AP · Rates of change (average and over equal intervals) · 14 min read · Updated 2026-05-10

Rates of change (average and over equal intervals) — AP Precalculus Study Guide

For: AP Precalculus candidates sitting AP Precalculus.

Covers: Average rate of change formula, difference quotients, calculation over equal intervals, first and higher-order differences for polynomials, the constant nth difference property for degree n polynomials, and contextual interpretation of average rates of change.

You should already know: Function evaluation and notation. Limit definition of the derivative. Basic polynomial algebra.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Precalculus style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Rates of change (average and over equal intervals)?

Average rate of change measures how much a function changes per unit change in input across a specified interval, unlike instantaneous rate of change which describes change at a single point as the limit of smaller and smaller intervals. Per the AP Precalculus Course and Exam Description (CED), this topic is Unit 1 Topic 1.1, making up roughly 2-3% of total exam score, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections. It is also a foundational skill embedded in almost all other Unit 1 topics, so mastery is critical even for questions that do not explicitly ask for average rate of change. When we calculate average rate of change over equal intervals (intervals with identical width $Δ x$ ), we can use the properties of finite differences to identify the degree of an unknown polynomial from tabular data, a unique AP Precalculus skill not covered in earlier algebra courses. Synonyms for average rate of change include difference quotient, slope of the secant line, and mean rate of change.

2. Average Rate of Change (General Definition)

For any function $f (x)$ defined at both endpoints of the interval $[a, b]$ , the average rate of change between $x = a$ and $x = b$ is defined as the total change in the output of the function divided by the total change in the input. The formula for average rate of change is: $Average rate of change = \frac{f ( b ) - f ( a )}{b - a} = \frac{Δ f}{Δ x}$ Geometrically, this is exactly the slope of the secant line connecting the two points $(a, f (a))$ and $(b, f (b))$ on the graph of $f$ . The sign of the average rate of change tells you whether the function is increasing (positive) or decreasing (negative) on average across the interval; a value of zero means the starting and ending output values are equal, not that the function never changes on the interval. Unlike the instantaneous rate of change (which requires a limit to calculate), average rate of change can always be computed directly with basic algebra from two points.

Worked Example

For $f (x) = 2 x^{2} - 3 x + 1$ , calculate the average rate of change from $x = 1$ to $x = 4$ .

Evaluate $f (x)$ at the left endpoint: $f (1) = 2 (1)^{2} - 3 (1) + 1 = 2 - 3 + 1 = 0$ .
Evaluate $f (x)$ at the right endpoint: $f (4) = 2 (4)^{2} - 3 (4) + 1 = 32 - 12 + 1 = 21$ .
Calculate the change in output and change in input: $Δ f = f (4) - f (1) = 21 - 0 = 21$ , $Δ x = 4 - 1 = 3$ .
Divide to get the final result: $\frac{Δ f}{Δ x} = \frac{21}{3} = 7$ . The average rate of change over $[1, 4]$ is $7$ .

Exam tip: If you reverse the order of subtraction for both numerator and denominator, you will get the same result ( $\frac{f ( a ) - f ( b )}{a - b} = \frac{f ( b ) - f ( a )}{b - a}$ ). Always match the order of subtraction: if you subtract the left endpoint from the right endpoint in the denominator, do the same in the numerator to avoid sign errors.

3. Average Rate of Change Over Equal Intervals

Equal intervals are consecutive subintervals of the x-axis that all have the same width $h = Δ x$ . Calculating average rate of change over equal intervals allows for direct comparison of how quickly a function is changing across different regions of its domain, because the interval width is constant. A key shortcut for this case: since $h$ is the same for all intervals, the average rate of change $\frac{f ( x + h ) - f ( x )}{h}$ is directly proportional to the difference $f (x + h) - f (x)$ . This means you can compare the magnitude of average rates just by comparing the differences, skipping the division by $h$ to save time on exams. This shortcut only works for equal intervals; it never works for unequal interval widths.

Worked Example

The height of a projectile $t$ seconds after launch is given by $h (t) = - 16 t^{2} + 48 t + 64$ , for $0 \leq t \leq 4$ , where height is measured in feet. Calculate the average rate of change of height over each 1-second equal interval $[0, 1], [1, 2], [2, 3], [3, 4]$ , and interpret the results.

Confirm all intervals have equal width $h = 1$ , so the shortcut applies if we only need to compare magnitudes.
Calculate $h (t)$ at each endpoint: $h (0) = 64$ , $h (1) = - 16 (1) + 48 (1) + 64 = 96$ , $h (2) = - 16 (4) + 48 (2) + 64 = 96$ , $h (3) = - 16 (9) + 48 (3) + 64 = 64$ , $h (4) = - 16 (16) + 48 (4) + 64 = 0$ .
Calculate average rate of change for each interval:
- $[0, 1] : \frac{96 - 64}{1} = 32$ ft/s
- $[1, 2] : \frac{96 - 96}{1} = 0$ ft/s
- $[2, 3] : \frac{64 - 96}{1} = - 32$ ft/s
- $[3, 4] : \frac{0 - 64}{1} = - 64$ ft/s
Interpretation: The magnitude of the average rate increases after the projectile reaches its peak height between $t = 1$ and $t = 2$ , meaning the projectile speeds up as it falls, which matches physical expectations.

Exam tip: For FRQ questions asking for comparison of average rates over equal intervals, you can use the shortcut of comparing differences directly to save calculation time, but always write the full average rate calculation if the question asks for explicit values.

4. Higher-Order Differences and the Constant nth Difference Property

When working with polynomials over equal intervals, we can use finite differences to find the degree of an unknown polynomial from a table of values. A first difference $Δ^{1} f (x)$ is just the difference between consecutive function values: $Δ^{1} f (x) = f (x + h) - f (x)$ . A second difference is the difference between consecutive first differences: $Δ^{2} f (x) = Δ^{1} f (x + h) - Δ^{1} f (x)$ , and so on for higher-order differences. The core property tested on the AP exam is: For any polynomial of degree $n$ , the nth differences over equal intervals of any width $h > 0$ are constant. The value of the constant nth difference is given by: $Constant n th difference = n! \cdot h^{n} \cdot a_{n}$ where $a_{n}$ is the leading coefficient of the polynomial, and $n!$ is $n$ factorial ( $n! = n \times (n - 1) \times ... \times 1$ ). This property lets you find both the degree and the leading coefficient of an unknown polynomial from a table of values.

Worked Example

The table below gives values of a polynomial $p (x)$ at equal 1-unit intervals. What is the degree of $p (x)$ , and what is its leading coefficient?

$x$	0	1	2	3	4	5
$p (x)$	3	2	7	24	59	118

Calculate first differences: $2 - 3 = - 1$ , $7 - 2 = 5$ , $24 - 7 = 17$ , $59 - 24 = 35$ , $118 - 59 = 59$ . First differences are $[- 1, 5, 17, 35, 59]$ , which are not constant, so $p (x)$ is not degree 1.
Calculate second differences: $5 - (- 1) = 6$ , $17 - 5 = 12$ , $35 - 17 = 18$ , $59 - 35 = 24$ . Second differences are $[6, 12, 18, 24]$ , which are not constant, so $p (x)$ is not degree 2.
Calculate third differences: $12 - 6 = 6$ , $18 - 12 = 6$ , $24 - 18 = 6$ . Third differences are constant at 6, so $p (x)$ is degree 3.
Use the constant nth difference formula to find leading coefficient: $6 = 3! (1)^{3} a_{3} ⟹ 6 = 6 a_{3} ⟹ a_{3} = 1$ .

$p (x)$ is a degree 3 polynomial with leading coefficient 1.

Exam tip: You need at least two identical nth differences to confirm they are constant, which means you need at least $n + 2$ points to confirm a polynomial is degree $n$ . If you only have one nth difference, you cannot confirm it is constant, so you cannot determine the degree.

5. Common Pitfalls (and how to avoid them)

Wrong move: Calculating $\frac{f ( a ) - f ( b )}{b - a}$ when finding average rate of change over $[a, b]$ . Why: Students mix up the order of numerator and denominator subtractions, leading to a sign error. Correct move: Always match the order of subtraction: if you subtract the left x from the right x in the denominator, subtract the left function value from the right function value in the numerator.
Wrong move: Claiming a polynomial is degree $n$ after calculating one nth difference, with fewer than $n + 2$ total points. Why: Students assume one value is enough to confirm "constant" difference, but you need at least two matching values to confirm constancy. Correct move: Only confirm a degree $n$ when you have at least two identical nth differences, from a table with at least $n + 2$ points.
Wrong move: Comparing average rates of change over unequal intervals by comparing $f (x + h) - f (x)$ directly. Why: Students get used to the shortcut for equal intervals and forget it does not apply to unequal widths. Correct move: Always divide by the change in x to get average rate, regardless of interval size, when comparing rates across unequal intervals.
Wrong move: Interpreting a zero average rate of change over $[a, b]$ to mean the function is constant at every point on the interval. Why: Students confuse average rate over an interval with instantaneous rate at every point. Correct move: Remember that a zero average rate only means the starting and ending values of $f$ are equal; the function can change anywhere in between.
Wrong move: Forgetting to scale the nth difference by $h^{n}$ when calculating leading coefficient for intervals of width $h \neq = 1$ . Why: Students memorize the formula for unit intervals and omit the $h^{n}$ term for non-unit steps. Correct move: Always write the full formula $constant n th difference = n! h^{n} a_{n}$ , and plug in $h$ explicitly before solving for $a_{n}$ .

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

For $f (x) = x^{3} - 4 x^{2} + 5$ , what is the average rate of change over the interval $[- 1, 3]$ ? A) $- 4$ B) $- 1$ C) $0$ D) $1$

Worked Solution: First evaluate $f$ at both endpoints of the interval: $f (- 1) = (- 1)^{3} - 4 (- 1)^{2} + 5 = - 1 - 4 + 5 = 0$ , and $f (3) = 3^{3} - 4 (3)^{2} + 5 = 27 - 36 + 5 = - 4$ . Calculate the change in x: $Δ x = 3 - (- 1) = 4$ , and the change in f: $Δ f = f (3) - f (- 1) = - 4 - 0 = - 4$ . Divide to get average rate of change: $\frac{Δ f}{Δ x} = \frac{- 4}{4} = - 1$ . The correct answer is B.

Question 2 (Free Response)

The table below gives values of a polynomial function $f (x)$ at equal 2-unit intervals.

$x$	0	2	4	6	8	10
$f (x)$	$- 3$	$1$	$17$	$53$	$117$	$217$

(a) Calculate the average rate of change of $f (x)$ over each consecutive equal interval. (b) Use difference tables to find the degree of $f (x)$ . (c) Find the leading coefficient of $f (x)$ .

Worked Solution: (a) Each interval has width $Δ x = 2$ , so average rate of change for $[x, x + 2]$ is $\frac{f ( x + 2 ) - f ( x )}{2}$ :

$[0, 2] : \frac{1 - ( - 3 )}{2} = 2$
$[2, 4] : \frac{17 - 1}{2} = 8$
$[4, 6] : \frac{53 - 17}{2} = 18$
$[6, 8] : \frac{117 - 53}{2} = 32$
$[8, 10] : \frac{217 - 117}{2} = 50$ Average rates of change are $[2, 8, 18, 32, 50]$ .

(b) First differences are $[4, 16, 36, 64, 100]$ (not constant, so not degree 1). Second differences are $[12, 20, 28, 36]$ (not constant, so not degree 2). Third differences are $[8, 8, 8]$ (constant). $f (x)$ is a degree 3 polynomial.

(c) Use the constant nth difference formula: $8 = 3! (2)^{3} a_{3} ⟹ 8 = 48 a_{3} ⟹ a_{3} = \frac{1}{6}$ . The leading coefficient of $f (x)$ is $\frac{1}{6}$ .

Question 3 (Application / Real-World Style)

A coffee shop tracks its daily revenue over 5 business days, fitting a polynomial model $R (d) = - 20 d^{3} + 150 d^{2} - 120 d + 200$ , where $d = 1$ is Monday, $d = 2$ is Tuesday, ..., $d = 5$ is Friday, and $R (d)$ is revenue in dollars on day $d$ . Intervals between consecutive days are equal (1 day apart). Calculate the average rate of change of revenue from Monday to Friday, and identify which daily interval has the largest average increase in revenue.

Worked Solution: First evaluate $R (d)$ at each day: $R (1) = 210$ , $R (2) = 400$ , $R (3) = 650$ , $R (4) = 840$ , $R (5) = 850$ . The average rate of change from Monday to Friday is $\frac{R ( 5 ) - R ( 1 )}{5 - 1} = \frac{850 - 210}{4} = 160$ dollars per day. For daily intervals (width 1), the average rate of change equals $R (d + 1) - R (d)$ : the daily rates are $190$ (Monday to Tuesday), $250$ (Tuesday to Wednesday), $190$ (Wednesday to Thursday), and $10$ (Thursday to Friday). The largest average increase in revenue occurs between Tuesday and Wednesday, with an average increase of $250$ per day, meaning the coffee shop's revenue grows fastest mid-week under this model.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Average Rate of Change (General)	$\frac{f ( b ) - f ( a )}{b - a} = \frac{Δ f}{Δ x}$	Applies to any function over any interval $[a, b]$ ; equals slope of the secant line between $(a, f (a))$ and $(b, f (b))$
Average Rate Over Equal Intervals	$\frac{f ( x + h ) - f ( x )}{h}$	$h$ is constant interval width; you can skip dividing by $h$ to compare relative magnitudes of rates
First Difference	$Δ^{1} f (x) = f (x + h) - f (x)$	Numerator of average rate of change over equal interval width $h$
nth Difference	$Δ^{n} f (x) = Δ^{n - 1} f (x + h) - Δ^{n - 1} f (x)$	Higher differences are differences of the previous order of differences
Constant nth Difference Property	Constant nth difference $= n! h^{n} a_{n}$	Applies only to degree $n$ polynomials and equal intervals of width $h$ ; $a_{n}$ is the leading coefficient
Polynomial Degree Identification	Degree $= n$ if nth differences are constant	Requires at least 2 identical nth differences to confirm constancy; needs at least $n + 2$ total points

8. What's Next

Mastery of average rates of change over equal intervals is a non-negotiable prerequisite for all remaining topics in Unit 1: Polynomial and Rational Functions. Next, you will use the constant nth difference property to construct polynomial models from tabular data, and use average rates of change to analyze the increasing/decreasing behavior and concavity of higher-degree polynomials. Without mastering the skills in this chapter, you will not be able to correctly identify polynomial degrees or construct models from data, both common FRQ tasks on the AP exam. More broadly, this topic lays the conceptual foundation for the derivative, which you will explore when comparing instantaneous and average rates across other function families later in the course. Follow-up topics to study next: Polynomial construction from tables Comparing average and instantaneous rates Polynomial concavity and monotonicity Equations of secant lines

← Back to topic

Stuck on a specific question?
Snap a photo or paste your problem — Ollie (our AI tutor) walks through it step-by-step with diagrams.
Try Ollie free →

Rates of change (average and over equal intervals) — AP Precalculus Study Guide

1. What Is Rates of change (average and over equal intervals)?

2. Average Rate of Change (General Definition)

Worked Example

3. Average Rate of Change Over Equal Intervals

Worked Example

4. Higher-Order Differences and the Constant nth Difference Property

Worked Example

5. Common Pitfalls (and how to avoid them)

6. Practice Questions (AP Precalculus Style)

Question 1 (Multiple Choice)

Question 2 (Free Response)

Question 3 (Application / Real-World Style)

7. Quick Reference Cheatsheet

8. What's Next

More study guides