Mathematical-Reasoning · Unit 22 · Video 4 · Interactive Practice

Domain and Codomain Decide: $x^2 + 1$ and $e^{-x}$

IKey Formulas

Statement	Name	What it depends on
For all $s, t \in A$ : $f(s) = f(t) \Rightarrow s = t$	$f : A \to B$ is an injection	The domain $A$ only; changing $B$ cannot make two inputs share or stop sharing an output
For all $y \in B$ there is $x \in A$ with $f(x) = y$	$f : A \to B$ is a surjection	The codomain $B$ , and the domain $A$ through the range
$x^2 + 1$ : neither on $\mathbb{R} \to \mathbb{R}$ ; onto only on $\mathbb{R} \to T$ ; one-to-one only on $\mathbb{Z}^* \to \mathbb{N}$	One formula, three verdicts	$T = \{y \in \mathbb{R} \mid y \ge 1\}$ , $\mathbb{Z}^* = \{x \in \mathbb{Z} \mid x \ge 0\}$
$g(x) = e^{-x}$ , $g : \mathbb{R} \to \mathbb{R}^+$ ; for $b > 0$ , $g(-\ln b) = b$	A bijection	$b > 0$ makes $\ln b$ defined; for $f : \mathbb{R} \to \mathbb{R}$ with the same rule, $-1$ is never an output

Key Insight: A function is its rule together with its domain and its codomain. Change either set and you have a different function, whose verdicts must be checked again.

IIVisualization 1 — One Formula, Twelve Choices of Sets

Ten of the twelve choices of domain $A$ and codomain $B$ turn $x^2 + 1$ into a different function with its own verdicts; with domain $\mathbb{R}$ or $[0, \infty)$ and codomain $\mathbb{N}$ , the rule gives outputs outside $\mathbb{N}$ , so it defines no function at all.

Domain A Codomain B Height y 5

IIIVisualization 2 — Solving $e^{-a} = b$ Backward

The input $a = -\ln b$ exists exactly when $b > 0$ , so the codomain decides whether $e^{-x}$ is onto.

Codomain Target b 4

💡 The horizontal line at any height meets the curve at most once, which suggests one-to-one; the proof takes $\ln$ of both sides of $e^{-a} = e^{-b}$ and is the same for $f$ and $g$ , because they share the domain $\mathbb{R}$ .

IVQuiz Questions

Problem 1 · Absolute Value on ℝ

Given: $h : \mathbb{R} \to \mathbb{R}$ , $h(x) = |x - 2|$ . Which statement is true?

✅ Correct!

h(0) = h(4) = 2

breaks one-to-one, and

-1

is in the codomain but

|x - 2| \ge 0

never equals it.

❌ Not quite. Two inputs the same distance from

2

share an output:

h(0) = |-2| = 2

and

h(4) = |2| = 2

, with

0 \ne 4

❌ Not quite. The codomain is all of

\mathbb{R}

, and an absolute value is never negative, so

-1

is never an output.

❌ Not quite. A bijection needs both properties, and

h

has neither:

h(0) = h(4)

, and

h(x) = -1

has no solution.

Show solution

Not an injection. One pair of inputs is enough: $h(0) = |0 - 2| = 2$ and $h(4) = |4 - 2| = 2$ , but $0 \ne 4$ .

Not a surjection. For every real $x$ , $|x - 2| \ge 0$ . So $-1 \in \mathbb{R}$ , the codomain, is not $h(x)$ for any $x$ , and $-1$ is not in the range.

So $h$ is neither. (With codomain $[0, \infty)$ instead, $h$ would be onto, since $h(2 + y) = y$ for $y \ge 0$ , but still not one-to-one.)

Problem 2 · Restricting the Domain

Given: $k : [3, \infty) \to \mathbb{R}$ , $k(x) = x^2 - 6x$ . Which statement is true?

✅ Correct! On

[3, \infty)

k(s) = k(t)

forces

s = t

; but

k(x) = (x - 3)^2 - 9 \ge -9

, so

-10

is never an output.

❌ Not quite. Restricting the domain fixed one-to-one, but the codomain is still

\mathbb{R}

, and

k(x) = (x - 3)^2 - 9 \ge -9

misses

-10

❌ Not quite. On all of

\mathbb{R}

k(2) = k(4) = -8

, but

2

is not in the domain

[3, \infty)

. And the range is only

[-9, \infty)

❌ Not quite. The pair

2, 4

that shares an output on

\mathbb{R}

is gone: for

s, t \ge 3

the inputs never share an output. Only onto fails.

Show solution

Injection. Let $s, t \in [3, \infty)$ with $k(s) = k(t)$ . Then $s^2 - 6s = t^2 - 6t$ , so

s^2 - t^2 - 6(s - t) = (s - t)(s + t - 6) = 0.

Since $s \ge 3$ and $t \ge 3$ , $s + t - 6 \ge 0$ , with equality only when $s = t = 3$ . Either way $s = t$ : if $s + t - 6 \ne 0$ then $s - t = 0$ , and if $s + t - 6 = 0$ then $s = t = 3$ .

Not a surjection. Completing the square, $k(x) = (x - 3)^2 - 9 \ge -9$ for every $x$ . So $-10 \in \mathbb{R}$ is not in the range.

The domain decided the injection verdict; the codomain $\mathbb{R}$ still contains numbers below $-9$ that are never outputs.

Problem 3 · Choose the Codomain

Given: the rule $k(x) = x^2 - 6x$ on the domain $[3, \infty)$ from Problem 2. For which codomain $C$ is $k : [3, \infty) \to C$ a bijection, and which input does an arbitrary $y \in C$ come from?

Which codomain?

Which input x gives k(x) = y?

✅ Correct! With

C = [-9, \infty)

, every

y \in C

has

y + 9 \ge 0

, so

x = 3 + \sqrt{y + 9}

is a real number at least

3

, and

k(x) = y

❌ Not quite.

k(3) = 9 - 18 = -9 \notin [0, \infty)

, so the rule does not even define a function from

[3, \infty)

[0, \infty)

❌ Not quite.

k(3) = -9

must lie in the codomain, and the open interval

(-9, \infty)

leaves it out, so this is not a function.

❌ Not quite. That is Problem 2: with codomain

\mathbb{R}

, the value

-10

is never reached.

❌ Not quite. This does solve

(x - 3)^2 = y + 9

, but it is at most

3

, so for

y > -9

it is not in the domain

[3, \infty)

❌ Not quite. Check forward with

y = 7

x = \sqrt{16} - 3 = 1

, which is not in the domain, and

k(1) = -5 \ne 7

❌ Not quite. Completing the square gives

(x - 3)^2 = y + 9

, not

y - 9

; also

\sqrt{y - 9}

is undefined for

y < 9

Show solution

The codomain. $k(x) = (x - 3)^2 - 9$ . For $x \ge 3$ , $(x - 3)^2$ takes every value in $[0, \infty)$ , so the range is $[-9, \infty)$ . A bijection needs codomain = range, and every output (including $k(3) = -9$ ) must lie in the codomain. So $C = [-9, \infty)$ .

Work backward. Let $y \in C$ . From $(x - 3)^2 - 9 = y$ we get $(x - 3)^2 = y + 9$ , and $y + 9 \ge 0$ , so $x - 3 = \pm\sqrt{y + 9}$ . The domain needs $x \ge 3$ , so take $x = 3 + \sqrt{y + 9}$ .

Verify forward.

k\big(3 + \sqrt{y + 9}\big) = \big(\sqrt{y + 9}\big)^2 - 9 = y + 9 - 9 = y.

Together with the injection from Problem 2 (which depends only on the domain), $k : [3, \infty) \to [-9, \infty)$ is a bijection.

Problem 4 · Reciprocals on Two Domains

Given: $p : (0, \infty) \to (0, \infty)$ and $q : \mathbb{R} - \{0\} \to \mathbb{R}$ , both defined by the rule $\dfrac{1}{x}$ . Which statement is true?

✅ Correct! Both are one-to-one;

p

reaches every

y > 0

via

x = \tfrac{1}{y}

, while

q

misses

0

❌ Not quite. The codomain of

q

contains

0

, and

\tfrac{1}{x} = 0

has no solution, so

q

is not onto.

❌ Not quite. Unlike

x^2

, the reciprocal never merges a negative input with a positive one:

\tfrac{1}{s} = \tfrac{1}{t}

gives

s = t

on all of

\mathbb{R} - \{0\}

❌ Not quite.

p

is onto: for

y \in (0, \infty)

, the input

x = \tfrac{1}{y}

is in

(0, \infty)

and

p(\tfrac{1}{y}) = y

Show solution

Both are injections. If $\tfrac{1}{s} = \tfrac{1}{t}$ with $s, t \ne 0$ , multiply both sides by $st$ to get $t = s$ . This works on either domain.

$p$ is a surjection. Let $y \in (0, \infty)$ . Work backward: $\tfrac{1}{x} = y$ gives $x = \tfrac{1}{y}$ , which is defined and positive because $y > 0$ . Verify: $p\big(\tfrac{1}{y}\big) = \dfrac{1}{1/y} = y$ . So $p$ is a bijection.

$q$ is not a surjection. $0 \in \mathbb{R}$ , but $\tfrac{1}{x} = 0$ would need $1 = 0 \cdot x = 0$ . So $0$ is in the codomain but not in the range.

Every other real $y \ne 0$ is reached by $x = \tfrac{1}{y}$ , so changing the codomain of $q$ to $\mathbb{R} - \{0\}$ would make it a bijection too.

Solved: 0 / 4

Domain and Codomain Decide: x2+1x^2 + 1x2+1 and e−xe^{-x}e−x

IKey Formulas

IIVisualization 1 — One Formula, Twelve Choices of Sets

IIIVisualization 2 — Solving e−a=be^{-a} = be−a=b Backward

IVQuiz Questions

Domain and Codomain Decide: $x^2 + 1$ and $e^{-x}$

IIIVisualization 2 — Solving $e^{-a} = b$ Backward