Unit 5.3 – Determining Intervals on Which a Function Is Increasing or Decreasing

AP® Calculus AB & BC | Using the First Derivative to Analyze Function Behavior

Why This Matters: Understanding when a function is increasing or decreasing is fundamental to analyzing function behavior! The first derivative test tells us: if \(f'(x) > 0\), the function is rising; if \(f'(x) < 0\), it's falling. This concept is essential for optimization, curve sketching, and real-world applications. By analyzing the sign of \(f'(x)\), we can determine where functions grow or shrink, identify local extrema, and understand the overall shape of graphs!

📈 Definitions: Increasing and Decreasing Functions

FORMAL DEFINITIONS

Increasing Function

A function \(f\) is increasing on an interval \(I\) if for any two numbers \(x_1\) and \(x_2\) in \(I\):

\[ \text{If } x_1 < x_2, \text{ then } f(x_1) < f(x_2) \]

In words: As you move from left to right, the function values get larger (graph goes up).

Decreasing Function

A function \(f\) is decreasing on an interval \(I\) if for any two numbers \(x_1\) and \(x_2\) in \(I\):

\[ \text{If } x_1 < x_2, \text{ then } f(x_1) > f(x_2) \]

In words: As you move from left to right, the function values get smaller (graph goes down).

Constant Function

A function \(f\) is constant on an interval \(I\) if for any two numbers \(x_1\) and \(x_2\) in \(I\):

\[ f(x_1) = f(x_2) \]

In words: The function stays at the same height (horizontal line).

📝 Important Terminology:

"Increasing" = "rising" = "going up" = positive slope
"Decreasing" = "falling" = "going down" = negative slope
Strictly increasing: \(f(x_1) < f(x_2)\) (no flat parts)
Non-decreasing: \(f(x_1) \leq f(x_2)\) (can have flat parts)

🎯 The Increasing/Decreasing Test (First Derivative Test)

Increasing/Decreasing Test

Let \(f\) be a function that is continuous on \([a, b]\) and differentiable on \((a, b)\):

If \(f'(x) > 0\) for all \(x\) in \((a, b)\)

\[ \text{Then } f \text{ is INCREASING on } [a, b] \]

Positive derivative → Function is rising

If \(f'(x) < 0\) for all \(x\) in \((a, b)\)

\[ \text{Then } f \text{ is DECREASING on } [a, b] \]

Negative derivative → Function is falling

If \(f'(x) = 0\) for all \(x\) in \((a, b)\)

\[ \text{Then } f \text{ is CONSTANT on } [a, b] \]

Zero derivative → Function is flat

🔑 The Key Connection:

Derivative Sign and Function Behavior
Sign of \(f'(x)\)	Function Behavior	Graph Direction	Slope
\(f'(x) > 0\)	Increasing	↗ Rising	Positive
\(f'(x) < 0\)	Decreasing	↘ Falling	Negative
\(f'(x) = 0\)	Constant or Critical Point	→ Horizontal	Zero
\(f'(x)\) undefined	Possible discontinuity	Corner/Cusp/Vertical	Undefined

💡 Memory Trick:

Positive = Pointing UP → \(f'(x) > 0\) means increasing
Negative = Nosedive DOWN → \(f'(x) < 0\) means decreasing
Zero = Flat/Level → \(f'(x) = 0\) means horizontal (at that instant)

🔍 Step-by-Step Procedure: Finding Intervals of Increase/Decrease

The Complete Method:

Find the derivative \(f'(x)\)
Find all critical points: Solve \(f'(x) = 0\) and find where \(f'(x)\) is undefined
Create a number line and mark all critical points and any discontinuities
Choose test points in each interval between critical points
Evaluate \(f'(\text{test point})\) to determine the sign in each interval
Determine behavior:
- If \(f'(x) > 0\) in the interval → \(f\) is increasing
- If \(f'(x) < 0\) in the interval → \(f\) is decreasing
Write your answer using interval notation

📝 Key Point: Critical points divide the number line into intervals. The derivative doesn't change sign within an interval (unless there's another critical point). So test ONE point in each interval to determine the sign throughout!

📊 Sign Charts (Number Line Analysis)

Example Sign Chart Format:

f'(x):    +++++++++    0    ----------    0    +++++++++
          increasing   c₁    decreasing   c₂    increasing
              
x-axis:  ←-----------|-----------|----------→
                     c₁          c₂
                     
Behavior: Increasing | Decreasing | Increasing
Extrema:            Max          Min

How to Read a Sign Chart:

+ + + region: \(f'(x) > 0\) → function is increasing
− − − region: \(f'(x) < 0\) → function is decreasing
Zero mark (0): Critical point where \(f'(x) = 0\)
Sign changes: Indicate local extrema
- + to − → Local maximum
- − to + → Local minimum
- No change → Not an extremum

💡 Pro Tip: Always draw a sign chart! It's the clearest way to visualize where \(f'(x)\) is positive or negative, and it helps identify local extrema. AP® graders love to see organized sign charts in FRQs!

🏔️ First Derivative Test for Local Extrema

FIRST DERIVATIVE TEST

Let \(c\) be a critical point of \(f\) (where \(f'(c) = 0\) or \(f'(c)\) is undefined). Then:

Local Maximum at \(c\)

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.

Visual: Function rises to peak, then falls → hilltop

Local Minimum at \(c\)

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.

Visual: Function falls to valley, then rises → valley bottom

NOT an Extremum at \(c\)

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

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

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

Visual: Could be an inflection point with horizontal tangent

Sign Change Patterns:

First Derivative Test Summary
Sign Pattern	Visual	Result
+ + + → 0 → − − −	↗ peak ↘	Local Maximum
− − − → 0 → + + +	↘ valley ↗	Local Minimum
+ + + → 0 → + + +	↗ → ↗	No extremum (inflection or horizontal tangent)
− − − → 0 → − − −	↘ → ↘	No extremum (inflection or horizontal tangent)

📖 Comprehensive Worked Examples

Example 1: Finding Intervals of Increase and Decrease

Problem: Find the intervals on which \(f(x) = x^3 - 3x^2 - 9x + 5\) is increasing and decreasing.

Solution:

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

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

Step 2: Find critical points (solve \(f'(x) = 0\))

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

Divide by 3:

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

Factor:

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

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

Step 3: Create a number line with critical points

Intervals:    (-∞, -1)    |    (-1, 3)    |    (3, ∞)
                          -1              3

Step 4: Choose test points and evaluate \(f'(\text{test point})\)

Interval	Test Point	\(f'(\text{test})\)	Sign	Behavior
\((-\infty, -1)\)	\(x = -2\)	\(3(-2)^2 - 6(-2) - 9 = 12 + 12 - 9 = 15\)	+	Increasing
\((-1, 3)\)	\(x = 0\)	\(3(0)^2 - 6(0) - 9 = -9\)	−	Decreasing
\((3, \infty)\)	\(x = 4\)	\(3(4)^2 - 6(4) - 9 = 48 - 24 - 9 = 15\)	+	Increasing

Step 5: Create sign chart

f'(x):    + + + + +    0    − − − − −    0    + + + + +
          increasing  -1    decreasing   3    increasing
              
Extrema:             MAX                MIN

Answer:
• Increasing: \((-\infty, -1) \cup (3, \infty)\)
• Decreasing: \((-1, 3)\)
• Local maximum at \(x = -1\)
• Local minimum at \(x = 3\)

Example 2: Function with Undefined Derivative

Problem: Find intervals where \(f(x) = x^{2/3}(x - 5)\) is increasing and decreasing.

Solution:

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

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

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

Step 2: Find critical points

\(f'(x) = 0\): Numerator = 0 → \(x - 2 = 0\) → \(x = 2\)
\(f'(x)\) undefined: Denominator = 0 → \(x^{1/3} = 0\) → \(x = 0\)

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

Step 3: Test intervals

Interval	Test Point	Sign of \(f'(x)\)	Behavior
\((-\infty, 0)\)	\(x = -1\)	\(\frac{5(-3)}{3(-1)} = \frac{-15}{-3} = +\)	Increasing
\((0, 2)\)	\(x = 1\)	\(\frac{5(-1)}{3(1)} = -\)	Decreasing
\((2, \infty)\)	\(x = 8\)	\(\frac{5(6)}{3(2)} = +\)	Increasing

Step 4: Sign chart

f'(x):    + + + +    DNE    − − − −    0    + + + +
          incr.      0      decr.      2    incr.
              
Extrema:          MAX               MIN

Answer:
• Increasing: \((-\infty, 0) \cup (2, \infty)\)
• Decreasing: \((0, 2)\)
• Local max at \(x = 0\) (derivative undefined, but sign changes)
• Local min at \(x = 2\)

Example 3: Function with No Local Extrema at Critical Point

Problem: Analyze \(f(x) = x^3\) and determine if it has local extrema.

Solution:

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

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

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

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

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: Analyze sign change

f'(x):    + + + + +    0    + + + + +
          increasing   0    increasing
              
NO SIGN CHANGE → No extremum at x = 0!

Answer:
• Increasing: \((-\infty, \infty)\) (entire domain)
• Decreasing: Never
• No local extrema at \(x = 0\) because \(f'(x)\) doesn't change sign
• Point \((0, 0)\) is an inflection point with horizontal tangent

Example 4: Piecewise Function

Problem: Find intervals of increase/decrease for \(f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 2x & \text{if } x > 1 \end{cases}\)

Solution:

Step 1: Find \(f'(x)\) in each piece

For \(x < 1\): \(f'(x) = 2x\)
For \(x > 1\): \(f'(x) = 2\)
At \(x = 1\): Check continuity and differentiability
- \(f(1^-) = 1^2 = 1\)
- \(f(1^+) = 2(1) = 2\)
- NOT continuous at \(x = 1\) → NOT differentiable there

Step 2: Find critical points in each piece

For \(x < 1\): \(2x = 0\) → \(x = 0\)
For \(x > 1\): \(f'(x) = 2 > 0\) always (no critical points)
At \(x = 1\): Discontinuity (check separately)

Step 3: Test intervals

Interval	Test Point	\(f'(\text{test})\)	Sign	Behavior
\((-\infty, 0)\)	\(x = -1\)	\(2(-1) = -2\)	−	Decreasing
\((0, 1)\)	\(x = 0.5\)	\(2(0.5) = 1\)	+	Increasing
\((1, \infty)\)	\(x = 2\)	\(2\)	+	Increasing

Answer:
• Increasing: \((0, 1) \cup (1, \infty)\) or \((0, \infty)\) excluding \(x = 1\)
• Decreasing: \((-\infty, 0)\)
• Local min at \(x = 0\)
• Discontinuity at \(x = 1\) (jump)

Example 5: Trigonometric Function

Problem: Find intervals of increase/decrease for \(f(x) = \sin(x) + \cos(x)\) on \([0, 2\pi]\).

Solution:

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

\[ f'(x) = \cos(x) - \sin(x) \]

Step 2: Find critical points

\[ \cos(x) - \sin(x) = 0 \]

\[ \cos(x) = \sin(x) \]

\[ \tan(x) = 1 \]

In \([0, 2\pi]\): \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\)

Step 3: Test intervals

Interval	Test	\(f'(\text{test})\)	Sign	Behavior
\((0, \frac{\pi}{4})\)	\(x = 0\)	\(\cos(0) - \sin(0) = 1 - 0 = 1\)	+	Increasing
\((\frac{\pi}{4}, \frac{5\pi}{4})\)	\(x = \pi\)	\(\cos(\pi) - \sin(\pi) = -1 - 0 = -1\)	−	Decreasing
\((\frac{5\pi}{4}, 2\pi)\)	\(x = 2\pi\)	\(\cos(2\pi) - \sin(2\pi) = 1 - 0 = 1\)	+	Increasing

Answer:
• Increasing: \([0, \frac{\pi}{4}] \cup [\frac{5\pi}{4}, 2\pi]\)
• Decreasing: \([\frac{\pi}{4}, \frac{5\pi}{4}]\)
• Local max at \(x = \frac{\pi}{4}\)
• Local min at \(x = \frac{5\pi}{4}\)

⚠️ Special Cases and Considerations

Important Cases to Watch:

Endpoints: When analyzing on a closed interval \([a, b]\), check behavior at endpoints separately
Discontinuities: Function behavior may change at points of discontinuity
Vertical asymptotes: \(f'(x)\) may not be defined; analyze each side separately
Corners/cusps: \(f'(x)\) undefined but function continuous; can still have extrema
Inflection points: Where \(f'(x) = 0\) but no sign change; not an extremum

📝 Interval Notation Reminder:

Use open intervals \((a, b)\) when stating where function is increasing/decreasing
Use closed intervals \([a, b]\) when including endpoints (if function defined and continuous there)
Union symbol \(\cup\): Combines multiple intervals (e.g., \((-\infty, 2) \cup (5, \infty)\))
Don't include critical points in the intervals (they're boundaries, not part of the behavior)

💡 Tips, Tricks & Strategies

✅ Essential Problem-Solving Tips:

Always factor \(f'(x)\): Makes finding zeros much easier!
Choose easy test points: Use integers when possible (0, 1, −1, etc.)
Don't just plug in critical points: Test points from BETWEEN critical points
Draw the sign chart: Visual organization prevents mistakes
Check your intervals: Make sure you've covered the entire domain
Remember the connection: Sign changes in \(f'\) indicate extrema
Use proper notation: Open/closed intervals as appropriate
State both increasing AND decreasing: Don't forget either one!

🎯 The Complete Strategy:

Step-by-Step Process
Step	Action	Why
1	Find \(f'(x)\)	Need derivative to analyze behavior
2	Find critical points	Divide domain into test intervals
3	Draw number line	Visualize intervals clearly
4	Test each interval	Determine sign of \(f'\) in each region
5	Create sign chart	Organize results visually
6	Write intervals	Answer the question with proper notation
7	Identify extrema	Look for sign changes

🔥 Quick Checks:

Polynomial degree n: At most \(n - 1\) critical points
Even function: If \(f(-x) = f(x)\), behavior is symmetric about y-axis
Odd function: If \(f(-x) = -f(x)\), behavior is symmetric about origin
Periodic function: Behavior repeats every period
Reality check: Does your answer make sense with the graph?

❌ Common Mistakes to Avoid

Mistake 1: Confusing where \(f'(x) = 0\) with where \(f(x) = 0\) (zeros vs critical points)
Mistake 2: Testing AT critical points instead of BETWEEN them
Mistake 3: Forgetting to check where \(f'(x)\) is undefined
Mistake 4: Including critical points inside interval notation (use them as boundaries)
Mistake 5: Assuming \(f'(x) = 0\) always means an extremum (could be inflection point!)
Mistake 6: Not factoring \(f'(x)\) before solving (makes it much harder!)
Mistake 7: Mixing up positive/negative: \(f' > 0\) means INCREASING, not decreasing!
Mistake 8: Forgetting to state BOTH increasing AND decreasing intervals
Mistake 9: Using wrong inequality: \(f'(x) > 0\) means \(>\), not \(\geq\)
Mistake 10: Not considering the domain restrictions of the original function

📝 Practice Problems

Set A: Finding Intervals

Find intervals of increase/decrease: \(f(x) = x^3 - 12x + 1\)
Find intervals of increase/decrease: \(f(x) = x^4 - 4x^3\)
Find intervals of increase/decrease: \(f(x) = \frac{x^2}{x - 1}\)

Answers:

Inc: \((-\infty, -2) \cup (2, \infty)\); Dec: \((-2, 2)\)
Inc: \((3, \infty)\); Dec: \((-\infty, 0) \cup (0, 3)\); Note: \(x = 0\) is inflection point
Inc: \((-\infty, 0) \cup (2, \infty)\); Dec: \((0, 1) \cup (1, 2)\); Note: \(x = 1\) is vertical asymptote

Set B: Identifying Extrema

Use First Derivative Test to classify critical points: \(f(x) = x^3 - 6x^2 + 9x + 1\)
Use First Derivative Test to classify critical points: \(f(x) = x^4 - 2x^2\)

Answers:

Critical points at \(x = 1, 3\); Local max at \(x = 1\), local min at \(x = 3\)
Critical points at \(x = -1, 0, 1\); Local max at \(x = 0\), local min at \(x = \pm 1\)

Set C: Conceptual Questions

If \(f'(x) = (x - 2)^2(x + 1)\), where is \(f\) increasing?
Can a function be both increasing and decreasing on the same interval? Explain.
If \(f\) is increasing on \((a, b)\) and decreasing on \((b, c)\), what can you conclude about \(x = b\)?

Answers:

\((-1, \infty)\) (sign changes from − to + at \(x = -1\), but not at \(x = 2\))
No—a function has only one behavior (up or down) at any given point
\(x = b\) is a local maximum (function rises then falls)

✏️ AP® Exam Success Tips

What AP® Graders Look For:

Show derivative explicitly: Write \(f'(x) = \ldots\)
Show critical point work: Solving \(f'(x) = 0\) with algebra
Include sign chart or number line: Visual evidence of analysis
Test points shown: Show at least one test point per interval
State conclusions clearly: "f is increasing on \((\ldots)\)"
Justify extrema: Cite sign changes: "f' changes from + to −, so local max"
Use proper interval notation: Parentheses vs brackets matter
Check domain restrictions: Note discontinuities or undefined points

Common FRQ Formats:

"Find the intervals on which f is increasing/decreasing"
"Use the First Derivative Test to classify each critical point"
"Justify your answer" (must show sign analysis)
"At what values of x does f have a local extremum?"
"Determine whether f has a local maximum, local minimum, or neither at x = c"
"Given a graph of f', determine where f is increasing"

💯 Earning Full Credit:

Derivative (1 pt): Correctly find and simplify \(f'(x)\)
Critical points (1 pt): Find all critical points
Sign analysis (2 pts): Determine sign of \(f'\) in each interval
Conclusion (1 pt): State intervals of increase/decrease
Justification (1 pt): For extrema, show sign change

⚡ Quick Reference Card

Increasing/Decreasing Quick Reference
Concept	Key Fact/Formula
Increasing Test	\(f'(x) > 0\) on interval → \(f\) increasing on interval
Decreasing Test	\(f'(x) < 0\) on interval → \(f\) decreasing on interval
Critical Points	\(f'(c) = 0\) or \(f'(c)\) undefined
Local Max	\(f'\) changes from + to − at \(c\)
Local Min	\(f'\) changes from − to + at \(c\)
No Extremum	\(f'\) doesn't change sign at \(c\)
Test Point Method	Pick one point in each interval between critical points
Sign Chart	Visual tool: + + + \| 0 \| − − − shows behavior

📋 Complete Visual Summary

The Complete Process Flowchart

START: Given function \(f(x)\)

↓

Find \(f'(x)\)

↓

Find Critical Points

Solve \(f'(x) = 0\) and find where \(f'(x)\) undefined

↓

Draw Number Line

Mark all critical points

↓

Test Each Interval

Pick test points, evaluate \(f'(\text{test})\)

↓

Create Sign Chart

Show + + + and − − − regions

↓

RESULT:

• State intervals of increase/decrease

• Identify local extrema from sign changes

🔗 Connections to Other Topics

Topic 5.3 Connects To:

Topic 5.1 (MVT): MVT proves why \(f' > 0\) implies increasing
Topic 5.2 (Extrema): Increasing/decreasing analysis identifies local max/min
Topic 5.4 (Concavity): Second derivative determines how fast function increases/decreases
Topic 5.5 (Optimization): Finding where \(f\) increases/decreases solves max/min problems
Topic 5.7 (Curve Sketching): Inc/dec intervals determine graph shape
Unit 6 (Integrals): If \(f\) is increasing, then \(F'(x) = f(x) > 0\)
Real applications: Profit increasing, population growing, velocity positive

Master the Increasing/Decreasing Test! The sign of the first derivative determines function behavior: \(f'(x) > 0\) means increasing (rising), and \(f'(x) < 0\) means decreasing (falling). To find intervals: (1) find \(f'(x)\), (2) find critical points, (3) test intervals between critical points, (4) create a sign chart, (5) write intervals in proper notation. The First Derivative Test uses sign changes to identify extrema: + to − gives local max, − to + gives local min, no change means no extremum. Always draw sign charts for visual organization! On the AP® exam, show all work: derivative, critical points, test points, sign analysis, and justified conclusions. This topic is essential for optimization, curve sketching, and understanding function behavior in calculus and real-world applications! 🎯✨