Second derivative test — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Core statement of the second derivative test for local extrema, conditions for applicability, resolution of inconclusive cases, comparison to the first derivative test, and use of the test for justifying optimization extrema.

You should already know: How to compute first and second derivatives of elementary, composite, and implicit functions. How to locate critical points of a differentiable function. The definition of local and absolute extrema.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Second derivative test?

The second derivative test is a fast, concavity-based method to classify interior critical points of a twice-differentiable function as local maxima, local minima, or neither. As part of Unit 5: Analytical Applications of Differentiation, it accounts for roughly 25% of the unit’s exam weight (per the AP CED), so it contributes 4-5% of your total AP exam score. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, most often paired with optimization, curve sketching, or implicit differentiation problems. Unlike the first derivative test, which requires checking the sign of the first derivative on either side of the critical point, the second derivative test only requires evaluating the second derivative at the critical point itself, making it much faster for functions with simple second derivatives. It does have limitations: it only works for critical points where the first derivative is zero and the second derivative exists and is non-zero. When the second derivative is zero or undefined, the test is inconclusive, and you must fall back to the first derivative test.

2. Core Statement and Conditions for the Second Derivative Test

To use the second derivative test, we start with two key prerequisites: we have a function (f(x)) that is twice differentiable on an open interval containing (c), and (c) is an interior critical point of (f) such that (f'(c) = 0) (the test cannot be used for critical points where (f'(c)) is undefined). The test then follows three rules:

1. If (f''(c) < 0), then (f(x)) has a local maximum at (x=c)
2. If (f''(c) > 0), then (f(x)) has a local minimum at (x=c)
3. If (f''(c) = 0), the test is inconclusive The intuition for this rule comes directly from the definition of concavity: if (f''(c) < 0), the function is concave down at (c), meaning it curves downward around the critical point, forming a peak (local maximum). If (f''(c) > 0), the function is concave up, curving upward around the critical point to form a valley (local minimum). When (f''(c) = 0), the concavity may or may not change at (c), so we cannot draw a conclusion.

Worked Example

Problem: Classify all local extrema of (f(x) = x^3 - 6x^2 + 9x + 2) using the second derivative test.

1. Find the first derivative and critical points: $$f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$$ Setting (f'(x) = 0) gives critical points at (x=1) and (x=3), both with (f'(c)=0) so eligible for the test.
2. Compute the second derivative: $$f^{\prime \prime }(x)=6x-12$$
3. Evaluate (f'') at each critical point:
   • (f''(1) = 6(1) - 12 = -6 < 0), so (f) has a local maximum at (x=1)
   • (f''(3) = 6(3) - 12 = 6 > 0), so (f) has a local minimum at (x=3)
4. Find the extremum values: (f(1) = 6) and (f(3) = 2), so the local maximum is at ((1,6)) and the local minimum is at ((3,2)).

Exam tip: Always confirm (f'(c) = 0) before applying the second derivative test. The test only works for critical points where the first derivative is zero, not for inflection point candidates or endpoints.

3. Resolving Inconclusive Second Derivative Test Cases

When the second derivative test returns (f''(c) = 0), the test gives no information about whether (c) is an extremum. This is because the second derivative being zero at (c) just means the curvature is zero there; it does not tell us if the first derivative changes sign. For example:

• (f(x) = x^4): (f'(0) = 0), (f''(0) = 0), and (x=0) is a local minimum
• (f(x) = -x^4): (f'(0) = 0), (f''(0) = 0), and (x=0) is a local maximum
• (f(x) = x^3): (f'(0) = 0), (f''(0) = 0), and (x=0) is not an extremum The only reliable approach when the test is inconclusive is to fall back to the first derivative test: check the sign of (f'(x)) for points just left and just right of (c). If the sign changes from positive to negative, it is a local maximum; if it changes from negative to positive, it is a local minimum; if there is no sign change, there is no extremum at (c).

Worked Example

Problem: Classify the critical point at (x=3) for (f(x) = x^4 - 8x^3 + 18x^2) using the second derivative test if possible; if not, use an appropriate alternative.

1. Confirm (x=3) is a critical point: (f'(x) = 4x^3 - 24x^2 + 36x = 4x(x-3)^2), so (f'(3) = 0), so it is an eligible critical point.
2. Compute the second derivative: (f''(x) = 12x^2 - 48x + 36 = 12(x-1)(x-3))
3. Evaluate at (x=3): (f''(3) = 12(2)(0) = 0), so the second derivative test is inconclusive.
4. Use the first derivative test:
   • Left of 3, (x=2): (f'(2) = 4(2)(-1)^2 = 8 > 0)
   • Right of 3, (x=4): (f'(4) = 4(4)(1)^2 = 16 > 0) (f'(x)) does not change sign at (x=3), so there is no local extremum at (x=3).

Exam tip: Never automatically assume an inconclusive second derivative test means no extremum, or that it means the point is automatically an inflection point. Always run the first derivative test to confirm.

4. Using the Second Derivative Test for Optimization Justification

On AP Calculus FRQ, optimization problems require you to justify that the critical point you found is actually the maximum or minimum you are looking for. The second derivative test is the fastest way to earn full justification points, especially for applied problems with only one interior critical point. For an applied optimization problem, the domain of the objective function is almost always an open interval ((a,b)), where the value of the objective function at the endpoints is non-optimal (e.g., volume of a box is zero when (x=0) or (x=\text{max})). If there is only one interior critical point (c), and the second derivative test confirms it is a local maximum, then it must be the absolute maximum of the function on the interval, since the function tends to a lower value at the endpoints. The same logic applies for local minimum being the absolute minimum.

Worked Example

Problem: A rectangular open-top box is made from a 12 cm by 12 cm sheet of cardboard by cutting out squares of side length (x) from each corner and folding up the sides. Use the second derivative test to confirm the value of (x) that maximizes the volume of the box.

1. Write the objective volume function: After cutting and folding, the base of the box is ((12-2x)) by ((12-2x)) and the height is (x), so: $$V(x)=x(12-2x{)}^2,0<x<6$$
2. Find the first derivative and critical points: $$V^{\prime }(x)=(12-2x{)}^2-4x(12-2x)=12(6-x)(2-x)$$ The only interior critical point on ((0,6)) is (x=2) ((x=6) is an endpoint).
3. Compute the second derivative: (V''(x) = 24x - 96)
4. Evaluate at (x=2): (V''(2) = 48 - 96 = -48 < 0), so (V(x)) has a local maximum at (x=2). Since (V(0)=V(6)=0) and (V(2)=128), this is the absolute maximum volume.

Exam tip: On FRQ optimization problems, justifying your extremum with the second derivative test is faster than testing endpoints or checking first derivative sign changes, and it earns full justification points if done correctly.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Applying the second derivative test to a critical point where (f'(c) \neq 0), or to a critical point where (f'(c)) is undefined. Why: Students confuse the requirement that (f'(c)=0) with just being any critical point, and often accidentally test inflection point candidates. Correct move: Always verify that (f'(c) = 0) before evaluating (f''(c)) for the second derivative test; only eligible critical points are tested.
• Wrong move: Concluding there is no extremum when the second derivative test is inconclusive ((f''(c)=0)). Why: Students assume an inconclusive result means no extremum, but many functions have extrema at points where (f''(c)=0) (e.g., (f(x)=x^4) at (x=0)). Correct move: Whenever (f''(c) = 0) or (f''(c)) does not exist, immediately fall back to the first derivative test to check for a sign change of (f').
• Wrong move: Claiming a point where (f''(c) = 0\ is automatically an inflection point. Why: Students confuse the second derivative test's inconclusive result with the inflection point condition, which requires more than just (f''(c)=0). Correct move: To confirm an inflection point, check that (f'') changes sign around (c); never use (f''(c)=0) alone as sufficient proof.
• Wrong move: Using the second derivative test to classify an endpoint extremum on a closed interval. Why: Students forget endpoints cannot be local extrema (they only have one side in the domain), so the test does not apply. Correct move: Use the Extreme Value Theorem to compare function values at all critical points and endpoints to find absolute extrema on closed intervals; never use the second derivative test for endpoints.
• Wrong move: Evaluating (f(c)) (the original function) instead of (f''(c)) at the critical point to classify the extremum. Why: Students mix up steps when working quickly on multi-step problems, especially on MCQ. Correct move: After finding critical points from (f'(x)=0), explicitly write out (f''(x)) and evaluate it at each critical point before classifying.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

The function (h(x) = e^{2x} - 6x) has a critical point at (x = \frac{\ln 3}{2}). What is the correct classification of this critical point per the second derivative test? A) Local maximum, because (h''\left(\frac{\ln 3}{2}\right) < 0) B) Local minimum, because (h''\left(\frac{\ln 3}{2}\right) > 0) C) Local maximum, because (h''\left(\frac{\ln 3}{2}\right) > 0) D) The second derivative test is inconclusive for this point

Worked Solution: First confirm (x = \frac{\ln 3}{2}) is a critical point: (h'(x) = 2e^{2x} - 6). Substituting (x = \frac{\ln 3}{2}) gives (h'\left(\frac{\ln 3}{2}\right) = 2e^{\ln 3} - 6 = 6 - 6 = 0), so the point is eligible. Next compute the second derivative: (h''(x) = 4e^{2x}). Evaluate at the critical point: (h''\left(\frac{\ln 3}{2}\right) = 4e^{\ln 3} = 4(3) = 12 > 0). A positive second derivative at a critical point means a local minimum. The correct answer is B.

Question 2 (Free Response)

Let (f(x) = x^4 - 8x^3 + 18x^2). (a) Find all critical points of (f(x)). (b) Use the second derivative test to classify each critical point as a local maximum, local minimum, or state that the test is inconclusive. (c) If the test is inconclusive for any critical point, use the first derivative test to classify it.

Worked Solution: (a) Compute the first derivative: $$f^{\prime }(x)=4x^3-24x^2+36x=4x(x-3{)}^2$$ Setting (f'(x) = 0) gives critical points at (x=0) and (x=3).

(b) Compute the second derivative: $$f^{\prime \prime }(x)=12x^2-48x+36=12(x-1)(x-3)$$ Evaluate at each critical point:

• (f''(0) = 12(-1)(-3) = 36 > 0): (f) has a local minimum at (x=0)
• (f''(3) = 12(2)(0) = 0): the second derivative test is inconclusive for (x=3)

(c) For (x=3), check the sign of (f'(x)) on either side:

• Left of 3: (x=2), (f'(2) = 4(2)(-1)^2 = 8 > 0)
• Right of 3: (x=4), (f'(4) = 4(4)(1)^2 = 16 > 0) (f'(x)) does not change sign at (x=3), so there is no local extremum at (x=3).

Question 3 (Application / Real-World Style)

A small company producing portable bluetooth speakers has a weekly profit function given by (P(q) = -0.001q^3 + 0.9q^2 - 100q - 500), where (q) is the number of speakers produced per week ((q \geq 0)) and (P(q)) is profit in US dollars. Use the second derivative test to find the production level that maximizes weekly profit.

Worked Solution: First find the marginal profit (first derivative): (P'(q) = -0.003q^2 + 1.8q - 100). Set (P'(q) = 0) and multiply through by 1000: (-3q^2 + 1800q - 100000 = 0), or (3q^2 - 1800q + 100000 = 0). Use the quadratic formula: $$q=\frac{1800\pm \sqrt{1800^2-4(3)(100000)}}{2(3)}=\frac{1800\pm 1428.3}{6}$$ This gives critical points at (q \approx 62) and (q \approx 538). Next compute the second derivative: (P''(q) = -0.006q + 1.8). Evaluate at each critical point: (P''(62) \approx 1.43 > 0) (local minimum), (P''(538) \approx -1.43 < 0) (local maximum). Interpretation: Producing approximately 538 speakers per week gives the company the maximum possible weekly profit.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the second derivative test is a critical prerequisite for curve sketching, where you combine first and second derivative information to draw the graph of a function, a skill tested on nearly every AP Calculus exam. It is also the foundation for justifying extrema in applied optimization, which is a common FRQ topic. Without mastering the conditions for the second derivative test and how to resolve inconclusive cases, you will lose easy justification points on FRQ, which make up half your total exam score. This topic also connects to higher-order derivative tests for Taylor polynomials later in the AP BC curriculum, where you use higher-order derivatives to classify extrema when lower-order derivatives are zero.

Follow-on topics: Curve sketching with derivatives First derivative test Applied optimization Concavity and inflection points