5.7 Using the Second Derivative Test to Determine Extrema

The Second Derivative Test tells you whether a critical point is a local maximum 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, 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 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 has only one critical point on an interval and that point is a local extremum, 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 looks like a hill (max), concave up 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.

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 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. That points to an important limitation of the 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 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, 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. On a closed interval, 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
• 5.1 Using the Mean Value Theorem
• 5.2 Extreme Value Theorem, Global vs Local Extrema, and Critical Points
• 5.3 Determining Intervals on Which a Function is Increasing or Decreasing
• 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
• 5.11 Solving Optimization Problems