Second Derivative Test- Complete Lesson and Examples

What Is the Second Derivative Test?

The Second Derivative Test is a calculus tool that tells you whether a critical point is a local maximum, local minimum, or neither. You find the second derivative, evaluate it at your critical points, and read the sign.

That's it. Three steps.

When to Use It

Use this test when you already have critical points from the first derivative and you want to classify them quickly. It's faster than the First Derivative Test in most cases because you skip the sign chart.

You need two things to be true:

• The function must be differentiable near your critical point (second derivative exists)
• The second derivative must not equal zero at the critical point

If the second derivative equals zero at a critical point, the test tells you nothing. Move on.

The Rules

For a critical point where f'(c) = 0:

• If f''(c) < 0 → local maximum
• If f''(c) > 0 → local minimum
• If f''(c) = 0 → test is inconclusive

How To Apply It

Step-by-Step Process

1. Find the first derivative f'(x)
2. Set f'(x) = 0 and solve to find critical points
3. Find the second derivative f''(x)
4. Evaluate f''(x) at each critical point
5. Apply the rules above

Examples

Example 1: Finding a Local Maximum

Find the local extrema of f(x) = -2x² + 8x - 3

Step 1: First derivative

f'(x) = -4x + 8

Step 2: Set equal to zero

-4x + 8 = 0
4x = 8
x = 2

Step 3: Second derivative

f''(x) = -4

Step 4: Evaluate at x = 2

f''(2) = -4

Step 5: Apply the test

f''(2) < 0, so x = 2 is a local maximum. The point is (2, 5).

Example 2: Finding a Local Minimum

Find the local extrema of f(x) = x³ - 12x + 2

Step 1: First derivative

f'(x) = 3x² - 12

Step 2: Set equal to zero

3x² - 12 = 0
x² = 4
x = ±2

Step 3: Second derivative

f''(x) = 6x

Step 4: Evaluate at both critical points

f''(2) = 6(2) = 12 > 0 → local minimum at x = 2
f''(-2) = 6(-2) = -12 < 0 → local maximum at x = -2

Step 5: Get the points

Local minimum: (2, -14)
Local maximum: (-2, 18)

Example 3: Inconclusive Test

Find the local extrema of f(x) = x⁴

Step 1: First derivative

f'(x) = 4x³

Step 2: Set equal to zero

4x³ = 0
x = 0

Step 3: Second derivative

f''(x) = 12x²

Step 4: Evaluate at x = 0

f''(0) = 12(0)² = 0

Step 5: The test fails. f''(0) = 0 means inconclusive. But you know from the original function that x = 0 is actually a local minimum. The Second Derivative Test just can't tell you that here.

Second Derivative Test vs First Derivative Test

Here's when to use each one:

Why the Test Works

Think about it this way: the second derivative measures the concavity of a function.

When f''(c) < 0, the graph is concave down at that point. A concave down curve looks like an upside-down bowl. The peak of an upside-down bowl is a local maximum.

When f''(c) > 0, the graph is concave up at that point. A concave up curve looks like a right-side-up bowl. The bottom of a right-side-up bowl is a local minimum.

Common Mistakes

• Forgetting to find critical points first. You need f'(c) = 0 before you can use the test.
• Evaluating f'(x) instead of f''(x). Read the problem. It says second derivative.
• Assuming zero means "no extremum." Zero means inconclusive, not "neither." You still need to check another method.
• Skipping the f'' = 0 case. Always evaluate the second derivative at your critical points. Don't assume it's nonzero.

Getting Started Checklist

Before you start any problem:

• ✓ Differentiate once to get f'(x)
• ✓ Solve f'(x) = 0 for critical points
• ✓ Differentiate again to get f''(x)
• ✓ Plug each critical point into f''(x)
• ✓ Read the sign and classify

Practice Problem

Find all local extrema of f(x) = 3x⁴ - 4x³ - 12x² + 7

Answer below when you're ready to check.

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Solution:

f'(x) = 12x³ - 12x² - 24x = 12x(x² - x - 2) = 12x(x - 2)(x + 1)

Critical points: x = 0, x = 2, x = -1

f''(x) = 36x² - 24x - 24

f''(0) = -24 < 0 → local maximum at (0, 7)

f''(2) = 36(4) - 48 - 24 = 72 > 0 → local minimum at (2, -25)

f''(-1) = 36(1) + 24 - 24 = 36 > 0 → local minimum at (-1, 2)