MATHEMATICS · LESSON 01 OF 10

Functions you need for calculus

The function language that every calculus lesson uses: domains and ranges, piecewise rules, transformations, composition, inverses, exponentials and logarithms, and trigonometric functions in radians.

What this lesson explains

Calculus studies how functions change. Before you can find a limit or a derivative you must be able to read a function: which inputs are allowed, what its graph looks like, how it is built from simpler functions, and when it has an inverse. Most errors in first-year calculus are not calculus errors at all; they are domain, algebra or radian errors made along the way.

Before you begin

Interval notation

[a, b] includes both ends; (a, b) excludes them; [a, b) includes a only. ∪ joins pieces: (−∞, 2) ∪ (2, ∞) means every real number except 2.

Exponent rules

a^maⁿ = a^m+n; a^m / aⁿ = a^m−n; (a^m)ⁿ = a^mn; a⁻ⁿ = 1 / aⁿ; a^1/n = the nth root of a (for even n this needs a ≥ 0). They combine powers of the same base only: x² + x³ is not x⁵.

The idea, made visible

A function assigns exactly one output to each allowed input. The domain is the set of allowed inputs; the range is the set of outputs the function actually produces. A graph represents a function of x exactly when every vertical line meets it at most once (the vertical-line test).

A formula restricts its own domain in three common ways: a denominator cannot be zero; an even root (such as √u) needs u ≥ 0; a logarithm ln u needs u > 0 (strictly). A piecewise function uses different rules on different parts of the domain; the inequality attached to each piece decides which rule applies at a boundary point.

y = −2(x − 3)² + 1: move the parent parabola 3 right and 1 up, stretch it by 2 and flip it. The point (1, 1) goes to (1 + 3, −2·1 + 1) = (4, −1).

Transformations build new graphs from a known “parent” graph. For y = a·f(b(x − h)) + k: changes outside f act on outputs (a stretches and, if negative, reflects; k shifts up or down); changes inside act on inputs (h shifts right by h; b scales horizontally by the factor 1/|b|, a compression when |b| > 1, and a negative b also reflects in the y-axis). The safest check is to track one known point.

Composition feeds one function’s output into another: (f ∘ g)(x) = f(g(x)). Order matters, and the domain must survive both stages: x must be allowed in g, and g(x) must be allowed in f. Recognising compositions is exactly what the chain rule needs later.

An inverse function f⁻¹ reverses f: f⁻¹(f(x)) = x. It exists only if f is one-to-one (every horizontal line meets the graph at most once); otherwise the domain must be restricted, as with x² on x ≥ 0 or sin x on [−π/2, π/2]. The domain of f⁻¹ is the range of f.

Exponential functions b^x (b > 0, b ≠ 1) have the variable in the exponent; the logarithm log_b x is the inverse: log_b x = y means b^y = x, and it needs x > 0. The natural base e ≈ 2.71828 and ln x = log_e x are the ones calculus uses. Trigonometric functions in calculus use radians: θ = arc length ÷ radius, so 180° = π rad.

Parent functions used throughout calculus
function	domain	range	key feature
x²	all reals	[0, ∞)	even; vertex at origin
x³	all reals	all reals	odd; one-to-one
√x	[0, ∞)	[0, ∞)	starts at origin
1/x	x ≠ 0	y ≠ 0	asymptotes x = 0, y = 0
e^x	all reals	(0, ∞)	always positive; inverse of ln x
ln x	(0, ∞)	all reals	ln 1 = 0; undefined for x ≤ 0
sin x	all reals	[−1, 1]	odd; period 2π
cos x	all reals	[−1, 1]	even; period 2π

Key terms

Function: A rule giving exactly one output f(x) for each input x in its domain.
Domain and range: Domain: all allowed inputs. Range: all outputs actually produced.
Even and odd functions: Even: f(−x) = f(x) (symmetric about the y-axis, e.g. x², cos x). Odd: f(−x) = −f(x) (symmetric about the origin, e.g. x³, sin x). Most functions are neither.
Composition: (f ∘ g)(x) = f(g(x)): apply g first, then f.
One-to-one; inverse: f is one-to-one if different inputs give different outputs. Then f⁻¹ exists, with f⁻¹(f(x)) = x on the domain of f.
Logarithm: log_b x = y ⇔ b^y = x, for b > 0, b ≠ 1, x > 0. ln x is log_e x.
Radian: The angle subtended by an arc equal in length to the radius. 2π rad = 360°.

The formulas and what they mean

Symbols, meanings and units
Symbol	Meaning	Unit
f(x)	output of f at input x	output units
f ∘ g	composition, g first then f	—
f⁻¹	inverse function (not 1/f)	—
A, B, h, k	amplitude, frequency factor, horizontal and vertical shifts in y = A sin(B(x − h)) + k	output units; per unit of x; units of x; output units

Domain restrictions

denominator ≠ 0; √u: u ≥ 0; ln u: u > 0

Conditions and limits: Apply every restriction present in the formula, then combine them.

Transformation form

y = a·f(b(x − h)) + k

Conditions and limits: a: vertical stretch/reflection; k: vertical shift; h: horizontal shift (right for h > 0); b: horizontal scale 1/|b|. Factor b out first: f(2x − 6) = f(2(x − 3)) has h = 3, not 6.

Logarithm laws

ln(ab) = ln a + ln b; ln(a / b) = ln a − ln b; ln(a^r) = r ln a; e^{ln x} = x

Conditions and limits: Only for positive a, b, x. Check solutions of log equations against the original domain.

Trigonometric identities and graphs

sin² x + cos² x = 1; tan x = sin x / cos x; period of A sin(Bx) = 2π / ∣B∣, amplitude ∣A∣

Conditions and limits: Angles in radians. tan x is undefined where cos x = 0.

Inverse trigonometric ranges

arcsin x ∈ [−π/2, π/2]; arccos x ∈ [0, π]; arctan x ∈ (−π/2, π/2)

Conditions and limits: arcsin and arccos need −1 ≤ x ≤ 1. They return one principal angle, not every angle with that value.

Why it works: Why an inside shift moves the graph the “wrong” way

Compare y = √x with y = √x + 4.

√x starts at x = 0, where the radicand is 0.
Starting point of the parent graph.
√x + 4 starts where x + 4 = 0, that is at x = −4.
The same output now happens 4 units earlier.
So the whole graph moves 4 units left.
Every output is reached at an input 4 smaller.

Inside changes act on inputs, so they undo themselves: to get the same output you need x + 4 to equal the old x. Test a landmark point instead of memorising a sign rule.

A first worked example

Reading a function before calculus

Find the domain: list every denominator, even root and logarithm, write each restriction, and combine them.
Identify the structure: which parent function, which transformations, which composition (inside and outside)?
For an equation with logs or roots, solve algebraically, then check every candidate in the original equation’s domain.
For trigonometry, work in radians, find the reference angle, then fix the sign from the quadrant.

A domain with a root and a denominator

Problem. Find the domain of f(x) = √x + 3 / x − 2, and evaluate f(−3).

Root: x + 3 ≥ 0, so x ≥ −3.
An even root needs a non-negative radicand; equality is allowed.
Denominator: x − 2 ≠ 0, so x ≠ 2.
Division by zero is undefined.
Combine: [−3, 2) ∪ (2, ∞).
Keep inputs that satisfy both conditions.
f(−3) = √0 / −5 = 0.
The endpoint −3 is allowed.

Result: Domain [−3, 2) ∪ (2, ∞); f(−3) = 0.

What it means: With a logarithm instead, ln(x + 3), the endpoint −3 would be excluded, because ln needs a strictly positive argument.

A different case

Composition in both orders

Problem. For f(x) = √x and g(x) = x − 3, find (f ∘ g)(x) and (g ∘ f)(x) with their domains.

(f ∘ g)(x) = f(x − 3) = √x − 3; domain x ≥ 3.
The input reaching the root is x − 3, which must be ≥ 0.
(g ∘ f)(x) = g(√x) = √x − 3; domain x ≥ 0.
Now the root acts first, on x itself.

Result: √x − 3 on [3, ∞) and √x − 3 on [0, ∞): different functions.

What it means: Order matters. Recognising which function is inside is the key step of the chain rule.

More worked cases

Each case below uses a different skill. Every step and result is shown.

A transformation traced by a point

Problem. Describe y = −2(x − 3)² + 1 as a transformation of y = x², and give its range.

Inside: x − 3 shifts the graph 3 right.
h = 3.
Outside: × (−2) stretches vertically by 2 and reflects in the x-axis; + 1 shifts up 1.
a = −2, k = 1.
Vertex (0, 0) → (3, 1); the parabola now opens downward.
Track the landmark point.

Result: Vertex (3, 1), opening downward; range (−∞, 1].

What it means: The maximum value 1 occurs at x = 3 — the kind of fact that optimization later finds with derivatives.

An inverse with its domain

Problem. Find the inverse of f(x) = √x − 1, x ≥ 1, and state its domain.

Write y = √x − 1 and solve: y² = x − 1, so x = y² + 1.
Solve for the input.
Exchange letters: f⁻¹(x) = x² + 1.
The inverse takes outputs back to inputs.
Domain of f⁻¹ = range of f = [0, ∞).
A square root never produces negative outputs.

Result: f⁻¹(x) = x² + 1 for x ≥ 0.

What it means: Without the restriction x ≥ 0, x² + 1 would not be one-to-one and could not be the inverse.

A logarithm equation with a domain check

Problem. Solve ln(x − 1) + ln(x + 1) = ln 8.

Domain: x − 1 > 0 and x + 1 > 0, so x > 1.
Every logarithm argument must be positive.
Combine: ln((x − 1)(x + 1)) = ln 8, so x² − 1 = 8.
ln a + ln b = ln(ab); ln is one-to-one.
x² = 9, so x = 3 or x = −3.
Algebraic candidates.
Only x = 3 satisfies x > 1.
Reject −3: ln(−4) is undefined.

Result: x = 3.

What it means: Algebra can create candidates that the original equation does not allow. Always check.

Trigonometric values from one ratio

Problem. If sin θ = 5/13 and θ is in quadrant II, find cos θ and tan θ.

cos² θ = 1 − (5/13)² = 144/169, so |cos θ| = 12/13.
Pythagorean identity gives the size.
In quadrant II cosine is negative: cos θ = −12/13.
The quadrant gives the sign.
tan θ = sin θ / cos θ = −5/12.
Definition of tangent.

Result: cos θ = −12/13, tan θ = −5/12.

What it means: The identity fixes magnitudes; only the quadrant can fix signs.

Reading a sinusoidal graph

Problem. For y = 3 cos(2x) − 1, give the amplitude, period, midline, maximum and minimum.

Amplitude |A| = 3; midline y = −1.
A = 3, k = −1.
Period = 2π/2 = π.
B = 2.
Maximum −1 + 3 = 2 (at x = 0); minimum −1 − 3 = −4 (at x = π/2).
cos 0 = 1 and cos π = −1.

Result: Amplitude 3, period π, midline y = −1, maximum 2, minimum −4.

What it means: Oscillating quantities (vibrations, alternating current) are described this way.

Common misunderstandings

Misunderstanding: Including a zero argument in a logarithm domain.
Correct idea: ln u needs u > 0 strictly; √u allows u = 0.
Misunderstanding: Reading f(x − 3) as a shift to the left.
Correct idea: x − 3 shifts right by 3; test the point where the inside equals the old landmark.
Misunderstanding: Writing f⁻¹(x) = 1/f(x).
Correct idea: The inverse function reverses f; 1/f is a reciprocal, a different thing.
Misunderstanding: Combining x² + x³ into x⁵.
Correct idea: Exponent rules apply to products and quotients of the same base, not to sums.
Misunderstanding: Using degrees in calculus.
Correct idea: Calculus formulas for sin, cos and tan assume radians.
Misunderstanding: Assuming arcsin gives every solution.
Correct idea: arcsin returns one principal angle in [−π/2, π/2]; other solutions come from symmetry and periodicity.

Keep in mind

denominator ≠ 0; √u: u ≥ 0; ln u: u > 0Domain restrictions
y = a·f(b(x − h)) + kTransformation form
ln(ab) = ln a + ln b; ln(a / b) = ln a − ln b; ln(a^r) = r ln a; e^{ln x} = xLogarithm laws
sin² x + cos² x = 1; tan x = sin x / cos x; period of A sin(Bx) = 2π / ∣B∣, amplitude ∣A∣Trigonometric identities and graphs
arcsin x ∈ [−π/2, π/2]; arccos x ∈ [0, π]; arctan x ∈ (−π/2, π/2)Inverse trigonometric ranges

Scope of this lesson

This is a review of prerequisites, not a full precalculus course. Hyperbolic functions and detailed identities are not included.

Next: Limits. A limit describes the value a function approaches near a point, not the value at the point.

Original study text. Sources and credits.

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