AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/5/5-1)
Unit 5 · Topic 5.1
5.1 Using the Mean Value Theorem
The Mean Value Theorem (MVT) says that a smooth enough function must, at some moment, be changing at exactly its average rate. If you average 60 mph on a trip, at some instant your speedometer read exactly 60. It is an existence theorem, and on the exam you must check both of its conditions.
Key terms
- Mean Value Theorem
- average rate of change
- instantaneous rate of change
- continuous
- differentiable
- secant line
The theorem, with both conditions
If f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one number c in (a, b) such that f′(c) = (f(b) − f(a)) / (b − a).
The left side is an instantaneous rate (a tangent slope). The right side is the average rate of change over the interval (a secant slope). So the theorem says some tangent line is parallel to the secant line through the endpoints.
Why each condition matters
Drop either condition and the conclusion can fail:
- Continuous on [a, b]: no jumps or holes, including at the endpoints. A jump could let the function “teleport” and skip the average rate entirely.
- Differentiable on (a, b): no corners, cusps or vertical tangents inside. At a corner, the slope can jump right past the average value. For example, |x| on [−1, 2] has average rate 1/3, but its slopes are only −1 and 1.
- Differentiability isn't needed at the endpoints, only strictly between them.
Rolle's Theorem: a special case
If f(a) = f(b), the average rate of change is 0, so MVT guarantees a c in (a, b) with f′(c) = 0: a horizontal tangent. This special case is called Rolle's Theorem. Picture a ball thrown up and caught at the same height: at some moment its vertical velocity is 0.
Using MVT with tables
Most exam questions give a table of values for a differentiable function and ask whether there must be a time when the rate equals some value. Compute the average rate of change over an interval from the table. If it equals the target value and the function is differentiable, MVT says yes.
Since differentiable implies continuous, “f is differentiable” covers both conditions. Say so in your answer.
MVT vs. IVT
The Intermediate Value Theorem (1.16) is about the function's values: f(c) = N. The Mean Value Theorem is about the derivative's values: f′(c) = average rate. If a question asks whether f itself reaches a number, use IVT. If it asks whether the rate of change reaches a number, use MVT.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Finding the value of c
Let f(x) = x³ − 3x on [0, 2]. Verify that MVT applies, and find every value of c that it guarantees.
Show the solutionHide the solution
- Step 1: f is a polynomial, so it's continuous on [0, 2] and differentiable on (0, 2).
- Step 2: Average rate: f(0) = 0 and f(2) = 8 − 6 = 2, so (2 − 0)/(2 − 0) = 1.
- Step 3: Solve f′(c) = 1: 3c² − 3 = 1, so c² = 4/3 and c = ±2/√3.
- Step 4: Only c = 2/√3 ≈ 1.155 is in (0, 2).
Answer: c = 2/√3 ≈ 1.155
- Example 2
MVT with a table
A car's position, in miles, is a differentiable function s(t) with t in hours. s(1) = 40 and s(3) = 160. Must there be a time between t = 1 and t = 3 when the car's velocity is exactly 60 mph? Justify.
Show the solutionHide the solution
- Step 1: s is differentiable, so it's also continuous on [1, 3] and differentiable on (1, 3).
- Step 2: Average rate of change: (160 − 40)/(3 − 1) = 120/2 = 60 mph.
- Step 3: MVT guarantees a c in (1, 3) with s′(c) = 60.
Answer: Yes. Because s is differentiable and (s(3) − s(1))/(3 − 1) = 60, the Mean Value Theorem guarantees a time c in (1, 3) when s′(c) = 60 mph.
- Example 3
Trap: a corner breaks MVT
For f(x) = |x| on [−1, 2], does MVT guarantee a c with f′(c) = 1/3?
Show the solutionHide the solution
- Step 1: Average rate: (f(2) − f(−1))/(2 − (−1)) = (2 − 1)/3 = 1/3.
- Step 2: Check the conditions: f is continuous on [−1, 2], but it has a corner at x = 0, which is inside (−1, 2). So f is not differentiable on (−1, 2).
- Step 3: In fact f′(x) is −1 for x < 0 and 1 for x > 0. It never equals 1/3.
Answer: No. MVT doesn't apply because f isn't differentiable at x = 0, and no such c exists.
Common mistakes
- Skipping the conditions. Free-response scoring requires you to state that f is continuous and differentiable (or just differentiable) on the interval.
- Using MVT when the question is about function values. That's IVT.
- Giving a c outside the open interval (a, b).
On the exam
- The classic free-response form: a table of values for a differentiable function and the question “Must there be a value of t, for a < t < b, such that f′(t) = k? Justify.” Compute the average rate and cite MVT by name.
- If the question gives a table but doesn't say the function is differentiable, you can't use MVT. Look for that statement.
Connected topics
- Unit 11.16 Working with the Intermediate Value Theorem (IVT)
- Unit 22.1 Defining Average and Instantaneous Rates of Change at a Point
- Unit 22.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
- Unit 55.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
Videos
Check yourself
4 questions on 5.1 Using the Mean Value Theorem. Pick an answer to see if you got it, and why.
Let f(x) = x³ − 3x. What value of c satisfies the conclusion of the Mean Value Theorem for f on the interval [0, 3]?
For which of the following functions are the hypotheses of the Mean Value Theorem satisfied on the given interval?
On a straight highway, a car passes mile marker 40 at 1:00 p.m. and mile marker 70 at 1:20 p.m. Assume the car's position is a differentiable function of time. Which of the following must be true?
Let f(x) = x + 1/x. Then f(1/2) = f(2) = 5/2. What value of c in the open interval (1/2, 2) satisfies the conclusion of the Mean Value Theorem for f on [1/2, 2] ?
0 of 4 answered