Mathematical-Reasoning · Unit 20 · Video 3 · Interactive Practice

Images and Preimages: The Formal Definition of a Function

IKey Formulas

Formula	Name	Meaning
$f : A \to B$	Function (mapping) from $A$ to $B$	A rule giving each $x \in A$ exactly one element of $B$ ; $A = \text{dom}(f)$ , $B = \text{codom}(f)$
$f(x) = y$	Image and preimage	$y$ is the image of $x$ under $f$ ; $x$ is a preimage of $y$ for $f$
$\{x \in A \mid f(x) = y\}$	Set of preimages of $y$	Solve $f(x) = y$ over the domain; the set may be empty, have one element, or more
$x^2 - 5x = 2$	Same formula, different domain	On $\mathbb{R}$ : two solutions, $\frac{5 \pm \sqrt{33}}{2}$ ; on $\mathbb{Z}$ : none, so the set is $\emptyset$

Key Insight: A function is a rule together with its domain and codomain. Every input has exactly one image; an output may have several preimages, exactly one, or none, and two inputs sharing an output does not break the definition.

IIVisualization 1 — The Image Versus the Preimages

Under $g : \mathbb{R} \to \mathbb{R}$ , $g(x) = x^2 - 2$ , each input has exactly one image, while an output can have two, one or no preimages.

Input x −2 Output y 5

IIIVisualization 2 — Same Formula, Different Function

On $\mathbb{R}$ and on $\mathbb{Z}$ the formula $x^2 - 5x$ defines different functions, and their preimage sets can differ.

Output y 2

$f : \mathbb{R} \to \mathbb{R}$ , $f(x) = x^2 - 5x$ (the whole curve)

$g : \mathbb{Z} \to \mathbb{Z}$ , $g(m) = m^2 - 5m$ (only the squares)

IVVisualization 3 — Arrow Diagrams and the Definition

A diagram defines a function from $A$ to $B$ exactly when each element of $A$ sends out one arrow.

VQuiz Questions

Problem 1 · Computing an Image

Given: $h : \mathbb{R} \to \mathbb{R}$ defined by $h(x) = x^2 + 3x - 1$ . Find the image of $-4$ under $h$ .

✅ Correct!

h(-4) = 16 - 12 - 1 = 3

, the one output

h

assigns to

-4

❌ Not quite. The middle term is

3 \cdot (-4) = -12

, not

+12

h(-4) = 16 - 12 - 1

❌ Not quite.

(-4)^2 = 16

, not

-16

: the square of a negative number is positive.

❌ Not quite. The middle term is

3x

, so at

x = -4

it is

3 \cdot (-4) = -12

, not

-4

Show solution

The image of $-4$ under $h$ is $h(-4)$ , found by substituting into the formula:

h(-4) = (-4)^2 + 3(-4) - 1 = 16 - 12 - 1 = 3.

So the image of $-4$ under $h$ is $3$ . Every input has exactly one image, so this is the image of $-4$ .

Problem 2 · Preimages Depend on the Domain

Given: $q : \mathbb{N} \to \mathbb{N}$ defined by $q(n) = n^2 + n$ , where $\mathbb{N} = \{1, 2, 3, \ldots\}$ . Find the set of preimages of $6$ for $q$ .

✅ Correct!

q(2) = 4 + 2 = 6

, and

2

is the only natural number with

q(n) = 6

❌ Close, but check the domain.

-3

does solve

n^2 + n = 6

, but

-3 \notin \mathbb{N}

, so it is not in the domain of

q

and cannot be a preimage.

❌ Not quite.

42 = q(6)

is the image of

6

. A preimage of

6

is an input

n

with

q(n) = 6

❌ Not quite.

q(2) = 2^2 + 2 = 6

, so

6

has at least one preimage.

Show solution

The preimages of $6$ are the $n \in \mathbb{N}$ with $q(n) = 6$ :

n^2 + n = 6 \iff n^2 + n - 6 = 0 \iff (n + 3)(n - 2) = 0 \iff n = -3 \text{ or } n = 2.

Only $2$ lies in the domain $\mathbb{N}$ , and $q(2) = 4 + 2 = 6$ . So the set of preimages of $6$ for $q$ is $\{2\}$ . For the function from $\mathbb{Z}$ to $\mathbb{Z}$ given by the same formula, the set would be $\{-3, 2\}$ .

Problem 3 · One Formula, Two Functions

Given: $F : \mathbb{R} \to \mathbb{R}$ and $G : \mathbb{Z} \to \mathbb{Z}$ , both defined by the formula $x^2 - 3x$ . Find the set of preimages of $1$ for each function.

Set of preimages of $1$ for $F$

Set of preimages of $1$ for $G$

✅ Correct!

F

has two preimages of

1

, and neither is an integer, so

G

, whose inputs must be integers, has none.

❌ Not quite.

\{0, 3\}

solves

x^2 - 3x = 0

, so it is the set of preimages of

0

, not of

1

❌ Not quite.

x^2 - 3x - 1 = 0

has discriminant

9 + 4 = 13 > 0

, so it has two real solutions, and every real number is in the domain of

F

❌ Check the sign. For

x^2 - 3x - 1 = 0

the quadratic formula gives

x = \frac{3 \pm \sqrt{13}}{2}

; these options solve

x^2 + 3x - 1 = 0

❌ Check the domain. These two numbers are not integers, since

3 < \sqrt{13} < 4

, so they are not in the domain

\mathbb{Z}

G

❌ Not quite.

G(3) = 9 - 9 = 0

, not

1

. Rounding

\frac{3 + \sqrt{13}}{2} \approx 3.30

does not produce a preimage.

❌ Not quite. Solve

x^2 - 3x = 1

over

\mathbb{R}

❌ Not quite. Solve

m^2 - 3m = 1

over

\mathbb{Z}

Show solution

For $F$ . Solve $x^2 - 3x = 1$ , that is $x^2 - 3x - 1 = 0$ . It does not factor over the integers, so use the quadratic formula:

x = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}.

Both are real, so the set of preimages of $1$ for $F$ is $\left\{\frac{3 - \sqrt{13}}{2}, \frac{3 + \sqrt{13}}{2}\right\}$ , about $-0.30$ and $3.30$ .

For $G$ . An integer preimage would have to be one of the same two numbers. Since $9 < 13 < 16$ , we have $3 < \sqrt{13} < 4$ , so

3 < \frac{3 + \sqrt{13}}{2} < \frac{7}{2} \quad \text{and} \quad -\frac{1}{2} < \frac{3 - \sqrt{13}}{2} < 0.

Neither number is an integer, so no $m \in \mathbb{Z}$ has $G(m) = 1$ , and the set of preimages of $1$ for $G$ is $\emptyset$ .

Problem 4 · Reading the Definition

Given: $f : \{-2, -1, 0, 1, 2\} \to \mathbb{Z}$ defined by $f(x) = x^2$ . Which statement is true?

✅ Correct!

f(-2) = f(2) = 4

and no element of the domain squares to

3

; neither fact affects whether

f

is a function.

❌ Not quite. Two inputs sharing an output is allowed. The definition requires each input to have exactly one output, and

f(-1) = 1

and

f(1) = 1

are single outputs.

❌ Not quite. The definition places no requirement on elements of the codomain: some may be the output of no input.

❌ Check the domain.

4

is not in

\{-2, -1, 0, 1, 2\}

, so

f(4)

is not defined;

4

is an output of

f

, not an input.

Show solution

The images are $f(-2) = 4$ , $f(-1) = 1$ , $f(0) = 0$ , $f(1) = 1$ , $f(2) = 4$ . Each of the five inputs has exactly one output in $\mathbb{Z}$ , so $f$ is a function.

The preimages of $4$ are the domain elements with $x^2 = 4$ , namely $-2$ and $2$ , so the set is $\{-2, 2\}$ . No element of the domain has square $3$ , so the set of preimages of $3$ is $\emptyset$ . Statement (a) is true.

Statements (b) and (c) describe an output with two preimages and an output with none; the definition allows both. Statement (d) fails because $4$ is not in the domain, so it has no image under $f$ .

Solved: 0 / 4