AP Calculus AB, AP exams, 15–18%

Unit 5: Analytical Applications of Differentiation

Unit 5 carries one of the largest published weightings in AB. It covers the Mean Value Theorem and Extreme Value Theorem, the first and second derivative tests, concavity and points of inflection, curve sketching, and optimization. Justification language matters more here than anywhere else in the course.

Position: 5 of 8 units
Published weighting: 15–18% in the official framework
Topics covered: 6 topics in this guide

What is covered

Topics inside this unit.

These topics describe the scope students most often review for this part of AP Calculus AB. Treat them as a study checklist rather than an exhaustive list of what a single exam form can sample.

Mean Value and Extreme Value Theorems: Stating hypotheses correctly and applying each theorem to guarantee a value on an interval.
Increasing and decreasing behavior: Using the sign of the first derivative to describe intervals of increase and decrease.
First and second derivative tests: Classifying critical points as local maxima, minima, or neither, and knowing when each test applies.
Concavity and points of inflection: Using the second derivative to determine concavity and to locate where it changes.
Curve sketching: Assembling derivative information into a complete graph, including asymptotes and endpoints.
Optimization: Setting up a model, reducing it to one variable, and justifying that a candidate is the extremum.

Study tips

How to practice Analytical Applications.

Section-specific guidance that follows from how this material is usually tested.

Memorize the exact justification phrasing for each test; the rubric wants the derivative sign change stated explicitly.
Work optimization problems all the way through the justification step, not just to the candidate value.
Practice reconstructing a function's behavior from a derivative graph, which is a recurring multiple-choice pattern.

Where this fits

Unit 5: Analytical Applications of Differentiation is 5 of 8 units in the AP Calculus AB guide, and it carries a published weighting of 15–18% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus AB guide Official exam information

Sample Analytical Applications questions

Practice questions, not yet expert-reviewed

Original practice questions written to the published framework for this section. No expert has reviewed them, and they are not taken from any official exam. Try each one before you open the answer.

Question 1Foundational
At what x-value does f(x) = x^3 - 3x have a local minimum?
1. Ax = -1
2. Bx = 0
3. Cx = 1
4. Dx = 3
Show answer
Answer: C. x = 1
Setting f'(x) = 3x^2 - 3 = 0 gives critical points at x = -1 and x = 1. Since f''(x) = 6x is positive at x = 1 and negative at x = -1, the graph is concave up at x = 1, so that is the local minimum and x = -1 is the local maximum.
Key insight: Critical points come from the first derivative; the second derivative sorts them into maxima and minima.
Question 2Foundational
At what x-value does the graph of f(x) = x^3 - 6x^2 + 5 have a point of inflection?
1. Ax = -2
2. Bx = 0
3. Cx = 2
4. Dx = 4
Show answer
Answer: C. x = 2
The second derivative is f''(x) = 6x - 12, which is zero at x = 2 and changes from negative to positive there, so concavity changes and the point is an inflection point. The value x = 4 solves f'(x) = 3x^2 - 12x = 0 and locates a critical point, not an inflection point.
Key insight: An inflection point needs a sign change in f'', not merely a zero of f''.
Question 3Intermediate
For f(x) = x^2 on the closed interval [0, 4], what value of c satisfies the conclusion of the Mean Value Theorem?
1. Ac = 1
2. Bc = 2
3. Cc = 4
4. Dc = 8
Show answer
Answer: B. c = 2
The average rate of change is (f(4) - f(0))/(4 - 0) = 16/4 = 4, and f'(c) = 2c, so 2c = 4 and c = 2. Answering c = 8 sets f'(c) equal to f(4) - f(0) = 16 without dividing by the interval length, and c = 4 confuses the endpoint with the guaranteed interior value.
Key insight: The Mean Value Theorem sets the instantaneous rate equal to the average rate over the interval.
Question 4Intermediate
What is the absolute maximum value of f(x) = x^3 - 3x^2 on the closed interval [-1, 1]?
1. A-4
2. B-2
3. C0
4. D1
Show answer
Answer: C. 0
The candidates are the endpoints and the critical points in the interval: f'(x) = 3x^2 - 6x = 0 gives x = 0 and x = 2, and only x = 0 lies in [-1, 1]. Comparing f(-1) = -4, f(0) = 0, and f(1) = -2 shows the absolute maximum value is 0.
Key insight: On a closed interval, compare function values at every critical point and both endpoints.
Question 5Advanced
A farmer encloses a rectangular pen along a straight river, fencing only the three sides away from the river with a total of 100 meters of fence. What is the largest possible area of the pen?
1. A625 square meters
2. B1250 square meters
3. C2000 square meters
4. D2500 square meters
Show answer
Answer: B. 1250 square meters
If the two sides perpendicular to the river each have length y and the side parallel to it has length x, then x + 2y = 100 and A = xy = 100y - 2y^2. Setting A'(y) = 100 - 4y = 0 gives y = 25, x = 50, and A = 1250 square meters. The value 625 is the largest area when all four sides are fenced, and 2500 comes from taking x = y = 50 without checking the constraint.
Key insight: Reduce the model to one variable using the constraint before differentiating.

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Unit 5: Analytical Applications of Differentiation