Calculate and interpret the average rate of change of a function over an interval, and estimate it from a graph

Learning goal: Calculate the average rate of change of a function over a specified interval when the function is given symbolically or as a table, interpret the result with units in terms of the quantities, and estimate the rate of change from a graph by reading coordinates.

The definition. The average rate of change of a function f over the interval from x = a to x = b is the change in output divided by the change in input: [f(b) − f(a)] / (b − a). It is the slope of the line segment joining the two points (a, f(a)) and (b, f(b)) on the graph, a segment called a secant. Its units are output units per input unit: meters per second, dollars per year, cells per hour. Two conditions matter. The interval must be specified, because for anything other than a linear function the answer depends on which interval you choose. And the order of subtraction must match in numerator and denominator: f(b) − f(a) over b − a, never f(b) − f(a) over a − b.

Why a linear function is the special case. For f(x) = 3x + 2, the average rate of change from 1 to 4 is (14 − 5)/(4 − 1) = 9/3 = 3, and from 10 to 20 it is (62 − 32)/10 = 3. A linear function has the same average rate of change over every interval, and that constant is its slope. This is exactly why a table with constant first differences signals a linear model. For every other kind of function the average rate of change varies from interval to interval, which is why the word average is in the name: the function may be changing quickly at one end of the interval and slowly at the other, and the quotient reports the single constant rate that would have produced the same overall change.

A fully worked symbolic example. Let f(x) = x². Over the interval from 1 to 4, f(4) − f(1) = 16 − 1 = 15 and 4 − 1 = 3, so the average rate of change is 15/3 = 5. Over 0 to 2 it is (4 − 0)/2 = 2, and over 2 to 4 it is (16 − 4)/2 = 6. The rate over the whole interval 0 to 4 is (16 − 0)/4 = 4, which lies between the rates on the two halves, as an average should. Geometrically the secant from (1, 1) to (4, 16) rises 15 units over a run of 3 units; the curve itself lies below that secant between the two points, rising slowly near x = 1 and quickly near x = 4.

From a table. When a function is presented as a table, the average rate of change over an interval uses the two rows at its endpoints and ignores the rows in between. Suppose a savings account is modeled by A(t) = 1000(1.06)ᵗ dollars after t years and a table lists A(0) = 1000.00, A(5) = 1338.23 and A(10) = 1790.85, rounded to the cent. The average rate of change over the first five years is (1338.23 − 1000)/5 = 67.65 dollars per year; over the next five years it is (1790.85 − 1338.23)/5 = 90.52 dollars per year; over the full ten years it is (1790.85 − 1000)/10 = 79.08 dollars per year. The interpretation is that the account grew, on average, by 67.65 dollars for each year of the first five years and by 90.52 dollars for each year of the second five. The rate increases because exponential growth adds more each year as the balance grows. Note that the rate over the full interval is not the average of the two half-interval rates in general; here it is, only because the two halves have equal width.

Interpreting sign and units. A negative average rate of change means the output decreased over the interval. For the ball with height h(t) = −5t² + 20t + 25, the average rate of change from t = 2 to t = 5 is (0 − 45)/(5 − 2) = −15 meters per second: on average the ball lost 15 meters of height each second during those three seconds. From t = 0 to t = 2 the rate is (45 − 25)/2 = 10 meters per second upward. From t = 1 to t = 3 the rate is (40 − 40)/2 = 0. The ball certainly moved during that interval, up to 45 meters and back down, but its height at the two endpoints was the same, so the average rate of change is zero. This is the misconception to confront: a zero average rate of change does not mean nothing happened, only that the net change over the interval was zero. The average rate of change reports the net change per unit of input, not the amount of motion.

Estimating from a graph. When only a graph is available, choose the two points on the curve above the endpoints of the interval, read their coordinates as accurately as the grid allows, and compute the quotient. The result is an estimate, and its accuracy depends on how finely the axes are marked. If a graph of a town's population appears to pass through (2000, 4200) and (2010, 5100), the estimated average rate of change is (5100 − 4200)/10 = 90 people per year; if the true values were 4,180 and 5,120, the true rate would be 94 people per year, so reading to the nearest gridline gave an estimate within a few units. Reporting an estimated rate as approximately 90 people per year, rather than exactly, is the honest statement. To estimate the rate at a single moment rather than over an interval, draw the tangent line, the line through that point whose steepness matches the curve's there, and estimate its slope (a tangent line can cross the curve, as the tangent to y = x³ at x = 0 does, and at a sharp corner there is none); that idea belongs to later courses, but the average rate over a short interval around the point is a good approximation of it.

Worked symbolic calculation: Let p(x) = x³ − 4x. Over the interval from −1 to 2: p(2) = 8 − 8 = 0 and p(−1) = −1 + 4 = 3, so the average rate of change is (0 − 3)/(2 − (−1)) = −3/3 = −1. Over the interval from 2 to 3: p(3) = 27 − 12 = 15, so the rate is (15 − 0)/(3 − 2) = 15. The function fell, on average, 1 unit per unit of x over the first interval and rose 15 units per unit of x over the second. The denominator 2 − (−1) is 3, not 1; forgetting to subtract a negative correctly is the most common arithmetic slip here.

Worked table calculation with interpretation: A hypothetical hiker records altitude, in meters, every 30 minutes: 400 at 0 minutes, 520 at 30, 610 at 60, 610 at 90, 700 at 120. Over the first hour the average rate of change is (610 − 400)/60 = 3.5 meters per minute. Over the interval from 60 to 90 minutes it is (610 − 610)/30 = 0 meters per minute: the net change in altitude was zero. The hiker may have rested, walked on level ground, or climbed and come back down by the same amount; the average rate cannot tell which. Over the whole two hours it is (700 − 400)/120 = 2.5 meters per minute. The rows in the middle of each interval do not enter the calculation; only the endpoints do.

Worked estimate from a graph: A graph of a car's distance from home, with gridlines every 10 minutes and every 5 kilometers, shows the curve passing close to (10, 5) and (40, 35). The estimated average rate of change is (35 − 5)/(40 − 10) = 30/30 = 1 kilometer per minute, which is 60 kilometers per hour. Because the coordinates were read from the picture, each could be off a little; if each reading is off by as much as 1 kilometer or 1 minute, the rise lies between 28 and 32 and the run between 28 and 32, so the true rate lies between about 28/32 ≈ 0.88 and 32/28 ≈ 1.14 kilometers per minute. The estimate should be reported as approximately 1 kilometer per minute.

Representation task: For f(x) = x², make a table for x = 0, 1, 2, 3, 4, plot the points and draw the curve. Draw the secant segment from (1, 1) to (4, 16) and label its rise 15, its run 3 and its slope 5. Then compute and record in a table the average rate of change over each of the intervals [0, 1], [1, 2], [2, 3] and [3, 4], and write one sentence about the pattern you see and what it says about the shape of the graph.

Independent performance: Choose a function given symbolically, such as an exponential model of money or population, and a function given only as a table of at least six real or hypothetical measurements. For each, calculate the average rate of change over three different intervals, including one interval over which the rate is negative or zero if the function allows it, and write an interpretation of each result in a full sentence with units. Then sketch a graph of one of the functions and estimate the average rate of change over an interval by reading coordinates from the sketch, comparing the estimate with the exact calculation. Success criteria: every quotient has the output change in the numerator and the input change in the denominator with matching order; units are stated as output units per input unit; each interpretation says what happened to the quantity over that interval; and the estimate from the graph is reported as approximate, with the exact value alongside.