Rate of change and average rate of change

Explain it like I am five

A rate of change tells you how much one quantity changes for each single unit of another. A car going 68 miles per hour adds 68 miles for every hour that passes, and a rat population growing by 40 rats per week adds 40 rats each week. The word 'per' means 'for each' and signals a rate. Gas prices in the table below rose in some years and fell in others. If you use only the values at the start and the end of a stretch of years, you get the average rate of change. It works like splitting a restaurant bill evenly: people ate different amounts, but everyone pays the total divided by the number of people.

Each column keeps one year directly above its gasoline price.

Reminder

Dollars to cents and rounding. Multiply dollars by 100: 0.196 dollars = 19.6 cents. In 0.195714..., the fourth decimal digit is 7, so the three-place rounded value is 0.196.
Signed subtraction. Smaller minus bigger is negative: 2.41 − 2.84 = −0.43. Subtracting a negative adds its opposite: 2 − (−1) = 3.
Division by zero. A quotient needs a nonzero denominator: 2012 − 2012 = 0, so two identical inputs cannot give this average rate.

Why it works. A total change hides how fast it happened: a $1.37 rise in gas prices over one year is very different from the same rise over seven years. Dividing by the change in input shares the total out equally, one share per unit of input, which is why the result is called an average: if the price had risen by exactly that share every year, it would land on the same final price. The units are output units per input unit because you divided dollars per gallon by years, as miles divided by hours gives miles per hour.

RuleAverage rate of change = = = , with ≠ and both outputs defined. Use the same endpoint order on top and bottom. Units are output units per input unit.

The same idea, five ways

Say it

Say 'delta y over delta x', or 'change in output per unit change in input'.

Write it

The average rate of change compares the net output change with the input change between two distinct endpoints.

In math

$\frac{Δ y}{Δ x}$
$\frac{Δ f}{Δ x}$
$f( x sub 2 ) − f( x sub 1 ) over x sub 2 − x sub 1$ , $x sub 2$ ≠ $x sub 1$

The difference between two odometer readings divided by the time between them.

See it

The move from 2005 to 2012 is Δx = 7 years.

The same idea, other ways

Sharing one total

A restaurant bill divided by the number of people gives the equal share per person. In the same way, a price increase divided by the years gives the equal increase per year that would produce the same final price.

total change ÷ input change

= average change per unit

Sharing the net change evenly is what the average measures.

With the gas numbers

Δ is the Greek capital letter delta, pronounced DEL-tuh and read 'change in'. Δy = 3.68 − 2.31 = 1.37 dollars per gallon. Δx = 2012 − 2005 = 7 years. Dividing gives $137 over 700$ dollars per gallon per year. Δf is another name for the output change; it does not change the function into a different function.

Each column keeps one year directly above its gasoline price.

The point name tags

$x sub 1$ , read 'x one' or 'x sub one', names the input of point 1. $x sub 2$ names the input of point 2. A subscript is a name tag, not a power and not multiplication. $y sub 2$ and f( $x sub 2$ ) name the same output. On this table, $x sub 2$ = 2012 and $y sub 2$ = f(2012) = 3.68.

point 1: (2005, 2.31)

point 2: (2012, 3.68)

: input of point 2

: x multiplied by x

A small lower name tag identifies the point; a raised 2 means squaring.

As rise divided by run

On f(x) = $x^{2}$ , going from x = 1 to x = 3 raises the output from 1 to 9. That is rise 8 over run 2, giving $8 over 2$ = 4. A straight ruler through the two points climbs 4 for each unit right, even though the curve changes steepness.

The dashed ruler connects two curve points; it is not another piece of the function.

Year	Average price of a gallon of gas
2005	$2.31
2006	$2.62
2007	$2.84
2008	$3.30
2009	$2.41
2010	$2.84
2011	$3.58
2012	$3.68

.1A negative average, even with a rise in the middle

The average compares the two ends. A price can rise in the middle and still end lower than it began. The negative sign records the net drop, rather than claiming the price fell every year.

From 2007 to 2008 the price rises by 3.30 − 2.84 = 0.46 dollars per gallon.
From 2007 to 2009 the net change is −0.43 dollars, so the average is negative.

Each column keeps one year directly above its gasoline price.

Worked exampleAverage rate of change of a used truck's resale value

A dealership tracks the typical resale value of one model of pickup truck as it ages. Let t be the truck's age in years and V(t) be its resale value in dollars. The table gives these values.

Age t (years): 1, 3, 6, 8
Resale value V(t) ($): 38,400, 31,200, 22,800, 18,600

(a) Find the average rate of change of V on the interval [1, 6]. Give the units and say what the sign means.
(b) Find the average rate of change of V on [6, 8], and compare it with part (a).

Table of the pickup truck's resale value at ages 1, 3, 6 and 8 years, with the endpoints t = 1 and t = 6 marked for part (a).

Input: t, the age in years. Output: V(t), the resale value in dollars.The units of the rate are output units per input unit, so both units are needed first.
For (a), let $x sub 1$ = 1 with $y sub 1$ = V(1) = 38,400, and $x sub 2$ = 6 with $y sub 2$ = V(6) = 22,800.Working left to right, the smaller input is $x sub 1$ . Each value stays paired with its own age from the table.
Δy = 22,800 − 38,400 = −15,600 and Δx = 6 − 1 = 5.Both differences use the same order, second endpoint minus first, so the sign is meaningful.
Δx = 5 ≠ 0, so the average rate of change = $−15,600 over 5$ = −3,120.The average rate of change is $\frac{Δ y}{Δ x}$ , and it is defined only when Δx is not zero.
The average rate of change on [1, 6] is −3,120 dollars per year. The value fell by an average of $3,120 per year from age 1 to age 6.The inputs increase, so a negative result means the output fell overall.
For (b), let $x sub 1$ = 6 with $y sub 1$ = 22,800, and $x sub 2$ = 8 with $y sub 2$ = 18,600. Then Δy = 18,600 − 22,800 = −4,200 and Δx = 8 − 6 = 2.This uses the same left-to-right order on the new interval.
Δx = 2 ≠ 0, so the average rate of change = $−4,200 over 2$ = −2,100 dollars per year.Divide Δy by Δx and attach dollars per year.
Compare the two rates: −2,100 is closer to zero than −3,120. The value still falls on [6, 8], but by about $1,020 less per year than on [1, 6].Both rates are negative, so both are net falls. The one with the smaller size is the slower fall.

Answer

(a) −3,120 dollars per year on [1, 6]. The resale value fell an average of $3,120 each year. (b) −2,100 dollars per year on [6, 8]. The value is still falling, but more slowly in those later years.

Check Start at V(1) = 38,400 and apply the rate over 5 years: 38,400 + 5(−3,120) = 38,400 − 15,600 = 22,800 = V(6). For (b): 22,800 + 2(−2,100) = 18,600 = V(8). Reversing the order also agrees: = = −3,120.

Work to write

Input t in years, output V(t) in dollars
(a) $x sub 1$ = 1, $y sub 1$ = 38,400; $x sub 2$ = 6, $y sub 2$ = 22,800
Δy = 22,800 − 38,400 = −15,600; Δx = 6 − 1 = 5
$−15,600 over 5$ = −3,120 dollars per year
Negative: the value fell an average of $3,120 per year on [1, 6]
(b) $18,600 − 22,800 over 8 − 6$ = $−4,200 over 2$ = −2,100 dollars per year
The value falls more slowly on [6, 8] than on [1, 6]

.2Rats per week

Count rats added for each week that passes.

The source example is growth of 40 rats per week.
Output: population change in rats. Input: elapsed weeks.

80 rats in 2 weeks

40 rats per week

The output unit is rats and the input unit is weeks.

Worked exampleA population rate

This asks how many rats are added for each week. A population grows by 80 rats over 2 weeks. Find its average rate.

80 rats in 2 weeks

40 rats per week

The output unit is rats and the input unit is weeks.

Change in output is 80 rats; change in input is 2 weeks.The problem gives both changes, so there are no endpoint outputs to subtract.
$80 over 2$ = 40 rats per week.Dividing the total addition into two equal weekly shares gives the amount per week.

Answer

40 rats per week.

Check 2 · 40 = 80 rats.

.3Miles per hour

A travel rate shares a distance over elapsed hours.

The source example is 68 miles per hour.
Output: miles traveled. Input: hours.

136 miles in 2 hours

68 miles per hour

The fraction bar reads as per.

Worked exampleA travel rate

This asks how many miles are covered per hour. A car covers 136 miles in 2 hours. Find its average rate.

136 miles in 2 hours

68 miles per hour

The fraction bar reads as per.

$136 over 2$ = 68 miles per hour.Miles are the output change and hours are the input change, so divide miles by hours.

Answer

68 miles per hour.

Check 2 · 68 = 136 miles.

.4Miles per gallon

You can compare distance with fuel instead of time.

The source example is 27 miles per gallon.
Output: miles traveled. Input: gallons used.

54 miles using 2 gallons

27 miles per gallon

The input can be gallons instead of time.

Worked exampleA fuel-use rate

This asks how many miles are covered for each gallon, rather than for each hour. A car covers 54 miles using 2 gallons. Find its rate.

54 miles using 2 gallons

27 miles per gallon

The input can be gallons instead of time.

$54 over 2$ = 27 miles per gallon.Distance is the output and fuel used is the input, so time does not belong in this denominator.

Answer

27 miles per gallon.

Check 2 gallons · 27 miles per gallon = 54 miles.

.5Amperes per volt

You can compare two measured electrical quantities without learning a physics formula.

The source example is current increasing by 0.125 amperes for each extra volt.
Output: current change in amperes. Input: voltage change in volts.

0.25 amperes per 2 volts

0.125 amperes per volt

Read unfamiliar unit names in the same output-per-input order.

Worked exampleAn electrical rate

This asks how much the current increases for each extra volt. The current increases by 0.25 amperes when voltage increases by 2 volts. Find the rate.

0.25 amperes per 2 volts

0.125 amperes per volt

Read unfamiliar unit names in the same output-per-input order.

$0.25 over 2$ = 0.125 amperes per volt.An ampere measures electrical current and a volt measures voltage. Dividing the current change by the voltage change gives the requested comparison.

Answer

0.125 amperes per volt.

Check 2 · 0.125 = 0.25 amperes.

.6Dollars per quarter

The output can fall while the input time increases.

The source example is a college account decreasing by $4,000 per quarter.
A quarter of a year is three months.

−$8,000 in 2 quarters

−$4,000 per quarter

The minus sign records a loss, and a quarter here means three months.

Worked exampleA falling account balance

This asks how much money changes per quarter. An account loses $8,000 over 2 quarters. Find the average rate.

−$8,000 in 2 quarters

−$4,000 per quarter

The minus sign records a loss, and a quarter here means three months.

Change in money = −8,000 dollars per gallon.A loss is a negative change in the output balance.
$−8,000 over 2$ = −4,000 dollars per quarter.Divide the loss by the positive time span. Here a quarter means three months.

Answer

−$4,000 per quarter.

Check 2 · (−4,000) = −8,000 dollars per gallon.

Strategy: step by step

Name the input and output, including their units.
For a left-to-right calculation, call the smaller input $x sub 1$ and the larger input $x sub 2$ ; keep each output with its input.
Find Δy = $y sub 2$ − $y sub 1$ and Δx = $x sub 2$ − $x sub 1$ , using the same order.
Check Δx ≠ 0, then divide Δy by Δx.
Attach output units per input unit. With inputs in increasing order, positive means a net rise and negative a net fall.

Strategy

Find an average rate from the information given

Is the information a table?

YesRead the output directly beneath each chosen input.

NoLook for a graph or formula.

↓

Is it a graph?

YesRead the two heights at the chosen inputs; lesson 1 shows how.

NoIf neither graph nor table was given, evaluate the formula; otherwise keep the outputs already read.

↓

Are the two inputs equal?

YesStop: the denominator is 0, so this average rate is undefined.

NoSubtract in matching order and divide.

↓

Is an endpoint a letter or is the input step h?

YesKeep the endpoint symbolic; lessons 3 and 4 show those forms.

NoCompute the numerical rate.

Find two distinct input values and their outputs.
Read a table column, read a graph height, or evaluate a formula, depending on what is given.
Subtract the two outputs and the two inputs in matching order.
Divide and state units; keep exact fractions until a rounded answer is requested.

Worked exampleAverage rate of change of a regional gas price, 2014 to 2021

A regional energy office records the average price of one gallon of regular gasoline each year. Let P(t) be the price in dollars per gallon in year t. The table gives these values.

Year t: 2014, 2016, 2019, 2021
Price P(t) ($/gal): 3.36, 2.14, 2.60, 3.01

(a) Find the average rate of change of the gas price from 2014 to 2021. Give units and say what the sign means.
(b) Find the average rate of change from 2016 to 2021. Explain why the signs in (a) and (b) differ, even though both intervals end in 2021.

Average regional gas price P(t) in dollars per gallon for the years 2014, 2016, 2019 and 2021. The endpoints 2014 and 2021 used in part (a) are marked.

Input: t, the year, in years. Output: P(t), the price, in dollars per gallon ($/gal).The units of the answer are output units per input unit, so both units are needed first.
(a) Let $x sub 1$ = 2014 with $y sub 1$ = P(2014) = 3.36, and $x sub 2$ = 2021 with $y sub 2$ = P(2021) = 3.01.For a left-to-right calculation the smaller input is $x sub 1$ . Each price stays paired with its own year.
Δy = 3.01 − 3.36 = −0.35 and Δx = 2021 − 2014 = 7.Both differences use the same order, the later value minus the earlier value.
Δx = 7 ≠ 0, so the average rate of change is $−0.35 over 7$ = −0.05.Average rate of change = $\frac{Δ y}{Δ x}$ . Division is allowed because Δx is not zero.
(a) The rate is −0.05 dollars per gallon per year. The price fell by a net 5 cents per gallon per year from 2014 to 2021.The inputs are in increasing order, so a negative rate means a net fall. The price did not fall every year: it rose from 2016 to 2021. The rate describes only the overall change between the endpoints.
(b) Let $x sub 1$ = 2016 with $y sub 1$ = 2.14, and $x sub 2$ = 2021 with $y sub 2$ = 3.01. Then Δy = 3.01 − 2.14 = 0.87 and Δx = 2021 − 2016 = 5.This is the same left-to-right method on the new interval, keeping each price with its year.
Δx = 5 ≠ 0, so the rate is $0.87 over 5$ = 0.174 dollars per gallon per year, a net rise of about 17.4 cents per gallon per year.The rate is positive with the inputs in increasing order, so the price rose overall on [2016, 2021].
The signs differ because the starting prices differ. The 2014 price of $3.36 is above the 2021 price of $3.01. The 2016 price of $2.14 is below it.An average rate of change depends only on the two endpoint values, not on the values in between.

Answer

(a) = = −0.05 dollars per gallon per year, a net fall of 5 cents per gallon per year from 2014 to 2021. (b) = = 0.174 dollars per gallon per year, a net rise of about 17.4 cents per gallon per year from 2016 to 2021.

Check Start at the first endpoint price and add the rate times the number of years. (a) 3.36 + 7(−0.05) = 3.36 − 0.35 = 3.01 = P(2021) ✓. (b) 2.14 + 5(0.174) = 2.14 + 0.87 = 3.01 = P(2021) ✓. Reversing the endpoint order in (a) gives = = −0.05, the same result, as it should be.

Work to write

Input t in years; output P(t) in dollars per gallon
(a) $x sub 1$ = 2014, $y sub 1$ = 3.36; $x sub 2$ = 2021, $y sub 2$ = 3.01
Δy = 3.01 − 3.36 = −0.35, Δx = 2021 − 2014 = 7
$−0.35 over 7$ = −0.05 dollars per gallon per year
Negative: net fall of 5 cents per gallon per year from 2014 to 2021
(b) $3.01 − 2.14 over 2021 − 2016$ = $0.87 over 5$ = 0.174 dollars per gallon per year, a net rise
The signs differ because P(2014) > P(2021) but P(2016) < P(2021)

(a) $3.01 − 3.36 over 2021 − 2014$ = $−0.35 over 7$ = −0.05 dollars per gallon per year, a net fall of 5 cents per gallon per year from 2014 to 2021. (b) $3.01 − 2.14 over 2021 − 2016$ = $0.87 over 5$ = 0.174 dollars per gallon per year, a net rise of about 17.4 cents per gallon per year from 2016 to 2021.

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: A population rate

This asks how many rats are added for each week. A population grows by 80 rats over 2 weeks. Find its average rate.

80 rats in 2 weeks

40 rats per week

The output unit is rats and the input unit is weeks.

Change in output is 80 rats; change in input is 2 weeks.The problem gives both changes, so there are no endpoint outputs to subtract.
$80 over 2$ = 40 rats per week.Dividing the total addition into two equal weekly shares gives the amount per week.

Answer

40 rats per week.

Check 2 · 40 = 80 rats.

Rung 2Average yearly change in gas prices from a table, 2016 to 2023

A state consumer office reports the average price of one gallon of regular gasoline in selected years. Let P(t) be the price in dollars per gallon in year t. The table gives these values.

Year t: 2016, 2018, 2020, 2023
Price P(t) ($/gal): 2.27, 2.85, 2.24, 3.53

(a) Find the average rate of change of P from 2016 to 2023. Give units and say what the sign means.
(b) Find the average rate of change of P from 2018 to 2020. Give units and say what the sign means.

Average price of regular gasoline in 2016, 2018, 2020 and 2023, with the endpoints 2016 and 2023 marked for part (a).

Input: t, the year. Output: P(t), the price in dollars per gallon. The rate will be in dollars per gallon per year.The units of an average rate of change are output units per input unit, so both quantities and their units must be named first.
(a) Let $x sub 1$ = 2016 with P(2016) = 2.27, and $x sub 2$ = 2023 with P(2023) = 3.53.For a left-to-right calculation the smaller input is $x sub 1$ . Each price stays with its own year.
Δy = P(2023) − P(2016) = 3.53 − 2.27 = 1.26 and Δx = 2023 − 2016 = 7.Both differences use the same order, later minus earlier, so the quotient has the correct sign.
Δx = 7 ≠ 0, so the average rate of change = $1.26 over 7$ = 0.18.Average rate of change = $\frac{Δ y}{Δ x}$ , which is defined only when the inputs differ.
The rate is 0.18 dollars per gallon per year. It is positive, so the price rose overall: on average it went up 18 cents per gallon each year from 2016 to 2023.The inputs are in increasing order, so a positive rate means a net rise in the output.
(b) Let $x sub 1$ = 2018 with P(2018) = 2.85, and $x sub 2$ = 2020 with P(2020) = 2.24. Then Δy = 2.24 − 2.85 = −0.61 and Δx = 2020 − 2018 = 2.These are the same steps with the new endpoints, again taking later minus earlier on both top and bottom.
Δx = 2 ≠ 0, so the average rate of change = $−0.61 over 2$ = −0.305 dollars per gallon per year.The difference in price is divided by the difference in years.
The rate is negative, so the price fell overall from 2018 to 2020: on average it dropped about 30.5 cents per gallon each year.With the inputs in increasing order, a negative rate means a net fall in the output.

Answer

(a) 0.18 dollars per gallon per year. The price rose by 18 cents per gallon per year on average from 2016 to 2023. (b) −0.305 dollars per gallon per year. The price fell by about 30.5 cents per gallon per year on average from 2018 to 2020.

Check Start at each first price and add the rate times the number of years. (a) 2.27 + 7(0.18) = 2.27 + 1.26 = 3.53, which matches P(2023). (b) 2.85 + 2(−0.305) = 2.85 − 0.61 = 2.24, which matches P(2020). The signs also agree with the table: the price went up from 2016 to 2023 and down from 2018 to 2020.

Work to write

Input t = year; output P(t) in dollars per gallon
(a) $x sub 1$ = 2016, P = 2.27; $x sub 2$ = 2023, P = 3.53
$3.53 − 2.27 over 2023 − 2016$ = $1.26 over 7$ = 0.18
0.18 dollars per gallon per year; positive, so the price rose on average
(b) $2.24 − 2.85 over 2020 − 2018$ = $−0.61 over 2$ = −0.305
−0.305 dollars per gallon per year; negative, so the price fell on average

Rung 3Average rate of change of a fuel price from 2015 to 2019

A small delivery company keeps a fuel log of the average price it paid for one gallon of regular gasoline each year. Let P(t) be the price in dollars per gallon in year t. The table gives these values.

Year t: 2015, 2017, 2018, 2019
Price P(t) ($/gal): 2.62, 2.48, 2.71, 3.02

Find the average rate of change of P over the four years from 2015 to 2019. Give units and say what the sign means.

The company's fuel-log prices for 2015, 2017, 2018 and 2019. The endpoint years 2015 and 2019 are the two rows used in the average rate of change.

Input: t, the year (years). Output: P(t), the price of gasoline (dollars per gallon).The units of the rate of change are output units per input unit, so both must be named first.
Let $x sub 1$ = 2015 with $y sub 1$ = P(2015) = 2.62. Let $x sub 2$ = 2019 with $y sub 2$ = P(2019) = 3.02.For a left-to-right calculation the smaller input is $x sub 1$ . Each output stays with its own input. The 2017 and 2018 values are not endpoints of the interval, so they are not used.
Δy = $y sub 2$ − $y sub 1$ = 3.02 − 2.62 = 0.40 dollars per gallon. Δx = $x sub 2$ − $x sub 1$ = 2019 − 2015 = 4 years.The same endpoint order (2019 first, then 2015) is used on top and bottom, so the signs are consistent.
Δx = 4 ≠ 0, so the average rate of change = $\frac{Δ y}{Δ x}$ = $0.40 over 4$ = 0.10.The rule $\frac{Δ y}{Δ x}$ is defined only when the inputs differ. Here they do.
Average rate of change = 0.10 dollars per gallon per year. It is positive.The inputs are in increasing order, so a positive rate means a net rise in price from 2015 to 2019. The dip in 2017 does not change the net result, because only the endpoints enter the calculation.

Answer

The average rate of change of P from 2015 to 2019 is 0.10 dollars per gallon per year. On average the price rose by 10 cents per gallon each year. The positive sign means a net increase over the four years.

Check Start at 2.62 and add 0.10 for each of the 4 years: 2.62 + 0.10 · 4 = 2.62 + 0.40 = 3.02, which matches P(2019).

Work to write

Input t in years; output P(t) in dollars per gallon
$x sub 1$ = 2015, $y sub 1$ = 2.62; $x sub 2$ = 2019, $y sub 2$ = 3.02
Δy = 3.02 − 2.62 = 0.40
Δx = 2019 − 2015 = 4
$\frac{Δ y}{Δ x}$ = $0.40 over 4$ = 0.10
0.10 dollars per gallon per year
Positive: the price rose overall from 2015 to 2019

Rung 4How fast a game console's price falls

A retailer tracks the price of a new handheld game console. Let t be the number of months since launch and P(t) be the price in dollars. The table gives these values.

Months since launch t: 2, 6, 14, 20
Price P(t) ($): 449, 425, 377, 335

(a) Find the average rate of change of P on the interval [2, 14]. Give units and say what the sign means.
(b) Find the average rate of change of P on [14, 20]. Over which interval did the price fall faster on average?

Table of the console's price at 2, 6, 14 and 20 months after launch. The highlighted entries at t = 2 and t = 14 are the endpoints used in part (a).

Input: t, the time since launch in months. Output: P(t), the price in dollars.The units of the rate are output units per input unit. Here that is dollars per month.
For (a), let $x sub 1$ = 2 and $y sub 1$ = P(2) = 449. Let $x sub 2$ = 14 and $y sub 2$ = P(14) = 377.Going left to right, the smaller input is $x sub 1$ . Each price stays paired with its own month.
Δy = 377 − 449 = −72 and Δx = 14 − 2 = 12.Both differences subtract in the same order: second endpoint minus first.
Δx = 12 ≠ 0, so the average rate of change = $−72 over 12$ = −6.The average rate of change is $\frac{Δ y}{Δ x}$ , and it is defined only when Δx ≠ 0.
The rate on [2, 14] is −6 dollars per month. The sign is negative, so the price had a net fall. On average it dropped $6 each month.The inputs are in increasing order, so a negative rate means a net decrease.
For (b), let $x sub 1$ = 14, $y sub 1$ = 377, $x sub 2$ = 20 and $y sub 2$ = 335. Then Δy = 335 − 377 = −42 and Δx = 20 − 14 = 6.These are the same steps on the new interval, keeping the same endpoint order on top and bottom.
Δx = 6 ≠ 0, so the rate = $−42 over 6$ = −7 dollars per month.Divide Δy by Δx and attach dollars per month.
Compare: |−7| > |−6|, so the price fell faster on average over [14, 20].Both rates are negative. The rate with the larger size shows the steeper average drop.

Answer

(a) −6 dollars per month: on average the price fell $6 per month from month 2 to month 14. (b) −7 dollars per month. The price fell faster on average over [14, 20].

Check Start at the first price and add the rate times the length of the interval. Each result should match the table.
(a) 449 + (−6)(12) = 449 − 72 = 377 = P(14). ✓
(b) 377 + (−7)(6) = 377 − 42 = 335 = P(20). ✓

Work to write

Input t in months, output P(t) in dollars
[2, 14]: $x sub 1$ = 2, $y sub 1$ = 449; $x sub 2$ = 14, $y sub 2$ = 377
Δy = 377 − 449 = −72, Δx = 14 − 2 = 12
Average rate of change = $−72 over 12$ = −6 dollars per month
Negative, so the price had a net fall of $6 per month on average
[14, 20]: $335 − 377 over 20 − 14$ = $−42 over 6$ = −7 dollars per month
The price fell faster on average over [14, 20]

(a) −6 dollars per month: on average the price fell $6 per month from month 2 to month 14. (b) −7 dollars per month. The price fell faster on average over [14, 20].

Rung 5Rung 5: A rate from clock times across an hour

A faucet adds 30 gallons from 7:52 p.m. to 8:04 p.m. Find the average gallons added per minute; first find how many minutes actually passed.

The walk measures elapsed minutes: 8 minutes to the hour plus 4 afterward makes a 12-minute run.

From 7:52 p.m. to 8:00 p.m. takes 60 − 52 = 8 minutes.An hour contains 60 minutes, so count forward to the next hour.
From 8:00 p.m. to 8:04 p.m. takes 4 more minutes. Total elapsed time: 8 + 4 = 12 minutes.Both spans belong to the same continuous elapsed-time interval; add them rather than subtracting clock digits as decimals.
Average rate = $30 over 12$ = $5 over 2$ = 2.5 gallons per minute.Gallons are the output change and elapsed minutes the input change; divide top and bottom by 6 to reduce.

Answer

2.5 gallons per minute.

Check 12 · 2.5 = 30 gallons, recovering the amount added. Count to the next hour, then add the minutes after it.

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

✗ Not this: Δx means Δ multiplied by x, and means x squared.

Δx is one name for the input change. The small lower 2 identifies the second point; the raised 2 in means x · x.

✓ Instead: For the gas example, Δx = 2012 − 2005 = 7; = 2012.

✗ Not this: A negative average means the price fell at every intermediate year.

The 2007-to-2009 average is −0.215, but the price rises 0.46 dollars from 2007 to 2008.

✓ Instead: An average compares only the endpoint outputs; it can hide the path.

✗ Not this: Use = ≈ 5.1 as the price change per year.

That division has units years per dollar, so it answers a different question.

✓ Instead: Use = dollars per gallon per year.

✗ Not this: A rate must be measured per unit of time.

Miles per gallon and amperes per volt use gallons and volts as inputs.

✓ Instead: Always divide output units by input units, whatever the input measures.

Tips and tricks

Remember 'output over input' and 'same trip, same order'. Write the units as $\frac{d o l l a r s}{y e a r}$ and read the bar as 'per'.
A rate is also negative if output rises while input falls: reversing both endpoint orders changes both signs and leaves the ratio unchanged.
A subscript names a point. A superscript tells you a power.
If both outputs match and the inputs differ, the average rate is 0; the function may still move up and down between those endpoints.

Trap. Subtracting in different orders on top and bottom flips the sign. Point 2 minus point 1 must be used for both output and input changes.