---
title: "AP Calculus 5.7: Second Derivative Test for Extrema"
description: "Review AP Calculus second derivative test, including critical points, concavity, local maxima, local minima, inconclusive cases, and absolute extrema."
canonical: "https://fiveable.me/ap-calc/unit-5/using-second-derivative-test-determine-extrema/study-guide/g8rSveVmLeEuKLK4T7ep"
type: "study-guide"
subject: "AP Calculus AB/BC"
unit: "Unit 5 – Analytical Applications of Differentiation"
lastUpdated: "2026-06-09"
---

# AP Calculus 5.7: Second Derivative Test for Extrema

## Summary

Review AP Calculus second derivative test, including critical points, concavity, local maxima, local minima, inconclusive cases, and absolute extrema.

## Guide

The Second Derivative Test tells you whether a [critical point](/ap-calc/key-terms/critical-point "fv-autolink") is a [local maximum](/ap-calc/key-terms/local-maximum "fv-autolink") or a local minimum by checking the concavity there. Find critical points using the first derivative, plug them into the second derivative, and read the sign: negative means a local max, and positive means a local min. For AP Calculus, use the First Derivative Test if the second derivative is 0 or undefined at the critical point.

## Why This Matters for the AP Calculus Exam

This topic is part of analyzing [function behavior](/ap-calc/unit-5/connecting-a-function-its-first-derivative-and-its-second-derivative/study-guide/NQXfonM48ssVKKRHQ7Te "fv-autolink"), one of the most tested ideas in AP Calculus. On the exam you will need to find extrema, justify why a point is a max or min, and connect a [function](/ap-calc/unit-1/defining-continuity-at-point/study-guide/JbsR9iQfAzCznNOCG6JK "fv-autolink") to its first and second derivatives. The Second Derivative Test gives you a fast, clean way to classify critical points, and it shows up both in multiple-choice questions and in free-response work where you have to support your conclusion with reasoning.

There is also a powerful shortcut here: when a [continuous function](/ap-calc/key-terms/continuous-function "fv-autolink") has only one critical point on an interval and that point is a [local extremum](/ap-calc/key-terms/local-extremum "fv-autolink"), it is also the absolute extremum on that interval. That idea connects directly to optimization problems later in the unit.

## Key Takeaways

- A critical point is where $f'(x) = 0$ or where $f'(x)$ does not exist.
- Plug each critical point into $f''(x)$: a negative result means a local maximum, a positive result means a local minimum.
- [Concave down](/ap-calc/key-terms/concave-down "fv-autolink") looks like a hill (max), [concave up](/ap-calc/key-terms/concave-up "fv-autolink") looks like a bowl (min).
- If $f''(c) = 0$ or $f''(c)$ does not exist, the Second Derivative Test is inconclusive; use the First Derivative Test instead.
- If a continuous function has exactly one critical point on an interval and it is a local extremum, that point is also the absolute extremum on that interval.
- Refer to $f$, $f'$, and $f''$ by name when you justify, not "it" or "the function."

## Finding Critical Points First

Before you can use the Second Derivative Test, you need critical points. A **critical point** is where the first derivative equals zero or fails to exist.

Take this function:

$$f(x)=\frac{2}{3}x^3-\frac{5}{2}x^2-3x$$

Compute the first derivative:

$$f'(x)=2x^{2}-5x-3$$

Set it equal to zero and factor:

$$0=2x^2-5x-3$$

$$0=(2x+1)(x-3)$$

Solve like a normal equation. The two $x$ values are your critical points:

$$x=-\frac{1}{2}, \quad x=3$$

The second derivative tells you about concavity: $f''(x) > 0$ means concave up, $f''(x) < 0$ means concave down. The Second Derivative Test uses exactly that idea to classify each critical point.

## The Second Derivative Test

By checking concavity at a critical point, you can tell whether it is a local minimum or a local maximum.

### Steps

1. Find the critical points of $f(x)$ using $f'(x)$.
2. Plug each critical point into $f''(x)$.
3. Read the sign of the result to decide min or max.

> **Key Idea:**
> - If $f''(c)$ is positive, you have a local **minimum** (concave up, like a bowl).

> - If $f''(c)$ is negative, you have a local **maximum** (concave down, like a hill).

### Walkthrough

Continue with the example above.

**Step 1: Critical points from $f'(x)$.** Already found: $x = -0.5$ and $x = 3$.

**Step 2: Plug into $f''(x)$.** First find the second derivative:

$$f''(x)=4x-5$$

Then evaluate at each critical point:

$$f''(-\tfrac{1}{2})=4(-\tfrac{1}{2})-5=-7$$

$$f''(3)=4(3)-5=7$$

**Step 3: Classify.** Since $f''(-0.5) = -7 < 0$, the function is concave down there, so $x = -0.5$ gives a local maximum. Since $f''(3) = 7 > 0$, the function is concave up there, so $x = 3$ gives a local minimum.

Here is the intuition: a maximum sits at the peak of a hill, which is a concave-down surface. A minimum sits at the bottom of a bowl, which is a concave-up surface. So concave down at a critical point $c$ points to a maximum, and concave up at $c$ points to a minimum.

## Practice Problems

For each function, classify the critical points as local maxima or minima.

### Example 1

$$f(x)=4\sin(x), \quad 0<x<2\pi$$

First and second derivatives:

$$f'(x)=4\cos(x)$$

$$f''(x)=-4\sin(x)$$

Set $f'(x) = 0$ to find critical points:

$$0=4\cos(x)$$

$$0=\cos(x)$$

[Cosine](/ap-calc/unit-2/derivatives-cos-x-sinx-ex-ln-x/study-guide/SbmDK3t2kYI0u2wLz9Hv "fv-autolink") equals zero at:

$$x=\frac{\pi}{2}, \quad x=\frac{3\pi}{2}$$

Plug into the second derivative:

$$f''(\tfrac{\pi}{2})=-4\sin(\tfrac{\pi}{2})=-4 \cdot 1=-4$$

$$f''(\tfrac{3\pi}{2})=-4\sin(\tfrac{3\pi}{2})=-4 \cdot -1=4$$

Since $f''(\tfrac{\pi}{2}) = -4 < 0$, $x = \tfrac{\pi}{2}$ is a local maximum of $f(x)$. Since $f''(\tfrac{3\pi}{2}) = 4 > 0$, $x = \tfrac{3\pi}{2}$ is a local minimum of $f(x)$.

### Example 2

$$f(x)=247\ln(x^2)$$

First and second derivatives:

$$f'(x)=\frac{247}{x}$$

$$f''(x)=-\frac{247}{x^2}$$

Set $f'(x) = 0$:

$$0=\frac{247}{x}$$

This has no [solution](/ap-calc/unit-7/verifying-solutions-for-differential-equations/study-guide/s2nX7AhIBDxGIWwlL82x "fv-autolink"). That points to an important limitation of the test:

> If $f'(c) = 0$ and $f''(c) = 0$, or if $f''(c)$ does not exist, the Second Derivative Test is **inconclusive**. There may be a point of inflection, or there may be an extremum you cannot pin down with this test. In those cases, fall back on the First Derivative Test.

## How to Use This on the AP Calculus Exam

### MCQ

Many multiple-choice questions give you a function or its derivatives and ask you to classify a critical point. Find the critical point with $f'$, then check the sign of $f''$. If $f''$ at the point is negative, it is a max; if positive, it is a min. Watch for cases where $f''$ is zero or undefined, since the test gives no answer there.

### Free Response

When a free-response question asks you to justify whether a point is a relative max, relative min, or neither, state your reasoning clearly. For example: "$f$ has a [relative maximum](/ap-calc/key-terms/relative-maximum "fv-autolink") at $x = c$ because $f'(c) = 0$ and $f''(c) < 0$, so $f$ is concave down there." Always name $f$, $f'$, and $f''$ instead of saying "it." Vague justifications lose the connection between the derivative and the conclusion.

### Common Trap

If the Second Derivative Test comes back inconclusive, do not guess. Switch to the First Derivative Test and analyze the sign of $f'$ on each side of the critical point.

## Common Misconceptions

- **A positive second derivative does not mean a maximum.** Positive $f''$ means concave up, which is a local minimum. Many students flip this. Bowl up means min, hill means max.
- **Inconclusive does not mean "no extremum."** When $f''(c) = 0$ or does not exist, the test simply gives no information. There could still be a max, a min, or a point of inflection. Use the First Derivative Test to decide.
- **A critical point is not always an extremum.** The first derivative being zero only makes a point a candidate. You still have to test it.
- **Don't forget critical points where $f'$ is undefined.** Critical points include places where $f'$ does not exist, like corners or cusps, not just where $f' = 0$.
- **The single-critical-point shortcut needs all its conditions.** It only works when the function is [continuous](/ap-calc/key-terms/continuous "fv-autolink"), has exactly one critical point on the interval, and that point is a local extremum. Then it is also the absolute extremum on that interval.
- **The Second Derivative Test does not find [endpoints](/ap-calc/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ "fv-autolink").** On a [closed interval](/ap-calc/key-terms/closed-interval "fv-autolink"), absolute extrema can occur at endpoints, which this test does not check. Compare endpoint values separately.

## Related AP Calculus Guides

- [Unit 5 Overview: Analytical Applications of Differentiation](/ap-calc/unit-5/review/study-guide/22AdFpcITnvM0bXKRl55)
- [5.1 Using the Mean Value Theorem](/ap-calc/unit-5/using-mean-value-theorem/study-guide/79sP2PXcyvRvBsjb3HRq)
- [5.2 Extreme Value Theorem, Global vs Local Extrema, and Critical Points](/ap-calc/unit-5/extreme-value-theorem-global-vs-local-extrema-critical-points/study-guide/xcQI1ZzNbmWJ5uRNiFCo)
- [5.3 Determining Intervals on Which a Function is Increasing or Decreasing](/ap-calc/unit-5/determining-intervals-on-which-function-is-increasing-or-decreasing/study-guide/Y2fgjyl7H1dKPI2YsB4Y)
- [5.4 Using the First Derivative Test to Determine Relative (Local) Extrema](/ap-calc/unit-5/using-first-derivative-test-to-determine-relative-local-extrema/study-guide/BjnQNCShz0uQiZGhSl2g)
- [5.11 Solving Optimization Problems](/ap-calc/unit-5/solving-optimization-problems/study-guide/u2Y3MpOG6kkTtbLH38S7)

## Vocabulary

- **absolute maximum**: The highest value of a function over its entire domain or a specified interval.
- **absolute minimum**: The lowest value of a function over its entire domain or a specified interval.
- **continuous**: A function that has no breaks, jumps, or holes in its graph over a given interval.
- **critical point**: A point in the domain of a function where the derivative is zero or undefined, which are candidates for local and absolute extrema.
- **global extremum**: The absolute maximum or minimum value of a function over its entire domain or a specified interval.
- **relative maximum**: A point where a function reaches a highest value in a neighborhood around that point.
- **relative minimum**: A point where a function reaches a lowest value in a neighborhood around that point.
- **second derivative**: The derivative of the first derivative, denoted f'', which describes the concavity of a function and indicates where it is concave up or concave down.

## FAQs

### What is the Second Derivative Test?

The Second Derivative Test classifies a critical point by checking concavity there. If f double prime is positive, the point is a local minimum; if f double prime is negative, it is a local maximum.

### What do I do before using the Second Derivative Test?

First find critical points using the first derivative. Critical points occur where f prime equals zero or where f prime does not exist.

### How do I know whether a critical point is a max or min?

Evaluate the second derivative at the critical point. Positive means concave up and a local minimum; negative means concave down and a local maximum.

### When is the Second Derivative Test inconclusive?

The test is inconclusive when f double prime at the critical point equals zero or does not exist. In that case, use the First Derivative Test or another justification.

### Can the Second Derivative Test find absolute extrema?

Not by itself on a closed interval. However, if a continuous function has exactly one critical point on an interval and that point is a local extremum, that point is also the absolute extremum on the interval.

### How should I justify the Second Derivative Test on AP Calculus FRQs?

Name the function and derivatives clearly. For example, say f has a local maximum at x = c because f prime of c is zero and f double prime of c is negative, so f is concave down there.

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