1. Section 3.2 Day 1 Mean Value Theorem
AP Calculus AB

2. Learning Targets
Define & apply the Mean Value Theorem

3. Mean Value Theorem: Definition
If 𝑓 is continuous on the closed interval [𝑎, 𝑏] and differentiable on the open interval (𝑎, 𝑏), then there exists a number 𝑐 in (𝑎, 𝑏) such that
𝑓 ′ 𝑐 = 𝑓 𝑏 −𝑓 𝑎 𝑏−𝑎

4. Mean Value Theorem: Conceptual
Geometrically: The tangent line is parallel to the secant line at some point
Application: The IROC is equal to the AROC at some point

5. Example 1
Determine all the numbers 𝑐 which will satisfy the conclusions of the Mean Value Theorem for the following function: 𝑓 𝑥 = 𝑥 3 +2 𝑥 2 −𝑥 on [−1, 2]
1. Confirm that 𝑓(𝑥) is continuous and differentiable on the interval
2. 𝑓 ′ 𝑐 = 𝑓 2 −𝑓 −1 2− −1
3. 3 𝑐 2 +4𝑐−1=4
4. Use calculator: 𝑐=0.7863

6. Example 2
Show that the function 𝑓 𝑥 = 𝑥 2 satisfies the hypotheses of the Mean Value Theorem on the interval 0, 2 . Then, find the value of 𝑐 that satisfies the Mean Value Theorem
1. Confirm 𝑓(𝑥) is continuous and differentiable on the interval
2. 𝑓 ′ 𝑐 = 𝑓 2 −𝑓 0 2−0
3. 2𝑐=2
4. 𝑐=1

7. Example 3
The functions 𝑓 and 𝑔 are differentiable for all real numbers. Also, ℎ 𝑥 =𝑓 𝑔 𝑥 −6
Explain why there must be a value 𝑐 for <𝑐<3 such that ℎ ′ 𝑐 =−5.

8. Example 3: Answer ℎ 3 −ℎ 1 3−1 = −7−3 3−1 =−5
ℎ 3 −ℎ 1 3−1 = −7−3 3−1 =−5
Since ℎ is continuous and differentiable, by the Mean Value Theorem, there exists a value 𝑐, 1 <𝑐<3, such that ℎ ′ 𝑐 =−5

9. Example 4
Find a value 𝑐, that satisfies the Mean Value Theorem for the function 𝑓 𝑥 =4 𝑥 2 −𝑥−6 on the interval [1, 3]
A) 15
B) 2
C) -6
D) 8 B

10. Example 5
A trucker handed in a ticket at a toll both showing that in 2 hours she had covered miles on a toll road with speed limit 65 mph. Does the trucker deserve a ticket?
A) Yes
B) No
C) Cannot be determined

11. Example 6
Determine the value of 𝑐 that satisfies the Mean Value Theorem for 𝑓 𝑥 = 1 𝑥−1 on the interval 0, 3
A) 1/2
B) 0
C) 1
D) Cannot be determined

12. Exit Ticket for Feedback
Let 𝑓 𝑥 = 𝑥 4 −2 𝑥 2 .
Find all the values of x where 𝑓 ′ 𝑥 =0 on the interval (−2, 2)