Second derivative test — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: The statement of the second derivative test for local maxima and minima, application to critical points, resolution of inconclusive cases, and use in AP-style optimization problems.

You should already know: How to find critical points of a differentiable function; How to compute first and second derivatives using basic differentiation rules; The definition of local maxima and local minima.

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 AB 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 an analytical method to classify critical points of a twice-differentiable function as local maxima, local minima, or inconclusive. Per the AP Calculus AB Course and Exam Description (CED), this topic falls in Unit 5 (Analytical Applications of Differentiation), which accounts for 15–18% of the total AP exam score. The second derivative test appears in both multiple-choice (MCQ) and free-response (FRQ) sections: it is commonly tested as a standalone MCQ classification question, or as a justification step in an FRQ optimization problem.

Unlike the first derivative test, which requires analyzing the sign change of the first derivative around a critical point, the second derivative test only requires evaluating the second derivative at the critical point itself, making it faster when the second derivative is easy to compute. Some sources refer to this test as the second derivative test for local extrema, but it is almost always shortened to "second derivative test" on the AP exam. It has clear limitations that you must recognize, most notably that it cannot produce a definitive result when the second derivative is zero or undefined at a critical point.

2. Formal Statement and Geometric Intuition for the Second Derivative Test

Recall that a critical point of a function $f(x)$ is a point $x=c$ in the domain of $f$ where $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined. The second derivative test only applies to critical points where $f^{\prime }(c)=0$ and $f^{\prime \prime }(c)$ exists; if $f^{\prime }(c)$ is undefined, you must use the first derivative test to classify it regardless of the second derivative.

The formal statement of the test is: if $f$ is twice differentiable at $x=c$ and $f^{\prime }(c)=0$, then:

1. If $f^{\prime \prime }(c)<0$, then $f$ has a local maximum at $x=c$
2. If $f^{\prime \prime }(c)>0$, then $f$ has a local minimum at $x=c$
3. If $f^{\prime \prime }(c)=0$ or $f^{\prime \prime }(c)$ does not exist, the test is inconclusive

The intuition behind this rule comes from concavity: the second derivative tells us the direction the graph curves around the critical point (where the tangent line is horizontal). If $f^{\prime \prime }(c)<0$, the function is concave down at $c$, meaning the graph curves downward around the horizontal tangent, forming a peak (a local maximum). If $f^{\prime \prime }(c)>0$, the function is concave up, curving upward around the horizontal tangent to form a valley (a local minimum).

Worked Example

Problem: Consider $f(x)=x^3-6x^2+9x+2$. Find all critical points where $f^{\prime }(c)=0$ and classify them using the second derivative test.

Solution:

1. Compute the first derivative and find critical points: $f^{\prime }(x)=3x^2-12x+9=3(x-1)(x-3)$. Setting $f^{\prime }(x)=0$ gives critical points at $x=1$ and $x=3$, both in the domain of $f$.
2. Compute the second derivative: $f^{\prime \prime }(x)=6x-12$.
3. Evaluate $f^{\prime \prime }$ at $x=1$: $f^{\prime \prime }(1)=6(1)-12=-6<0$. By the second derivative test, $f$ has a local maximum at $x=1$.
4. Evaluate $f^{\prime \prime }$ at $x=3$: $f^{\prime \prime }(3)=6(3)-12=6>0$. By the second derivative test, $f$ has a local minimum at $x=3$.
5. Final result: Local maximum at $(1,6)$, local minimum at $(3,2)$.

Exam tip: On the AP exam, you will almost always lose points if you do not explicitly connect the sign of $f^{\prime \prime }(c)$ to your classification. Always state the result of the test explicitly, do not just give the final classification.

3. Inconclusive Cases and Resolution

When $f^{\prime \prime }(c)=0$ at a critical point (where $f^{\prime }(c)=0$), the second derivative test cannot give a definitive answer. This does not mean there is no extremum at $x=c$: it only means we need a different method to classify the point. For example, $f(x)=x^4$ has a critical point at $x=0$, $f^{\prime \prime }(0)=0$, but $x=0$ is still a local minimum. In contrast, $f(x)=x^3$ also has $f^{\prime }(0)=0$ and $f^{\prime \prime }(0)=0$, but $x=0$ is neither a maximum nor a minimum.

When the test is inconclusive, the only valid approach on the AP exam is to fall back to the first derivative test: check the sign of $f^{\prime }(x)$ on intervals on either side of $c$. If $f^{\prime }(x)$ 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, it is not an extremum. You will sometimes be explicitly asked to resolve an inconclusive second derivative test case on both MCQ and FRQ sections.

Worked Example

Problem: Given $f(x)=x^4-4x^3$, find all critical points and classify any that are inconclusive by the second derivative test.

Solution:

1. First derivative: $f^{\prime }(x)=4x^3-12x^2=4x^2(x-3)$. Critical points where $f^{\prime }(x)=0$ are $x=0$ and $x=3$.
2. Second derivative: $f^{\prime \prime }(x)=12x^2-24x=12x(x-2)$.
3. Classify with the second derivative test first: $f^{\prime \prime }(3)=12(3)(1)=36>0$, so we confirm a local minimum at $x=3$.
4. Evaluate at $x=0$: $f^{\prime \prime }(0)=12(0)(-2)=0$, so the second derivative test is inconclusive here.
5. Use the first derivative test to resolve: For $x<0$, $f^{\prime }(-1)=4(1)(-4)=-16<0$. For $0<x<3$, $f^{\prime }(1)=4(1)(-2)=-8<0$. There is no sign change of $f^{\prime }$ around $x=0$, so $x=0$ is neither a local maximum nor a local minimum.

Exam tip: If an FRQ asks you to classify all critical points, leaving an inconclusive point unclassified will cost you at least one point. Always explicitly state that the test is inconclusive and show your work for the first derivative test resolution.

4. Using the Second Derivative Test for Absolute Extrema

The second derivative test can be used as a first step to find absolute extrema of a continuous function on a closed interval $[a,b]$. By the Extreme Value Theorem, absolute extrema on a closed interval occur either at critical points inside the interval or at the endpoints of the interval. After using the second derivative test to classify all local extrema among the critical points, you still must evaluate $f(x)$ at all critical points inside the interval and at both endpoints, then compare the function values to find the absolute maximum and minimum.

This workflow is faster than using the first derivative test for classification, because you only need to evaluate the second derivative at critical points rather than checking the sign of the first derivative on either side of each critical point. It is the most common approach used for optimization FRQs on the AP exam.

Worked Example

Problem: Find the absolute maximum and absolute minimum of $f(x)=2x^3-9x^2+12x$ on the closed interval $[0,4]$, using the second derivative test to classify critical points.

Solution:

1. First derivative: $f^{\prime }(x)=6x^2-18x+12=6(x-1)(x-2)$. Critical points inside $[0,4]$ are $x=1$ and $x=2$.
2. Second derivative: $f^{\prime \prime }(x)=12x-18$.
3. Classify critical points with the second derivative test: $f^{\prime \prime }(1)=12(1)-18=-6<0$, so local maximum at $x=1$. $f^{\prime \prime }(2)=12(2)-18=6>0$, so local minimum at $x=2$.
4. Evaluate $f(x)$ at all candidates (critical points + endpoints): $f(0)=0$, $f(1)=5$, $f(2)=4$, $f(4)=32$.
5. Compare values: The smallest value is $0$ at $x=0$, the largest is $32$ at $x=4$. So absolute minimum is $0$ at $x=0$, absolute maximum is $32$ at $x=4$.

Exam tip: If the absolute extremum occurs at an endpoint, do not try to classify it as a local maximum or minimum, and do not apply the second derivative test to it. Endpoints are only candidates for absolute extrema.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Classifying a critical point where $f^{\prime }(c)$ is undefined using the second derivative test. Why: Students forget that the test requires $f^{\prime }(c)=0$ (and $f^{\prime \prime }(c)$ exists) to work, and critical points can also occur where $f^{\prime }(c)$ is undefined. Correct move: Always use the first derivative test to classify any critical point where $f^{\prime }(c)$ does not exist.
• Wrong move: Concluding that $x=c$ is not a local extremum when the second derivative test is inconclusive ($f^{\prime \prime }(c)=0$). Why: Students assume "inconclusive" means no extremum, but many extrema (like $x=0$ for $f(x)=x^4$) have $f^{\prime \prime }(c)=0$ and are still extrema. Correct move: Always fall back to the first derivative test to check for a sign change of $f^{\prime }(x)$ around $c$ when the test is inconclusive.
• Wrong move: Stating that $f^{\prime \prime }(c)<0$ means $f$ has a local minimum at $x=c$ (or vice versa). Why: Students mix up the sign rules, confusing concavity direction with the type of extremum. Correct move: Use geometric intuition to remember: concave down ($f^{\prime \prime }<0$) makes a hill (local maximum), concave up ($f^{\prime \prime }>0$) makes a valley (local minimum).
• Wrong move: Setting $f^{\prime \prime }(x)=0$ to find critical points for the second derivative test. Why: Students confuse critical points (found from the first derivative) with inflection points (found from the second derivative). Correct move: Always find critical points by setting $f^{\prime }(x)=0$ first, then plug those $x$-values into $f^{\prime \prime }(x)$ for the test.
• Wrong move: Stopping after classifying local extrema and not checking endpoints when finding absolute extrema on a closed interval. Why: Students assume the absolute maximum must be the largest local maximum, but the Extreme Value Theorem allows absolute extrema to occur at endpoints. Correct move: Always list all candidates: all critical points inside the interval plus the two endpoints, then evaluate $f$ at all candidates.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

The function $f(x)$ is twice differentiable for all real $x$, and has a critical point at $x=2$ with $f^{\prime }(2)=0$. If $f^{\prime \prime }(x)=3x^2-10x+8$, which of the following is true about $f(x)$ at $x=2$? A) $f(x)$ has a local maximum at $x=2$ B) $f(x)$ has a local minimum at $x=2$ C) $f(x)$ has an inflection point at $x=2$ D) The second derivative test is inconclusive at $x=2$

Worked Solution: To apply the second derivative test, we first evaluate $f^{\prime \prime }$ at the critical point $x=2$. Substituting gives $f^{\prime \prime }(2)=3(2{)}^2-10(2)+8=12-20+8=0$. By definition, if $f^{\prime \prime }(c)=0$ at a critical point $c$, the second derivative test is inconclusive. While $f^{\prime \prime }(2)=0$ is a necessary condition for an inflection point, it is not sufficient (we need a sign change of $f^{\prime \prime }$ to confirm an inflection point, so option C is incorrect). The correct answer is D.

Question 2 (Free Response)

Let $f(x)=\frac{1}{4}{x}^4-2x^2+3$. (a) Find all critical points of $f(x)$ on $(-\infty ,\infty )$. (b) Classify each critical point using the second derivative test. Justify your answer. (c) Find the absolute minimum value of $f(x)$ on the interval $[-3,1]$. Justify your answer.

Worked Solution: (a) First derivative: $f^{\prime }(x)=x^3-4x=x(x-2)(x+2)$. Setting $f^{\prime }(x)=0$ gives critical points at $x=-2$, $x=0$, and $x=2$. All are in the domain of $f(x)$, so these are all critical points. (b) Second derivative: $f^{\prime \prime }(x)=3x^2-4$. Evaluate at each critical point:

• $f^{\prime \prime }(-2)=3(-2{)}^2-4=8>0$, so by the second derivative test, $f$ has a local minimum at $x=-2$.
• $f^{\prime \prime }(0)=3(0{)}^2-4=-4<0$, so by the second derivative test, $f$ has a local maximum at $x=0$.
• $f^{\prime \prime }(2)=3(2{)}^2-4=8>0$, so by the second derivative test, $f$ has a local minimum at $x=2$. (c) Candidates for absolute minimum on $[-3,1]$ are critical points inside the interval ($x=-2$, $x=0$) and endpoints ($x=-3$, $x=1$). Evaluate $f$ at each candidate: $f(-3)=\frac{21}{4}=5.25$, $f(-2)=-1$, $f(0)=3$, $f(1)=\frac{5}{4}=1.25$. The smallest value is $-1$, so the absolute minimum value of $f(x)$ on $[-3,1]$ is $-1$.

Question 3 (Application / Real-World Style)

A bakery determines that the daily profit $P(x)$ (in dollars) from selling $x$ loaves of sourdough bread is given by $P(x)=-0.005x^3+0.75x^2+2x-500$, for $0\le x\le 200$. Use the second derivative test to find how many loaves the bakery should sell to maximize its daily profit, and confirm that your result is a maximum.

Worked Solution:

1. First derivative: $P^{\prime }(x)=-0.015x^2+1.5x+2$. Set $P^{\prime }(x)=0$ and solve with the quadratic formula, giving one positive critical point at $x\approx 101.3$ (the other solution is negative, so it is discarded, as you cannot sell negative loaves).
2. Second derivative: $P^{\prime \prime }(x)=-0.03x+1.5$.
3. Evaluate at the critical point: $P^{\prime \prime }(101.3)=-0.03(101.3)+1.5\approx -1.54<0$.
4. By the second derivative test, the critical point is a local maximum, and since it is the only local maximum on the interval $[0,200]$, it is the absolute maximum.

Interpretation: The bakery will maximize its daily profit when it sells approximately 101 loaves of sourdough bread, for a maximum daily profit of approximately $\$1830$.

7. Quick Reference Cheatsheet

8. What's Next

Mastering the second derivative test is a critical prerequisite for all optimization problems, which are the most heavily tested application of derivatives in Unit 5 on the AP exam. Next, you will extend the concepts of concavity and extrema to identify inflection points and intervals of concavity, which are regularly tested in both MCQ and FRQ sections. You will also apply the second derivative test to constrained optimization problems, where you find the maximum or minimum of a real-world function subject to a domain constraint. Without correctly classifying extrema using the second derivative test, you will not be able to correctly justify your answers to these optimization problems, which frequently require explicit justification for full credit.

• Finding critical points
• First derivative test for extrema
• Concavity and inflection points
• Optimization problems