AP Calculus BC, AP exams, 8–11%

Unit 5: Analytical Applications of Differentiation

Unit 5 covers the Mean Value and Extreme Value Theorems, the first and second derivative tests, concavity, curve sketching, and optimization. The justification language required here is the same language rewarded across the whole free-response section.

Position: 5 of 10 units
Published weighting: 8–11% 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 BC. 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 on a closed interval.
Increasing and decreasing behavior: Using the first derivative sign to describe intervals of increase and decrease.
First and second derivative tests: Classifying critical points and knowing when each test applies or is inconclusive.
Concavity and inflection points: Second derivative sign analysis and locating changes in concavity.
Curve sketching: Combining derivative information into a complete graph with asymptotes and endpoints.
Optimization: Building a one-variable model and justifying that the candidate is the required extremum.

Study tips

How to practice Analytical Applications.

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

Write justifications in rubric language: name the derivative, state its sign change, and draw the conclusion.
Finish optimization problems with the justification, not just the candidate value.
Practice reading a derivative graph and describing the original function's behavior from it.

Where this fits

Unit 5: Analytical Applications of Differentiation is 5 of 10 units in the AP Calculus BC guide, and it carries a published weighting of 8–11% in the official framework. Use the full guide to balance it against everything else you have to review.

Back to the AP Calculus BC 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
On what open interval is f(x) = x^3 - 3x^2 decreasing?
1. A(0, 2)
2. B(-∞, 0)
3. C(2, ∞)
4. D(0, ∞)
Show answer
Answer: A. (0, 2)
The derivative f'(x) = 3x^2 - 6x = 3x(x - 2) is negative exactly when x lies strictly between the roots 0 and 2, so f decreases on (0, 2) and increases outside it. Reporting (2, ∞) reads the sign chart backwards.
Key insight: Decreasing corresponds to f' < 0, which for a product of two factors happens between the roots.
Question 2Foundational
How many points of inflection does the graph of f(x) = x^4 - 4x^3 have?
1. A0
2. B1
3. C2
4. D3
Show answer
Answer: C. 2
The second derivative is f''(x) = 12x^2 - 24x = 12x(x - 2), which is zero at x = 0 and x = 2 and changes sign at both, so there are two points of inflection. Counting only x = 2 misses the sign change at the origin.
Key insight: Every zero of f'' is a candidate, but only those with a sign change are inflection points.
Question 3Intermediate
Does the Mean Value Theorem apply to f(x) = |x - 1| on the closed interval [0, 3], and why?
1. ANo, because f is not differentiable at x = 1
2. BNo, because f is not continuous at x = 1
3. CYes, because f is continuous on the interval
4. DYes, and the guaranteed value is c = 1
Show answer
Answer: A. No, because f is not differentiable at x = 1
The Mean Value Theorem requires continuity on the closed interval and differentiability on the open interval. This function is continuous everywhere, but the corner at x = 1 destroys differentiability there, so the hypotheses fail even though continuity holds.
Key insight: State both hypotheses separately: continuity on [a, b] and differentiability on (a, b).
Question 4Intermediate
A twice-differentiable function f satisfies f'(2) = 0 and f''(2) = -5. What does the second derivative test conclude about x = 2?
1. Af has a local maximum at x = 2
2. Bf has a local minimum at x = 2
3. Cf has a point of inflection at x = 2
4. DThe test is inconclusive
Show answer
Answer: A. f has a local maximum at x = 2
A critical point where the second derivative is negative sits on a concave-down piece of the graph, so it is a local maximum. The test is inconclusive only when f''(c) = 0, and a nonzero f'' rules out an inflection point at that x-value.
Key insight: Negative second derivative means concave down, so a critical point there is a local maximum.
Question 5Advanced
What is the minimum value of f(x) = x + 4/x for x > 0?
1. A2
2. B4
3. C8
4. DThere is no minimum value
Show answer
Answer: B. 4
Setting f'(x) = 1 - 4/x^2 = 0 gives x = 2 for x > 0, and f''(x) = 8/x^3 is positive there, confirming a minimum. The minimum value is f(2) = 2 + 2 = 4. Answering 2 reports the x-value at which the minimum occurs rather than the minimum value itself.
Key insight: A question asking for the minimum value wants f(c), not the location c.

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Unit 5: Analytical Applications of Differentiation