Second Derivative Test

AP Calculus AB· AP Calculus AB CED — Analytical Applications of Differentiation· 14 min read

1. Core Definition and Purpose★★☆☆☆⏱ 3 min

The second derivative test is an analytical method to classify critical points of a twice-differentiable function as local maxima, local minima, or inconclusive. For AP Calculus AB, this topic falls in Unit 5, which makes up 15–18% of your total exam score, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections.

Unlike the first derivative test, which requires analyzing the sign change of $f^{'} (x)$ 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. It has clear limitations you must recognize for the exam.

2. Formal Statement and Geometric Intuition★★☆☆☆⏱ 4 min

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

📘 Definition

Second Derivative Test for Local Extrema

If $f$ is twice differentiable at $x = c$ and $f^{'} (c) = 0$ :

If $f^{''} (c) < 0$ , $f$ has a local maximum at $x = c$
If $f^{''} (c) > 0$ , $f$ has a local minimum at $x = c$
If $f^{''} (c) = 0$ or $f^{''} (c)$ does not exist, the test is inconclusive

The intuition comes from concavity: if $f^{''} (c) < 0$ , the function is concave down at $c$ , curving downward around the horizontal tangent to form a peak (local maximum). If $f^{''} (c) > 0$ , the function is concave up, curving upward around the horizontal tangent to form a valley (local minimum).

📐 Worked Example

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

1
Compute first derivative and find critical points
$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$
2
Setting $f^{'} (x) = 0$ gives critical points at $x = 1$ and $x = 3$ , both in the domain of $f$ .
3
Compute the second derivative
$f''(x) = 6x - 12$
4
Evaluate $f^{''}$ at $x = 1$
$f''(1) = 6(1) - 12 = -6 < 0$
5
By the second derivative test, $f$ has a local maximum at $x = 1$ .
6
Evaluate $f^{''}$ at $x = 3$
$f''(3) = 6(3) - 12 = 6 > 0$
7
By the second derivative test, $f$ has a local minimum at $x = 3$ . Final result: local maximum at $(1, 6)$ , local minimum at $(3, 2)$ .

Exam tip:

Always explicitly connect the sign of $f^{''} (c)$ to your classification to earn full credit on FRQs.

3. Inconclusive Cases and Resolution★★★☆☆⏱ 3 min

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When $f^{''} (c) = 0$ at a critical point where $f^{'} (c) = 0$ , the second derivative test cannot give a definitive answer. This does not mean there is no extremum at $x = c$ : for example, $f (x) = x^{4}$ has $f^{''} (0) = 0$ but $x = 0$ is a local minimum, while $f (x) = x^{3}$ also has $f^{''} (0) = 0$ but $x = 0$ is not an extremum.

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 $f^{'} (x)$ on intervals on either side of $c$ .

📐 Worked Example

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

1
First derivative
$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$
2
Critical points where $f^{'} (x) = 0$ are $x = 0$ and $x = 3$ .
3
Second derivative
$f''(x) = 12x^2 - 24x = 12x(x - 2)$
4
Classify with the second derivative test first
$f''(3) = 12(3)(1) = 36 > 0$
5
This confirms a local minimum at $x = 3$ .
6
Evaluate at $x = 0$
$f''(0) = 12(0)(-2) = 0$
7
The test is inconclusive here. Check the sign of $f^{'} (x)$ around $x = 0$ : for $x < 0$ , $f^{'} (- 1) = - 16 < 0$ . For $0 < x < 3$ , $f^{'} (1) = - 8 < 0$ . No sign change means $x = 0$ is neither a maximum nor minimum.

Exam tip:

Leaving an inconclusive point unclassified will cost you at least one point on FRQs.

4. Second Derivative Test for Absolute Extrema★★★☆☆⏱ 4 min

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By the Extreme Value Theorem, absolute extrema of a continuous function on a closed interval $[a, b]$ occur either at critical points inside the interval or at the endpoints. After using the second derivative test to classify local extrema, you still must compare function values at all candidates to find the absolute maximum and minimum.

This workflow is faster than using the first derivative test for classification, and is the most common approach for optimization FRQs on the AP exam.

📐 Worked Example

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

1
First derivative
$f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$
2
Critical points inside $[0, 4]$ are $x = 1$ and $x = 2$ .
3
Second derivative
$f''(x) = 12x - 18$
4
Classify critical points
$f''(1) = 12(1) - 18 = -6 < 0 \implies \text{local maximum at } x=1$
5
$f''(2) = 12(2) - 18 = 6 > 0 \implies \text{local minimum at } x=2$
6
Evaluate $f (x)$ at all candidates (critical points + endpoints): $f (0) = 0$ , $f (1) = 5$ , $f (2) = 4$ , $f (4) = 32$ .
7
Compare values: absolute minimum is $0$ at $x = 0$ , absolute maximum is $32$ at $x = 4$ .

Exam tip:

Never apply the second derivative test to endpoints, they are only candidates for absolute extrema.

5. Check Your Understanding★★★☆☆⏱ 3 min

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Test your knowledge with this AP-style multiple choice question:

The function $f (x)$ is twice differentiable for all real $x$ , and has a critical point at $x = 2$ with $f^{'} (2) = 0$ . If $f^{''} (x) = 3 x^{2} - 10 x + 8$ , 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 $x = 2$
Reveal answer
3 —
$f^{''} (2) = 3 (2)^{2} - 10 (2) + 8 = 0$ , so the test is inconclusive. $f^{''} (c) = 0$ is not sufficient to confirm an inflection point, so option C is incorrect.

6. Common Pitfalls

Wrong move:

Classifying a critical point where $f^{'} (c)$ is undefined using the second derivative test.

Why:

The test requires $f^{'} (c) = 0$ and $f^{''} (c)$ exists to work, and critical points can occur where $f^{'} (c)$ is undefined.

Correct move:

Always use the first derivative test to classify any critical point where $f^{'} (c)$ does not exist.

Wrong move:

Concluding $x = c$ is not an extremum when the second derivative test is inconclusive ( $f^{''} (c) = 0$ ).

Why:

Inconclusive means the test cannot give a result, not that there is no extremum. Many extrema have $f^{''} (c) = 0$ .

Correct move:

Always fall back to the first derivative test to check for a sign change of $f^{'} (x)$ around $c$ .

Wrong move:

Mixing up sign rules: stating $f^{''} (c) < 0$ means a local minimum (or vice versa).

Why:

Confusion between concavity direction and the type of extremum is a very common exam mistake.

Correct move:

Use geometric intuition: concave down ( $f^{''} < 0$ ) is a hill (local maximum), concave up ( $f^{''} > 0$ ) is a valley (local minimum).

Wrong move:

Setting $f^{''} (x) = 0$ to find critical points for the second derivative test.

Why:

Students confuse critical points (found from first derivative) with inflection points (found from second derivative).

Correct move:

Always find critical points by setting $f^{'} (x) = 0$ first, then plug those $x$ -values into $f^{''} (x)$ .

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 at endpoints.

Correct move:

Always list all candidates: all critical points inside the interval plus both endpoints, then evaluate $f$ at all candidates.

7. Quick Reference Cheatsheet

Category	Rule / Statement	Notes
Local Maximum Rule	If $f^{'} (c) = 0$ and $f^{''} (c) < 0$ → local max at $x = c$	Only applies when $f^{'} (c) = 0$ and $f^{''} (c)$ exists
Local Minimum Rule	If $f^{'} (c) = 0$ and $f^{''} (c) > 0$ → local min at $x = c$	Same requirements as local maximum rule
Inconclusive Case	If $f^{'} (c) = 0$ and $f^{''} (c) = 0$ or $f^{''} (c)$ undefined → test inconclusive	Never conclude no extremum; always use first derivative test
Critical Point Step	Find critical points by solving $f^{'} (x) = 0$	Do not use $f^{''} (x) = 0$ to find test candidates
Undefined $f^{'} (c)$	Critical points with undefined $f^{'} (c)$ cannot use SDT	Use first derivative test for these points
Absolute Extrema Candidates	All critical points inside interval + interval endpoints	SDT classifies local extrema; always compare function values
Geometric Mnemonic	$f^{''} < 0$ = concave down = hill = local max; $f^{''} > 0$ = concave up = valley = local min	Use this to fix sign confusion errors

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Classify critical point with SDT
2022 · FRQ
Justify extremum in optimization

What's Next

Mastering the second derivative test is a critical prerequisite for all optimization problems, the most heavily tested application of derivatives in Unit 5 of AP Calculus AB. 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 problems, which frequently require explicit justification for full credit.