AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/2/2-1)
Unit 2 · Topic 2.1
2.1 Defining Average and Instantaneous Rates of Change at a Point
The average rate of change on an interval is the slope of a secant line. Let the interval shrink toward one point and, if the slopes settle on a value, you get the instantaneous rate of change: the slope of the tangent line. This is the doorway to the derivative.
Key terms
- average rate of change
- secant line
- difference quotient
- instantaneous rate of change
- tangent line
Average rate of change and the difference quotient
On [a, b], the average rate of change of f is (f(b) − f(a)) / (b − a), the slope of the secant line through (a, f(a)) and (b, f(b)).
It's often written with a step size h instead of a second endpoint. If the interval runs from a to a + h, the average rate of change is (f(a + h) − f(a)) / h. This expression is called a difference quotient. h can be positive (the interval goes right from a) or negative (it goes left).
A second form uses a moving point x instead of a step h: (f(x) − f(a)) / (x − a) is the slope between (a, f(a)) and (x, f(x)). Both forms describe the same secant slope, just labeled differently, and you should recognize both.
From average to instantaneous
The instantaneous rate of change at x = a is the limit of the difference quotient as the step shrinks: lim (h→0) (f(a + h) − f(a)) / h, if that limit exists. Geometrically, the secant lines through (a, f(a)) pivot toward the tangent line, and the secant slopes approach the tangent slope.
Notice that substituting h = 0 directly gives 0/0. That's exactly why you need the limit tools from Unit 1.
Seeing the pattern numerically
For f(x) = x³ at a = 1, the difference quotient with h = 0.1, 0.01 and 0.001 gives 3.31, 3.0301 and 3.003001. The values are closing in on 3, so the instantaneous rate of change at x = 1 appears to be 3. In 2.2 you'll compute it exactly.
| h | Interval | Secant slope (f(1 + h) − f(1)) / h |
|---|---|---|
| 0.1 | [1, 1.1] | 3.31 |
| 0.01 | [1, 1.01] | 3.0301 |
| 0.001 | [1, 1.001] | 3.003001 |
Units and meaning
Both rates have units of output per input. If s(t) is position in meters and t is time in seconds, the average rate of change is average velocity in m/s, and the instantaneous rate is velocity at a moment, also in m/s.
Words in a question tell you which one is wanted. “Over the interval,” “between,” or “from t = 2 to t = 5” mean average. “At t = 3” or “when the radius is 4” means instantaneous.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Secant slope for a polynomial
Find the average rate of change of f(x) = x³ on [1, 3].
Show the solutionHide the solution
- Step 1: f(3) = 27 and f(1) = 1.
- Step 2: Average rate of change = (f(3) − f(1)) / (3 − 1) = (27 − 1) / 2.
Answer: 13. On average, f increases by 13 units of output per unit of x on [1, 3].
- Example 2
Average rate for a square root, with interpretation
A tank's water depth is d(t) = √t feet, t hours after filling starts. Find the average rate of change of d from t = 4 to t = 9 and interpret it.
Show the solutionHide the solution
- Step 1: d(9) = 3 and d(4) = 2.
- Step 2: Average rate = (3 − 2) / (9 − 4) = 1/5.
- Step 3: Units: feet per hour.
Answer: 1/5 foot per hour: between hours 4 and 9, the water depth rose by an average of 0.2 foot per hour.
- Example 3
Trap: dividing by the wrong number
Let f(x) = x². A student writes the average rate of change of f on [3, 3.5] as (f(3 + 0.5) − f(3)) / 3.5. Find the correct value.
Show the solutionHide the solution
- Step 1: The bottom of a difference quotient is the change in x, which is the width of the interval: 3.5 − 3 = 0.5. It is not the right endpoint.
- Step 2: f(3.5) = 12.25 and f(3) = 9, so the change in f is 3.25.
- Step 3: Average rate = 3.25 / 0.5 = 6.5. The student's version gives 3.25/3.5 ≈ 0.929, which is far too small.
Answer: 6.5
Common mistakes
- Plugging h = 0 into the difference quotient and writing 0/0 as the answer. The instantaneous rate is a limit.
- Mixing up the order of subtraction on the top and bottom.
- Answering an “at t = 3” question with an average rate over some interval.
On the exam
- Average rate of change shows up constantly in free response, often from tables. Show the expression with numbers, like (85 − 76)/(5 − 2), and give units.
- Questions that ask which expression equals the slope of a secant line test whether you can recognize a difference quotient.
Connected topics
Videos
Check yourself
4 questions on 2.1 Defining Average and Instantaneous Rates of Change at a Point. Pick an answer to see if you got it, and why.
What is the average rate of change of f(x) = x³ − 2x on the interval [1, 3]?
What is the average rate of change of f(x) = sin x on the interval [0, π/2] ?
For f(x) = x², the average rate of change of f over the interval [1, b] is 6, where b > 1. What is the value of b ?
The function f is differentiable, and f(3) = 5. The line tangent to the graph of f at x = 3 passes through the point (1, 1). What is the value of f′(3)?
0 of 4 answered