Unit 5.4 – Using the First Derivative Test to Determine Relative (Local) Extrema

AP® Calculus AB & BC | The Most Powerful Tool for Finding Local Maxima and Minima

Why This Matters: The First Derivative Test is THE primary method for identifying and classifying local extrema (relative maxima and minima) in calculus! By analyzing how the sign of \(f'(x)\) changes at critical points, we can definitively determine whether each critical point is a local max, local min, or neither. This test is essential for optimization problems, curve sketching, and understanding function behavior. It's more reliable than just finding where \(f'(x) = 0\) because it actually tests the nature of each critical point!

📍 Review: Critical Points and Local Extrema

KEY DEFINITIONS (REVIEW)

Critical Point

A number \(c\) in the domain of \(f\) is a critical point if:

\[ f'(c) = 0 \quad \text{OR} \quad f'(c) \text{ does not exist} \]

Local (Relative) Maximum

\(f(c)\) is a local maximum if \(f(c) \geq f(x)\) for all \(x\) near \(c\).

Visual: A hilltop or peak in the neighborhood of \(c\).

Local (Relative) Minimum

\(f(d)\) is a local minimum if \(f(d) \leq f(x)\) for all \(x\) near \(d\).

Visual: A valley or trough in the neighborhood of \(d\).

📝 Important Connection: Local extrema can only occur at:

Critical points (interior points where \(f'(c) = 0\) or \(f'(c)\) undefined)
Endpoints of the domain (if applicable)

However, not every critical point is an extremum! That's why we need the First Derivative Test.

🎯 The First Derivative Test

First Derivative Test for Local Extrema

Let \(c\) be a critical point of a continuous function \(f\). Then:

Case 1: Local Maximum

If \(f'(x)\) changes from positive to negative at \(c\):

\[ f'(x) > 0 \text{ for } x < c \quad \text{and} \quad f'(x) < 0 \text{ for } x > c \]

Then \(f(c)\) is a LOCAL MAXIMUM

Sign Change:    + + + +  →  0  →  − − − −
Behavior:       Rising      Peak    Falling
Result:            LOCAL MAXIMUM at c

Case 2: Local Minimum

If \(f'(x)\) changes from negative to positive at \(c\):

\[ f'(x) < 0 \text{ for } x < c \quad \text{and} \quad f'(x) > 0 \text{ for } x > c \]

Then \(f(c)\) is a LOCAL MINIMUM

Sign Change:    − − − −  →  0  →  + + + +
Behavior:       Falling    Valley   Rising
Result:            LOCAL MINIMUM at c

Case 3: No Extremum

If \(f'(x)\) does NOT change sign at \(c\):

\[ f'(x) > 0 \text{ on both sides} \quad \text{OR} \quad f'(x) < 0 \text{ on both sides} \]

Then \(f(c)\) is NOT a local extremum

Sign Pattern:   + + + +  →  0  →  + + + +  OR  − − − −  →  0  →  − − − −
Result:         No extremum (possibly inflection point)

🔑 The Key Insight:

The First Derivative Test is all about sign changes:

Sign Change Patterns and Results
Sign of \(f'\) Before \(c\)	At \(c\)	Sign of \(f'\) After \(c\)	Conclusion
+ (positive)	0 or undefined	− (negative)	Local Max
− (negative)	0 or undefined	+ (positive)	Local Min
+ (positive)	0 or undefined	+ (positive)	No extremum
− (negative)	0 or undefined	− (negative)	No extremum

💡 Memory Tricks:

"Peak to valley": + to − = Maximum (peak), − to + = Minimum (valley)
"Happy to sad": Positive (↗ happy) to negative (↘ sad) = Maximum
"Low to high": Negative (low) to positive (high) = Minimum at the bottom
"No change, no extremum": Same sign on both sides = Not an extremum

📋 Step-by-Step Procedure: Using the First Derivative Test

The Complete 7-Step Process:

Find \(f'(x)\) (take the derivative)
Find all critical points:
- Solve \(f'(x) = 0\)
- Find where \(f'(x)\) is undefined (but \(f(x)\) is defined)
Create a number line marking all critical points
Choose test points in each interval between critical points
Evaluate the sign of \(f'(\text{test point})\) in each interval
Create a sign chart showing where \(f'(x)\) is positive or negative
Apply the First Derivative Test:
- If sign changes from + to − → Local Max
- If sign changes from − to + → Local Min
- If no sign change → No extremum

📝 Pro Tip: Always organize your work with a clear sign chart! It makes the sign changes obvious and helps prevent errors. AP® graders look for this organizational tool in FRQs.

📊 Sign Chart Template

Standard Sign Chart Format:

Critical Points:        c₁         c₂         c₃

Number Line:    ←-------|---------|---------|-------→
                        c₁        c₂        c₃

Test Points:      t₁       t₂       t₃       t₄
                (pick)   (pick)   (pick)   (pick)

Sign of f'(x):   + + +     − − −     + + +     − − −

f(x) behavior:   ↗ Inc    ↘ Dec    ↗ Inc    ↘ Dec

Conclusion:           MAX       MIN       MAX
                      at c₁     at c₂     at c₃

How to Read the Sign Chart:

Top row: Mark all critical points on a number line
Test points: Choose values between critical points (and beyond endpoints)
Sign of \(f'(x)\): Evaluate \(f'(\text{test point})\) to get + or −
Behavior: + means increasing (↗), − means decreasing (↘)
Extrema: Look for sign changes at critical points

📖 Comprehensive Worked Examples

Example 1: Classic Polynomial Function

Problem: Use the First Derivative Test to find and classify all local extrema of \(f(x) = x^3 - 3x^2 - 9x + 5\).

Solution:

Step 1: Find \(f'(x)\)

\[ f'(x) = 3x^2 - 6x - 9 \]

Step 2: Find critical points

Set \(f'(x) = 0\):

\[ 3x^2 - 6x - 9 = 0 \]

Divide by 3:

\[ x^2 - 2x - 3 = 0 \]

Factor:

\[ (x - 3)(x + 1) = 0 \]

\[ x = -1 \quad \text{or} \quad x = 3 \]

Check if \(f'(x)\) is undefined anywhere: No (polynomial is differentiable everywhere)

Critical points: \(x = -1\) and \(x = 3\)

Step 3: Set up number line and choose test points

Interval	Test Point	Sign Calculation
\((-\infty, -1)\)	\(x = -2\)	\(f'(-2) = 3(4) - 6(-2) - 9 = 12 + 12 - 9 = 15 > 0\)
\((-1, 3)\)	\(x = 0\)	\(f'(0) = 3(0) - 6(0) - 9 = -9 < 0\)
\((3, \infty)\)	\(x = 4\)	\(f'(4) = 3(16) - 6(4) - 9 = 48 - 24 - 9 = 15 > 0\)

Step 4: Create sign chart

Critical Points:          -1              3

f'(x) sign:      + + + +  0  − − − −  0  + + + +

f(x) behavior:    ↗ Inc     ↘ Dec     ↗ Inc

Test points:   -2       0       4
               +        −       +

Sign changes:        +→−           −→+
                     MAX           MIN

Step 5: Apply First Derivative Test

At \(x = -1\): \(f'\) changes from + to − → LOCAL MAXIMUM
- Calculate: \(f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10\)
At \(x = 3\): \(f'\) changes from − to + → LOCAL MINIMUM
- Calculate: \(f(3) = (3)^3 - 3(3)^2 - 9(3) + 5 = 27 - 27 - 27 + 5 = -22\)

Final Answer:
• Local maximum: \(f(-1) = 10\) at \(x = -1\)
• Local minimum: \(f(3) = -22\) at \(x = 3\)

Example 2: Function with Undefined Derivative

Problem: Use the First Derivative Test to classify extrema of \(f(x) = x^{2/3}(x - 5)\).

Solution:

Step 1: Expand and find \(f'(x)\)

\(f(x) = x^{5/3} - 5x^{2/3}\)

\[ f'(x) = \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} = \frac{5x^{2/3}}{3} - \frac{10}{3x^{1/3}} \]

Common denominator:

\[ f'(x) = \frac{5x - 10}{3x^{1/3}} = \frac{5(x - 2)}{3x^{1/3}} \]

Step 2: Find critical points

\(f'(x) = 0\): Numerator = 0 → \(5(x - 2) = 0\) → \(x = 2\)
\(f'(x)\) undefined: Denominator = 0 → \(3x^{1/3} = 0\) → \(x = 0\)
Check: Is \(f(0)\) defined? Yes: \(f(0) = 0^{2/3}(0 - 5) = 0\) ✓

Critical points: \(x = 0\) and \(x = 2\)

Step 3: Test intervals

Interval	Test Point	Sign of \(f'(x) = \frac{5(x-2)}{3x^{1/3}}\)	Result
\((-\infty, 0)\)	\(x = -1\)	\(\frac{5(-3)}{3(-1)} = \frac{-15}{-3} = +\)	Positive
\((0, 2)\)	\(x = 1\)	\(\frac{5(-1)}{3(1)} = \frac{-5}{3} = -\)	Negative
\((2, \infty)\)	\(x = 8\)	\(\frac{5(6)}{3(2)} = \frac{30}{6} = +\)	Positive

Step 4: Sign chart

Critical Points:          0              2

f'(x) sign:      + + + + DNE − − − −  0  + + + +

f(x) behavior:    ↗ Inc    ↘ Dec     ↗ Inc

Sign changes:        +→−           −→+
                     MAX           MIN

Step 5: Apply First Derivative Test and calculate values

At \(x = 0\): \(f'\) changes from + to − → LOCAL MAXIMUM
- \(f(0) = 0\)
At \(x = 2\): \(f'\) changes from − to + → LOCAL MINIMUM
- \(f(2) = 2^{2/3}(2 - 5) = 2^{2/3}(-3) \approx -4.76\)

Final Answer:
• Local maximum: \(f(0) = 0\) at \(x = 0\) (derivative undefined)
• Local minimum: \(f(2) = -3\cdot 2^{2/3}\) at \(x = 2\)

Example 3: Critical Point That Is NOT an Extremum

Problem: Use the First Derivative Test to analyze \(f(x) = x^3\).

Solution:

Step 1: Find \(f'(x)\) and critical points

\[ f'(x) = 3x^2 \]

Set \(f'(x) = 0\):

\[ 3x^2 = 0 \quad \Rightarrow \quad x = 0 \]

Critical point: \(x = 0\)

Step 2: Test intervals around \(x = 0\)

Interval	Test Point	\(f'(\text{test})\)	Sign
\((-\infty, 0)\)	\(x = -1\)	\(3(-1)^2 = 3\)	+
\((0, \infty)\)	\(x = 1\)	\(3(1)^2 = 3\)	+

Step 3: Sign chart

Critical Point:          0

f'(x) sign:      + + + +  0  + + + +

f(x) behavior:    ↗ Inc     ↗ Inc

Sign change:         NONE!
Result:           NO EXTREMUM

Step 4: Apply First Derivative Test

At \(x = 0\): \(f'\) does NOT change sign (positive on both sides)

→ NO local extremum at \(x = 0\)

Note: \((0, 0)\) is an inflection point with horizontal tangent

Final Answer:
• No local extrema
• \(x = 0\) is a critical point but NOT an extremum
• This shows why we can't just assume \(f'(c) = 0\) means an extremum exists!

Example 4: Multiple Extrema

Problem: Find and classify all local extrema of \(f(x) = x^4 - 4x^3\).

Solution:

Step 1: Find \(f'(x)\) and critical points

\[ f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) \]

Set \(f'(x) = 0\):

\[ 4x^2(x - 3) = 0 \]

\[ x = 0 \quad \text{or} \quad x = 3 \]

Critical points: \(x = 0\) and \(x = 3\)

Step 2: Test intervals

Interval	Test Point	\(f'(x) = 4x^2(x-3)\)	Sign
\((-\infty, 0)\)	\(x = -1\)	\(4(1)(-4) = -16\)	−
\((0, 3)\)	\(x = 1\)	\(4(1)(-2) = -8\)	−
\((3, \infty)\)	\(x = 4\)	\(4(16)(1) = 64\)	+

Step 3: Sign chart

Critical Points:          0              3

f'(x) sign:      − − − −  0  − − − −  0  + + + +

f(x) behavior:    ↘ Dec     ↘ Dec     ↗ Inc

Sign changes:        NO           −→+
                   CHANGE         MIN

Step 4: Apply First Derivative Test

At \(x = 0\): \(f'\) stays negative (− to −) → NO extremum
At \(x = 3\): \(f'\) changes from − to + → LOCAL MINIMUM
- \(f(3) = 3^4 - 4(3)^3 = 81 - 108 = -27\)

Final Answer:
• Local minimum: \(f(3) = -27\) at \(x = 3\)
• No extremum at \(x = 0\) (inflection point with horizontal tangent)

Example 5: Rational Function

Problem: Find and classify all local extrema of \(f(x) = \frac{x^2}{x^2 - 4}\).

Solution:

Step 1: Find \(f'(x)\) using quotient rule

\[ f'(x) = \frac{(x^2 - 4)(2x) - x^2(2x)}{(x^2 - 4)^2} = \frac{2x^3 - 8x - 2x^3}{(x^2 - 4)^2} = \frac{-8x}{(x^2 - 4)^2} \]

Step 2: Find critical points

\(f'(x) = 0\): \(-8x = 0\) → \(x = 0\)
\(f'(x)\) undefined: \((x^2 - 4)^2 = 0\) → \(x = \pm 2\)
Check: Is \(f(\pm 2)\) defined? No! Vertical asymptotes.
Only critical point: \(x = 0\)

Note: \(x = \pm 2\) are NOT critical points (not in domain)

Step 3: Test intervals (must consider asymptotes)

Interval	Test Point	\(f'(x) = \frac{-8x}{(x^2-4)^2}\)	Sign
\((-\infty, -2)\)	\(x = -3\)	\(\frac{-8(-3)}{25} = \frac{24}{25} > 0\)	+
\((-2, 0)\)	\(x = -1\)	\(\frac{-8(-1)}{9} = \frac{8}{9} > 0\)	+
\((0, 2)\)	\(x = 1\)	\(\frac{-8(1)}{9} = -\frac{8}{9} < 0\)	−
\((2, \infty)\)	\(x = 3\)	\(\frac{-8(3)}{25} = -\frac{24}{25} < 0\)	−

Step 4: Sign chart

                        -2     0     2
                        VA     c     VA

f'(x) sign:  + + + | + + +  0  − − − | − − −

f(x) behavior: Inc | Inc    Dec | Dec

Sign change:         +→−
                     MAX at x=0

Step 5: Apply First Derivative Test

At \(x = 0\): \(f'\) changes from + to − → LOCAL MAXIMUM

\(f(0) = \frac{0}{-4} = 0\)

Final Answer:
• Local maximum: \(f(0) = 0\) at \(x = 0\)
• Domain: \((-\infty, -2) \cup (-2, 2) \cup (2, \infty)\)
• Vertical asymptotes at \(x = \pm 2\) (not critical points)

⚠️ Special Cases and Important Notes

Important Cases to Watch:

Multiple critical points with \(f'(c) = 0\):
- Factor \(f'(x)\) completely to find all zeros
- Example: \(f'(x) = (x-1)^2(x-3)\) has critical points at \(x = 1, 3\)
- Test each one separately!
Critical point where derivative is undefined:
- Can still have extrema (corners, cusps)
- Example: \(f(x) = |x|\) at \(x = 0\) is a local (and absolute) minimum
Repeated roots in \(f'(x)\):
- \(f'(x) = (x-2)^2\): Sign doesn't change at \(x = 2\) (no extremum)
- \(f'(x) = (x-2)^3\): Sign changes at \(x = 2\) (possible extremum)
Endpoints:
- First Derivative Test applies to interior critical points only
- Endpoints can be absolute extrema but not local extrema
Discontinuities:
- Function must be continuous for First Derivative Test to work
- At jump discontinuities, extrema may not exist

📝 Terminology Reminder:

"Local" = "Relative": These terms are interchangeable
Local extrema: Maxima and minima in a neighborhood
Absolute extrema: Highest/lowest over entire domain (may also be local)
Critical point: An x-value (input), not a point on the graph
Critical value: The function value at a critical point: \(f(c)\)

🔄 First Derivative Test vs Other Methods

When to Use Each Method:

Comparison of Extrema Tests
Method	When to Use	Pros	Cons
First Derivative Test	Always works for continuous functions	• Always reliable • Works when \(f''(c)\) doesn't exist • Shows inc/dec behavior	• Requires testing intervals • More work than 2nd derivative test
Second Derivative Test (Topic 5.6)	When \(f'(c) = 0\) and \(f''(c)\) exists	• Quick for simple functions • No intervals to test	• Fails when \(f''(c) = 0\) • Doesn't work if \(f'(c)\) undefined • Must compute \(f''\)
Closed Interval Method (Topic 5.2)	Finding absolute extrema on \([a,b]\)	• Finds absolute max/min • Systematic approach	• Requires closed interval • Compares many values

💡 Which Test to Use?

Default choice: First Derivative Test (always works!)
Quick check: Second Derivative Test (if applicable)
If 2nd derivative test fails: Fall back to First Derivative Test
For absolute extrema on \([a,b]\): Closed Interval Method
On AP exam: Use whichever method the question specifies!

💡 Tips, Tricks & Strategies

✅ Essential Problem-Solving Tips:

Always factor \(f'(x)\) completely: Makes finding zeros easier and reveals repeated roots
Organize with sign charts: Prevents confusion and errors
Test BETWEEN critical points: Not at them!
Check domain restrictions: Don't call points critical if they're not in the domain
Calculate function values: Find \(f(c)\) for each extremum (that's the actual max/min value)
State conclusions clearly: "Local max of ___ at x = ___"
Watch for repeated roots: \((x-a)^n\) in \(f'(x)\) affects sign changes
Don't assume: Not every critical point is an extremum—test them all!

🎯 The Ultimate Strategy Checklist:

☑ Find and simplify \(f'(x)\)
☑ Factor \(f'(x)\) completely
☑ Solve \(f'(x) = 0\) for all zeros
☑ Find where \(f'(x)\) is undefined (check if \(f(x)\) is defined there)
☑ List all critical points
☑ Draw number line with critical points marked
☑ Choose test points in each interval
☑ Evaluate \(f'(\text{test point})\) for sign
☑ Create organized sign chart
☑ Look for sign changes
☑ Apply First Derivative Test to classify each critical point
☑ Calculate \(f(c)\) for each extremum
☑ Write final answer with proper notation

🔥 Quick Checks for Common Functions:

Polynomial \(f(x) = ax^n\): Has at most \(n-1\) critical points
Product \(f(x) = g(x) \cdot h(x)\): Use product rule, expect critical points from both factors
Quotient \(f(x) = \frac{g(x)}{h(x)}\): Watch for vertical asymptotes (not critical points!)
Composition \(f(g(x))\): Chain rule may create complex \(f'(x)\)
Trig functions: Remember periodic nature—many critical points possible

❌ Common Mistakes to Avoid

Mistake 1: Assuming \(f'(c) = 0\) automatically means an extremum (need sign change!)
Mistake 2: Testing AT critical points instead of BETWEEN them
Mistake 3: Forgetting to check where \(f'(x)\) is undefined
Mistake 4: Including points where \(f(x)\) is undefined as critical points
Mistake 5: Confusing sign of \(f'(x)\) with sign of \(f(x)\)
Mistake 6: Not factoring \(f'(x)\) before solving \(f'(x) = 0\)
Mistake 7: Saying "max at \((c, f(c))\)" instead of "max of \(f(c)\) at \(x = c\)"
Mistake 8: Stopping after finding critical points without testing them
Mistake 9: Forgetting that repeated roots affect sign changes
Mistake 10: Not organizing work with a sign chart (increases errors)
Mistake 11: Confusing local extrema with absolute extrema
Mistake 12: Testing derivative sign at the critical point itself (where it's 0 or undefined!)

📝 Practice Problems

Set A: Basic Applications

Use First Derivative Test: \(f(x) = x^3 - 3x + 2\)
Use First Derivative Test: \(f(x) = x^4 - 8x^2\)
Use First Derivative Test: \(f(x) = x + \frac{1}{x}\)

Answers:

Local max at \(x = -1\), local min at \(x = 1\)
Local max at \(x = 0\), local min at \(x = \pm 2\)
Local min at \(x = 1\) (domain: \(x \neq 0\))

Set B: More Challenging

Use First Derivative Test: \(f(x) = x^{2/3}(2-x)\)
Use First Derivative Test: \(f(x) = \frac{x^2-1}{x^2+1}\)
Explain why \(f(x) = x^4\) has no local extrema despite having \(f'(0) = 0\)

Answers:

Local max at \(x = 0\) (undefined deriv), local min at \(x = \frac{4}{5}\)
Local min at \(x = 0\) (test intervals around asymptotes at none—no asymptotes!)
\(f'(x) = 4x^3\) doesn't change sign at \(x=0\) (positive on both sides); no extremum

Set C: Conceptual Questions

Can the First Derivative Test ever fail to classify a critical point? Explain.
If \(f'(x) = (x-2)^2(x+1)\), identify all local extrema.
True or False: If \(f\) has a local maximum at \(c\), then \(f'(c) = 0\). Explain.

Answers:

No—it always works for continuous functions. It tells you whether there's an extremum based on sign changes.
Local min at \(x = -1\) (sign changes − to +); no extremum at \(x = 2\) (no sign change due to even power)
False—\(f'\) might not exist at \(c\) (e.g., corner or cusp). Correct: If local max at \(c\) AND \(f'(c)\) exists, then \(f'(c) = 0\)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Derivative shown explicitly: Write \(f'(x) = \ldots\)
Critical points identified: Both where \(f'(x) = 0\) and undefined
Sign chart or number line: Visual organization of analysis
Test points shown: Display calculations: \(f'(\text{test}) = \ldots\)
Sign analysis stated: "\(f'(x) > 0\) for \(x < c\) and \(f'(x) < 0\) for \(x > c\)"
Justification given: "Since \(f'\) changes from + to −, there is a local max at \(x = c\)"
Function values calculated: Find \(f(c)\) for each extremum
Proper language: "Local maximum" not just "maximum"

Common FRQ Formats:

"Use the First Derivative Test to determine the x-coordinates of all local extrema"
"Justify your answer" (must show sign analysis!)
"Classify each critical point as a local max, local min, or neither"
"At what x-values does f have a local maximum? Justify."
"Use the First Derivative Test to determine whether f has a relative extremum at x = c"
"Given f'(x), determine all local extrema of f"

💯 Earning Full Credit (Typical 5-Point Question):

1 point: Finding \(f'(x)\) correctly
1 point: Finding all critical points
2 points: Sign analysis of \(f'(x)\) in intervals (1 pt for attempting, 1 pt for correct)
1 point: Correct classification with justification

Key: Partial credit often given for correct method even with arithmetic errors!

⚡ Quick Reference Card

First Derivative Test Quick Reference
Concept	Key Fact
Critical Point	\(f'(c) = 0\) or \(f'(c)\) undefined (and \(f(c)\) exists)
Local Max Test	\(f'\) changes from + to − at \(c\)
Local Min Test	\(f'\) changes from − to + at \(c\)
No Extremum	\(f'\) doesn't change sign at \(c\)
Test Point Method	Pick one value in each interval between critical points
Sign Chart	+ + + \| 0 \| − − − shows max; − − − \| 0 \| + + + shows min
Always Works?	YES—for continuous functions (unlike 2nd derivative test)
Key Question	"Does the sign of \(f'(x)\) change at \(c\)?"

🗺️ First Derivative Test Flowchart

Decision Flowchart for Classifying Critical Points

START: Found critical point at \(x = c\)

↓

Test \(f'\) just LEFT of \(c\)

Is \(f'(x)\) positive or negative?

↓

Test \(f'\) just RIGHT of \(c\)

Is \(f'(x)\) positive or negative?

↓

Compare Signs:

Left: + and Right: − → LOCAL MAXIMUM
Left: − and Right: + → LOCAL MINIMUM
Left: + and Right: + → NO EXTREMUM
Left: − and Right: − → NO EXTREMUM

↓

If extremum exists: Calculate \(f(c)\)

State: "Local [max/min] of \(f(c)\) at \(x = c\)"

🔗 Connections to Other Topics

Topic 5.4 Connects To:

Topic 5.2 (Critical Points): Provides the candidates to test
Topic 5.3 (Inc/Dec): Sign of \(f'\) shows increasing/decreasing behavior
Topic 5.5 (Optimization): Identifies max/min for applied problems
Topic 5.6 (2nd Derivative Test): Alternative method for some cases
Topic 5.7 (Curve Sketching): Extrema are key features of graphs
Unit 6 (FTC): Understanding behavior of antiderivatives
Real applications: Maximizing profit, minimizing cost, optimal design

Master the First Derivative Test! This is THE most reliable method for identifying and classifying local extrema. The key is sign analysis: if \(f'(x)\) changes from positive to negative at critical point \(c\), there's a local maximum; if it changes from negative to positive, there's a local minimum; if there's no sign change, there's no extremum. Always use the 7-step process: (1) find \(f'(x)\), (2) find critical points, (3) create number line, (4) choose test points, (5) determine signs, (6) make sign chart, (7) apply test. Organize work with clear sign charts, test between critical points (not at them), and state conclusions with proper justification. This method works for ALL continuous functions—even when second derivative test fails. Essential for optimization and curve sketching! 🎯✨