AP Precalculus · Unit 1: Polynomial and Rational Functions · Lesson 1.2

Deep Dive: Rates of Change

🔬 Deep Dive
This is your textbook for this topic. Take your time. Read it more than once.
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Average Rate of Change

Knowing that a function increases is only half the story. How fast does it increase? The average rate of change of a function over an interval of its domain is the ratio of the change in the output values to the change in the input values over that interval.

AROC = [f(b) − f(a)] / (b − a)

It is also the constant rate of change that yields the same change in the output as the function actually produced on that interval. A cyclist who speeds up and slows down can be replaced, for bookkeeping, by one who rides at a single steady speed and covers the same distance in the same time. That steady speed is the average rate of change, and it is the slope of the line through the two points (the secant line).

Pick a function and drag the two inputs. The dashed line through the two points is the constant-rate stand-in: a function that changes at one steady rate and still goes from the first output to the second. Its slope is the average rate of change.

input a1.0
input b3.0
0123450510152025xf(x)
change in output: f(3.00) − f(1.00) = 9.00 − (1.00) = 8.00change in input: 3.00 − 1.00 = 2.00average rate of change = 8.00 ÷ 2.00 = 4.00A steady rate of 4.00 over 2.00 input units gives the same total change, 8.00.
⚠️Subtract in the same order on top and bottom: f(b) − f(a) over b − a. The rate is not f(b)/b, not the average of the two outputs, and it is not the change in output alone. Always divide by the change in input.
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Rates from a Table

A table gives you the outputs at certain inputs, so choose two columns and apply the same formula. Only the two endpoints of your interval matter. Columns in between don't change the average rate over the whole interval.

A cyclist's distance D (miles) from home is recorded at uneven times t (hours). Choose two columns. The rate divides by the change in t, not by how many columns apart they are.

t (h)01346
D (mi)012303850
from columnt = 0
to columnt = 3
change in D: 30 − 0 = 30 mileschange in t: 3 − 0 = 3 hoursaverage rate of change = 30 ÷ 3 = 10.00 miles per hour⚠ Common slip: dividing by the number of steps (2) gives 15.00, which is wrong because the times are not evenly spaced.
🔑Tables are often unevenly spaced. Count the change in the input row, not the number of columns. Include units: miles ÷ hours gives miles per hour.
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Rate of Change at a Point

The rate of change at a point quantifies how fast the output would change were the input to change at that exact point. You can't use the formula on an interval of length zero, but you can approximate the rate by using the average rate of change over small intervals containing the point, if such values exist.

We can't divide by an interval of length 0, so we estimate the rate of change at the point with the average rate over a small interval that contains it. Start with [a, a + h] and shrink h to watch the averages settle. (For x², the two-sided interval happens to give exactly the same value for every h, so try both one-sided modes too.)

interval size h1
-0.500.511.522.5302468xf(x)
hintervalaverage rate of change
1[1.00, 2.00]3.000
0.5[1.00, 1.50]2.500
0.25[1.00, 1.25]2.250
0.1[1.00, 1.10]2.100
0.05[1.00, 1.05]2.050
0.01[1.00, 1.01]2.010
current: [1.00, 2.00] → 3.000As h shrinks, the values settle near 2.00: that is the rate of change at x = 1.
💡Estimates can come from the right of the point, the left, or both sides. The two-sided interval [a − h, a + h] is usually the most balanced, and when you have a table, use the smallest interval around the point that the table allows.
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Comparing Two Points

The rates of change at two points can be compared by estimating each one with an average rate over a sufficiently small interval containing that point, then comparing the two numbers. Use intervals of the same small size around each point so the comparison is fair.

f(x) = x³ − 3x. Each point's rate is estimated with an interval of length 0.2 centered on it. Move the points and compare. The dashed lines show the slope each estimate found.

point p0.0
point q2.0
-2-1012-8-6-4-202468xf(x)pq
rate near p ≈ -2.99 (decreasing)rate near q ≈ 9.01 (increasing)greater rate (more positive): qchanging faster (larger size, ignoring sign): q
⚠️"Greater" and "faster" are different questions. A rate of −9 is less than a rate of 3, but the output is changing faster at the point with rate −9, because only the size matters for speed. The sign tells the direction.
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Positive and Negative Rates

A positive rate of change indicates that as one quantity increases or decreases, the other quantity does the same. They move together. A negative rate of change indicates that as one quantity increases, the other decreases. They move in opposite directions.

The graph shows the temperature T (°F) of a room t hours after the heater is switched on. Pick any interval and read the sign of the average rate of change.

start t0.0
end t2.0
012345678404550556065707580t (hours)T (°F)
(62.0 − 50.0) ÷ (2.0 − 0.0) = 6.00 °F per hourPositive rate: the quantities move in the same direction. As time increases, temperature increases. Looking backward in time, as time decreases, temperature decreases too.
⚠️Try start t = 0 and end t = 8. The average rate is positive (50 → 52 °F), yet the temperature fell during the last four hours. An average rate only compares the two endpoints, so a positive average does not mean the function increases at every point of the interval.
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Rates in Context

Rates of change quantify how two quantities vary together. In a real situation the rate always carries units: output units per input unit. A complete description names the interval, the quantity, the value of the rate, and the units.

Choose a situation and an interval. The sentence below is the pattern to copy on the exam: name the interval, the quantity, the rate, and the units.

start t2
end t5
024681000.511.522.533.5t (min)V (m³)
“On the interval from t = 2 to t = 5 minutes, the average rate of change of V is 0.36 cubic meters per minute.”Positive: as t increases, V increases.
🔑Sentence frame: “On the interval from ___ to ___ (input units), the average rate of change of ___ is ___ (output units per input unit).” Add whether the output increases or decreases to show you understand the sign.
← Back to Lesson 1.2Try Guided Practice →Ready for 1.3? Rates of Change in Linear and Quadratic Functions shows how these rates behave for two important families.