AP Calc Unit 5 Review: Analyzing Functions (AB and BC)

AP Calculus Unit 5, Analytical Applications of Differentiation, teaches you to read a function through its derivatives. The first derivative says where a function rises and falls. The second derivative says how it bends. Together they find maximums and minimums, which is what optimization is all about. It is one of the largest units on the AB exam.

What Unit 5 covers

College Board splits the unit into these topics. Each row links to a lesson where you can practice it with new numbers every time.

The key ideas

The Mean Value Theorem guarantees a matching slope. If f is continuous on [a, b] and differentiable on (a, b), at least one point has a tangent line parallel to the secant line. Read the Mean Value Theorem and the Extreme Value Theorem pages for the exact conditions.

f' controls direction, f'' controls shape. f' > 0 means increasing and f' < 0 means decreasing. f'' > 0 means concave up and f'' < 0 means concave down. A point of inflection is where f'' changes sign.

Optimization is extrema with a story. Write the quantity to maximize or minimize as a function of one variable, find its critical points, and justify that your answer is a max or min, using the first derivative test, the second derivative test, or the candidates test.

Sample questions

1. Mean Value Theorem. For f(x) = x^2 on [1, 3], the average rate of change is \frac{9 - 1}{3 - 1} = 4. Setting f'(c) = 2c = 4 gives c = 2, which is inside the interval.

2. Relative extrema. For f(x) = x^3 - 3x^2 - 9x + 2:

f' changes from positive to negative at x = -1, so f(-1) = 7 is a relative max. It changes from negative to positive at x = 3, so f(3) = -25 is a relative min. Since f''(x) = 6x - 6 changes sign at x = 1, there is a point of inflection at (1, -9).

3. Candidates test. On [-2, 4], check the endpoints and critical points: f(-2) = 0, f(-1) = 7, f(3) = -25 and f(4) = -18. The absolute max is 7 at x = -1 and the absolute min is -25 at x = 3.

4. Optimization. A farmer has 100 m of fence to enclose a rectangle along a river, fencing only three sides. With width x, the area is A(x) = x(100 - 2x). Then A'(x) = 100 - 4x = 0 at x = 25, and A''(x) = -4 < 0, so this is a max. The best area is 25 \times 50 = 1250 m².

How to study Unit 5

• Practice with the graph of f'. Many free-response questions give only f'. Read increasing from where f' is above the axis, and concave up from where f' is rising.
• Justify with sign changes. Write "because f' changes from negative to positive at x = 3". Graders want the reason, not just the answer.
• Never skip the endpoints. On a closed interval the absolute max is often at an endpoint.
• Check MVT conditions first. State continuity and differentiability before you conclude anything.
• Work through problem sets. Try the optimization practice problems, curve sketching problems and Mean Value Theorem problems, each with full solutions. The guide to curve sketching with derivatives pulls the whole unit together.

For the full course map, see the AP Calculus AB guide or the AP Calculus BC guide.

AP and Advanced Placement are registered trademarks of College Board, which is not affiliated with PracticeCalculus and does not endorse this guide. Unit names and weights come from College Board's course framework; check it for any changes.