Average rate of change from a table or graph
A quantity rose fast, then slowly, then fell a little. What one number says how fast it changed between two chosen moments?
Where you are 4 in Average rate of change
When you reach for it
In rate of change one number described a whole relationship, because the rate was the same between any two inputs: a constant rate, and a straight graph. Most quantities are not like that. A sprinter speeds up and then holds her pace, a city grows for decades and shrinks for one, a snowman melts fast at first and slowly later. Ask “how fast did it change?” and the honest reply is “between which two moments?”
The average rate of change answers once the two moments are chosen. You reach for it when you know a function’s value at two inputs, as two rows of a table or two points on a graph, and want one number for how much the output changed per unit of input across that whole stretch. When the function is given by a rule, the same quotient comes from evaluating the rule: average rate of change from a formula.
The procedure
The definition. Take a function f and two different inputs a and b where f has values. The average rate of change of f over the interval from a to b is the change in output divided by the change in input:
The Greek capital \Delta (delta) is read “change in”. In function notation f(b) is the output at input b, so the top is the end value minus the start value.
Finding the two values. In a table the two inputs pick out two rows; on a graph they pick out two points, (a, f(a)) and (b, f(b)), whose heights you read from the vertical axis. Nothing else enters the calculation: the two ends are compared and everything between them is ignored.
Subtracting in the same order. Top and bottom describe the same trip, from the first point to the second, so both subtract the first from the second. Doing both the other way round, \displaystyle \frac{f(a) - f(b)}{a - b}, flips the sign of the top and of the bottom, and the two flips cancel. Reversing only one of them flips the answer and turns a rise into a fall.
Keeping the units. The result is measured in output units per input unit: meters per second, people per year. A positive average says the output ended higher than it started, a negative one lower, and zero says it ended where it started, which is not the same as staying put.
It is the slope of the secant line. A straight line through two points of a graph is a secant line of the graph. By the slope formula its slope is rise over run between the two points, which is exactly \displaystyle \frac{f(b) - f(a)}{b - a} (Figure 1). A steeper secant line means a larger average rate; a secant line falling to the right means a negative one.
Why it is called an average. Suppose the output had changed steadily from f(a) to f(b), the same amount for every unit of input. That steady change is the secant line, and its rate is the quotient above. So the average rate of change is the one constant rate that gives the same total change over the same interval, as an average speed is the steady speed that covers the same distance in the same time. It says nothing about any single moment inside the interval.
A line gives one answer; a curve gives many. If f(x) = mx + c, then for any two inputs
The start value c cancels, and so does the width: every interval gives m. When the graph bends, secant lines over different intervals point in different directions, so the average depends on the interval and must be quoted with it. Two intervals with different averages show a function is not linear; two with the same average do not show that it is (Example 5).
Worked examples
A sprinter’s first six seconds
Briony’s distance d(t), in meters, t seconds after the start of a 100-meter race (made-up numbers):
| t (seconds) | d(t) (meters) |
|---|---|
| 0 | 0 |
| 1 | 5 |
| 2 | 13 |
| 3 | 22 |
| 4 | 32 |
| 5 | 42 |
| 6 | 52 |
Find the average rate of change of d (a) over the first two seconds, (b) from t = 3 to t = 6, and (c) over all six seconds. Say what each means.
Solution(a) Use the rows for t = 0 and t = 2:
Meters over seconds gives 6.5 meters per second, Briony’s average speed over those two seconds. She ran 5 meters in the first second and 8 in the second, never 6.5; 6.5 is the steady speed that covers the same 13 meters in the same 2 seconds.
(b) From t = 3 to t = 6:
so 10 meters per second. The table shows exactly 10 meters in each of these seconds, so here the average also describes every second.
(c) Over all six seconds:
about 8.67 meters per second. Three intervals, three answers: d is not linear, so “her average speed” means nothing until you say over which seconds.
A city that grew, shrank and grew again
The population P(t), in thousands, of a made-up city in year t:
| year t | P(t) (thousands) |
|---|---|
| 1960 | 120 |
| 1970 | 150 |
| 1980 | 172 |
| 1990 | 180 |
| 2000 | 176 |
| 2010 | 190 |
| 2020 | 216 |
(a) Find the average rate of change of P from 1990 to 2000, in people per year. (b) Do the same from 1960 to 2020. (c) Dagmar finds the average rate from 1980 to 2000 is positive and says the city grew all through those twenty years. Is she right?
Solution(a) Later minus earlier, top and bottom:
That is -0.4 thousand people per year. Multiplying by \displaystyle \frac{1000 \text{ people}}{1 \text{ thousand}}, which equals one (a conversion factor), gives -400 people per year: on average the city lost 400 people a year.
(b) Over the whole table:
that is, 1600 people per year. This one positive number covers sixty years that include the shrinking decade of part (a).
(c) From 1980 to 2000:
so 200 people per year. But the city rose to 180 thousand in 1990 and then fell to 176 thousand, so Dagmar is not right. A positive average says only that the city ended larger than it began; it cannot see the rise and fall between.
A snowman melting, read from a graph
Figure 2 graphs the height h(t), in centimeters, of Ambrose-Kai’s snowman t hours after the sun came out. Find the average rate of change of h from t = 0 to t = 4 and from t = 4 to t = 12, and compare them.
Each marked point sits on a grid crossing, so the readings are exact: h(0) = 150, h(4) = 100 and h(12) = 60.
So the snowman lost 12.5 centimeters an hour on average over the first four hours, and only 5 an hour over the next eight. Each answer is the slope of a dashed secant line, and faster melting shows as the steeper secant. Neither number is the melting rate during any particular hour.
The secant line as a steady snowman
The snowman’s graph also passes through (8, 70). Find the equation of the secant line through the points at t = 0 and t = 8, say what it describes, and compare it with the snowman at t = 4.
SolutionThe slope of the secant line is the average rate of change over the interval:
The line passes through (0, 150), so it is y = 150 - 10t. Check: at t = 8 it gives 150 - 80 = 70.
The line is a snowman that starts 150 centimeters tall and melts a steady 10 centimeters an hour, so it too is 70 centimeters tall at t = 8: same start, same end. At t = 4 the line gives 150 - 40 = 110, but h(4) = 100, so the real snowman is 10 centimeters shorter: it melted faster than the average early and more slowly later.
Same over every interval, or not?
Weekly viewers, in millions, of two shows on a streaming service (made-up numbers):
| week | Show A | Show B |
|---|---|---|
| 1 | 2.0 | 1.5 |
| 2 | 3.2 | 1.8 |
| 3 | 4.0 | 2.1 |
| 4 | 4.4 | 2.4 |
| 5 | 4.5 | 2.7 |
| 6 | 4.4 | 3.0 |
(a) For each show, find the average rate of change from week 1 to week 3 and from week 3 to week 5. (b) Which show’s viewers could be a linear function of the week? (c) Zofia notices that the snowman’s average rate from t = 4 to t = 8 equals its average from t = 0 to t = 12. Does that make its height linear?
Solution(a) Show A: \displaystyle \frac{4.0 - 2.0}{3 - 1} = 1.0 and \displaystyle \frac{4.5 - 4.0}{5 - 3} = 0.25 million viewers per week. Show B: \displaystyle \frac{2.1 - 1.5}{3 - 1} = 0.3 and \displaystyle \frac{2.7 - 2.1}{5 - 3} = 0.3 million viewers per week.
(b) A linear function has the same average over every interval. Show A has two different averages, so it is not linear: it gained fast and then leveled off. Show B gains exactly 0.3 million every week, so over any n weeks it gains 0.3n million and the average is always 0.3. Its table fits the line V = 1.2 + 0.3w exactly.
(c) \displaystyle \frac{70 - 100}{4} = -7.5 and \displaystyle \frac{60 - 150}{12} = -7.5: they agree. But Example 3 found -12.5 from t = 0 to t = 4, so the height is not linear. Two equal averages do not prove a function is linear. A line needs every interval to agree; one disagreement rules a line out, and no number of chosen agreements rules it in.
Working backwards from an average
From week 6 to week 10, Show A’s average rate of change was -0.35 million viewers per week. How many viewers did it have in week 10?
SolutionWrite V(w) for Show A’s viewers, in millions, in week w. The average is the total change divided by the width, so the total change is the average times the width:
With V(6) = 4.4 from the table, that gives V(10) = 4.4 - 1.4 = 3.0 million viewers. Nothing says how the loss was spread over the four weeks; only the total is fixed.
Boundaries
- Two different inputs. If b = a the bottom is 0 and the quotient is undefined. A rate at a single instant is a different question.
- Values at both ends. f(a) and f(b) must both be known, so from a table both inputs must be rows of it.
- One rate for every interval only for a line. For any function that is not linear, the average belongs to an interval and means nothing without it.
- Graph readings. A point between grid lines gives an estimated height, and an estimated rate.
- Matching units. Both inputs must be in the same unit, and both outputs; convert minutes to hours, say, before subtracting.
Where this goes next
Comparing averages over equal intervals is how linear and exponential growth are told apart. Shrinking the interval leads toward calculus: as the second point slides toward the first, the averages settle on a rate at an instant.
Exercises
From a table
Cato’s distance d(t), in meters, t seconds after the start of a 100-meter race (made-up numbers):
| t (seconds) | d(t) (meters) |
|---|---|
| 0 | 0 |
| 2 | 11 |
| 4 | 28 |
| 6 | 48 |
| 8 | 66 |
| 10 | 84 |
| 12 | 100 |
- Find the average rate of change of d from t = 0 to t = 4, with its units.
- Find the average rate of change of d from t = 4 to t = 8.
- Find the average rate of change of d from t = 8 to t = 12. Compared with Exercise 2, did Cato speed up or slow down, on average?
- Find Cato’s average speed over the whole race, to two decimal places.
- Over which of the six two-second intervals in the table was Cato’s average speed greatest, and what was it?
With function notation
The population Q(t), in thousands, of a made-up town in year t:
| year t | Q(t) (thousands) |
|---|---|
| 1950 | 40 |
| 1960 | 46 |
| 1970 | 55 |
| 1980 | 61 |
| 1990 | 58 |
| 2000 | 52 |
| 2010 | 57 |
- Find \displaystyle \frac{Q(1980) - Q(1950)}{1980 - 1950} and say what it means in people per year.
- Find the average rate of change of Q from 1980 to 2000, in people per year.
- Find the average rate of change of Q from 1950 to 2010, in people per year, to the nearest person.
- Which decade had the greatest average rate of change, and what was it in people per year?
From a graph
Exercises 10 to 13 use the graph of g in Figure 4.
- Find the average rate of change of g from x = 0 to x = 2.
- Find the average rate of change of g from x = 3 to x = 6.
- Find the average rate of change of g from x = 1 to x = 5. What does the secant line over this interval look like?
- Find the equation of the secant line through the points of the graph at x = 0 and x = 3.
Secant lines and steady change
- A function f is linear, with f(2) = 7 and f(5) = 19. Find its average rate of change from x = 2 to x = 5, its average rate of change from x = 10 to x = 30, and f(10).
- A show had 1.8 million viewers in week 3, and its average rate of change from week 3 to week 7 was -0.25 million viewers per week. How many viewers did it have in week 7?
- Yannick-Lee’s snowman is 120 cm tall at noon and 84 cm tall at 3 pm. (a) Find its average rate of change over those three hours. (b) How tall would a snowman following the secant line, melting steadily at that rate from the same start, be at 1:30 pm?
Reasoning
- Dagmar says: “Cato’s average speed over the whole race was about 8.33 meters per second, so six seconds in he was running at 8.33 meters per second.” Use the table to explain why this does not follow.
- Show that \displaystyle \frac{f(a) - f(b)}{a - b} = \frac{f(b) - f(a)}{b - a}, so the order of the two points does not matter. Then explain what goes wrong if only the top is reversed.
- In Figure 4 the average rate of change of g from x = 1 to x = 5 is 0, yet g is not constant on that interval. Explain how both can be true, and say what a zero average rate does tell you.
Exercise 20 uses two functions given only by this table:
| x | F(x) | G(x) |
|---|---|---|
| 0 | 1 | 1 |
| 0.5 | 2 | 4 |
| 1 | 3 | 3 |
| 1.5 | 4 | 2 |
| 2 | 5 | 5 |
| 3 | 7 | 7 |
- (a) Show that F and G each have an average rate of change of 2 from 0 to 1, from 1 to 2 and from 2 to 3. (b) Zofia concludes that both are linear. Show that she is wrong about one of them.
Answers
- 7 meters per second: \displaystyle \frac{28 - 0}{4 - 0} = \frac{28}{4}.
- 9.5 meters per second: \displaystyle \frac{66 - 28}{8 - 4} = \frac{38}{4}.
- 8.5 meters per second: \displaystyle \frac{100 - 66}{12 - 8} = \frac{34}{4}. That is less than 9.5, so on average Cato slowed down over the last four seconds.
- About 8.33 meters per second: \displaystyle \frac{100 - 0}{12 - 0} = \frac{25}{3}.
- From t = 4 to t = 6, at 10 meters per second: \displaystyle \frac{48 - 28}{2} = 10. The six two-second averages are 5.5, 8.5, 10, 9, 9 and 8.
- 0.7 thousand per year: \displaystyle \frac{61 - 40}{30} = \frac{21}{30}. From 1950 to 1980 the town grew by 700 people per year on average.
- -450 people per year: \displaystyle \frac{52 - 61}{20} = -0.45 thousand per year. On average the town lost 450 people a year.
- About 283 people per year: \displaystyle \frac{57 - 40}{60} = \frac{17}{60} \approx 0.2833 thousand per year.
- 1960 to 1970, at 900 people per year: \displaystyle \frac{55 - 46}{10} = 0.9 thousand. The six decades give 0.6, 0.9, 0.6, -0.3, -0.6 and 0.5 thousand per year.
- -4: g(0) = 5 and g(2) = -3, so \displaystyle \frac{-3 - 5}{2 - 0} = -4.
- 3: g(3) = -4 and g(6) = 5, so \displaystyle \frac{5 - (-4)}{6 - 3} = \frac{9}{3}.
- 0: g(1) = 0 and g(5) = 0, so the change in output is 0. The secant line through (1, 0) and (5, 0) is horizontal; it is the x-axis itself.
- y = -3x + 5: the slope is \displaystyle \frac{-4 - 5}{3 - 0} = -3, and the line passes through (0, 5).
- 4, 4 and 39: \displaystyle \frac{19 - 7}{5 - 2} = 4; a linear function has the same average rate over every interval; f(10) = 7 + 4 \times 8 = 39.
- 0.8 million: the total change is -0.25 \times 4 = -1, and 1.8 - 1 = 0.8.
- (a) -12 cm per hour: \displaystyle \frac{84 - 120}{3} = -12. (b) 102 cm: 120 - 12 \times 1.5 = 102. The real snowman need not be that tall at 1:30 pm; only its two ends are known.
- The 8.33 is the steady speed that covers 100 meters in 12 seconds; it compares only the start and the finish. The table shows Cato averaging 10 meters per second from t = 4 to t = 6 and 9 from t = 6 to t = 8, faster than 8.33 around the six-second mark. An average over an interval does not fix the speed at any one moment inside it.
- f(a) - f(b) = -(f(b) - f(a)) and a - b = -(b - a); dividing, the two factors of -1 cancel. Reversing only the top multiplies the quotient by -1, which turns a rise into a fall.
- A zero average rate says only that g(5) = g(1): the output ended where it started. In between, g fell to -4 at x = 3 and rose back to 0; the fall and the rise cancel in the total change, and the average rate sees only the total.
- (a) For F: \displaystyle \frac{3 - 1}{1} = 2, \displaystyle \frac{5 - 3}{1} = 2 and \displaystyle \frac{7 - 5}{1} = 2. G has the same values at 0, 1, 2 and 3, so the same three averages. (b) From 0 to 0.5, G has average \displaystyle \frac{4 - 1}{0.5} = 6, not 2, so G is not linear; every row of F fits y = 2x + 1. Those three averages use only the values at 0, 1, 2 and 3; a line needs the same average over every pair of inputs.