One-to-one means each output identifies one input

Explain it like I am five

Imagine a coat check where every ticket number belongs to one coat, and every coat has its own ticket number. You can start at either side and recover one partner. A one-to-one function has that extra property. It already gives one output for each input, as every function must. It also keeps different inputs from sharing an output. If two account numbers show the same balance, the balance cannot tell you which account you meant. The forward rule can still be a function, but the backward search is ambiguous. Always name the accepted inputs before deciding whether the extra property holds.

The grade table has a unique partner in either direction.

Reminder

Domain and range. Domain lists allowed inputs; range lists outputs actually reached. Restricting $x^{2}$ to x ≥ 0 removes the negative-input duplicates.

Why it works. A function's definition controls the forward direction only. It allows two inputs to point to the same output. The one-to-one condition adds a separate restriction on the backward direction. If an output has two possible inputs, knowing that output cannot recover a unique original input. If every output the function actually produces has one input, reversing the pairs gives a function from the original range back to the original domain.

RuleA one-to-one function gives different outputs to different inputs. Equivalently, each value in its range comes from exactly one input. Check function status first, then check for shared outputs.

The same idea, five ways

Say it

Different inputs give different outputs

Write it

Every output actually produced identifies exactly one input.

In math

f(a) = f(b) only when a = b
For $x^{2}$ : f(−2) = f(2) = 4
For $x^{3}$ : f(−2) = −8 and f(2) = 8

Every coat has its own ticket, so either partner identifies the other.

See it

Each input has one arrow out, and each listed output has one arrow in.

The same idea, other ways

As a coat check

Different tickets need different coats. If two tickets point to the same coat, the coat alone cannot identify which ticket to return.

As a backward machine

A one-to-one function can be read backward from every output it actually produces without having to choose between two inputs.

Each listed output receives an arrow from only one letter.

.1Square versus cube

Squaring is like removing the direction from a walk before measuring its size. Input 2 and input −2 both give 4. You can still calculate one square for either input, so the square rule is a function. The result loses which side of zero you started on. Cubing multiplies three copies instead of two. Input 2 gives 8, while input −2 gives −8, so the direction survives. More generally, as an input increases, its cube increases. Every real output has one cube root, so the cube rule lets you recover the original input.

$x^{2}$ is a function but is not one-to-one on (−∞, ∞).
$x^{3}$ is a one-to-one function on (−∞, ∞).
If the square rule is restricted to x ≥ 0, it becomes one-to-one. The domain matters.
For positive inputs, multiplying a larger positive number three times gives a larger cube. For negative inputs, the farther-left number has a larger positive magnitude, so cubing that magnitude and restoring the minus gives a smaller cube. Across zero the signs differ. Thus a bigger real input always gives a bigger cube, and two different inputs cannot have the same cube.
The toolkit rule $x^{3}$ is one-to-one. Other cubic functions may share outputs, as the counterexample below shows.

Two different inputs reach the same height on the square curve.

Reminder

Signed powers. (−2 $) squared$ = 4 but (−2 $) cubed$ = −8: two negative factors give positive, and the third makes negative.

The same idea, five ways

Say it

Negative two squared and two squared both equal four; negative two cubed equals negative eight.

Write it

Squaring loses the input’s sign, while the toolkit cube rule lets each output identify one input.

In math

(−2 $) squared$ = $2 squared$ = 4
(−2 $) cubed$ = −8
$2 cubed$ = 8
$x^{3}$ = y gives x = $\sqrt[3]{y}$
For x ≥ 0, $x^{2}$ = y gives x = $\sqrt{y}$

A distance may hide which side you stood on; a signed direction preserves that choice.

See it

Two different inputs reach the same height on the square curve.

Worked exampleA repeated square and a unique negative cube

First decide whether the square outputs for inputs −1 and 1 are distinct. Then solve $x^{3}$ = −27, meaning find the input whose cube output is −27.

The repeated output defeats one-to-one but still permits a function.

A negative cube output identifies one negative input.

What it asks. Compare two squares, then recover the input of a negative cube.

Plan. Multiply the signed factors and take one cube root for = −27.

Two negative factors give positive; three give negative.
A cube root has one real answer, with no ±.

Compute (−1 $) squared$ = 1 and $1 squared$ = 1; the outputs are equal.Squaring multiplies two copies, and two negative copies have a positive product.
Write x = −3 since (−3 $) cubed$ = (−3) × (−3) × (−3) = −27.A cube root finds the single real input that produces the given cube.

Answer

The two square outputs are both 1, so they are not distinct.
$x^{3}$ = −27 gives x = −3.

Check For the square, reverse output 1 and recover both −1 and 1. For the cube, multiply the three copies of −3 and recover −27.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: The square rule is not a function because f(−2) = f(2).

A function may have shared outputs. The definition forbids one input from having two outputs.

✓ Instead: is a function, but it is not one-to-one on all real inputs.

✗ Not this: Every cubic function is one-to-one because is.

The cubic v(x) = − 4x gives v(−2) = 0, v(0) = 0 and v(2) = 0.

✓ Instead: The toolkit cubic is one-to-one. A different cubic may fail; three inputs of v share output 0.

Tips and tricks

A single repeated output disproves one-to-one. To prove it for all real inputs, use the rule's behavior rather than a short table.
(−1 $) squared$ = $1 squared$ = 1, but (−1 $) cubed$ = −1 and $1 cubed$ = 1. The square loses the sign; the cube keeps it.
Output 4 under the square rule leaves two possible inputs, −2 and 2. Output 8 under the cube rule leaves only input 2.
As an everyday comparison: Squaring is like removing the direction from a walk before measuring its size. Input 2 and input −2 both give 4. You can still calculate one square for either input, so the square rule is a function. The result loses which side of zero you started on. Cubing multiplies three copies instead of two. Input 2 gives 8, while input −2 gives −8, so the direction survives. More generally, as an input increases, its cube increases. Every real output has one cube root, so the cube rule lets you recover the original input.
With the worked values: For the square, reverse output 1 and recover both −1 and 1. For the cube, multiply the three copies of −3 and recover −27.

.2Radius and area

Recall the circle-area rule from the refresher: A = π $r^{2}$ , where the radius r is a positive length. Picture enlarging a wheel. A radius of 2 gives area 4π, and a radius of 3 gives area 9π. Increasing a positive radius always increases its area. Therefore one area cannot belong to two different positive radii. You can work backward by dividing the area by π, then taking the positive square root. The positive-radius domain matters because a negative algebraic candidate is not a permitted length.

Circle: the set of points at one fixed distance from a center. Radius: that distance. Area: the space inside.
Area rule: A(r) = π $r^{2}$ , with domain r > 0 and range A > 0.
Backward rule: r = $\sqrt{\frac{A}{π}}$ , the positive radius. Dividing by π and then taking the square root undoes the forward operations.
A larger positive radius has a larger square, and multiplying by the same positive π keeps that order. Thus every positive area identifies one positive radius.

The radius runs from center to edge. The shaded interior reminds you that area measures the space inside; the area of the whole circle is π.

Reminder

Square root versus squared equation. $square root of 25$ = 5. Solving $r^{2}$ = 25 gives r = 5 or r = −5 before checking the domain.
Circle area. A = π $r^{2}$ , with positive radius r. For r = 2, the area is 4π square units.

The same idea, five ways

Say it

Area equals pi times radius squared; radius is the positive square root of the whole quotient area divided by pi.

Write it

Each positive circle area recovers exactly one allowed positive radius.

In math

A(r) = π $r^{2}$
r > 0
A > 0
r = $\sqrt{\frac{A}{π}}$
A = 25π gives r = 5

A larger wheel covers more area; the covered area recovers one positive radius.

See it

The radius runs from center to edge. The shaded interior reminds you that area measures the space inside; the area of the whole circle is π.

Worked exampleA positive radius gives a unique circle area

Find the area for radius 1, meaning evaluate the area rule at r = 1. Then find the positive radius for area 25π, meaning solve π $r^{2}$ = 25π for the allowed input r.

The recovered positive radius returns the requested area.

What it asks. Find an area from a radius, then find a positive radius from an area.

Plan. Use A = π forward. Divide by π and take the positive root to go backward.

Keep π for an exact answer.
r > 0 excludes the negative square-root candidate.
Check by returning the radius to A = π $r^{2}$ .

A(1) = π × $1 squared$ = π.The area formula squares the given radius and multiplies by π.
Divide π $r^{2}$ = 25π by π on both sides to get $r^{2}$ = 25.π is nonzero, so equal division preserves the equation and cancels the shared factor.
The squared equation allows 5 and −5, but keep r = 5.The radius domain is r > 0, so −5 is not an allowed input.
More generally, any positive area A gives the unique positive radius r = $\sqrt{\frac{A}{π}}$ . Therefore the area rule is one-to-one for r > 0.Dividing by π and taking the positive square root recovers one allowed input from every output in the range.

Answer

Radius 1 gives area π square units.
Area 25π square units gives radius 5 units.
The circle-area function A = π $r^{2}$ is one-to-one for r > 0.

Check Substitute the recovered radius: π × = 25π. The area and radius are positive, as the stated domains require.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Area 25π gives two radii, 5 and −5, so area is not one-to-one.

The algebraic squared equation has two real solutions, but −5 is outside the positive-radius domain.

✓ Instead: The single allowed radius is 5, so the area rule is one-to-one for r > 0.

Tips and tricks

Write r > 0 before working backward. The domain tells you which square-root candidate to keep.
Leave π in an exact area answer; a decimal for π makes the answer approximate.
For positive radii 1 and 2, the areas are π and 4π. Increasing the radius increases the covered space.
Forward: square the positive radius, then multiply by π. Backward: divide the area by π, then choose the positive square root.
As an everyday comparison: Recall the circle-area rule from the refresher: A = π $r^{2}$ , where the radius r is a positive length. Picture enlarging a wheel. A radius of 2 gives area 4π, and a radius of 3 gives area 9π. Increasing a positive radius always increases its area. Therefore one area cannot belong to two different positive radii. You can work backward by dividing the area by π, then taking the positive square root. The positive-radius domain matters because a negative algebraic candidate is not a permitted length.
With the worked values: Substitute the recovered radius: π × $5 squared$ = 25π. The area and radius are positive, as the stated domains require.

.3A letter grade identifies one grade-point value

Imagine labels attached to four storage boxes. Each label names one box, and each box has its own label. The letter-grade table works that way: A gives 4 points, B gives 3, C gives 2, and D gives 1. You can look in either direction and recover one partner. Other grading rules can group different percents into one letter, so do not decide from the word grade alone. Inspect the actual pairs and state which direction you mean.

For this table, letters A, B, C and D give grade points 4, 3, 2 and 1.
Every letter has one output and every output has one letter input, so the table gives a one-to-one function.
Output 3 identifies B because the output row has exactly one 3.

Each output appears once. The output 3 identifies input B.

Reminder

Shared outputs. 101 and 205 can both give 90 without breaking a function. They do break one-to-one.

The same idea, five ways

Say it

A gives four points, B gives three, C gives two, and D gives one.

Write it

Each letter has one point value, and each listed point value identifies one letter.

In math

A → 4
B → 3
C → 2
D → 1
3 → B

Each storage box has its own unique label, so either partner identifies the other.

See it

Each output appears once. The output 3 identifies input B.

Worked exampleRecover a letter from its grade points

In the table, A gives 4 points, B gives 3, C gives 2 and D gives 1. Which letter gives 3 grade points? Is this table one-to-one?

Each output appears once. The output 3 identifies input B.

What it asks. Read backward from output 3 and decide whether any output has two letters.

Plan. Find 3 in the output row, read up, then inspect every listed output.

Look above the single 3.
Each of 4, 3, 2 and 1 appears once.

The column above output 3 has input B.A backward lookup finds the input paired with the given output.
The letter-to-points function is one-to-one.Each letter has one point value and each listed point value appears once.

Answer

3 grade points identifies B.
The table is one-to-one.

Check The other outputs identify A, C and D respectively. No output is shared.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Every grading rule is one-to-one.

Some rules put many different percents into one letter. The later grade-band example compares 81 and 88.

✓ Instead: The letter-to-points table here is one-to-one. Inspect other grading rules separately.

Tips and tricks

Before reversing a table, check whether its output row contains repeated values.
The number of partners, rather than the context word grade, decides.
As an everyday comparison: Imagine labels attached to four storage boxes. Each label names one box, and each box has its own label. The letter-grade table works that way: A gives 4 points, B gives 3, C gives 2, and D gives 1. You can look in either direction and recover one partner. Other grading rules can group different percents into one letter, so do not decide from the word grade alone. Inspect the actual pairs and state which direction you mean.
With the worked values: The other outputs identify A, C and D respectively. No output is shared.

.4Account numbers can share a balance

Picture two bank accounts holding the same amount. A snapshot gives one balance for each account number, so account number to balance is a function. But the balance alone does not identify which account it came from. The two accounts share an output, so the function is not one-to-one.

Account 101 has $90, account 205 has $90, and account 309 has $140.
At this moment, balance is a function of account number. It is not one-to-one because 101 and 205 share 90.
In the reverse direction, balance 90 has two account outputs, so account number is not a function of balance for these data.

Two 90s point back to two different account numbers.

The same idea, five ways

Say it

Account one hundred one and account two hundred five each have a ninety-dollar balance.

Write it

A shared balance leaves one output per account but prevents that balance from identifying one account.

In math

101 → 90
205 → 90
309 → 140
90 → 101 and 90 → 205

Two coats can have the same price, so the price alone need not name the coat.

See it

Two 90s point back to two different account numbers.

Worked exampleCheck the two directions for balances

Use the account table. Is balance a function of account number? Is account number a function of balance? Is the forward rule one-to-one?

The shared output 90 makes the backward question ambiguous.

What it asks. Check both directions and the forward rule's one-to-one property.

Plan. Read from account to balance first, then count partners for output 90.

Each account column has one balance.
Output 90 appears under two different accounts.

Balance is a function of account number.Each listed account has one balance at the chosen moment.
Account number is not a function of balance for these accounts.Balance 90 corresponds to both 101 and 205.
Account number to balance is not one-to-one.Two different accounts share the same output 90.

Answer

Balance is a function of account number.
Account number is not a function of balance.
The forward function is not one-to-one.

Check The single balance in each column proves the first claim. The two columns holding 90 disprove the other two claims.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Shared balances mean account number to balance is not a function.

Each account still has exactly one balance. Shared outputs are allowed in a function.

✓ Instead: It is a function, but not one-to-one.

Tips and tricks

Name the direction before counting partners.
Write: accounts 101 and 205 both give 90, so the function is not one-to-one.
As an everyday comparison: Picture two bank accounts holding the same amount. A snapshot gives one balance for each account number, so account number to balance is a function. But the balance alone does not identify which account it came from. The two accounts share an output, so the function is not one-to-one.
With the worked values: The single balance in each column proves the first claim. The two columns holding 90 disprove the other two claims.

.5A grade band groups several inputs

Imagine placing several scores in one labeled box. Under a rule where every whole-number percent from 80 through 89 earns B, the letter identifies a group rather than one exact score. Each percent earns one letter, so percent to letter is a function. The shared letter cannot recover one percent, so it is not one-to-one.

For this rule, every percent from 80 through 89 earns B.
Inputs 81 and 88 both give output B.
Giving one grade per score establishes function status. Sharing B defeats one-to-one.

Two different percent inputs share the same letter output.

The same idea, five ways

Say it

Eighty-one and eighty-eight both give the letter B.

Write it

The grade band groups different percent inputs under one shared letter output.

In math

81 → B
88 → B
(81, B) and (88, B)

One labeled box can hold several different scores.

See it

Two different percent inputs share the same letter output.

Worked exampleCan B identify one percent?

Every percent from 80 through 89 earns B. Does B tell you whether the percent was 81 or 88? Is this percent-to-letter rule one-to-one?

One shared letter prevents unique backward recovery.

What it asks. Decide whether output B recovers one exact percent.

Plan. Apply the stated band to both inputs and compare their letters.

81 and 88 both lie in the stated band.
Different inputs with one shared output defeat one-to-one.

81 gives B, and 88 gives B.Both scores lie from 80 through 89.
B cannot tell which score was used, so the rule is not one-to-one.The same output comes from two different inputs.

Answer

B does not distinguish 81 from 88.
The percent-to-letter function is not one-to-one.

Check Looking backward from B finds both listed inputs, so the backward question has more than one answer.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: One letter per percent means one-to-one.

That checks only the forward function requirement.

✓ Instead: Also check whether different percent inputs share a letter.

Tips and tricks

A function may deliberately group inputs.
For one-to-one, each produced output needs one input partner.
As an everyday comparison: Imagine placing several scores in one labeled box. Under a rule where every whole-number percent from 80 through 89 earns B, the letter identifies a group rather than one exact score. Each percent earns one letter, so percent to letter is a function. The shared letter cannot recover one percent, so it is not one-to-one.
With the worked values: Looking backward from B finds both listed inputs, so the backward question has more than one answer.

.6One-to-one from a table

Look at the output row like a row of coat tags. A tag repeated under different inputs cannot tell you which input to return. In the g table, output 6 belongs to inputs 2 and 4; output 8 belongs to inputs 1 and 5. Every input still has one output, so g is a function. Its repeated outputs mean it is not one-to-one.

g(2) = g(4) = 6, and g(1) = g(5) = 8.
The listed relation is a function because each listed input has exactly one output. Distinct input labels identify separate columns; the decisive condition is that no input is assigned different outputs.
In contrast, the letter-grade table has no shared outputs and is one-to-one.

The output 6 appears under inputs 2 and 4. The output 8 also repeats, under inputs 1 and 5.

The same idea, five ways

Say it

g of two and g of four both equal six.

Write it

Each listed input has one output, but the repeated output 6 prevents unique backward recovery.

In math

g(2) = g(4) = 6
g(1) = g(5) = 8
2 ≠ 4
(2, 6) and (4, 6)

A repeated coat tag cannot tell you which of two tickets to return.

See it

The output 6 appears under inputs 2 and 4. The output 8 also repeats, under inputs 1 and 5.

Worked exampleCheck the g table for one-to-one

Use the pictured g table. Is g a function? Is it one-to-one? Name two inputs sharing an output.

The output 6 appears under inputs 2 and 4. The output 8 also repeats, under inputs 1 and 5.

What it asks. Check input partners first, then output partners.

Plan. Look for conflicting outputs at one input. Then inspect the repeated 6s and 8s.

No input has two outputs.
Both 2 and 4 give 6.

g is a function.Each listed input has exactly one output.
g(2) = g(4) = 6, so g is not one-to-one.Inputs 2 and 4 are different but share one output.
g(1) = g(5) = 8 supplies a second repeated output.The 8s are under two different input columns.

Answer

g is a function.
g is not one-to-one.
Inputs 2 and 4 share 6; inputs 1 and 5 share 8.

Check Reading down each column gives one output. Reading up from either repeated output gives two inputs, which checks the two definitions separately.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Repeating output 6 makes g fail the function definition.

Only one input with conflicting outputs would break a function.

✓ Instead: The repeated output breaks one-to-one, while g remains a function.

Tips and tricks

Function: inspect input partners. One-to-one: inspect output partners.
A shared output disproves one-to-one; the letter-grade table illustrates a table with no shared output.
As an everyday comparison: Look at the output row like a row of coat tags. A tag repeated under different inputs cannot tell you which input to return. In the g table, output 6 belongs to inputs 2 and 4; output 8 belongs to inputs 1 and 5. Every input still has one output, so g is a function. Its repeated outputs mean it is not one-to-one.
With the worked values: Reading down each column gives one output. Reading up from either repeated output gives two inputs, which checks the two definitions separately.

.7A horizontal line can meet three times

A curve can climb, turn and return to the same height. Three visits to height 0 are enough to show that the output does not identify one input. The cubic v(x) = $x^{3}$ − 4x gives output 0 at inputs −2, 0 and 2. It is still a function because its formula calculates one answer at each input, but it is not one-to-one.

v(−2) = −8 + 8 = 0; v(0) = 0; v(2) = 8 − 8 = 0.
The three graph points (−2, 0), (0, 0) and (2, 0) lie on the horizontal line y = 0.
The toolkit cubic $x^{3}$ is one-to-one. The cubic v(x) = $x^{3}$ − 4x is not.

For v(x) = − 4x, three different inputs share output 0. The horizontal line at height 0 has three intersections.

The same idea, five ways

Say it

v of negative two, v of zero, and v of two all equal zero.

Write it

Three different inputs share output 0, so one horizontal level meets the graph three times.

In math

v(x) = $x^{3}$ − 4x
v(−2) = v(0) = v(2) = 0
(−2, 0), (0, 0), (2, 0)
y = 0

Three trains may share one arrival time, so that time cannot name a unique train.

See it

For v(x) = − 4x, three different inputs share output 0. The horizontal line at height 0 has three intersections.

Worked exampleThree zero outputs from one cubic

For v(x) = $x^{3}$ − 4x, find v(−2), v(0) and v(2). What do those outputs tell you about one-to-one?

For v(x) = − 4x, three different inputs share output 0. The horizontal line at height 0 has three intersections.

What it asks. Find three outputs and see whether the same height identifies one input.

Plan. Use parentheses around each input, evaluate, and compare the results.

(−2 $) cubed$ = −8 and −4(−2) = 8.
One shared output is enough to disprove one-to-one.

v(−2) = (−2 $) cubed$ − 4(−2) = −8 + 8 = 0.A negative cube is negative, while the two negatives in −4(−2) make positive.
v(0) = $0 cubed$ − 4(0) = 0.Both terms are zero.
v(2) = $2 cubed$ − 4(2) = 8 − 8 = 0.Cube 2, then subtract four copies of 2.
v is not one-to-one.Three different inputs share output 0, so the horizontal line y = 0 meets its graph three times.

Answer

v(−2) = 0
v(0) = 0
v(2) = 0
v is not one-to-one.

Check Factor v(x) as x( − 4). Inputs −2, 0 and 2 each make a factor zero, confirming all three outputs.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: A horizontal line can meet a function only twice.

This function has three points at height 0. The function definition controls vertical lines, not horizontal ones.

✓ Instead: A horizontal line can meet a function any number of times. Two or more intersections defeat one-to-one.

Tips and tricks

Write the repeated equality, then the conclusion.
Do not extend a toolkit property to every formula of the same degree.
As an everyday comparison: A curve can climb, turn and return to the same height. Three visits to height 0 are enough to show that the output does not identify one input. The cubic v(x) = $x^{3}$ − 4x gives output 0 at inputs −2, 0 and 2. It is still a function because its formula calculates one answer at each input, but it is not one-to-one.
With the worked values: Factor v(x) as x( $x^{2}$ − 4). Inputs −2, 0 and 2 each make a factor zero, confirming all three outputs.

Strategy: step by step

State the domain and the direction of the rule.
Confirm that each input has one output.
Search for two different inputs with the same output. One such pair proves the function is not one-to-one.
If claiming one-to-one, explain why every output the function actually produces recovers exactly one allowed input. A few sample pairs alone do not prove it for an infinite domain.

Strategy

Check one-to-one after checking function status

Does one input have two different outputs?

YesIt is not a function. Stop before asking whether it is one-to-one.

NoNow check the outputs.

↓

Does the table have a shared output?

YesName two inputs with that same output. It is not one-to-one.

NoIf the table lists every allowed input, its outputs each identify one input.

↓

Is the rule given by a graph?

YesLook for a horizontal line meeting it twice or more.

NoFor a formula, a pair of repeated outputs disproves one-to-one. To prove it for all inputs, explain why an output can come from only one input.

State the domain and the direction of the rule.
Confirm that each input has one output.
Search for two different inputs with the same output. One such pair proves the function is not one-to-one.
If claiming one-to-one, explain why every output the function actually produces recovers exactly one allowed input. A few sample pairs alone do not prove it for an infinite domain.

Worked exampleOne repeated square and one recoverable cube

Decide whether f(x) = $x^{2}$ and g(x) = $x^{3}$ are one-to-one on all real inputs. The question asks whether every output they actually produce identifies exactly one input.

A function may give the same output to two inputs, which defeats one-to-one.

The cube rule gives distinct outputs to distinct inputs.

What it asks. Check whether a square output and a cube output each identify one input.

Plan. Find a repeated square. Then explain why each cube has one cube root.

Try inputs 2 and −2 for the square.
The toolkit cubic is $x^{3}$ . Its negative and positive inputs have different cubes.
A few distinct sample outputs alone do not prove the full rule is one-to-one.

f(2) = 4 and f(−2) = 4, so f is not one-to-one.Two distinct allowed inputs share one output.
An output y of g has the single input x = $\sqrt[3]{y}$ , so g is one-to-one.As the input goes up, its cube goes up. Every real output has one real cube root, so two different inputs cannot share a cube.

Answer

f(x) = $x^{2}$ : not one-to-one on all real inputs.
g(x) = $x^{3}$ : one-to-one on all real inputs.

Check The square output 4 recovers both −2 and 2. The cube output 8 recovers only 2; input −2 instead gives −8.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Any function can be reversed into a function.

Shared outputs become repeated backward inputs with different answers.

✓ Instead: A function reverses to a function on its range when it is one-to-one.

✗ Not this: An output never produced by the rule must have one input too.

One-to-one refers to outputs in the range; values the function never produces have no partner.

✓ Instead: For restricted to x ≥ 0, output −1 has no input. Every output the function actually produces identifies one input.

Tips and tricks

Remember one-to-one as one partner going forward and one partner coming back.
A counterexample needs only two distinct inputs with the same output.
On an exam, write a witness: g(2) = g(4) = 6, so g is not one-to-one.
A vertical line represents one input. A horizontal line represents one output.

Trap. Calling a function one-to-one after checking only that each input has one output. Also check that no two different inputs share an output, and state the domain.