Order changes the result, and units must fit

Explain it like I am five

Picture a store applying a discount and a coupon. A 64-dollar price falls to 48 dollars after 25% off. An 8-dollar coupon then leaves 40 dollars. Reverse the jobs: the coupon leaves 56 dollars, and 25% off that amount leaves 42 dollars. The order changes what the second job receives. Composition can work the same way. Commutative means switching two jobs always preserves the result; composition does not have that property in general. Meaning matters too. If a driving rule gives miles from hours, a fuel rule accepting miles can follow it. The miles fit its input slot, the way a matching key fits a lock.

Miles fit the fuel rule's input slot.

Reminder

Substitution. Replacing x with 3 − x in 2x + 1 gives 2(3 − x) + 1.
Distribution. 2(3 − x) = 6 − 2x, because 2 multiplies each term.
Subtracting parentheses. 3 − (2x + 1) = 3 − 2x − 1.
Composition order. G(m(2)) starts with m(2), then puts that answer into G.
Function equality. A formula accepting every x differs from the same formula restricted to x ≠ 1.
Discount fractions. 25% off leaves 75%, so multiply the price by 0.75. At 64 dollars, 0.75 × 64 = 48 dollars.
Divide a fraction by a whole number. $6 over 5$ ÷ 2 = $6 over 5$ × $1 over 2$ = $3 over 5$ . Likewise $96 over r squared$ ÷ 8 = $12 over r squared$ .

Why it works. The first function changes what the second function receives. Reversing them therefore changes the second input, which can change the final answer. Agreement at a single input is not enough to establish equal functions: they must have the same domain and agree at every input there. In an application, a unit tells you what a number measures. A number labeled gallons cannot fill an input slot labeled hours, even when the bare formulas permit the arithmetic.

RuleUsually f(g(x)) ≠ g(f(x)) as functions. Check both orders rather than assuming equality.
For a meaningful composition, the inner output must be an acceptable outer input, including its units.

The same idea, five ways

Say it

order matters; f after g may differ from g after f

Write it

Changing which job happens first can change the next input and the final output.

In math

f(g(x)) and g(f(x)) are usually different functions
2 + 3 = 3 + 2: addition is commutative
5 − 2 ≠ 2 − 5: subtraction is not commutative
hours → miles → gallons

A discount and a coupon act on different intermediate prices when reversed.

See it

Discount then coupon gives 40 dollars; the reverse order gives 42 dollars.

The same idea, other ways

As two jobs

The first job sets up the second. Adding a coat after washing an object is a different process from washing after adding the coat.

First job changes the second input

Switching jobs can switch the answer

Order affects what reaches the second step.

As labeled packages

The inner function hands over a package with a label. If the label says miles, the next function must accept miles. A number alone does not tell you its meaning.

The unit on each output matches the next input.

.1Compare both orders

Composition can happen to commute for a particular pair. You still need to check. Two functions can also happen to agree at one input without being equal everywhere.

One unequal output proves functions differ.
One equal output does not prove functions equal.

The solid + 1 and dashed (x + 1 agree at 0 but give different heights at 1.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Compare both orders

Write it

Composition can happen to commute for a particular pair. You still need to check. Two functions can also happen to agree at one input without being equal everywhere.

In math

One unequal output proves functions differ.
One equal output does not prove functions equal.

Composition can happen to commute for a particular pair. You still need to check. Two functions can also happen to agree at one input without being equal everywhere.

See it

The solid + 1 and dashed (x + 1 agree at 0 but give different heights at 1.

Worked exampleA coincidence is not function equality

For f(x) = x + 1 and g(x) = $x^{2}$ , compare both orders at 0 and at 1. You want to test whether one matching answer tells the whole story. Plan: use the quantity or input named in the question, write the first result, then finish the stated operation.

One matching input: not enough

One differing input: functions differ

Compare + 1 with + 2x + 1

Function equality concerns every allowed input.

At 0, f(g(0)) = f(0) = 1 and g(f(0)) = g(1) = 1.Both paths happen to end at 1 for this input.
At 1, f(g(1)) = f(1) = 2 and g(f(1)) = g(2) = 4.The first stage now sends different numbers to the second stage.

Answer

They agree at 0, but differ at 1, so they are different functions.

Check Their formulas are + 1 and (x + 1 = + 2x + 1. The extra 2x disappears at 0 but not at 1.

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

✗ Not this: Composition never commutes for any pair.

Two adding machines can commute: adding 1 then 2 has the same effect as adding 2 then 1.

✓ Instead: Composition is not commutative in general; particular pairs may commute.

Tips and tricks

Use a small test input to catch a difference, then use formulas and domains to establish equality.

.2Interpret a composition with units

Read the composition as a sentence. If s takes minutes and returns sit-ups, and c takes sit-ups and returns calories, c(s(t)) is calories burned during t minutes of sit-ups.

The inner s output counts sit-ups.
The outer c output counts calories.
The input and output letters are placeholders; the units explain their meaning.

The intermediate 48 is a sit-up count, not a time.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Interpret a composition with units

Write it

Read the composition as a sentence. If s takes minutes and returns sit-ups, and c takes sit-ups and returns calories, c(s(t)) is calories burned during t minutes of sit-ups.

In math

The inner s output counts sit-ups.
The outer c output counts calories.
The input and output letters are placeholders; the units explain their meaning.

Read the composition as a sentence. If s takes minutes and returns sit-ups, and c takes sit-ups and returns calories, c(s(t)) is calories burned during t minutes of sit-ups.

See it

The intermediate 48 is a sit-up count, not a time.

Worked exampleMinutes to sit-ups to calories

Use the illustrative models s(t) = 12t sit-ups and c(n) = $n over 4$ calories. Interpret and evaluate c(s(4)). You want the calories assigned by this model to the sit-ups completed in 4 minutes. Plan: use the quantity or input named in the question, write the first result, then finish the stated operation.

The function labels tell you what each stage measures.

s(4) = 12 × 4 = 48 sit-ups.s accepts minutes and returns the exercise count.
c(s(4)) = c(48) = $48 over 4$ = 12 calories.c accepts the count produced by s.

Answer

Interpretation: c(s(4)) is the modeled calories burned in 4 minutes of sit-ups.
Value: c(s(4)) = 12 calories.

Check c(s(t)) = = 3t, so at 4 minutes the combined model gives 3 × 4 = 12 calories.

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

✗ Not this: c(s(4)) counts the sit-ups done in 4 minutes.

The inner s output counts sit-ups. The outer c output counts calories.

✓ Instead: s(4) counts sit-ups; c(s(4)) counts the modeled calories burned doing those sit-ups.

Tips and tricks

Read inside to outside, naming each quantity: minutes, sit-ups, calories.
An interpretation question needs a sentence. A numerical value can be found only when the functions’ rules or values are supplied.

.3Check whether the reverse order makes sense

A numerical formula may accept a number that its application cannot interpret. In the driving story, fuel use must follow distance. Feeding gallons into a driving-time rule gives the wrong kind of input.

If m accepts hours and returns miles, and G accepts miles and returns gallons, G(m(h)) is meaningful.
m(G(h)) does not match these units; G cannot start with hours and m cannot take gallons.

The valid connection starts with time and finishes with fuel.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Check whether the reverse order makes sense

Write it

In math

If m accepts hours and returns miles, and G accepts miles and returns gallons, G(m(h)) is meaningful.
m(G(h)) does not match these units; G cannot start with hours and m cannot take gallons.

See it

The valid connection starts with time and finishes with fuel.

Worked exampleHours to distance to fuel

Let m(h) = 45h miles and G(d) = $d over 30$ gallons. Evaluate the meaningful composition at 2 hours and explain the order. You want gallons used for the distance driven in 2 hours. Plan: use the quantity or input named in the question, write the first result, then finish the stated operation.

Follow the units to choose the order.

m(2) = 45 × 2 = 90 miles.The mileage rule accepts hours.
G(m(2)) = G(90) = $90 over 30$ = 3 gallons.The fuel rule accepts miles.
Reject m(G(h)) as an interpretation for these models.G expects miles rather than hours, and m expects hours rather than G's gallon output.

Answer

G(m(2)) = 3 gallons.
The order is hours to miles to gallons.

Check The model uses 1 gallon per 30 miles; 3 gallons cover 3 × 30 = 90 miles, the distance obtained from 2 hours.

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

✗ Not this: m(G(2)) converts 2 hours into gallons.

G accepts miles, and its gallon output cannot enter m, which accepts hours.

✓ Instead: G(m(2)) converts hours to miles to gallons.

Tips and tricks

Write the input and output unit under each function name before deciding which name goes inside.

.4Both orders can be meaningful and can sometimes commute

When two rules both accept and return ordinary real numbers, you can often use either order. The outputs can still differ, as the earlier linear pair showed. Some particular pairs do give equal functions in both orders.

Both orders can be meaningful when the input and output quantities match.
A pair that commutes does not make composition commutative in general.
For f(x) = x + 1 and g(x) = x + 2, both compositions give x + 3 on all real inputs.

f(x) = x + 1; g(x) = x + 2

f(g(x)) = (x + 2) + 1 = x + 3

g(f(x)) = (x + 1) + 2 = x + 3

These two addition rules happen to commute.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Both orders can be meaningful and can sometimes commute

Write it

In math

Both orders can be meaningful when the input and output quantities match.
A pair that commutes does not make composition commutative in general.
For f(x) = x + 1 and g(x) = x + 2, both compositions give x + 3 on all real inputs.

See it

f(x) = x + 1; g(x) = x + 2

f(g(x)) = (x + 2) + 1 = x + 3

g(f(x)) = (x + 1) + 2 = x + 3

These two addition rules happen to commute.

Worked exampleA pair whose order does not change the function

Let f(x) = x + 1 and g(x) = x + 2, both with all real inputs. Find both compositions and compare them. You want to see a case where reversing two connected rules gives the same complete function. Plan: use the quantity or input named in the question, write the first result, then finish the stated operation.

Adding 2 then 1 gives 7.

Adding 1 then 2 also gives 7.

f(g(x)) = f(x + 2) = (x + 2) + 1 = x + 3.g adds 2 first, then f adds 1.
g(f(x)) = g(x + 1) = (x + 1) + 2 = x + 3.f adds 1 first, then g adds 2.
Both composite domains are all real numbers.Both component rules accept every real intermediate answer.

Answer

Both compositions equal x + 3 on (−∞, ∞), so this pair commutes.

Check At x = 4 the first path is 4 to 6 to 7 and the reverse path is 4 to 5 to 7. The symbolic calculations establish the agreement for every real input.

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

✗ Not this: This example proves that all function compositions commute.

It proves equality for this specific pair; the earlier pair gave different formulas.

✓ Instead: Check each pair and its domains separately.

Tips and tricks

Functions with the same input and output units can permit both orders without giving equal answers.

.5Distance to force to acceleration

A distance rule can return a force, and a force rule can return an acceleration. Acceleration measures how quickly speed or direction changes. To find acceleration from distance, hand the distance rule's force output to the acceleration rule. In this model, G names the force rule, F names its force input to a, and r names distance. The letters are labels for their quantities.

For the scaled practice model below, the meaningful order is a(G(r)).
The units pass from distance to force to acceleration.
a(G(r)) means the modeled acceleration of a planet at distance r from the sun.

Distance gives force 6; dividing that force by 8 gives acceleration in the model units.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Distance to force to acceleration

Write it

In math

For the scaled practice model below, the meaningful order is a(G(r)).
The units pass from distance to force to acceleration.
a(G(r)) means the modeled acceleration of a planet at distance r from the sun.

See it

Distance gives force 6; dividing that force by 8 gives acceleration in the model units.

Worked exampleChoose a meaningful gravity-model composition

In an invented scaled model, G(r) = $96 over r squared$ force units for r > 0 distance units, and a(F) = $F over 8$ acceleration units for F ≥ 0 force units. Interpret and evaluate a(G(4)). You want acceleration obtained from the force at distance 4. Plan: use the quantity or input named in the question, write the first result, then finish the stated operation.

The handoff is force, and the final number measures acceleration in this invented scaled model.

G(4) = $96 over 4 squared$ = $96 over 16$ = 6 force units.The distance input goes into the inner force model first.
a(G(4)) = a(6) = $6 over 8$ = $3 over 4$ acceleration units.The outer model accepts the force output and divides it by 8.
The reverse connection does not match the stated quantity labels.G expects distance, not the acceleration returned by a.

Answer

a(G(4)) = $3 over 4$ acceleration units.
It means the modeled acceleration at distance 4.

Check a(G(r)) = ÷ 8 = for r > 0. At 4 it gives = , agreeing with the two-stage route.

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

✗ Not this: Use G(a(r)) to obtain acceleration from distance.

a expects force, and G expects distance; that order mixes the quantity labels.

✓ Instead: Use a(G(r)).

Tips and tricks

Read the input and output names before manipulating an application formula.

.6A price example makes order visible

A percentage reduction changes with the price it receives. A coupon removes a fixed amount. This is why moving the coupon earlier changes the later percentage savings. Compare the numeric prices first, then write the two formulas.

On starting prices x ≥ 11 dollars, both composed prices are nonnegative and both store instructions make sense.
The reverse order in this example costs 2 dollars more.

The coupon is applied to the 48-dollar intermediate price.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

A price example makes order visible

Write it

In math

On starting prices x ≥ 11 dollars, both composed prices are nonnegative and both store instructions make sense.
The reverse order in this example costs 2 dollars more.

See it

The coupon is applied to the 48-dollar intermediate price.

Worked exampleA store’s two orders

Let d(x) = 0.75x apply 25% off a price, and c(x) = x − 8 apply an 8-dollar coupon. Compare c(d(64)) and d(c(64)), then find both formulas. You want the price in each order. Plan: keep a separate line for each discount or coupon.

The coupon is applied to the 48-dollar intermediate price.

Reversing the jobs applies 25% off to 56 dollars.

d(64) = 0.75 × 64 = 48; c(48) = 48 − 8 = 40.The percentage discount comes first in c(d(64)); the coupon acts on its remaining price.
c(64) = 64 − 8 = 56; d(56) = 0.75 × 56 = 42.In the reverse order, the percentage discount acts on the price after the coupon.
c(d(x)) = 0.75x − 8; d(c(x)) = 0.75(x − 8) = 0.75x − 6.Substitute the whole inner price; distribution gives 0.75 × 8 = 6.

Answer

Discount then coupon: 40 dollars.
Coupon then discount: 42 dollars.
c(d(x)) = 0.75x − 8.
d(c(x)) = 0.75x − 6.

Check The discount after the coupon saves 25% of 56, which is 14 dollars: 56 − 14 = 42. The discount before the coupon saves 16 dollars, giving 64 − 16 − 8 = 40.

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

✗ Not this: A discount and coupon always give the same final price in either order.

The percentage reduction depends on its current input price: it saves 16 dollars before the coupon but 14 afterward.

✓ Instead: For this 64-dollar item the two prices are 40 and 42 dollars.

Tips and tricks

Write the current price after each instruction. Do not apply both instructions separately to the original price.

.7Interpret a word-only composition

You can explain a composition even when no numerical formulas are given. Each function’s description tells you the kind of input and output. Follow those descriptions as you would follow labels on packages. The outside function names the final quantity.

Interpret means state what the expression measures in words.
A quantity description by itself does not supply a numerical rate.

Read the quantity at every stage before giving the interpretation.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Interpret a word-only composition

Write it

In math

Interpret means state what the expression measures in words.
A quantity description by itself does not supply a numerical rate.

See it

Read the quantity at every stage before giving the interpretation.

Worked exampleTwo orders of composition, and a pool hose with water cost

(a) Let f(x) = 3x − 4 and g(x) = $\sqrt{x}$ . Find f(g(x)) and g(f(x)), give the domain of each, and decide whether they are the same function. (b) A hose fills a pool at a steady rate, so after t minutes the pool holds V(t) = 12t gallons. The water company charges W(v) = 0.005v dollars for v gallons. Decide which composition, W(V(t)) or V(W(t)), is meaningful. Find a formula for the meaningful one and evaluate it at t = 50. Label every quantity with its units.

For f(g(x)), the inner function is g, because g acts on x first. For g(f(x)), the inner function is f.In a composition the function written closest to x is applied first. Its output becomes the input of the outer function.
f(g(x)) = f( $\sqrt{x}$ ) = 3 $\sqrt{x}$ − 4. The domain is x ≥ 0.Replace x in f with $\sqrt{x}$ . The inner root needs x ≥ 0, and f accepts any real number, so no further restriction is added.
g(f(x)) = g(3x − 4) = $square root of 3 x − 4$ . The domain needs 3x − 4 ≥ 0, so x ≥ $4 over 3$ .Replace x in g with 3x − 4. A square root needs a non-negative input, so the inner output must be at least 0.
Test x = 4, which is in both domains: f(g(4)) = 3·2 − 4 = 2, and g(f(4)) = $square root of 8$ ≈ 2.83. The values are 2 and $square root of 8$ , which differ.One allowed input that gives different outputs proves the two functions are different. The domains also differ, [0, ∞) against $[4 over 3$ , ∞), which is a second reason.
In part (b), label the units. V takes t in minutes and returns gallons. W takes v in gallons and returns dollars.A composition makes sense only when the inner output is an acceptable input for the outer function, units included.
W(V(t)): the starting input is t minutes, the intermediate output V(t) is in gallons, and the final output W is in dollars. Gallons is exactly what W expects, so this order is meaningful.The intermediate unit (gallons) matches W's input unit (gallons).
V(W(t)) would feed W's output, which is in dollars, into V, which expects minutes. This order is not meaningful.The intermediate unit (dollars) does not match V's input unit (minutes). The formula can be written, but it does not describe anything real.
W(V(t)) = 0.005(12t) = 0.06t dollars. At t = 50: V(50) = 600 gallons, so W(600) = 0.005·600 = 3 dollars.Substitute the inner formula into the outer one, then evaluate. The result is the cost of the water after t minutes.

Answer

(a) f(g(x)) = 3 − 4 with domain x ≥ 0. g(f(x)) = with domain x ≥ . They are not the same function, since at x = 4 they give 2 and . (b) W(V(t)) is the meaningful composition. W(V(t)) = 0.06t dollars, and W(V(50)) = 3 dollars, so 50 minutes of filling costs $3. V(W(t)) is not meaningful, because it would feed dollars into a function that expects minutes.

Check (a) Recompute at x = 9: f(g(9)) = 3·3 − 4 = 5, and g(f(9)) = ≈ 4.80. These still differ. (b) 50 minutes × 12 gallons per minute = 600 gallons, and 600 gallons × $0.005 per gallon = $3. Also 0.06 × 50 = 3, which agrees.

Work to write

f(g(x)) = 3 $\sqrt{x}$ − 4, domain x ≥ 0
g(f(x)) = $square root of 3 x − 4$ , domain x ≥ $4 over 3$
At x = 4: f(g(4)) = 2 and g(f(4)) = $square root of 8$ , so f(g(x)) ≠ g(f(x))
t (minutes) → V(t) (gallons) → W(V(t)) (dollars); gallons matches W's input
V(W(t)) is not meaningful: dollars cannot be input as minutes
W(V(t)) = 0.06t dollars
W(V(50)) = W(600) = 3 dollars

(a) f(g(x)) = 3 $\sqrt{x}$ − 4 with domain x ≥ 0. g(f(x)) = $square root of 3 x − 4$ with domain x ≥ $4 over 3$ . They are not the same function, since at x = 4 they give 2 and $square root of 8$ . (b) W(V(t)) is the meaningful composition. W(V(t)) = 0.06t dollars, and W(V(50)) = 3 dollars, so 50 minutes of filling costs $3. V(W(t)) is not meaningful, because it would feed dollars into a function that expects minutes.

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

✗ Not this: c(s(3)) means 3 calories.

The 3 is the time input to s, not the final output of c.

✓ Instead: It means calories burned during 3 minutes of sit-ups; the number of calories is not supplied.

Tips and tricks

Answer template: starting input, inner quantity, final outer quantity.

.8Units alone choose an order

No speed or fuel formula is needed to choose the meaningful order. The matching labels decide the connection, like matching a plug to its socket. Here hours enter f and miles leave it; those miles fit g, whose final output measures gallons.

For these definitions, g(f(x)) is meaningful.
Bare-number substitution does not prove that an applied interpretation matches its units.

The textbook’s names f and g connect through the shared unit miles.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Units alone choose an order

Write it

In math

For these definitions, g(f(x)) is meaningful.
Bare-number substitution does not prove that an applied interpretation matches its units.

See it

The textbook’s names f and g connect through the shared unit miles.

Worked exampleChoose the textbook’s meaningful notation

f(x) gives miles driven in x hours, and g(y) gives gallons used for driving y miles. Which is meaningful: f(g(y)) or g(f(x))? You want the order whose handoff has the unit the second rule expects. Plan: label each input and output before choosing an order.

The textbook’s names f and g connect through the shared unit miles.

f: hours → miles; g: miles → gallons.Each description names its input and output quantity.
Choose g(f(x)): f(x) supplies miles, which g accepts.The quantity passed between functions matches.
Reject f(g(y)): g(y) supplies gallons, while f needs hours.Gallons cannot fill the time slot in these definitions.

Answer

g(f(x)) is meaningful: gallons used for the miles driven in x hours.
f(g(y)) does not match these quantity definitions.

Check Trace the chosen path: hours → miles → gallons. The middle miles appear both as f’s output and g’s required input.

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

✗ Not this: f(g(y)) must be meaningful because f is written first.

The written first name is outer and must accept the inner output, here gallons.

✓ Instead: g(f(x)) has the matching hours → miles → gallons path.

Tips and tricks

Write hours → miles and miles → gallons, then join the repeated unit.

.9Evaluate asks for output; solve asks for input

Two questions can use the same machines while searching in opposite directions. Evaluate supplies a starting input and asks what comes out. Solve supplies a desired output and asks which starting input gets there. The equation records that desired output, like finding how long a trip must take to reach a chosen destination.

Evaluate c(s(4)) means start with time 4 and find calories.
Solve c(s(t)) = 45 means start with desired calories 45 and find time t.

The given final output 45 is reached by starting with 15 minutes.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

Evaluate asks for output; solve asks for input

Write it

In math

Evaluate c(s(4)) means start with time 4 and find calories.
Solve c(s(t)) = 45 means start with desired calories 45 and find time t.

See it

The given final output 45 is reached by starting with 15 minutes.

Worked exampleFind the time from a requested calorie output

In an invented practice model, s(t) = 18t sit-ups and c(n) = $n over 6$ calories. How many minutes give c(s(t)) = 45 calories? You are given the output 45 and must find the input time. Plan: form the combined formula, set it equal to the requested output, solve for time, then check both original functions.

The given final output 45 is reached by starting with 15 minutes.

c(s(t)) = c(18t) = $18 t over 6$ = 3t.s supplies the whole exercise count to c, and 18 ÷ 6 = 3.
Solve 3t = 45. Divide both sides by 3: t = 15.This locates the starting time whose final calorie output is 45; division undoes multiplication by 3.
Check s(15) = 18 × 15 = 270 sit-ups, then c(270) = $270 over 6$ = 45 calories.The proposed time must produce the requested final output in the original two-stage calculation.

Answer

The model reaches 45 calories after 15 minutes.

Check The combined rate is 3 calories per minute. Fifteen groups of 3 give 45, confirming the time separately.

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

✗ Not this: Evaluate c(s(45)) to answer when 45 calories are reached.

That treats 45 as minutes and asks a different output question.

✓ Instead: Solve c(s(t)) = 45; in the worked model the time is 15 minutes.

Tips and tricks

Given input: evaluate. Given output and a question asking when or which input: solve.

.10A fresh interpretation to copy as a pattern

A printer first turns time into a page count. Ink use then depends on that count. This is the same handoff idea with different quantities, so you can rebuild the interpretation by reading labels rather than memorizing one story.

w(p(t)) describes grams of ink used during t minutes of printing.

Pages form the matching handoff between the two functions.

Reminder

Function notation. The name chooses a recipe and the parentheses supply its input: f(3) is the f output at input 3.

The same idea, five ways

Say it

A fresh interpretation to copy as a pattern

Write it

In math

w(p(t)) describes grams of ink used during t minutes of printing.

See it

Pages form the matching handoff between the two functions.

Worked exampleMinutes to pages to ink

p(t) counts pages printed in t minutes, and w(n) counts grams of ink used for n pages. Interpret w(p(10)) and explain why p(w(10)) does not match these definitions. You want a sentence about quantities. Plan: read each expression from the inside outward.

Pages form the matching handoff between the two functions.

p(10) counts pages printed in 10 minutes.p accepts minutes and returns pages.
w(p(10)) counts the grams of ink used for those pages.w accepts pages and returns grams.
p(w(10)) would hand grams of ink to p, which needs minutes.The output of w cannot fill p’s time input slot under the stated definitions.

Answer

w(p(10)) means grams of ink used in 10 minutes of printing.
p(w(10)) does not match the stated quantity labels.

Check The correct route starts with minutes and ends with ink: minutes → pages → grams. The final w tells you that the answer is ink.

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

✗ Not this: w(p(10)) returns pages.

The outside w returns grams, while only the inside p returns pages.

✓ Instead: The final answer measures grams of ink.

Tips and tricks

The outside function names the final quantity in your answer.

Strategy: step by step

1. For each order, identify the function nearest the original input.
2. Evaluate or substitute that inner function into the other function.
3. Compare the resulting formulas and their domains. A different answer at one allowed input proves the functions differ.
4. For an application, label the starting input, intermediate output, and final output with units.
5. Check that the intermediate unit matches the outer function's expected input unit.

Strategy

Choose an order and compare it

Do the inner output quantity and outer input quantity match?

YesThis order is meaningful; calculate if formulas are supplied.

NoReject this applied interpretation even if bare-number arithmetic can be carried out.

↓

Do the two orders have different domains or a different output at an allowed input?

YesThey are different functions.

NoUse complete formulas and domains to determine whether they agree everywhere.

Write each function as its input quantity → its output quantity. A unit is a named measure such as hours or miles.
Start with the function accepting the given input quantity; put it inside the other name.
Pass its output to a function accepting that output quantity.
For pure-number functions, calculate both formulas and their domains. One different output proves they differ; equality requires the same outputs everywhere and the same domain.

Worked exampleTwo orders of composition, and a unit check for fuel use

(a) Let f(x) = 2x + 5 and g(x) = $x^{2}$ . Find f(g(x)) and g(f(x)). Decide whether they are the same function. (b) A delivery van drives at a steady speed, so the distance it covers in t hours is D(t) = 60t miles. The van uses fuel according to G(d) = $d over 30$ gallons for d miles driven. Find the meaningful composition that gives gallons used after t hours, and evaluate it at t = 3. Explain why the other order, D(G(d)), is not meaningful.

For f(g(x)), the inner function is g, because it acts on x first. For g(f(x)), the inner function is f.In a composition, the function written nearest the input is applied first.
Substitute g(x) = $x^{2}$ into f: f(g(x)) = 2( $x^{2}$ ) + 5 = 2 $x^{2}$ + 5.f doubles its input and adds 5, and here its input is $x^{2}$ .
Substitute f(x) = 2x + 5 into g: g(f(x)) = (2x + 5 $) squared$ = 4 $x^{2}$ + 20x + 25.g squares its input, and here its input is the whole expression 2x + 5.
Compare the two results at x = 1. f(g(1)) = 2(1) + 5 = 7 and g(f(1)) = (7 $) squared$ = 49. Both functions have domain all real numbers.A single allowed input that gives different outputs proves the functions are different.
In part (b), label the units. For D, the input t is in hours and the output is in miles. For G, the input d is in miles and the output is in gallons.A composition is meaningful only if the inner output can be used as the outer input.
Use the order G(D(t)). The chain is t hours → D(t) miles → G(D(t)) gallons. So G(D(t)) = $60 t over 30$ = 2t gallons.D gives miles, and miles is the unit G expects as its input.
Evaluate at t = 3. D(3) = 60(3) = 180 miles, and G(180) = $180 over 30$ = 6 gallons.First find the intermediate output, then feed it into the outer function.
Check the order D(G(d)). G(d) gives gallons, but D expects an input in hours.The intermediate unit (gallons) does not match the input unit D expects (hours), so this composition has no real-world meaning.

Answer

(a) f(g(x)) = 2 + 5 and g(f(x)) = 4 + 20x + 25. They are different functions, since at x = 1 the outputs are 7 and 49. (b) G(D(t)) = 2t gallons, and G(D(3)) = 6 gallons. D(G(d)) is not meaningful because G outputs gallons while D needs hours.

Check Part (a): at x = 0, f(g(0)) = 5 and g(f(0)) = 25. These still differ. Part (b): 3 hours at 60 miles per hour is 180 miles. At 30 miles per gallon, 180 miles uses 6 gallons. The formula 2t gives 2(3) = 6 and agrees.

Work to write

f(g(x)) = 2 $x^{2}$ + 5
g(f(x)) = (2x + 5 $) squared$ = 4 $x^{2}$ + 20x + 25
f(g(1)) = 7 ≠ 49 = g(f(1)), so f∘g ≠ g∘f
t (hours) → D(t) (miles) → G(D(t)) (gallons)
G(D(t)) = $60 t over 30$ = 2t
G(D(3)) = 6 gallons
D(G(d)) is not meaningful: G outputs gallons, but D needs hours

(a) f(g(x)) = 2 $x^{2}$ + 5 and g(f(x)) = 4 $x^{2}$ + 20x + 25. They are different functions, since at x = 1 the outputs are 7 and 49. (b) G(D(t)) = 2t gallons, and G(D(3)) = 6 gallons. D(G(d)) is not meaningful because G outputs gallons while D needs hours.

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

✗ Not this: Matching at x = 0 proves two compositions are equal.

Function equality requires matching domains and outputs at every allowed input.

✓ Instead: Compare formulas and domains; test another input to expose a possible difference.

✗ Not this: A fuel output of 3 gallons can become 3 hours without explanation.

The same numeral can measure different quantities, but changing its unit changes its meaning.

✓ Instead: Only hand an output to a function that accepts that output's quantity.

Tips and tricks

Write input units above the first arrow and output units above each following arrow.
Understand, then rebuild: the order from the parentheses or the units. Do not memorize which letter goes first; letters can change.

Trap. Treating a valid arithmetic calculation as proof that an application makes sense. Label the units as well as the numbers.