Unit 5.5 – Using the Candidates Test to Determine Absolute (Global) Extrema
AP® Calculus AB & BC | The Closed Interval Method for Finding Absolute Maximum and Minimum Values
Why This Matters: The Candidates Test (also called the Closed Interval Method) is a systematic procedure for finding the absolute (global) maximum and minimum values of a continuous function on a closed interval. This method is essential for optimization problems where you need to find the highest or lowest value a function can achieve. Unlike local extrema (which are "neighborhood champions"), absolute extrema are the "world champions"—the highest and lowest values over the entire interval. This is one of the most practical and frequently tested concepts in AP® Calculus!
🎯 Review: Absolute vs Local Extrema
KEY DEFINITIONS
\(f(c)\) is an absolute maximum of \(f\) on interval \(I\) if:
In words: \(f(c)\) is the highest value the function reaches on the entire interval.
\(f(d)\) is an absolute minimum of \(f\) on interval \(I\) if:
In words: \(f(d)\) is the lowest value the function reaches on the entire interval.
🔍 Absolute vs Local Extrema:
| Aspect | Absolute (Global) Extrema | Local (Relative) Extrema |
|---|---|---|
| Scope | Entire interval | Small neighborhood |
| Comparison | Compare to ALL points in interval | Compare to nearby points only |
| Uniqueness | May occur at multiple x-values | Can have many throughout interval |
| Where They Occur | Critical points OR endpoints | Critical points (interior only) |
| Analogy | "World champion" | "Neighborhood champion" |
📝 Important: Every absolute extremum is also a local extremum, but NOT every local extremum is absolute! An absolute max/min is simply the highest/lowest local extremum (including endpoints).
📐 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 at some \(c\) in \([a, b]\)
- An absolute minimum value at some \(d\) in \([a, b]\)
In words: Continuous functions on closed intervals MUST have both a highest point and a lowest point somewhere on that interval!
⚠️ Requirements for EVT:
- Continuous function: No breaks, jumps, or holes in \([a, b]\)
- Closed interval: Must include both endpoints: \([a, b]\) not \((a, b)\)
If either condition fails, absolute extrema may NOT exist!
🎯 The Candidates Test (Closed Interval Method)
The Candidates Test
To find the absolute maximum and minimum of a continuous function \(f\) on \([a, b]\):
The "Candidates" are:
- All critical points in the open interval \((a, b)\)
- Where \(f'(x) = 0\)
- Where \(f'(x)\) is undefined (but \(f(x)\) is defined)
- Both endpoints: \(x = a\) and \(x = b\)
- Find \(f'(x)\)
- Find all critical points in \((a, b)\)
- Evaluate \(f\) at all candidates:
- Each critical point
- Both endpoints \(a\) and \(b\)
- Compare all values:
- The largest value is the absolute maximum
- The smallest value is the absolute minimum
🔑 Key Insight:
Absolute extrema on a closed interval can ONLY occur at:
- Critical points (interior points where derivative is zero or undefined)
- Endpoints of the interval
That's why we check these "candidates" and compare their function values!
💡 Organizational Tip: Make a table with three columns:
| \(x\)-value (Candidate) | Type | \(f(x)\) Value |
|---|---|---|
| \(a\) | Left endpoint | \(f(a)\) |
| \(c_1, c_2, \ldots\) | Critical points | \(f(c_1), f(c_2), \ldots\) |
| \(b\) | Right endpoint | \(f(b)\) |
Then circle the largest and smallest values!
📋 Step-by-Step Procedure
The Complete Candidates Test Procedure:
- Verify EVT applies:
- Check that \(f\) is continuous on \([a, b]\)
- Confirm you have a closed interval
- Find the derivative \(f'(x)\)
- Find all critical points in \((a, b)\):
- Solve \(f'(x) = 0\)
- Find where \(f'(x)\) is undefined (but \(f(x)\) is defined)
- Verify each critical point is in the open interval \((a, b)\)
- Create a candidates table with:
- All critical points from step 3
- Both endpoints: \(x = a\) and \(x = b\)
- Evaluate \(f\) at each candidate
- Compare all values:
- Identify the largest value → Absolute maximum
- Identify the smallest value → Absolute minimum
- State your answer clearly:
- "Absolute maximum of [value] at \(x = \) [location]"
- "Absolute minimum of [value] at \(x = \) [location]"
📝 Important Note: You must evaluate the function at BOTH endpoints, even if they're not critical points. Endpoints are automatic candidates!
📖 Comprehensive Worked Examples
Example 1: Standard Polynomial Function
Problem: Find the absolute maximum and minimum values of \(f(x) = x^3 - 6x^2 + 9x + 1\) on \([0, 5]\).
Solution:
Step 1: Verify EVT applies
- \(f(x)\) is a polynomial → continuous everywhere, including \([0, 5]\) ✓
- \([0, 5]\) is a closed interval ✓
EVT guarantees absolute extrema exist!
Step 2: Find \(f'(x)\)
Step 3: Find critical points
Set \(f'(x) = 0\):
Divide by 3:
Factor:
Check: Are both in \((0, 5)\)? Yes! ✓
Check if \(f'(x)\) undefined anywhere: No (polynomial) ✓
Step 4: Create candidates table and evaluate
| \(x\)-value | Type | Calculation | \(f(x)\) |
|---|---|---|---|
| \(0\) | Left endpoint | \(0^3 - 6(0)^2 + 9(0) + 1\) | 1 |
| \(1\) | Critical point | \(1 - 6 + 9 + 1\) | 5 |
| \(3\) | Critical point | \(27 - 54 + 27 + 1\) | 1 |
| \(5\) | Right endpoint | \(125 - 150 + 45 + 1\) | 21 |
Step 5: Compare values
- Largest value: 21 at \(x = 5\)
- Smallest value: 1 at \(x = 0\) and \(x = 3\)
Final Answer:
• Absolute maximum: \(f(5) = 21\) at \(x = 5\)
• Absolute minimum: \(f(0) = f(3) = 1\) at \(x = 0\) and \(x = 3\)
(Note: The absolute minimum occurs at two different x-values!)
Example 2: Function with Undefined Derivative
Problem: Find the absolute extrema of \(f(x) = (x - 2)^{2/3}\) on \([1, 4]\).
Solution:
Step 1: Verify EVT applies
\(f(x) = (x - 2)^{2/3}\) is continuous everywhere (including at \(x = 2\)) ✓
Step 2: Find \(f'(x)\)
Step 3: Find critical points
- \(f'(x) = 0\): Numerator never equals 0, so no solutions
- \(f'(x)\) undefined: Denominator = 0 when \((x - 2)^{1/3} = 0\) → \(x = 2\)
- Check: Is \(f(2)\) defined? Yes: \(f(2) = 0\) ✓
- Is \(x = 2\) in \((1, 4)\)? Yes ✓
Critical point: \(x = 2\)
Step 4: Evaluate at all candidates
| \(x\)-value | Type | \(f(x) = (x-2)^{2/3}\) |
|---|---|---|
| \(1\) | Left endpoint | \((1-2)^{2/3} = (-1)^{2/3} = 1\) |
| \(2\) | Critical point | \((2-2)^{2/3} = 0\) |
| \(4\) | Right endpoint | \((4-2)^{2/3} = 2^{2/3} \approx 1.587\) |
Step 5: Compare
- Largest: \(2^{2/3} \approx 1.587\) at \(x = 4\)
- Smallest: \(0\) at \(x = 2\)
Final Answer:
• Absolute maximum: \(f(4) = 2^{2/3} \approx 1.587\) at \(x = 4\)
• Absolute minimum: \(f(2) = 0\) at \(x = 2\)
Example 3: Extrema at Endpoints Only
Problem: Find the absolute extrema of \(f(x) = 2x + 3\) on \([-1, 2]\).
Solution:
Step 1-2: Find \(f'(x)\)
(constant slope—it's a line!)
Step 3: Find critical points
- \(f'(x) = 2 \neq 0\) for all \(x\)
- \(f'(x)\) is never undefined
No critical points!
Step 4: Evaluate at endpoints only
| \(x\)-value | Type | \(f(x) = 2x + 3\) |
|---|---|---|
| \(-1\) | Left endpoint | \(2(-1) + 3 = 1\) |
| \(2\) | Right endpoint | \(2(2) + 3 = 7\) |
Final Answer:
• Absolute maximum: \(f(2) = 7\) at \(x = 2\)
• Absolute minimum: \(f(-1) = 1\) at \(x = -1\)
(For linear functions with non-zero slope, extrema always occur at endpoints!)
Example 4: Rational Function
Problem: Find the absolute extrema of \(f(x) = \frac{x}{x^2 + 1}\) on \([-2, 2]\).
Solution:
Step 1-2: Find \(f'(x)\) using quotient rule
Step 3: Find critical points
- \(f'(x) = 0\): Numerator = 0 → \(1 - x^2 = 0\) → \(x = \pm 1\)
- Both \(x = -1\) and \(x = 1\) are in \((-2, 2)\) ✓
- \(f'(x)\) is never undefined (denominator always positive)
Step 4: Evaluate at all candidates
| \(x\)-value | Type | \(f(x) = \frac{x}{x^2+1}\) |
|---|---|---|
| \(-2\) | Left endpoint | \(\frac{-2}{5} = -0.4\) |
| \(-1\) | Critical point | \(\frac{-1}{2} = -0.5\) |
| \(1\) | Critical point | \(\frac{1}{2} = 0.5\) |
| \(2\) | Right endpoint | \(\frac{2}{5} = 0.4\) |
Step 5: Compare
- Largest: \(0.5\) at \(x = 1\)
- Smallest: \(-0.5\) at \(x = -1\)
Final Answer:
• Absolute maximum: \(f(1) = \frac{1}{2}\) at \(x = 1\)
• Absolute minimum: \(f(-1) = -\frac{1}{2}\) at \(x = -1\)
Example 5: Trigonometric Function
Problem: Find the absolute extrema of \(f(x) = 2\sin(x) + \cos(2x)\) on \([0, \pi]\).
Solution:
Step 1-2: Find \(f'(x)\)
Factor:
Step 3: Find critical points in \((0, \pi)\)
Set \(f'(x) = 0\):
Case 1: \(\cos(x) = 0\) → \(x = \frac{\pi}{2}\) ✓
Case 2: \(1 - 2\sin(x) = 0\) → \(\sin(x) = \frac{1}{2}\) → \(x = \frac{\pi}{6}, \frac{5\pi}{6}\)
Check: \(\frac{\pi}{6} \approx 0.524\) and \(\frac{5\pi}{6} \approx 2.618\) both in \((0, \pi)\) ✓
Step 4: Evaluate at all candidates
| \(x\)-value | Type | \(f(x) = 2\sin(x) + \cos(2x)\) |
|---|---|---|
| \(0\) | Left endpoint | \(2(0) + \cos(0) = 1\) |
| \(\frac{\pi}{6}\) | Critical point | \(2(\frac{1}{2}) + \cos(\frac{\pi}{3}) = 1 + \frac{1}{2} = \frac{3}{2}\) |
| \(\frac{\pi}{2}\) | Critical point | \(2(1) + \cos(\pi) = 2 - 1 = 1\) |
| \(\frac{5\pi}{6}\) | Critical point | \(2(\frac{1}{2}) + \cos(\frac{5\pi}{3}) = 1 + \frac{1}{2} = \frac{3}{2}\) |
| \(\pi\) | Right endpoint | \(2(0) + \cos(2\pi) = 1\) |
Final Answer:
• Absolute maximum: \(f\left(\frac{\pi}{6}\right) = f\left(\frac{5\pi}{6}\right) = \frac{3}{2}\) at \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\)
• Absolute minimum: \(f(0) = f\left(\frac{\pi}{2}\right) = f(\pi) = 1\) at \(x = 0, \frac{\pi}{2}, \pi\)
⚠️ Special Cases and Important Notes
Important Cases to Consider:
- No critical points:
- If \(f'(x) \neq 0\) everywhere and no undefined points, extrema occur at endpoints only
- Example: Linear functions \(f(x) = mx + b\) with \(m \neq 0\)
- Critical point at boundary:
- If \(f'(a) = 0\) or \(f'(b) = 0\), the endpoint is both an endpoint AND a critical point
- Still only count it once in your table!
- Multiple extrema at same value:
- Absolute max/min can occur at multiple x-values
- List all locations where they occur
- Critical point outside interval:
- If you find \(f'(c) = 0\) but \(c \notin (a, b)\), DON'T include it
- Only candidates IN the interval count!
- Discontinuity in interval:
- If \(f\) is not continuous on \([a, b]\), EVT doesn't apply
- Extrema may not exist!
📝 Common Question: "What if the critical point is exactly at an endpoint?"
Answer: Count it as a candidate, but only list it once. The endpoint is automatically a candidate, so you don't need to list it twice.
🔄 Candidates Test vs First Derivative Test
When to Use Each Method:
| Aspect | Candidates Test (Topic 5.5) | First Derivative Test (Topic 5.4) |
|---|---|---|
| Purpose | Find absolute extrema | Classify local extrema |
| Domain | Requires closed interval \([a, b]\) | Works on any interval |
| Candidates | Critical points + endpoints | Critical points only (interior) |
| Method | Evaluate and compare values | Analyze sign changes of \(f'\) |
| Output | Highest and lowest values on \([a, b]\) | Nature of each critical point |
| Sign analysis | Not required | Essential |
| When to use | Optimization on closed intervals | Understanding function behavior |
💡 Key Distinction:
- Candidates Test: "What are the highest and lowest values on \([a, b]\)?" (Absolute)
- First Derivative Test: "Is this critical point a peak or valley?" (Local)
- Remember: Every absolute extremum is also a local extremum (except at endpoints)
💡 Tips, Tricks & Strategies
✅ Essential Problem-Solving Tips:
- Always check EVT first: Continuous + closed interval = guaranteed extrema
- Don't forget endpoints: They're automatic candidates—always evaluate at both!
- Verify critical points are in interval: Only include those in \((a, b)\)
- Make a table: Organize candidates, types, and values systematically
- Calculate carefully: Arithmetic errors are common—double-check evaluations
- Compare ALL values: Don't assume—actually compare to find largest and smallest
- State location AND value: "Max of [value] at \(x = \) [location]"
- List all occurrences: If max/min occurs at multiple x-values, list them all
🎯 The Ultimate Checklist:
- ☑ Verify \(f\) is continuous on \([a, b]\)
- ☑ Confirm you have a closed interval
- ☑ Find and simplify \(f'(x)\)
- ☑ Solve \(f'(x) = 0\) for critical points
- ☑ Find where \(f'(x)\) is undefined
- ☑ Check that critical points are in \((a, b)\)
- ☑ List all candidates: critical points + endpoints
- ☑ Evaluate \(f\) at each candidate
- ☑ Compare all values
- ☑ Identify largest (abs max) and smallest (abs min)
- ☑ State answers with proper notation
🔥 Quick Recognition Patterns:
- Linear functions: Extrema always at endpoints (unless horizontal line)
- Quadratic functions: One critical point (vertex), check it + endpoints
- Cubic functions: Up to 2 critical points, check all + endpoints
- Rational functions: Watch for vertical asymptotes (not in domain!)
- Trig functions: Many critical points possible—solve carefully
- Absolute value functions: Check corners (where derivative undefined)
❌ Common Mistakes to Avoid
- Mistake 1: Forgetting to evaluate at endpoints (THE most common error!)
- Mistake 2: Including critical points that are outside \((a, b)\)
- Mistake 3: Not checking whether EVT applies (continuity + closed interval)
- Mistake 4: Confusing absolute extrema with local extrema
- Mistake 5: Arithmetic errors when evaluating \(f\) at candidates
- Mistake 6: Assuming the extrema occur at critical points only (ignoring endpoints)
- Mistake 7: Not comparing all candidate values—jumping to conclusions
- Mistake 8: Stating only the x-location without the function value (or vice versa)
- Mistake 9: Including points where \(f(x)\) is undefined as candidates
- Mistake 10: Using this method on an open interval (it only works on closed intervals!)
- Mistake 11: Forgetting to find where \(f'(x)\) is undefined
- Mistake 12: Not listing all x-values when extremum occurs multiple times
📝 Practice Problems
Set A: Basic Applications
- Find absolute extrema of \(f(x) = x^3 - 3x + 2\) on \([0, 2]\)
- Find absolute extrema of \(f(x) = x^2 - 4x + 5\) on \([-1, 3]\)
- Find absolute extrema of \(f(x) = \frac{1}{x}\) on \([1, 4]\)
Answers:
- Max: 4 at \(x = 2\); Min: 0 at \(x = 1\)
- Max: 10 at \(x = -1\); Min: 1 at \(x = 2\)
- Max: 1 at \(x = 1\); Min: \(\frac{1}{4}\) at \(x = 4\)
Set B: More Challenging
- Find absolute extrema of \(f(x) = x^4 - 2x^2\) on \([-2, 2]\)
- Find absolute extrema of \(f(x) = x\sqrt{4 - x^2}\) on \([-2, 2]\)
- Find absolute extrema of \(f(x) = \sin(x) + \cos(x)\) on \([0, 2\pi]\)
Answers:
- Max: 8 at \(x = \pm 2\); Min: -1 at \(x = \pm 1\)
- Max: 2 at \(x = \sqrt{2}\); Min: -2 at \(x = -\sqrt{2}\)
- Max: \(\sqrt{2}\) at \(x = \frac{\pi}{4}\); Min: \(-\sqrt{2}\) at \(x = \frac{5\pi}{4}\)
Set C: Conceptual Questions
- Why must we always check endpoints in the Candidates Test?
- Can a function have an absolute maximum but no absolute minimum on a closed interval? Explain.
- If \(f'(x) > 0\) for all \(x\) in \([a, b]\), where do the absolute extrema occur?
Answers:
- Absolute extrema can occur at endpoints, not just critical points. Endpoints are automatic candidates!
- No—if continuous on closed \([a, b]\), EVT guarantees BOTH absolute max and min exist
- If always increasing: abs min at \(x = a\) (left), abs max at \(x = b\) (right)
✏️ AP® Exam Success Tips
What AP® Graders Look For:
- Clear identification of candidates: List all critical points and endpoints
- Organized table or list: Show all candidates with their function values
- Derivative shown: Write \(f'(x) = \ldots\) explicitly
- Critical point work: Show solving \(f'(x) = 0\)
- Function evaluations: Show calculations for each candidate
- Clear comparison: Explicitly identify largest and smallest values
- Proper answer format: "Absolute maximum of ___ at \(x = \)___"
- Include both location and value: Both the x-coordinate AND the function value
Common FRQ Formats:
- "Find the absolute maximum and absolute minimum values of f on [a, b]"
- "Justify your answer" (must show all candidates and comparisons)
- "At what value of x does f attain its absolute maximum on [a, b]?"
- "Find the x-coordinate of each point at which f attains an absolute extremum"
- "Show work leading to your answer" (must show derivative and evaluations)
- Context problems: "What is the maximum height/profit/area?"
💯 Earning Full Credit (Typical Scoring):
- 1 point: Finding \(f'(x)\) correctly
- 1 point: Finding all critical points in \((a, b)\)
- 1 point: Evaluating \(f\) at critical points AND endpoints
- 1 point: Correctly identifying absolute max and min with justification
- 1 point: Proper notation and complete answer
Key: Show ALL work—partial credit available for correct method even with errors!
📝 AP® Calculator Note: On calculator-allowed problems, you can use your calculator to find critical points (solving \(f'(x) = 0\)) and to evaluate function values. However, you must still show your setup and indicate what you're finding!
⚡ Quick Reference Card
| Concept | Key Information |
|---|---|
| Purpose | Find absolute (global) max and min on closed interval |
| Requirements | Continuous function on closed interval \([a, b]\) |
| Candidates | Critical points in \((a, b)\) + both endpoints |
| Critical Points | Where \(f'(x) = 0\) or \(f'(x)\) undefined (and \(f(x)\) exists) |
| Method | Evaluate \(f\) at all candidates, compare values |
| Absolute Max | Largest function value among all candidates |
| Absolute Min | Smallest function value among all candidates |
| EVT Guarantee | Both absolute max and min MUST exist |
| Don't Forget | Always check BOTH endpoints! |
🗺️ Candidates Test Flowchart
Complete Decision Flowchart
START: Find absolute extrema of \(f\) on \([a, b]\)
↓
Check EVT conditions
Is \(f\) continuous on \([a, b]\)?
Do you have a closed interval?
↓ YES
Find \(f'(x)\)
↓
Find ALL critical points in \((a, b)\)
• Solve \(f'(x) = 0\)
• Find where \(f'(x)\) undefined
↓
List ALL Candidates
• Critical points
• Left endpoint: \(x = a\)
• Right endpoint: \(x = b\)
↓
Evaluate \(f\) at Each Candidate
Create table with all values
↓
RESULT: Compare All Values
• Largest value → Absolute Maximum
• Smallest value → Absolute Minimum
State: "Abs max of ___ at \(x = \)___"
🔗 Connections to Other Topics
Topic 5.5 Connects To:
- Topic 5.1 (EVT): EVT guarantees absolute extrema exist on closed intervals
- Topic 5.2 (Critical Points): Critical points are candidates for extrema
- Topic 5.3 (Inc/Dec): Understanding function behavior helps predict where extrema occur
- Topic 5.4 (1st Derivative Test): Identifies local extrema; absolute extrema are special cases
- Topic 5.6-5.8 (Optimization): This method solves real-world optimization problems
- Real Applications: Maximum profit, minimum cost, optimal dimensions, best trajectory
- Applied Problems: Business, engineering, physics, economics—all use this!
Master the Candidates Test! This method systematically finds absolute (global) maximum and minimum values on closed intervals. The Extreme Value Theorem guarantees that continuous functions on \([a, b]\) MUST have both absolute max and min. The candidates are: (1) all critical points in \((a, b)\), and (2) both endpoints. The procedure: find \(f'(x)\), find critical points, evaluate \(f\) at all candidates (don't forget endpoints!), then compare values—largest is abs max, smallest is abs min. Always show organized work with a table listing candidates, types, and function values. This is THE method for optimization problems on closed intervals and is heavily tested on the AP® exam. Remember: absolute extrema can occur at critical points OR endpoints, so check everything! Essential for real-world applications! 🎯✨