# Second Derivative Test

> AP Calculus BC · Analytical Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u5-second-derivative-test/

This guide covers the second derivative test for classifying local extrema, its eligibility conditions, methods for resolving inconclusive cases, and justifying extrema in optimization problems for the AP Calculus BC exam.

**Prerequisites:** Computing first and second derivatives of composite and implicit functions; Locating critical points of differentiable functions; Definition of local and absolute extrema

## Learning objectives

- State the second derivative test for classifying local extrema
- Identify eligible critical points for the test
- Resolve inconclusive test cases
- Use the test to justify extrema in optimization problems
- Compare the second derivative test to the first derivative test

## Core Definition and Eligibility Conditions

The second derivative test is a fast, concavity-based method to classify interior critical points of a twice-differentiable function. It is faster than the first derivative test because you only evaluate the second derivative at the critical point, rather than checking the sign of the first derivative on both sides. It accounts for ~4-5% of your total AP Calculus BC exam score, appearing in both multiple-choice and free-response sections.

**Second Derivative Test** — If $f$ is twice differentiable on an open interval containing $c$, and $f'(c) = 0$: 1. $f''(c) < 0 \to$ local maximum at $x=c$; 2. $f''(c) > 0 \to$ local minimum at $x=c$; 3. $f''(c) = 0 \to$ test is inconclusive

*Notation:* For function $f(x)$ with critical point $c$

*Example:* Polynomials with simple second derivatives can be classified quickly using this test.

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

1. Find the first derivative and solve for 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 with $f'(c)=0$ so they are eligible for the test.
3. Compute the second derivative:

   $$f''(x) = 6x - 12$$
4. 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 $x=3$
5. The extremum values are $f(1) = 6$ 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. It only works for critical points where the first derivative is zero, not for endpoints or critical points where $f'(c)$ is undefined.

## Resolving Inconclusive Test Cases

When $f''(c) = 0$ at an eligible critical point, the second derivative test is inconclusive. This means we cannot draw a conclusion about whether an extremum exists, because the curvature is zero at the point, but concavity may or may not change. For example, $f(x)=x^4$, $f(x)=-x^4$, and $f(x)=x^3$ all have $f'(0)=0$ and $f''(0)=0$, but $x=0$ is a local minimum, local maximum, and no extremum respectively.

The only reliable approach to resolve an inconclusive test is to fall back to the first derivative test: check the sign of $f'(x)$ just left and right of $c$. A sign change indicates an extremum, while no sign change means no extremum.

**Worked example:** 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 an eligible critical point:

   $$f'(x) = 4x^3 - 24x^2 + 36x = 4x(x-3)^2 \\ f'(3) = 0$$
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. Check the sign of $f'(x)$ around $x=3$ using the first derivative test:  
- Left of 3 (at $x=2$): $f'(2) = 4(2)(-1)^2 = 8 > 0$  
- Right of 3 (at $x=4$): $f'(4) = 4(4)(1)^2 = 16 > 0$
5. Since $f'(x)$ does not change sign at $x=3$, there is no local extremum at $x=3$.

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

## Optimization Justification with the Second Derivative Test

On AP Calculus BC 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 most applied optimization problems, the objective function has a domain that is an open interval $(a,b)$, where the function value at the endpoints is non-optimal (e.g., volume of a box is zero when $x=0$ or at the maximum endpoint). If you have one interior critical point $c$, and the second derivative test confirms it is a local maximum, this must be the absolute maximum on the interval. The same logic applies for local minima becoming absolute minima.

**Worked example:** 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 with domain $0 < x < 6$:

   $$V(x) = x(12-2x)^2$$
2. Find the first derivative and critical points:

   $$V'(x) = (12-2x)^2 - 4x(12-2x) = 12(6-x)(2-x)$$
3. The only interior critical point on $(0,6)$ is $x=2$; $x=6$ is an endpoint and not eligible for the test.
4. Compute the second derivative:

   $$V''(x) = 24x - 96$$
5. Evaluate at $x=2$: $V''(2) = 48 - 96 = -48 < 0$, so $V(x)$ has a local maximum at $x=2$. Since $V(0) = V(6) = 0 < V(2) = 128$, this is the absolute maximum volume at $x=2$ cm.

> **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.

## Concept Check

**Check your understanding**

Test your understanding of the second derivative test with these AP-style questions:

1. The function $h(x) = e^{2x} - 6x$ has a critical point at $x = \frac{\ln 3}{2}$. What is the correct classification per the second derivative test?

   - Local maximum, because $h''\left(\frac{\ln 3}{2}\right) < 0$
   - Local minimum, because $h''\left(\frac{\ln 3}{2}\right) > 0$
   - Local maximum, because $h''\left(\frac{\ln 3}{2}\right) > 0$
   - The second derivative test is inconclusive for this point

   *Answer:* Local minimum, because $h''\left(\frac{\ln 3}{2}\right) > 0$

   *Why:* Correct! $h''(x) = 4e^{2x}$, so $h''\left(\frac{\ln 3}{2}\right) = 12 > 0$, which indicates a local minimum.

2. When the second derivative test gives $f''(c) = 0$ at a critical point $c$ where $f'(c)=0$, what should you do next?

   - Conclude there is no extremum at $c$
   - Conclude $c$ is an inflection point
   - Use the first derivative test to check for a sign change of $f'$
   - Conclude $c$ is a local extremum

   *Answer:* Use the first derivative test to check for a sign change of $f'$

   *Why:* Correct! An inconclusive second derivative test requires the first derivative test to resolve, because $f''(c)=0$ does not tell us if $f'$ changes sign around $c$.

## Common pitfalls

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

## Cheatsheet

| Category | Rule / Formula | Notes |
| --- | --- | --- |
| Second Derivative Test Core Statement | If $f'(c)=0$ and $f''(c)$ exists: <br> 1. $f''(c) < 0 \to$ local max <br> 2. $f''(c) > 0 \to$ local min <br> 3. $f''(c) = 0 \to$ inconclusive | Only for interior critical points with $f'(c)=0$; no endpoints or undefined $f'(c)$ |
| Inconclusive Case Procedure | Check sign of $f'(x)$ around $c$: <br> + to - = local max; - to + = local min; no change = no extremum | Always use this when $f''(c)=0$ |
| Optimization Justification | One interior critical point $c$ on $(a,b)$: <br> $f''(c) < 0 \to$ absolute max; $f''(c) > 0 \to$ absolute min | Valid for applied problems where endpoint values are non-optimal |
| Critical Point Eligibility | Eligible only if $f'(c) = 0$ | Critical points with undefined $f'(c)$ cannot be tested with this method |
| Inflection Point Rule | Inflection point requires $f''$ changes sign around $c$ | $f''(c) = 0$ is necessary but not sufficient; do not confuse with extremum classification |
| Closed Interval Extrema | Compare $f(c)$ at all critical points and endpoints | Never use the second derivative test for endpoint extrema |

## 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 BC exam. It is also the foundation for justifying extrema in applied optimization, a common free-response question topic. Failing to master the conditions for the test and how to resolve inconclusive cases can cost you 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, building on the core logic of the second derivative test.

- [Sketching graphs of f, f', f''](https://www.owlsprep.com/study/ap-calculus-bc-u5-sketching-graphs-of-f-f/)
- [Connecting f, f', f'' qualitatively](https://www.owlsprep.com/study/ap-calculus-bc-u5-connecting-f-f-f-qualitatively/)
- [Introduction to Optimization Problems](https://www.owlsprep.com/study/ap-calculus-bc-u5-introduction-to-optimization-problems/)

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