Mathematical-Reasoning · Unit 20 · Video 1 · Interactive Practice

What Is a Function? Exactly One Output for Each Input

IKey Formulas

Formula	Name	Meaning
$f\colon A \to B$	Function from $A$ to $B$	Each element of the domain $A$ is assigned exactly one element of the codomain $B$
$f(x) = x^2 \sin x$ , $\;f\!\left(\tfrac{\pi}{2}\right) = \tfrac{\pi^2}{4}$	Function notation	$f$ names the function; $f(x)$ is the output for the input $x$
$y^2 = x + 3 \;\Rightarrow\; y = \sqrt{x+3}$ or $y = -\sqrt{x+3}$	Two outputs	Every $x > -3$ gives two values of $y$ , so the equation does not define $y$ as a function of $x$
$y = \dfrac{x}{x-1}$ , domain $\{x \in \mathbb{R} \mid x \neq 1\}$	No output	$x = 1$ gives division by zero, so $1$ is left out of the domain

Key Insight: An input with two outputs rules an equation out; an input with no output is removed from the domain. Two different inputs sharing one output is allowed.

IIVisualization 1 — One Input, How Many Outputs?

An equation defines $y$ as a function of $x$ only if no input $x$ ever produces two values of $y$ .

Equation Input x 1

IIIVisualization 2 — Arrow Diagrams

An arrow diagram is a function exactly when one arrow leaves each element of the domain.

💡 Challenge: build a function in which every element of $B$ except one receives no arrow, then one in which four different elements of $B$ each receive an arrow.

IVQuiz Questions

Problem 1 · Which Equation Is a Function?

Given: real inputs $x$ and real outputs $y$ — which equation defines $y$ as a function of $x$ ?

✅ Correct! Squaring, multiplying and adding real numbers give exactly one result, so every real

x

produces exactly one

y

❌ Not quite. Put in

x = 5

: then

y^2 = 4

, so

y = 2

and

y = -2

both work. One input, two outputs.

❌ Not quite. This is the circle of radius

3

. At

x = 0

y^2 = 9

gives

y = 3

and

y = -3

: one input, two outputs.

❌ Not quite. At

x = 2

, both

y = 2

and

y = -2

satisfy

\lvert y \rvert = 2

. One input, two outputs.

Show solution

For each equation, ask whether a single input $x$ can produce two values of $y$ .

$y^2 = x - 1$ : at $x = 5$ , $y^2 = 4$ , so $y = \pm 2$ . Not a function.
$x^2 + y^2 = 9$ : at $x = 0$ , $y = \pm 3$ . Not a function.
$\lvert y \rvert = x$ : at $x = 2$ , $y = \pm 2$ . Not a function.
$y = 3x^2 - x + 4$ : for each real $x$ the right side is exactly one real number; for example $x = 2$ gives $y = 12 - 2 + 4 = 14$ . This one defines $y$ as a function of $x$ .

Answer: (d).

Problem 2 · Two Inputs, One Output

Given: (I) $y = x^2$ and (II) $y^2 = x$ , with real $x$ and real $y$ — which statement is correct?

✅ Correct! In (I) each input has exactly one output, even though

2

and

-2

share the output

4

. In (II) the input

4

has the two outputs

2

and

-2

❌ Not quite. Test (II) at

x = 4

y^2 = 4

gives

y = 2

and

y = -2

. One input with two outputs rules (II) out.

❌ Not quite. The roles are reversed. In (II) the input

x = 4

has two outputs,

y = 2

and

y = -2

; in (I) every input has exactly one output.

❌ Not quite. The definition limits how many outputs each input has. Two inputs sharing the output

4

is allowed, so (I) is a function.

Show solution

(I) $y = x^2$ . Each real $x$ has exactly one square, so each input has exactly one output. That $f(2) = f(-2) = 4$ does not matter: the requirement is about each input, not each output.

(II) $y^2 = x$ . At $x = 4$ , $y^2 = 4$ is satisfied by $y = 2$ and by $y = -2$ . One input has two outputs, so (II) does not define $y$ as a function of $x$ .

Answer: (b).

Problem 3 · Natural Domain and an Output

Given: $h(x) = \dfrac{\sqrt{x+3}}{x-2}$ . Its domain is the set of all real $x$ for which the formula produces a real number — find the domain and $h(6)$ .

What is the domain of $h$ ?

What is $h(6)$ ?

✅ Correct! The domain is

\{x \in \mathbb{R} \mid x \geq -3 \text{ and } x \neq 2\}

, and

h(6) = \tfrac{\sqrt{9}}{4} = \tfrac{3}{4}

❌ Not quite. Division by zero is excluded, but so is the square root of a negative number:

x = -7

gives

\sqrt{-4}

, which is not real.

❌ Not quite. The square root is fine for

x \geq -3

, but

x = 2

makes the denominator

0

, so

h(2)

is undefined.

❌ Not quite. At

x = -3

the formula gives

\tfrac{\sqrt{0}}{-5} = 0

, a real number, so

-3

belongs to the domain.

❌ Not quite. Two conditions are needed:

x + 3 \geq 0

for the root and

x - 2 \neq 0

for the quotient.

❌ Not quite.

\tfrac{9}{4}

forgets the square root:

\sqrt{6 + 3} = \sqrt{9} = 3

, not

9

❌ Not quite. The symbol

\sqrt{9}

means the nonnegative root,

3

. A function gives exactly one output, so

h(6)

is a single number.

❌ Not quite. Check the denominator:

6 - 2 = 4

, not

8

❌ Not quite. Substitute

x = 6

\sqrt{6 + 3} = 3

on top and

6 - 2 = 4

on the bottom.

Show solution

Domain. The formula produces a real number exactly when both pieces make sense:

x + 3 \geq 0 \quad\text{and}\quad x - 2 \neq 0, \qquad\text{that is,}\qquad x \geq -3 \text{ and } x \neq 2.

So the domain is $\{x \in \mathbb{R} \mid x \geq -3 \text{ and } x \neq 2\}$ . Note that $-3$ is included, since $h(-3) = \tfrac{0}{-5} = 0$ .

Output.

h(6) = \frac{\sqrt{6+3}}{6-2} = \frac{\sqrt{9}}{4} = \frac{3}{4}.

The square root symbol denotes the nonnegative root, so there is exactly one output.

Problem 4 · Arrow Diagrams and the Range

Given: $A = \{1, 2, 3, 4\}$ and $B = \{p, q, r, s\}$ . Each option lists the arrows of a diagram from $A$ to $B$ . For a function $f\colon A \to B$ , the range is $\{f(x) \mid x \in A\}$ , the set of outputs that actually occur.

Which diagram represents a function from $A$ to $B$ ?

What is the range of that function?

✅ Correct! Exactly one arrow leaves each element of

A

, so (c) is a function, and its only output is

p

: the range is

\{p\} \subseteq B

❌ Not quite. No arrow leaves

4

, so the input

4

has no output. Every element of the domain needs an output.

❌ Not quite. Two arrows leave

1

(to

p

and to

q

), so the input

1

has two outputs.

❌ Not quite. Two arrows leave

2

(to

r

and to

s

), so the input

2

has two outputs.

❌ Not quite. Count the arrows leaving each of

1, 2, 3, 4

: a function needs exactly one from each.

❌ Not quite.

\{p, q, r, s\}

is the codomain. The range contains only the outputs that occur, and here every arrow ends at

p

❌ Not quite.

\{1, 2, 3, 4\}

is the domain, the set of inputs. The range is a subset of the codomain

B

❌ Not quite. Every input has the output

p

, so

p

is in the range; the range of a function with a nonempty domain is never empty.

❌ Not quite. List

f(1), f(2), f(3), f(4)

and collect the distinct values.

Show solution

A diagram represents a function $A \to B$ exactly when one arrow leaves each element of $A$ .

(a): no arrow leaves $4$ . Not a function.
(b): two arrows leave $1$ . Not a function.
(c): one arrow leaves each of $1, 2, 3, 4$ . A function, even though all four arrows end at $p$ and $q$ , $r$ , $s$ receive none.
(d): two arrows leave $2$ . Not a function.

For (c), $f(1) = f(2) = f(3) = f(4) = p$ , so the range is

\{f(x) \mid x \in A\} = \{p\},

a proper subset of the codomain $B = \{p, q, r, s\}$ .

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