Mathematical-Reasoning · Unit 25 · Video 1 · Interactive Practice

Applying a Function to a Set: Forward and Backward

IKey Formulas

Set	Direction	How to compute it
$\{ f(x) \mid x \in A \}$	Forward, from $A \subseteq S$	Follow the arrow out of each element of $A$ and collect the outputs, each listed once
$\{ x \in S \mid f(x) \in C \}$	Backward, from $C \subseteq T$	Test every element of the domain and keep it when its output lies in $C$
$\{ x \in \mathbb{R} \mid x^2 = c \}$	Backward for $g(x) = x^2$	$\{-\sqrt{c}, \sqrt{c}\}$ if $c \gt 0$ , $\{0\}$ if $c = 0$ , $\emptyset$ if $c \lt 0$

Key Insight: Forward, different inputs can share an output, so the set of outputs can be smaller than $A$ . Backward, one target can catch several inputs or none, so the set of inputs can be larger than $C$ , or empty.

IIOne Arrow Diagram, Two Directions

The same four arrows answer both questions: which outputs a set of inputs reaches, and which inputs land in a set of outputs.

Direction

IIISquaring, Forward and Backward

For $g(x) = x^2$ , two inputs share each positive output, and a negative target catches nothing.

Direction

IVQuiz Questions

Problem 1 · Forward Through the Arrows

Given: $f: \{1, 2, 3, 4, 5\} \to \{p, q, r, s\}$ with $f(1) = q$ , $f(2) = r$ , $f(3) = q$ , $f(4) = p$ and $f(5) = r$ . Let $A = \{1, 3, 5\}$ . Find $\{ f(x) \mid x \in A \}$ .

✅ Correct!

1

and

3

both go to

q

, and

5

goes to

r

; the repeated

q

is listed once, so three inputs give two outputs.

❌ Not quite. That is

A

itself. The set

\{ f(x) \mid x \in A \}

collects outputs, so it is a subset of the codomain

\{p, q, r, s\}

❌ Not quite. Only arrows leaving elements of

A

count. No element of

A

is sent to

p

or to

s

Show solution

Follow the arrow out of each element of $A = \{1, 3, 5\}$ : $f(1) = q$ , $f(3) = q$ and $f(5) = r$ .

A set lists each element once, so $\{ f(x) \mid x \in A \} = \{q, r\}$ . The two inputs $1$ and $3$ share the output $q$ , which is why the set of outputs is smaller than $A$ .

Problem 2 · Backward to the Inputs

Given: the same $f$ as in Problem 1 and the set of outputs $C = \{q, s\}$ . Find $\{ x \in \{1, 2, 3, 4, 5\} \mid f(x) \in C \}$ .

✅ Correct! The inputs

1

and

3

land on

q

, and no input lands on

s

, so

s

contributes nothing.

❌ Not quite. An input is kept when its output is some element of

C

; it does not have to reach every element of

C

. Both

1

and

3

land on

q \in C

❌ Not quite.

f(4) = p

, and

p \notin C

, so

4

is dropped.

❌ Not quite.

f(2) = r

, and

r \notin C

, so

2

is dropped.

Show solution

Test each element of the domain against $C = \{q, s\}$ :

$f(1) = q \in C$ , keep $1$ ; $\ f(2) = r \notin C$ , drop $2$ ;
$f(3) = q \in C$ , keep $3$ ; $\ f(4) = p \notin C$ , drop $4$ ; $\ f(5) = r \notin C$ , drop $5$ .

No arrow reaches $s$ , so it adds no inputs. The answer is $\{1, 3\}$ .

Problem 3 · Squaring Backward

Given: $g: \mathbb{R} \to \mathbb{R}$ , $g(x) = x^2$ , and $C = \{4, 0, -9\}$ . Find $\{ x \in \mathbb{R} \mid g(x) \in C \}$ .

✅ Correct!

x^2 = 4

gives

\pm 2

x^2 = 0

gives

0

, and

x^2 = -9

has no real solution.

❌ Not quite.

(-2)^2 = 4

as well, so

-2

lands in

C

too.

❌ Not quite.

(\pm 3)^2 = 9

, but

9 \notin C

. The target is

-9

, and no real number squares to a negative number.

❌ Not quite. That squares the elements of

C

, which is the forward direction. Here we need the inputs whose squares lie in

C

Show solution

A real number $x$ qualifies when $x^2 = 4$ , $x^2 = 0$ or $x^2 = -9$ .

$x^2 = 4$ exactly when $x = 2$ or $x = -2$ ;
$x^2 = 0$ exactly when $x = 0$ ;
$x^2 = -9$ has no solution, since $x^2 \ge 0$ for every real $x$ .

So $\{ x \in \mathbb{R} \mid g(x) \in C \} = \{-2, 0, 2\}$ : three targets, three inputs.

Problem 4 · Counting Both Directions

Given: a function $f: S \to T$ with $S$ finite, a subset $A \subseteq S$ and a subset $C \subseteq T$ . Which statement is true for every such $f$ , $A$ and $C$ ?

✅ Correct! Each element of

A

contributes one output, and repeats are listed once, so there are never more outputs than elements of

A

❌ Not quite. Inputs can share an output: in the video,

B = \{a, d\}

has two elements but its set of outputs is

\{s\}

❌ Not quite. One target can catch several inputs: for

g(x) = x^2

, the four targets

1, 9, 15, -1

caught six real inputs.

❌ Not quite. A target that no arrow reaches catches nothing: in the video,

\{ x \in S \mid f(x) \in \{u\} \} = \emptyset

Show solution

(a) is true. If $A$ has $n$ elements, listing $f(x)$ for each $x \in A$ gives $n$ entries, some possibly equal; as a set it has at most $n$ elements.

(b) fails when two elements of $A$ share an output, as $a$ and $d$ share $s$ .

(c) fails when a target has several inputs: $\{ x \in \mathbb{R} \mid x^2 \in \{1, 9, 15, -1\} \} = \{-\sqrt{15}, -3, -1, 1, 3, \sqrt{15}\}$ has six elements, while the target set has four. (A finite example: the constant function on $\{1, 2\}$ sends both inputs into the one-element set $\{p\}$ .)

(d) fails when $C$ contains only values nothing reaches, such as $C = \{u\}$ .

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