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Analytical applications of differentiation practice problems AP Calculus, Unit 5

Written by STEM-Tutor's tutors · Updated October 2026 · 8 questions and 1 free-response question

The mean value theorem, increasing and decreasing intervals, the first and second derivative tests, concavity and optimisation. Try each question before you look: use the hint if you are stuck, then reveal the worked solution one step at a time.

On this page
  1. Key equations
  2. Warm-up questions
  3. Exam-style questions
  4. Free-response practice
  5. Common mistakes
  6. Keep going

Key equations for analytical applications of differentiation

f'(c) = [f(b) − f(a)]/(b − a)mean value theorem
f' changes + to −local maximum
f'' > 0concave up
Candidates testabsolute extrema occur at critical points or endpoints

Questions marked No calculator should be done by hand, as in Part A of the multiple-choice section. You can type fractions, such as 3/4. The full sheet is in our AP Calculus formula sheet.

Warm-up questions

Quick checks on the core ideas. If these feel easy, go straight to the exam-style questions.

Q1Warm-upNo calculator
Find the critical point of f(x) = x2 − 6x + 5.

Hint

Set f'(x) = 0.

Worked solution
  1. f'(x) = 2x − 6 = 0.
  2. x = 3.
Q2Warm-upNo calculator
If f''(x) > 0 on an interval, the graph of f there is:

Hint

The second derivative describes the bending.

Worked solution
  1. f'' > 0 means f' is increasing, so the slopes are getting larger.
  2. That is concave up. The answer is C.
Q3Warm-upNo calculator
For f(x) = x2 on [1, 3], find the value of c guaranteed by the mean value theorem.

Hint

Set f'(c) equal to the average rate of change.

Worked solution
  1. Average rate: (9 − 1)/(3 − 1) = 4.
  2. f'(c) = 2c = 4.
  3. c = 2.

Exam-style questions

Questions at the level of the AP Calculus exam. The ones marked Challenge combine more than one idea.

Q4Exam-styleNo calculator
Find the absolute maximum value of f(x) = x3 − 3x on [0, 2].

Hint

Check the critical points inside the interval and both endpoints.

Worked solution
  1. f'(x) = 3x2 − 3 = 0 gives x = 1 (in the interval).
  2. f(0) = 0, f(1) = −2, f(2) = 2.
  3. The absolute maximum is 2, at x = 2.
Q5Exam-styleNo calculator
Find the x-coordinate of the point of inflection of f(x) = x3 − 6x2 + 4.

Hint

Find where f'' changes sign.

Worked solution
  1. f''(x) = 6x − 12.
  2. f''(x) = 0 at x = 2, and it changes sign there.
  3. x = 2.
Q6Exam-styleNo calculator
A rectangle has a perimeter of 40 m. What is its largest possible area?
m²

Hint

Write the area in terms of one side length.

Worked solution
  1. Sides x and 20 − x, so A = x(20 − x) = 20x − x2.
  2. A'(x) = 20 − 2x = 0 at x = 10.
  3. A = 10 × 10 = 100 m², a square.
Q7Exam-styleNo calculator
f'(x) changes from positive to negative at x = c. Then f has:

Hint

Think about what f does before and after c.

Worked solution
  1. f' > 0 before c: f increases. f' < 0 after c: f decreases.
  2. So f has a local maximum at c. The answer is B.
Q8ChallengeNo calculator
Squares of side x are cut from the corners of a 12 cm by 12 cm sheet to make an open box. What is the largest possible volume?
cm³

Hint

V(x) = x(12 − 2x)2 for 0 < x < 6.

Worked solution
  1. V'(x) = (12 − 2x)2 − 4x(12 − 2x) = (12 − 2x)(12 − 6x).
  2. V'(x) = 0 at x = 2 (x = 6 gives zero volume).
  3. V(2) = 2 × 82 = 128 cm³.

Free-response practice

One original question in the style of the exam's free-response section. Answer it in full, with working, before you open the scoring guide.

No-calculator free-response style · 5 points · about 15 minutes

Let f(x) = x3 − 3x2 − 9x + 2 on the interval −2 ≤ x ≤ 4.

  1. (a) Find the intervals on which f is increasing. Justify your answer.
  2. (b) Find the absolute maximum and minimum values of f on the interval.
  3. (c) Find the x-coordinate of the point of inflection, and justify.

Write your full answer on paper first, then open the scoring guide and tick each point you earned.

Scoring guide and model answer

Part (a)

f'(x) = 3x2 − 6x − 9 = 3(x − 3)(x + 1). f' > 0 on (−2, −1) and (3, 4), so f increases there.

Part (b)

f(−2) = 0, f(−1) = 7, f(3) = −25, f(4) = −18. Maximum 7, minimum −25.

Part (c)

f''(x) = 6x − 6 changes sign at x = 1, so there is a point of inflection at x = 1.

Your mark: 0 of 5

Common mistakes in analytical applications of differentiation

Keep going

Revise the ideas behind these questions:

Then try real exam questions: AP Calculus past papers, with official scoring guidelines. · All AP Calculus units