Lesson 2.1: Function Definitions & Representations

What is a Function?

Each input maps to exactly one output. Learn domain, range, and how to represent functions in multiple ways with real meaning.

Learning Objectives

Define functions and identify domain/range
Use function notation f(x) and evaluate values

Translate between table, formula, and graph
Interpret contexts from representations

Definition & Notation

A function assigns every input in the domain exactly one output in the range. Notation: $y=f(x)$ .

Domain

All permissible x-values.

Range

All outputs f(x).

Function Value

$f(4)$ means output at $x=4$ .

Three Representations

Formula

Example: $y=2x+1$

Table

(x, y): (0,1), (1,3), (2,5)

Graph

Straight line rising with slope 2

Deeper Concepts

Function as Mapping

A function is a mapping $f: A\to B$ that assigns each $x\in A$ a unique $y\in B$ . The set $A$ is the domain; $f(A)\subseteq B$ is the range (image).

Set notation for domain restrictions (example rational function): $f(x)=\tfrac{1}{x-2},\quad \mathrm{Dom}(f)=\{x\in\mathbb{R},|,x\neq 2\}$

Piecewise definition: $f(x)=\begin{cases}x^2,& x\ge 0\\-x,& x<0\end{cases}$

One-to-one, Onto, Inverse

One-to-one (injective): $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ . Onto (surjective): $\forall y\in B,\exists x\in A: f(x)=y$ .

If $f$ is bijective, inverse $f^{-1}$ exists. Example: $f(x)=2x+1$ (on $\mathbb{R}$ ) is bijective with $f^{-1}(y)=\tfrac{y-1}{2}$ .

Composition preview: $(g\circ f)(x)=g(f(x))$ connects multi-stage processes.

Common Pitfalls

Confusing relation with function: a function cannot assign two different outputs to the same input.
Ignoring domain constraints: e.g., $\sqrt{x}$ requires $x\ge 0$ (over $\mathbb{R}$ ); denominators cannot be zero.
Mistaking range for codomain: range is actual outputs $f(A)$ , not the entire codomain.
Reading graphs: vertical line test fails → not a function of $x$ .

Worked Examples

1) Table → Formula

Given: x: 1,2,3; y: 4,7,10

Increase per step: +3 → $m=3$

Assume $y=3x+b$ ; plug $(1,4)$ → $b=1$

Formula: $y=3x+1$

2) Domain with Radicals and Fractions

$f(x)=\dfrac{\sqrt{x-1}}{x-3}$

Constraints: $x-1\ge 0\Rightarrow x\ge 1$ , and $x\neq 3$

Domain: $[1,3)\cup(3,\infty)$

3) Inverse Check (One-to-one on Restricted Domain)

$f(x)=x^2$ is not one-to-one on $\mathbb{R}$ , but is bijective on $[0,\infty)$ .

Inverse there: $f^{-1}(y)=\sqrt{y}$

Practice

1) Evaluate $f(x)=3x-2$ at $x=4$

Solution

f(4)=3\cdot 4-2=10

2) Identify domain of $y=\dfrac{1}{x-2}$

Solution

All real

x

except

x=2

3) Determine whether the relation is a function: $x=\pm \sqrt{y}$

Solution

Fails vertical line test (two x-values for one y). Not a function of

y

4) Find inverse on restricted domain: $f(x)=x^2,\ x\ge 0$

Solution

y=x^2\Rightarrow x=\sqrt{y}\ (x\ge 0);\ f^{-1}(y)=\sqrt{y}

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