Unit 5.7 – Using the Second Derivative Test to Determine Extrema
AP® Calculus AB & BC | A Quick Alternative to the First Derivative Test
Why This Matters: The Second Derivative Test provides a faster, more elegant way to classify critical points as local maxima or minima—when it works! Instead of testing intervals around critical points (as in the First Derivative Test), you simply evaluate the second derivative at the critical point. If \(f''(c) > 0\), you have a local minimum (concave up like a valley); if \(f''(c) < 0\), you have a local maximum (concave down like a peak). However, this test has an important limitation: it's inconclusive when \(f''(c) = 0\). Understanding when to use this test and when to fall back on the First Derivative Test is essential for efficient problem-solving in calculus!
🎯 The Second Derivative Test
Second Derivative Test for Local Extrema
Let \(f\) be a function such that \(f'(c) = 0\) and \(f''(c)\) exists. Then:
If \(f''(c) > 0\), then \(f\) has a local minimum at \(x = c\).
Reasoning: The function is concave up at \(c\) (shaped like ∪), so the horizontal tangent is at a valley bottom.
If \(f''(c) < 0\), then \(f\) has a local maximum at \(x = c\).
Reasoning: The function is concave down at \(c\) (shaped like ∩), so the horizontal tangent is at a hilltop.
If \(f''(c) = 0\) or \(f''(c)\) does not exist, then the test is INCONCLUSIVE.
Important: The critical point could be a max, min, or neither. You must use another method!
🔑 Quick Summary Table:
| Condition | Result | Shape | Visual |
|---|---|---|---|
| \(f'(c) = 0\) and \(f''(c) > 0\) | Local Minimum | Concave up | ∪ valley |
| \(f'(c) = 0\) and \(f''(c) < 0\) | Local Maximum | Concave down | ∩ peak |
| \(f'(c) = 0\) and \(f''(c) = 0\) | Inconclusive | Unknown | ??? Test fails |
| \(f'(c) = 0\) and \(f''(c)\) DNE | Inconclusive | Unknown | ??? Test fails |
💡 Memory Trick:
- Positive = Pointing Up: \(f''(c) > 0\) → concave up → minimum (valley)
- Negative = N for "iNverted": \(f''(c) < 0\) → concave down → maximum (peak)
- Zero = "Uh-oh": \(f''(c) = 0\) → test fails, use First Derivative Test
🧠 Why the Second Derivative Test Works
THE INTUITION
The second derivative \(f''(x)\) tells us about the concavity of the function—how it curves.
- At a critical point \(f'(c) = 0\): The tangent line is horizontal
- If \(f''(c) > 0\): Function is concave up (∪)
- Graph curves upward around \(c\)
- Function values are higher on both sides of \(c\)
- Therefore, \(c\) is at the bottom → local minimum
- If \(f''(c) < 0\): Function is concave down (∩)
- Graph curves downward around \(c\)
- Function values are lower on both sides of \(c\)
- Therefore, \(c\) is at the top → local maximum
📝 Connection to First Derivative: The second derivative tells us how \(f'(x)\) is changing. If \(f''(c) > 0\), then \(f'\) is increasing at \(c\), meaning slopes go from negative to positive—exactly what happens at a minimum!
📋 Step-by-Step Procedure
Using the Second Derivative Test:
- Find the first derivative \(f'(x)\)
- Find all critical points: Solve \(f'(x) = 0\)
- Note: Second Derivative Test only works for points where \(f'(c) = 0\)
- It does NOT work for points where \(f'(c)\) is undefined
- Find the second derivative \(f''(x)\)
- For each critical point \(c\) where \(f'(c) = 0\):
- Evaluate \(f''(c)\)
- Classify based on sign of \(f''(c)\):
- If \(f''(c) > 0\) → Local minimum
- If \(f''(c) < 0\) → Local maximum
- If \(f''(c) = 0\) → Inconclusive (use First Derivative Test)
- Calculate \(f(c)\) for each extremum (the actual max/min value)
⚠️ Important Limitations:
- Only works when \(f'(c) = 0\): Cannot use for corners or cusps
- Fails when \(f''(c) = 0\): Must use alternative method
- Requires \(f''\) to exist: Cannot use if second derivative is undefined
- Only classifies one point at a time: Doesn't tell you about intervals
📖 Comprehensive Worked Examples
Example 1: Standard Application
Problem: Use the Second Derivative Test to classify the critical points of \(f(x) = x^3 - 6x^2 + 9x + 1\).
Solution:
Step 1: Find \(f'(x)\)
Step 2: Find critical points
Set \(f'(x) = 0\):
Divide by 3:
Factor:
Step 3: Find \(f''(x)\)
Step 4: Apply Second Derivative Test at each critical point
At \(x = 1\):
Since \(f''(1) < 0\) → LOCAL MAXIMUM at \(x = 1\)
Calculate value: \(f(1) = 1 - 6 + 9 + 1 = 5\)
At \(x = 3\):
Since \(f''(3) > 0\) → LOCAL MINIMUM at \(x = 3\)
Calculate value: \(f(3) = 27 - 54 + 27 + 1 = 1\)
Final Answer:
• Local maximum: \(f(1) = 5\) at \(x = 1\)
• Local minimum: \(f(3) = 1\) at \(x = 3\)
Example 2: When the Test is Inconclusive
Problem: Use the Second Derivative Test to classify critical points of \(f(x) = x^4\).
Solution:
Step 1-2: Find \(f'(x)\) and critical points
Set \(f'(x) = 0\):
Step 3: Find \(f''(x)\)
Step 4: Apply Second Derivative Test at \(x = 0\)
Since \(f''(0) = 0\), the Second Derivative Test is INCONCLUSIVE!
Step 5: Use First Derivative Test instead
Test sign of \(f'(x) = 4x^3\) around \(x = 0\):
- For \(x < 0\): \(f'(-1) = -4 < 0\) (decreasing)
- For \(x > 0\): \(f'(1) = 4 > 0\) (increasing)
Since \(f'\) changes from − to +, LOCAL MINIMUM at \(x = 0\)
Final Answer:
• Local minimum: \(f(0) = 0\) at \(x = 0\)
• Note: Second Derivative Test failed; had to use First Derivative Test
Example 3: Multiple Critical Points
Problem: Use the Second Derivative Test to classify critical points of \(f(x) = x^4 - 4x^3 + 6\).
Solution:
Step 1-2: Find \(f'(x)\) and critical points
Step 3: Find \(f''(x)\)
Step 4: Test each critical point
At \(x = 0\):
INCONCLUSIVE at \(x = 0\)
Use First Derivative Test:
- \(f'(-1) = 4(1)(-4) = -16 < 0\)
- \(f'(1) = 4(1)(-2) = -8 < 0\)
- No sign change → NOT an extremum
At \(x = 3\):
Since \(f''(3) > 0\) → LOCAL MINIMUM at \(x = 3\)
\(f(3) = 81 - 108 + 6 = -21\)
Final Answer:
• No extremum at \(x = 0\) (Second Derivative Test inconclusive; First Derivative Test shows no extremum)
• Local minimum: \(f(3) = -21\) at \(x = 3\)
Example 4: Trigonometric Function
Problem: Use the Second Derivative Test to classify critical points of \(f(x) = \sin(x) + \cos(x)\) on \([0, 2\pi]\).
Solution:
Step 1-2: Find \(f'(x)\) and critical points
Set \(f'(x) = 0\):
In \([0, 2\pi]\): \(x = \frac{\pi}{4}, \frac{5\pi}{4}\)
Step 3: Find \(f''(x)\)
Step 4: Test each critical point
At \(x = \frac{\pi}{4}\):
Since \(f''\left(\frac{\pi}{4}\right) < 0\) → LOCAL MAXIMUM
\(f\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2}\)
At \(x = \frac{5\pi}{4}\):
Since \(f''\left(\frac{5\pi}{4}\right) > 0\) → LOCAL MINIMUM
\(f\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2}\)
Final Answer:
• Local maximum: \(f\left(\frac{\pi}{4}\right) = \sqrt{2}\) at \(x = \frac{\pi}{4}\)
• Local minimum: \(f\left(\frac{5\pi}{4}\right) = -\sqrt{2}\) at \(x = \frac{5\pi}{4}\)
Example 5: Rational Function
Problem: Use the Second Derivative Test for \(f(x) = \frac{x^2}{x - 1}\) on its domain.
Solution:
Step 1-2: Find \(f'(x)\) using quotient rule
Set \(f'(x) = 0\):
Step 3: Find \(f''(x)\)
Using quotient rule on \(f'(x) = \frac{x^2-2x}{(x-1)^2}\):
Simplify (factor out \((x-1)\)):
Step 4: Test critical points
At \(x = 0\):
Since \(f''(0) < 0\) → LOCAL MAXIMUM
\(f(0) = \frac{0}{-1} = 0\)
At \(x = 2\):
Since \(f''(2) > 0\) → LOCAL MINIMUM
\(f(2) = \frac{4}{1} = 4\)
Final Answer:
• Local maximum: \(f(0) = 0\) at \(x = 0\)
• Local minimum: \(f(2) = 4\) at \(x = 2\)
• Note: Domain is \(x \neq 1\) (vertical asymptote)
🔄 First Derivative Test vs Second Derivative Test
When to Use Each Method:
| Aspect | First Derivative Test | Second Derivative Test |
|---|---|---|
| What to analyze | Sign of \(f'\) around critical point | Sign of \(f''\) at critical point |
| Requirements | \(f\) continuous at \(c\) | \(f'(c) = 0\) and \(f''(c)\) exists |
| Works when | Always (for continuous functions) | Only when \(f''(c) \neq 0\) |
| Work required | Test multiple points in intervals | Evaluate one value: \(f''(c)\) |
| Handles corners/cusps | Yes | No |
| Can fail | No (always works) | Yes (when \(f''(c) = 0\)) |
| Advantage | 100% reliable | Faster when it works |
| Disadvantage | More work | Sometimes inconclusive |
| Best for | Default method; always use when in doubt | Quick check for smooth functions |
💡 Decision Guide:
- Use Second Derivative Test when:
- You have \(f'(c) = 0\) (horizontal tangent)
- \(f''\) is easy to compute
- You want a quick answer
- Use First Derivative Test when:
- \(f'(c)\) is undefined (corner, cusp)
- \(f''(c) = 0\) (Second Derivative Test fails)
- You want guaranteed results
- You need to know intervals of increase/decrease anyway
⚠️ When the Second Derivative Test Fails
Cases Where You CANNOT Use the Second Derivative Test:
- When \(f'(c)\) is undefined:
- Example: \(f(x) = |x|\) at \(x = 0\) (corner)
- The test requires \(f'(c) = 0\), not undefined
- When \(f''(c) = 0\):
- Example: \(f(x) = x^4\) at \(x = 0\)
- Could be max, min, or neither
- Must use First Derivative Test
- When \(f''(c)\) doesn't exist:
- Example: \(f(x) = x^{4/3}\) at \(x = 0\)
- Cannot evaluate \(f''(0)\)
📝 Important Examples:
| Function | Critical Point | Why It Fails | Actual Result |
|---|---|---|---|
| \(f(x) = x^3\) | \(x = 0\) | \(f''(0) = 0\) | No extremum (inflection point) |
| \(f(x) = x^4\) | \(x = 0\) | \(f''(0) = 0\) | Local minimum |
| \(f(x) = -x^4\) | \(x = 0\) | \(f''(0) = 0\) | Local maximum |
| \(f(x) = |x|\) | \(x = 0\) | \(f'(0)\) DNE | Local minimum (use 1st deriv test) |
💡 Tips, Tricks & Strategies
✅ Essential Problem-Solving Tips:
- Check requirements first: Verify \(f'(c) = 0\) before using the test
- Simplify derivatives: Factor \(f'(x)\) and \(f''(x)\) for easier evaluation
- Be ready to switch: If \(f''(c) = 0\), immediately use First Derivative Test
- Remember the signs: Positive = minimum, Negative = maximum
- Always find \(f(c)\): State the actual max/min value, not just \(x = c\)
- Double-check arithmetic: Sign errors are common when evaluating \(f''(c)\)
- Use for speed: When available, it's faster than First Derivative Test
- Know when to quit: Don't waste time if test is inconclusive
🎯 Quick Decision Flowchart:
Found critical point at \(x = c\)
↓
Is \(f'(c) = 0\)?
↓ YES → Continue | NO → Use First Derivative Test
Can you easily find \(f''(c)\)?
↓ YES → Continue | NO → Use First Derivative Test
Evaluate \(f''(c)\)
↓
- If \(f''(c) > 0\) → Local Min ✓
- If \(f''(c) < 0\) → Local Max ✓
- If \(f''(c) = 0\) → Use First Derivative Test
🔥 Quick Recognition Patterns:
- Polynomials: Second derivative is easy to find—good candidate for test
- Exponentials: \(f(x) = e^{g(x)}\) often has simple \(f''\)—use the test
- Trig functions: \(f''\) alternates between \(\sin\) and \(\cos\)—test works well
- Rational functions: \(f''\) can be messy—consider First Derivative Test
- Even powers at origin: \(x^{2n}\) often has \(f''(0) = 0\)—test fails
❌ Common Mistakes to Avoid
- Mistake 1: Using the test when \(f'(c)\) is undefined (only works when \(f'(c) = 0\))
- Mistake 2: Continuing when \(f''(c) = 0\) instead of switching to First Derivative Test
- Mistake 3: Confusing the signs: thinking positive = max (it's minimum!)
- Mistake 4: Not checking if \(f''(c)\) exists before evaluating
- Mistake 5: Assuming test always works (it doesn't when inconclusive)
- Mistake 6: Forgetting to calculate \(f(c)\) to state the actual extremum value
- Mistake 7: Arithmetic errors when evaluating \(f''(c)\)
- Mistake 8: Not simplifying \(f'(x)\) before taking the second derivative
- Mistake 9: Using test for global extrema (it only identifies local extrema)
- Mistake 10: Stopping at "test is inconclusive" without using alternative method
📝 Practice Problems
Set A: Standard Applications
- Use Second Derivative Test: \(f(x) = x^3 - 3x + 2\)
- Use Second Derivative Test: \(f(x) = 2x^3 - 9x^2 + 12x + 1\)
- Use Second Derivative Test: \(f(x) = x^4 - 8x^2 + 3\)
Answers:
- Local max at \(x = -1\), local min at \(x = 1\)
- Local max at \(x = 1\), local min at \(x = 2\)
- Inconclusive at \(x = 0\) (use 1st deriv test: local max); local min at \(x = \pm 2\)
Set B: When Test Fails
- Try Second Derivative Test on \(f(x) = x^5\). What happens?
- Can you use Second Derivative Test on \(f(x) = |x - 2|\) at \(x = 2\)? Explain.
Answers:
- \(f'(0) = 0\) but \(f''(0) = 0\) → inconclusive; use 1st deriv test (no extremum)
- No—\(f'(2)\) doesn't exist (corner point); Second Derivative Test requires \(f'(c) = 0\)
Set C: Conceptual Questions
- Why does the Second Derivative Test work? Explain using concavity.
- Give an example where Second Derivative Test gives a different answer than expected.
Answers:
- If \(f'(c) = 0\) (horizontal tangent) and \(f''(c) > 0\) (concave up), the graph curves upward, so \(c\) is at the bottom (minimum). If \(f''(c) < 0\) (concave down), \(c\) is at the top (maximum).
- \(f(x) = x^4\): \(f'(0) = 0\) and \(f''(0) = 0\), but it's actually a minimum (not inconclusive in nature, but test can't determine it)
✏️ AP® Exam Success Tips
What AP® Graders Look For:
- Both derivatives shown: Write \(f'(x) = \ldots\) and \(f''(x) = \ldots\)
- Critical points found correctly: Show solving \(f'(x) = 0\)
- Evaluation shown: Calculate \(f''(c)\) explicitly
- Conclusion justified: State "Since \(f''(c) > 0\), there is a local minimum at..."
- Inconclusive cases handled: If \(f''(c) = 0\), state test is inconclusive
- Function values calculated: Find \(f(c)\) for each extremum
- Proper terminology: "Local maximum/minimum" not just "max/min"
- Complete answer: Include both location (\(x = c\)) and value (\(f(c)\))
Common FRQ Formats:
- "Use the Second Derivative Test to classify the critical points of f"
- "Determine whether f has a relative maximum or minimum at x = c. Justify your answer."
- "Show that f has a local minimum at x = c"
- "Can the Second Derivative Test be used to determine the nature of the critical point? Explain."
- "Find and classify all critical points"
💯 Earning Full Credit:
- 1 point: Finding \(f'(x)\) and critical points correctly
- 1 point: Finding \(f''(x)\) correctly
- 1 point: Evaluating \(f''(c)\) at each critical point
- 1 point: Correct classification with justification
- Bonus: Handling inconclusive cases properly shows mastery
⚡ Quick Reference Card
| Concept | Key Information |
|---|---|
| When to Use | When \(f'(c) = 0\) and \(f''\) is easy to find |
| Local Minimum | \(f'(c) = 0\) and \(f''(c) > 0\) |
| Local Maximum | \(f'(c) = 0\) and \(f''(c) < 0\) |
| Inconclusive | \(f'(c) = 0\) and \(f''(c) = 0\) (or DNE) |
| Why It Works | Concavity: positive = ∪ (min), negative = ∩ (max) |
| Doesn't Work For | Corners, cusps (where \(f'(c)\) undefined) |
| Alternative | First Derivative Test (always reliable) |
| Advantage | Faster—only one evaluation needed |
🔗 Connections to Other Topics
Topic 5.7 Connects To:
- Topic 5.4 (1st Deriv Test): Alternative method; use when 2nd deriv test fails
- Topic 5.6 (Concavity): Second derivative determines concavity, which reveals extrema
- Topic 5.2 (Critical Points): Need to find critical points first before testing
- Topic 5.5 (Optimization): Quick way to verify max/min in applied problems
- Topic 5.8 (Curve Sketching): Helps identify key features of graphs
- Physics: Acceleration (2nd derivative) determines motion at critical velocity
- Economics: Marginal cost analysis using second derivatives
Master the Second Derivative Test! This elegant method uses concavity to quickly classify critical points: if \(f'(c) = 0\) and \(f''(c) > 0\), you have a local minimum (concave up ∪); if \(f''(c) < 0\), you have a local maximum (concave down ∩). However, when \(f''(c) = 0\), the test is inconclusive—you must switch to the First Derivative Test. This method only works when \(f'(c) = 0\) (not for corners or cusps). The key advantage is speed: evaluate one value instead of testing intervals. But remember: the First Derivative Test is your reliable backup—it always works! Know when to use each method: Second Derivative Test for quick checks on smooth functions, First Derivative Test for guaranteed results. Both derivatives shown, evaluation explicit, and justification clear earn full AP® credit! 🎯✨