Mathematical-Reasoning · Unit 20 · Video 4 · Interactive Practice

The Range of a Function: From Graphs and Arrow Diagrams

IKey Formulas

Formula	Name	What it says
$\operatorname{range}(f) = \{f(x) \mid x \in A\}$	Range (image of $f$ )	All outputs, collected from the domain side
$\operatorname{range}(f) = \{y \in B \mid y = f(x) \text{ for some } x \in A\}$	Same set, codomain side	The $y$ in $B$ that have at least one preimage
$\operatorname{range}(f) \subseteq \operatorname{codom}(f)$	Range inside codomain	Always true; equality depends on $f$
$g(x) = (x - 1)^2 - 2 \ge -2$	Completed square	$\operatorname{range}(g) = \{y \in \mathbb{R} \mid y \ge -2\}$

Key Insight: The definition of a function constrains only the domain side: each input sends exactly one arrow. A codomain point may receive several arrows or none, and the points that receive none are exactly the elements of the codomain outside the range.

IIReading the Range of a Parabola

A height $c$ is in the range of $g(x) = x^2 - 2x - 1$ exactly when the horizontal line $y = c$ meets the graph.

Height c 3

IIIInputs Meet Once, Outputs Meet Many Times

For $f(x) = A\sin(Bx)$ with $A, B > 0$ , a vertical line meets the graph once; a horizontal line can meet it many times or never.

Input x = a −1 Output y = c 2

💡 The window shows only $-3.8 \le x \le 3.8$ ; because $f$ repeats with period about $3.1$ , every output from $-A$ to $A$ has infinitely many preimages in $\mathbb{R}$ , and $\operatorname{range}(f) = [-A, A]$ since $-1 \le \sin(Bx) \le 1$ with both bounds reached.

IVBuild an Arrow Diagram

An arrow diagram from $\{1, 2, 3, 4\}$ to $\{a, b, c\}$ is a function exactly when each domain point sends one arrow.

VQuiz Questions

Problem 1 · The Last Digit of a Square

Given: $f: \{1, 2, \ldots, 10\} \to \{0, 1, \ldots, 9\}$ , where $f(n)$ is the last digit of $n^2$ (for example $f(7) = 9$ , since $7^2 = 49$ ). Find $\operatorname{range}(f)$ .

✅ Correct! The ten outputs are

1, 4, 9, 6, 5, 6, 9, 4, 1, 0

, which collect into six values;

2

3

7

and

8

have no preimage.

❌ Not quite. That is the codomain. No square ends in

2

3

7

8

, so those digits are outside the range.

❌ Not quite. That is the domain, the set of inputs. The range is a subset of the codomain

\{0, 1, \ldots, 9\}

; it does not even contain

10

❌ Not quite. The range holds the last digits of the squares, not the one-digit squares:

f(4) = 6

from

16

f(5) = 5

from

25

, and

f(10) = 0

from

100

Show solution

The squares of $1, 2, \ldots, 10$ are $1, 4, 9, 16, 25, 36, 49, 64, 81, 100$ , so

f(1), f(2), \ldots, f(10) = 1, 4, 9, 6, 5, 6, 9, 4, 1, 0.

Collecting the distinct values, $\operatorname{range}(f) = \{f(n) \mid n \in \{1, \ldots, 10\}\} = \{0, 1, 4, 5, 6, 9\}$ .

This is a proper subset of the codomain: $2$ , $3$ , $7$ and $8$ have no preimage. Several outputs have two preimages, for example $f(4) = f(6) = 6$ ; the definition of a function allows that.

Problem 2 · Filling the Codomain with One Change

Given: $f: \{1, 2, 3, 4, 5\} \to \{a, b, c, d\}$ with arrows $1 \to b$ , $2 \to d$ , $3 \to b$ , $4 \to a$ , $5 \to a$ . Find $\operatorname{range}(f)$ , and which single change gives a function whose range is the whole codomain?

What is the range of f?

Which change makes the range equal to the codomain?

✅ Correct!

b

had two preimages,

1

and

3

, so one of them can be spent on

c

without losing

b

: the new outputs

c, d, b, a, a

fill the codomain.

❌ Not quite.

\{a, b, c, d\}

is the codomain. No arrow lands on

c

, so

c

is not in the range.

❌ Not quite.

c

is the one codomain point with no preimage, so it is the only point outside the range.

❌ Not quite.

\{1, 2, 3, 4, 5\}

is the domain. The range is a subset of the codomain

\{a, b, c, d\}

❌ Not quite.

2

is the only preimage of

d

. The new outputs

b, c, b, a, a

reach

c

but lose

d

❌ Not quite. Then two arrows leave

5

, so the input

5

has two outputs and the diagram is no longer a function.

❌ Not quite. The new outputs

b, d, d, a, a

still miss

c

: the range stays

\{a, b, d\}

Show solution

The outputs are $f(1) = b$ , $f(2) = d$ , $f(3) = b$ , $f(4) = a$ , $f(5) = a$ , so $\operatorname{range}(f) = \{a, b, d\}$ , a proper subset of the codomain. The point $c$ has no preimage, $a$ and $b$ have two each, and $d$ has one.

A change that keeps a function must still send exactly one arrow from each input, so it moves an arrow rather than adding one; that rules out (b). The moved arrow must land on $c$ , and it must come from an input whose old target keeps another preimage.

(a) Moving $2$ to $c$ leaves $d$ with no preimage: range $\{a, b, c\}$ .
(c) Moving $3$ to $d$ never reaches $c$ : range $\{a, b, d\}$ .
(d) Moving $1$ to $c$ gives outputs $c, d, b, a, a$ : range $\{a, b, c, d\}$ , the whole codomain.

Problem 3 · Range and Preimages of a Parabola

Given: $h: \mathbb{R} \to \mathbb{R}$ , $h(x) = x^2 + 4x + 1$ . Find $\operatorname{range}(h)$ and the set of all preimages of $-2$ .

What is the range?

What are the preimages of −2?

✅ Correct!

h(x) = (x + 2)^2 - 3

, so the range starts at the vertex height

-3

, and

-2

sits above it with two preimages.

❌ Not quite.

1 = h(0)

is the

y

-intercept, not the lowest output:

h(-2) = -3 < 1

❌ Not quite.

-2

is the

x

-coordinate of the vertex. The range is read from heights:

h(-2) = -3

❌ Not quite.

\mathbb{R}

is the codomain.

(x + 2)^2 - 3 \ge -3

, so no value below

-3

is an output.

❌ Not quite.

-2 > -3

, so

-2

is in the range: solve

(x + 2)^2 - 3 = -2

❌ Not quite.

h(-2) = -3

, not

-2

. A preimage of

-2

is an input

x

with

h(x) = -2

❌ Not quite. Check the sign:

h(1) = 6

. From

(x + 2)^2 = 1

we get

x + 2 = \pm 1

, so

x = -1

x = -3

Show solution

Range. Complete the square:

h(x) = x^2 + 4x + 4 - 3 = (x + 2)^2 - 3.

A square is never negative, so $h(x) \ge -3$ for every $x$ , with equality at $x = -2$ . For any $y \ge -3$ , the input $x = -2 + \sqrt{y + 3}$ gives $h(x) = (\sqrt{y + 3})^2 - 3 = y$ . So $\operatorname{range}(h) = \{y \in \mathbb{R} \mid y \ge -3\}$ , a proper subset of the codomain $\mathbb{R}$ .

Preimages of $-2$ . Solve $h(x) = -2$ :

(x + 2)^2 - 3 = -2 \iff (x + 2)^2 = 1 \iff x + 2 = 1 \text{ or } x + 2 = -1.

So $x = -1$ or $x = -3$ . Check: $h(-1) = 1 - 4 + 1 = -2$ and $h(-3) = 9 - 12 + 1 = -2$ . The set of preimages is $\{-3, -1\}$ .

Problem 4 · Same Formula, Integer Domain

Given: $k: \mathbb{Z} \to \mathbb{Z}$ , $k(n) = n^2 - 2n - 1$ , the video's formula for $g$ with the domain and codomain changed to $\mathbb{Z}$ . Which set equals $\operatorname{range}(k)$ ? (Here $\mathbb{N} = \{1, 2, 3, \ldots\}$ .)

✅ Correct!

k(n) = (n - 1)^2 - 2

, and

m = n - 1

runs through all of

\mathbb{Z}

n

does.

❌ Not quite. That is the real answer cut down to integers, but

0

is in it and

(n - 1)^2 - 2 = 0

needs

(n - 1)^2 = 2

, which has no integer solution.

❌ Not quite. This misses

-2

k(1) = -2

, but

m^2 - 2 = -2

needs

m = 0 \notin \mathbb{N}

❌ Not quite.

\mathbb{Z}

is the codomain. Since

k(n) = (n - 1)^2 - 2 \ge -2

, the integer

-3

has no preimage.

Show solution

Complete the square: $k(n) = (n - 1)^2 - 2$ .

Every output has the form $m^2 - 2$ with $m \in \mathbb{Z}$ : take $m = n - 1$ , an integer.

Every $m^2 - 2$ with $m \in \mathbb{Z}$ is an output: the integer $n = m + 1$ gives $k(m + 1) = m^2 - 2$ .

So $\operatorname{range}(k) = \{m^2 - 2 \mid m \in \mathbb{Z}\}$ , whose smallest elements are $-2$ , $-1$ , $2$ , $7$ , $14$ , …

Over $\mathbb{R}$ the same formula had range $\{y \in \mathbb{R} \mid y \ge -2\}$ ; restricting the domain to $\mathbb{Z}$ removes the inputs that produced values like $0$ (from $x = 1 \pm \sqrt{2}$ ), so the range shrinks to a set with gaps. The $\mathbb{N}$ version omits $m = 0$ and with it the output $-2 = k(1)$ .

Solved: 0 / 4