Unit 5.2 – Extreme Value Theorem, Global vs Local Extrema, and Critical Points
AP® Calculus AB & BC | Finding Maximum and Minimum Values
Why This Matters: Understanding extreme values (maximum and minimum values) is essential for optimization problems in calculus! The Extreme Value Theorem guarantees that continuous functions on closed intervals must have both a highest and lowest point. Critical points are where these extrema can occur. This topic forms the foundation for solving real-world optimization problems like maximizing profit, minimizing cost, or finding the best design specifications!
🏔️ The Extreme Value Theorem (EVT)
Extreme Value Theorem
If a function \(f\) is continuous on a closed interval \([a, b]\), then \(f\) has:
- An absolute maximum value \(f(c)\) at some \(c\) in \([a, b]\)
- An absolute minimum value \(f(d)\) at some \(d\) in \([a, b]\)
In words: A continuous function on a closed interval must have both a highest point and a lowest point somewhere on that interval!
📝 Critical Conditions: The EVT requires BOTH conditions:
- Continuous function – no breaks, jumps, or holes
- Closed interval – must include both endpoints \([a, b]\)
If either condition fails, the theorem doesn't apply, and extrema may not exist!
⚠️ When EVT Does NOT Apply:
- Discontinuous function – e.g., \(f(x) = \frac{1}{x}\) on \([-1, 1]\) (discontinuous at \(x = 0\))
- Open interval – e.g., \(f(x) = x\) on \((0, 1)\) (no endpoints, so no max/min)
- Half-open interval – e.g., \(f(x) = x^2\) on \([0, 1)\) (no right endpoint)
- Unbounded interval – e.g., \(f(x) = x^2\) on \([0, \infty)\) (no upper bound)
📊 Global (Absolute) vs Local (Relative) Extrema
DEFINITIONS: TYPES OF EXTREMA
\(f(c)\) is an absolute maximum of \(f\) on an interval if:
Meaning: \(f(c)\) is the highest value the function reaches on the entire interval.
\(f(d)\) is an absolute minimum of \(f\) on an interval if:
Meaning: \(f(d)\) is the lowest value the function reaches on the entire interval.
\(f(c)\) is a local maximum of \(f\) if there exists an open interval \((a, b)\) containing \(c\) such that:
Meaning: \(f(c)\) is the highest point in the neighborhood around \(c\) (like a hilltop).
\(f(d)\) is a local minimum of \(f\) if there exists an open interval \((a, b)\) containing \(d\) such that:
Meaning: \(f(d)\) is the lowest point in the neighborhood around \(d\) (like a valley).
🔍 Key Differences: Global vs Local Extrema
| Aspect | Absolute/Global | Local/Relative |
|---|---|---|
| Scope | Entire interval | Small neighborhood |
| Comparison | Compare to ALL points | Compare to nearby points |
| Uniqueness | May not be unique (could tie) | Can have multiple local extrema |
| Visual | Highest/lowest peak overall | Hilltop or valley |
| Relationship | Every absolute extremum is also local | Not every local extremum is absolute |
💡 Memory Trick:
- Absolute/Global: Think "world champion" – the best overall!
- Local/Relative: Think "neighborhood champion" – the best in the area!
- Absolute is always local too: The world champion is also the neighborhood champion!
🎯 Critical Points (Critical Numbers)
DEFINITION: CRITICAL POINT
A number \(c\) in the domain of \(f\) is called a critical point (or critical number) if:
In words: Critical points occur where the derivative is zero (horizontal tangent) or where the derivative is undefined (corner, cusp, or vertical tangent).
📝 Important: Critical points are x-values (inputs), not points on the graph! When we say "\(c\) is a critical point," we mean the number \(c\), though sometimes we refer to the point \((c, f(c))\) on the graph.
Type 1: Stationary Points
- \(f'(c) = 0\) → horizontal tangent line
- The function "levels off" at this point
- Example: Top of a hill or bottom of a valley
Type 2: Singular Points
- \(f'(c)\) does not exist → derivative undefined
- Can be: corner, cusp, or vertical tangent
- Example: The point \(x = 0\) for \(f(x) = |x|\) (corner)
🌟 The Fundamental Connection:
If \(f\) has a local extremum (max or min) at \(c\), and \(f'(c)\) exists, then \(f'(c) = 0\).
This means: Local extrema can ONLY occur at critical points or endpoints!
⚠️ Critical Warning:
NOT every critical point is an extremum!
- If \(f'(c) = 0\), \(c\) might be a max, min, or neither (inflection point)
- Example: \(f(x) = x^3\) has \(f'(0) = 0\), but \(x = 0\) is NOT a local extremum!
- You must TEST critical points to determine their nature
🔍 How to Find Critical Points
Step-by-Step Process:
- Find the derivative \(f'(x)\)
- Solve \(f'(x) = 0\) to find stationary points
- Find where \(f'(x)\) is undefined (but \(f(x)\) is defined)
- Check that all critical points are in the domain of \(f\)
- List all critical points
📝 Key Point: A critical point must be in the domain of \(f\)! If \(f'(c)\) doesn't exist AND \(f(c)\) doesn't exist, then \(c\) is NOT a critical point.
Example: For \(f(x) = \frac{1}{x}\), even though \(f'(0)\) doesn't exist, \(x = 0\) is NOT a critical point because \(f(0)\) is undefined (not in domain).
📈 Finding Absolute Extrema on a Closed Interval
The Closed Interval Method
To find the absolute maximum and minimum of a continuous function \(f\) on \([a, b]\):
- Find all critical points of \(f\) in the open interval \((a, b)\)
- Evaluate \(f\) at all critical points
- Evaluate \(f\) at both endpoints: \(f(a)\) and \(f(b)\)
- Compare all values: The largest is the absolute maximum, the smallest is the absolute minimum
Remember: Absolute extrema on a closed interval occur at either critical points or endpoints!
💡 Pro Tip: Make a table to organize your work!
| \(x\)-value | Type | \(f(x)\) |
|---|---|---|
| \(a\) | Left endpoint | \(f(a)\) |
| \(c_1, c_2, \ldots\) | Critical points | \(f(c_1), f(c_2), \ldots\) |
| \(b\) | Right endpoint | \(f(b)\) |
Then identify the largest value (abs max) and smallest value (abs min)!
📖 Comprehensive Worked Examples
Example 1: Finding Critical Points
Problem: Find all critical points of \(f(x) = x^3 - 6x^2 + 9x + 1\).
Solution:
Step 1: Find \(f'(x)\)
Step 2: Solve \(f'(x) = 0\)
Divide by 3:
Factor:
Solutions:
Step 3: Check where \(f'(x)\) is undefined
\(f'(x)\) is a polynomial, so it's defined everywhere. No additional critical points.
Step 4: Verify domain
\(f(x)\) is a polynomial, so domain is all real numbers. Both \(x = 1\) and \(x = 3\) are in the domain. ✓
Answer: Critical points are \(x = 1\) and \(x = 3\)
Example 2: Critical Points with Undefined Derivative
Problem: Find all critical points of \(f(x) = x^{2/3}(x - 5)\).
Solution:
Step 1: Expand and find \(f'(x)\)
Expand: \(f(x) = x^{5/3} - 5x^{2/3}\)
Step 2: Simplify \(f'(x)\)
Factor out \(\frac{5}{3}x^{-1/3}\):
Step 3: Solve \(f'(x) = 0\)
Numerator = 0:
Step 4: Find where \(f'(x)\) is undefined
Denominator = 0 when:
Check: Is \(f(0)\) defined? \(f(0) = 0^{2/3}(0 - 5) = 0\) ✓
So \(x = 0\) is a critical point (singular point - vertical tangent).
Answer: Critical points are \(x = 0\) (vertical tangent) and \(x = 2\) (horizontal tangent)
Example 3: Finding Absolute Extrema on a Closed Interval
Problem: Find the absolute maximum and minimum values of \(f(x) = x^3 - 3x + 2\) on \([-2, 2]\).
Solution:
Step 1: Find critical points
\(f'(x) = 3x^2 - 3\)
Set \(f'(x) = 0\):
Both \(x = 1\) and \(x = -1\) are in \((-2, 2)\) ✓
Step 2: Evaluate \(f\) at critical points and endpoints
| \(x\) | Type | \(f(x)\) | Calculation |
|---|---|---|---|
| \(-2\) | Left endpoint | \(0\) | \((-2)^3 - 3(-2) + 2 = -8 + 6 + 2 = 0\) |
| \(-1\) | Critical point | \(4\) | \((-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4\) |
| \(1\) | Critical point | \(0\) | \((1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0\) |
| \(2\) | Right endpoint | \(4\) | \((2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4\) |
Step 3: Compare values
- Largest value: \(4\) (occurs at \(x = -1\) and \(x = 2\))
- Smallest value: \(0\) (occurs at \(x = -2\) and \(x = 1\))
Answer:
• Absolute maximum: \(f(-1) = 4\) and \(f(2) = 4\)
• Absolute minimum: \(f(-2) = 0\) and \(f(1) = 0\)
Note: This shows that absolute extrema need not be unique! Both the max and min occur at two different points.
Example 4: When EVT Doesn't Apply
Problem: Explain why EVT doesn't guarantee extrema for \(f(x) = \frac{1}{x - 1}\) on \([0, 2]\).
Solution:
Check EVT Conditions:
- Closed interval? YES – \([0, 2]\) is a closed interval ✓
- Continuous on \([0, 2]\)? NO ✗
\(f(x) = \frac{1}{x - 1}\) has a vertical asymptote at \(x = 1\), which is inside \([0, 2]\).
The function is discontinuous at \(x = 1\), so EVT does not apply!
What actually happens:
- As \(x \to 1^-\), \(f(x) \to -\infty\) (no minimum exists)
- As \(x \to 1^+\), \(f(x) \to +\infty\) (no maximum exists)
Conclusion: Because \(f\) is not continuous on \([0, 2]\), EVT doesn't apply, and indeed the function has no absolute maximum or minimum on this interval.
Example 5: Absolute Extrema with Trigonometric Functions
Problem: Find the absolute extrema of \(f(x) = 2\sin(x) - \cos(2x)\) on \([0, \pi]\).
Solution:
Step 1: Find \(f'(x)\)
Use identity \(\sin(2x) = 2\sin(x)\cos(x)\):
Step 2: Solve \(f'(x) = 0\)
Two cases:
- Case 1: \(\cos(x) = 0\) → \(x = \frac{\pi}{2}\) (in \([0, \pi]\)) ✓
- Case 2: \(1 + 2\sin(x) = 0\) → \(\sin(x) = -\frac{1}{2}\) → \(x = \frac{7\pi}{6}\) or \(\frac{11\pi}{6}\)
- Neither \(\frac{7\pi}{6}\) nor \(\frac{11\pi}{6}\) is in \([0, \pi]\) ✗
Critical point: \(x = \frac{\pi}{2}\)
Step 3: Evaluate at critical point and endpoints
- \(f(0) = 2\sin(0) - \cos(0) = 0 - 1 = -1\)
- \(f\left(\frac{\pi}{2}\right) = 2\sin\left(\frac{\pi}{2}\right) - \cos(\pi) = 2(1) - (-1) = 3\)
- \(f(\pi) = 2\sin(\pi) - \cos(2\pi) = 0 - 1 = -1\)
Answer:
• Absolute maximum: \(f\left(\frac{\pi}{2}\right) = 3\)
• Absolute minimum: \(f(0) = f(\pi) = -1\)
Example 6: Global vs Local Extrema
Problem: For \(f(x) = x^4 - 4x^3 + 4x^2\) on \([-1, 3]\), identify all local and global extrema.
Solution:
Step 1: Find critical points
\(f'(x) = 4x^3 - 12x^2 + 8x = 4x(x^2 - 3x + 2) = 4x(x - 1)(x - 2)\)
Set \(f'(x) = 0\): \(x = 0, 1, 2\) (all in \((-1, 3)\))
Step 2: Evaluate everywhere
| \(x\) | \(f(x)\) |
|---|---|
| \(-1\) | \(9\) |
| \(0\) | \(0\) |
| \(1\) | \(1\) |
| \(2\) | \(0\) |
| \(3\) | \(9\) |
Step 3: Identify extrema
- Absolute maximum: \(f(-1) = f(3) = 9\)
- Absolute minimum: \(f(0) = f(2) = 0\)
- Local maximum: \(f(1) = 1\) (peak between two valleys)
- Local minima: \(f(0) = f(2) = 0\) (also absolute minima)
Summary:
• Global max: \(9\) at \(x = -1, 3\)
• Global min: \(0\) at \(x = 0, 2\)
• Local max only: \(1\) at \(x = 1\) (not global)
📚 Key Theorems Summary
Three Essential Theorems
If \(f\) is continuous on \([a, b]\), then \(f\) has both an absolute max and absolute min on \([a, b]\).
If \(f\) has a local extremum at \(c\) and \(f'(c)\) exists, then \(f'(c) = 0\).
Contrapositive: If \(f'(c) \neq 0\), then \(f\) does NOT have a local extremum at \(c\).
If \(f\) has a local extremum at \(c\), then \(c\) is either:
- A critical point (\(f'(c) = 0\) or \(f'(c)\) undefined), OR
- An endpoint of the domain
Key insight: To find all possible extrema, check critical points and endpoints!
💡 Tips, Tricks & Strategies
✅ Essential Problem-Solving Tips:
- Always check EVT conditions first: Continuous + closed interval
- Critical points are x-values: Don't forget to find \(f(c)\) for the extremum value
- Include endpoints: They're just as important as critical points!
- Make a table: Organize critical points, endpoints, and function values
- Not all critical points are extrema: \(f'(c) = 0\) doesn't guarantee max or min
- Check the domain: Critical points must be where \(f\) is defined
- Absolute extrema can occur multiple times: Same y-value at different x-values
- Local extrema in interior: Must occur at critical points only
🎯 The Complete Strategy for Finding Extrema:
- Determine the domain and type of interval
- Find \(f'(x)\)
- Find critical points: Solve \(f'(x) = 0\) and find where \(f'(x)\) is undefined
- If closed interval: Evaluate \(f\) at critical points AND endpoints
- Compare values: Largest = abs max, smallest = abs min
- Identify local extrema: Look at the sign of \(f'\) around each critical point
📊 Quick Classification Guide:
| If \(f'\) changes from... | Then \(c\) is a... |
|---|---|
| Positive to negative | Local maximum |
| Negative to positive | Local minimum |
| Positive to positive | Neither (possible inflection) |
| Negative to negative | Neither (possible inflection) |
❌ Common Mistakes to Avoid
- Mistake 1: Forgetting to check endpoints when finding absolute extrema
- Mistake 2: Assuming every critical point is an extremum (could be inflection point!)
- Mistake 3: Saying "the critical point is \((c, f(c))\)" instead of "\(c\)" (critical points are x-values)
- Mistake 4: Including points where BOTH \(f\) and \(f'\) are undefined as critical points
- Mistake 5: Applying EVT to open intervals or discontinuous functions
- Mistake 6: Confusing "local max" with "absolute max" (they're different!)
- Mistake 7: Not simplifying \(f'(x)\) before solving \(f'(x) = 0\) (makes it harder!)
- Mistake 8: Forgetting to check if critical points are actually in the given interval
- Mistake 9: Stating extrema as x-values instead of y-values (extrema are function values!)
- Mistake 10: Thinking absolute extrema are unique (can occur at multiple points!)
📝 Practice Problems
Set A: Finding Critical Points
- Find all critical points of \(f(x) = x^4 - 8x^2 + 5\)
- Find all critical points of \(f(x) = x + \frac{1}{x}\)
- Find all critical points of \(f(x) = x^{2/3}(x - 4)\)
Answers:
- \(x = 0, \pm 2\)
- \(x = \pm 1\) (note: \(x = 0\) is NOT a critical point—not in domain!)
- \(x = 0\) (undefined derivative) and \(x = \frac{8}{5}\) (\(f' = 0\))
Set B: Finding Absolute Extrema
- Find absolute max and min of \(f(x) = x^2 - 4x + 1\) on \([0, 5]\)
- Find absolute max and min of \(f(x) = x^3 - 3x^2 - 9x + 5\) on \([-2, 4]\)
- Find absolute max and min of \(f(x) = \sin(x) + \cos(x)\) on \([0, 2\pi]\)
Answers:
- Abs max: \(6\) at \(x = 5\); Abs min: \(-3\) at \(x = 2\)
- Abs max: \(10\) at \(x = 3\); Abs min: \(-15\) at \(x = -1\)
- Abs max: \(\sqrt{2}\) at \(x = \frac{\pi}{4}\); Abs min: \(-\sqrt{2}\) at \(x = \frac{5\pi}{4}\)
Set C: Conceptual Questions
- Can a function have a local maximum at an endpoint? Explain.
- If \(f'(5) = 0\), must \(x = 5\) be a local extremum? Why or why not?
- Explain why EVT doesn't apply to \(f(x) = \frac{1}{x^2}\) on \((0, 1)\)
Answers:
- No—local extrema require the point to be interior (surrounded by nearby points on both sides)
- No—could be an inflection point. Example: \(f(x) = x^3\) has \(f'(0) = 0\) but no extremum at \(x = 0\)
- Two reasons: (1) Open interval (not closed), and (2) Has vertical asymptote at \(x = 0\) (not continuous)
✏️ AP® Exam Success Tips
What AP® Graders Look For:
- Show derivative work: Write out \(f'(x)\) explicitly
- Show equation solving: When finding \(f'(x) = 0\), show all algebraic steps
- Check and state domain: Verify critical points are in the domain/interval
- Create a clear table: Organize candidates (critical points + endpoints) and their values
- Label answers clearly: "Absolute maximum is ___ at x = ___"
- Justify EVT application: State "f is continuous on [a,b], so by EVT..."
- Use correct terminology: Absolute vs local, maximum vs minimum
- Include units: If the problem has context (cost, distance, etc.)
Common FRQ Formats:
- "Find all critical points of f"
- "Find the absolute maximum and minimum values on [a, b]"
- "Justify your answer" (requires EVT statement or showing f' changes sign)
- "Explain why the Extreme Value Theorem applies/doesn't apply"
- "Determine whether each critical point is a local max, local min, or neither"
- "Find where f has local extrema and justify" (requires First Derivative Test - Topic 5.3)
💯 Scoring Maximum Points:
- Critical points: Must show work solving \(f'(x) = 0\) (1 point)
- Endpoint evaluation: Must evaluate at BOTH endpoints (often 1 point)
- Comparison: Must compare all candidates to identify max/min (1 point)
- Justification: When asked, cite EVT or sign analysis (1 point)
- Proper notation: Use correct interval notation and function notation
⚡ Quick Reference Card
| Concept | Definition/Formula |
|---|---|
| Extreme Value Theorem | If \(f\) continuous on \([a,b]\) → has abs max and abs min |
| Absolute Maximum | \(f(c) \geq f(x)\) for ALL \(x\) in interval |
| Absolute Minimum | \(f(d) \leq f(x)\) for ALL \(x\) in interval |
| Local Maximum | \(f(c) \geq f(x)\) for all \(x\) NEAR \(c\) |
| Local Minimum | \(f(d) \leq f(x)\) for all \(x\) NEAR \(d\) |
| Critical Point | \(c\) where \(f'(c) = 0\) OR \(f'(c)\) undefined |
| Fermat's Theorem | Local extremum + \(f'\) exists → \(f'(c) = 0\) |
| Where Extrema Occur | Critical points OR endpoints |
| Closed Interval Method | Evaluate at critical points + endpoints, compare |
🗺️ Decision Flowchart
Finding Absolute Extrema Flowchart
START: Given \(f(x)\) on interval \(I\)
↓
Is \(I\) a closed interval [a, b]?
↓ YES → Continue | NO → EVT doesn't apply
Is \(f\) continuous on [a, b]?
↓ YES → EVT guarantees extrema exist!
Find \(f'(x)\) and solve \(f'(x) = 0\)
Find where \(f'(x)\) undefined
↓
Evaluate \(f\) at:
- All critical points in (a, b)
- Both endpoints: a and b
↓
RESULT:
Largest value = Absolute Maximum
Smallest value = Absolute Minimum
📋 Complete Concept Map
| Topic | Key Points |
|---|---|
| Extreme Value Theorem | Guarantees absolute max and min exist if continuous on closed interval |
| EVT Requirements | 1) Continuous function, 2) Closed interval [a, b] |
| Absolute Extrema | Highest/lowest value over ENTIRE interval |
| Local Extrema | Highest/lowest value in a NEIGHBORHOOD |
| Critical Points | \(f'(c) = 0\) or \(f'(c)\) undefined (but \(f(c)\) exists) |
| Stationary Point | Critical point where \(f'(c) = 0\) |
| Singular Point | Critical point where \(f'(c)\) undefined |
| Fermat's Theorem | Local extremum (interior) + derivative exists → \(f'(c) = 0\) |
| Where Extrema Can Occur | 1) Critical points, 2) Endpoints, 3) Discontinuities (not applicable if continuous) |
| Closed Interval Method | Find critical points → Evaluate at critical points + endpoints → Compare |
| Key Warning | Not every critical point is an extremum! Must test. |
🔗 Connections to Other Topics
Topic 5.2 Connects To:
- Topic 5.1 (MVT): MVT proves increasing/decreasing behavior used to classify extrema
- Topic 5.3 (First Derivative Test): Uses critical points to determine nature of extrema
- Topic 5.4 (Second Derivative Test): Alternative method to classify critical points
- Topic 5.5 (Optimization): Finding extrema IS optimization!
- Topic 5.7 (Curve Sketching): Critical points and extrema determine graph shape
- Unit 6 (Integration): Fundamental Theorem uses EVT concepts
- Real-world applications: Maximize profit, minimize cost, optimize design
Master Extrema and Critical Points! The Extreme Value Theorem guarantees that continuous functions on closed intervals have both absolute maximum and minimum values. These extrema occur at critical points (where \(f'(x) = 0\) or \(f'(x)\) is undefined) or at endpoints. Use the Closed Interval Method to find absolute extrema: (1) find critical points, (2) evaluate at critical points and endpoints, (3) compare values. Remember the difference between absolute (global—highest/lowest overall) and local (relative—highest/lowest nearby) extrema. Not every critical point is an extremum—you must test using sign analysis or second derivative test. On the AP® exam, always show derivative work, create organized tables, verify points are in the domain/interval, and use correct terminology. Understanding extrema is essential for optimization and real-world applications! 🎯✨