly), §1.1 (functions and their graphs), §1.2 (combining functions; shifting and scaling graphs), §1.3 (trigonometric functions). The primary source for this week's material and numbering.
- Pearson MyLab Math — the Week 1 homework set (10 questions, up to 3 attempts) is assigned directly from the sections above; see the Calendar for the due date.