The First Derivative Test
Using the derivative to classify critical points as maxima, minima, or neither
📊 The First Derivative Test
What is the First Derivative Test?
The First Derivative Test uses the sign of to determine whether a critical point is a local maximum, local minimum, or neither.
💡 Key Idea: If the derivative changes from positive to negative at a critical point, you have a local max. If it changes from negative to positive, you have a local min!
Understanding the Signs of the Derivative
What Tells Us
- : Function is increasing (going up)
- : Function is decreasing (going down)
- : Function has a horizontal tangent
Think of it like driving a car:
- : driving uphill ⬆️
- : driving downhill ⬇️
- : at the top or bottom of a hill
The First Derivative Test (Formal Statement)
Let be a critical point of (so or undefined).
Test the sign of on intervals around :
Case 1: Local Maximum
If changes from positive to negative at :
- for (increasing before)
- for (decreasing after)
- Then is a local maximum 📈➡️📉
Case 2: Local Minimum
If changes from negative to positive at :
- for (decreasing before)
- for (increasing after)
- Then is a local minimum 📉➡️📈
Case 3: Neither
If does not change sign at :
- Then is neither a max nor a min
- Could be an inflection point
Step-by-Step Process
How to Apply the First Derivative Test
Step 1: Find all critical points (solve and find where is undefined)
Step 2: Create a sign chart for
- Mark critical points on a number line
- Test the sign of in each interval
Step 3: Analyze the sign changes
- Positive → Negative = Local MAX
- Negative → Positive = Local MIN
- No change = Neither
Step 4: State your conclusions
Example 1: Complete Analysis
Classify the critical points of
Step 1: Find critical points
Critical points: and
Step 2: Create sign chart
Test intervals: , ,
Interval : Test ✓ POSITIVE
Interval : Test ✓ NEGATIVE
Interval : Test ✓ POSITIVE
Step 3: Sign chart
-1 3
++++ | ---- | ++++
▼ ▲
Step 4: Conclusions
At : goes from + to − → LOCAL MAXIMUM
At : goes from − to + → LOCAL MINIMUM
Calculate the values:
Answer:
- Local maximum of at
- Local minimum of at
Creating Sign Charts
Method 1: Test Points
Pick any value in each interval and substitute into .
Tip: Use easy numbers like
Method 2: Factored Form
If , analyze each factor's sign.
Example:
Interval (x+2) (x-1) Product
(-∞, -2) − − +
(-2, 1) + − −
(1, ∞) + + +
Increasing and Decreasing Intervals
The First Derivative Test also tells us where is increasing or decreasing:
From the Sign Chart
- Where : is increasing
- Where : is decreasing
Example from above:
- is increasing on and
- is decreasing on
Special Cases
Case 1: Inflection Point with Horizontal Tangent
If but the sign doesn't change, it's usually an inflection point.
Example: at
- , so
- for all (no sign change!)
- is an inflection point, not an extremum
Case 2: Multiple Critical Points in a Row
Handle each one separately using the test.
Case 3: Derivative Undefined
The test still works! Just check the sign change around the point.
Example: at
- for
- for
- Changes from − to + → local minimum at ✓
⚠️ Common Mistakes
Mistake 1: Testing AT the Critical Point
❌ Don't test where is the critical point ✅ Test points on either SIDE of
Mistake 2: Wrong Sign Interpretation
The sign of , not , determines increasing/decreasing!
Mistake 3: Forgetting to Check All Critical Points
Each critical point needs its own analysis.
Mistake 4: Not Using Factored Form
Factor completely to make sign analysis easier!
First Derivative Test vs. Other Methods
When to Use the First Derivative Test
✅ Always works (if derivative exists on either side) ✅ Good for understanding overall behavior ✅ Tells you increasing/decreasing intervals ✅ Works when second derivative is hard to compute
When to Use the Second Derivative Test (coming next!)
✅ Faster for simple cases ✅ Only need to evaluate at one point ❌ Fails if ❌ Doesn't give increasing/decreasing info
Quick Reference Chart
| Sign Change | Result | |-------------|--------| | + → − | Local MAX ⬆️⬇️ | | − → + | Local MIN ⬇️⬆️ | | + → + | Neither (increasing through) | | − → − | Neither (decreasing through) |
📝 Practice Strategy
- Find and factor completely
- Mark critical points on a number line
- Test one point in each interval (use easy numbers)
- Record signs carefully: + for positive, − for negative
- Look for sign changes: + to − is max, − to + is min
- State intervals: Where increasing? Where decreasing?
- Calculate values at extrema for complete answer
📚 Practice Problems
1Problem 1medium
❓ Question:
Use the First Derivative Test to find all local extrema of .
💡 Show Solution
Step 1: Find the derivative
Step 2: Find critical points
or
Step 3: Create sign chart
Test intervals: , ,
Test (in ): ✓ NEGATIVE
Test (in ): ✓ NEGATIVE
Test (in ): ✓ POSITIVE
Step 4: Analyze sign chart
0 3
---- | ---- | ++++
▲
At : − to − (no change) → NEITHER (inflection point)
At : − to + → LOCAL MINIMUM
Step 5: Calculate the minimum value
Answer: Local minimum of at . No local maximum. ( is an inflection point with horizontal tangent.)
2Problem 2hard
❓ Question:
Find the intervals where is increasing and decreasing. Identify all local extrema.
💡 Show Solution
Step 1: Find the derivative
Step 2: Find critical points
Set :
Also, is undefined at , but is also undefined, so is NOT a critical point (not in domain).
Critical points: and
Step 3: Create sign chart
Test intervals: , , ,
Note: is not in the domain but divides the intervals.
Test : ✓ POSITIVE
Test : ✓ NEGATIVE
Test : ✓ NEGATIVE
Test : ✓ POSITIVE
Step 4: Sign chart
-2 0 2
++++ | -- | -- | ++++
▼ ▲
Step 5: Classify critical points
At : + to − → LOCAL MAXIMUM
At : − to + → LOCAL MINIMUM
Calculate values:
Answer:
- Increasing on and
- Decreasing on and
- Local maximum of at
- Local minimum of at
3Problem 3expert
❓ Question:
Use the First Derivative Test on to find and classify all critical points.
💡 Show Solution
Step 1: Expand and find derivative
Step 2: Combine using common denominator
Step 3: Find critical points
: →
undefined: →
Critical points: and
Step 4: Test signs
Test (in ): ✓ POSITIVE
Test (in ): ✓ NEGATIVE
Test (in ): ✓ POSITIVE
Step 5: Sign chart
0 8/5
++++ | ---- | ++++
▼ ▲
At : + to − → LOCAL MAXIMUM
At : − to + → LOCAL MINIMUM
Step 6: Calculate values
Answer:
- Local maximum of at
- Local minimum of at
4Problem 4medium
❓ Question:
Use the first derivative test to classify the critical point at x = 2 for f(x) = x³ - 6x² + 9x.
💡 Show Solution
Step 1: Find f'(x): f(x) = x³ - 6x² + 9x f'(x) = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x - 1)(x - 3)
Step 2: Verify x = 2 is a critical point: f'(2) = 3(2 - 1)(2 - 3) = 3(1)(-1) = -3... Wait, this is NOT zero. Let me recalculate: Critical points where f'(x) = 0: 3(x - 1)(x - 3) = 0 x = 1 or x = 3 (not x = 2)
Actually, the problem asks about x = 2, but it's not a critical point. Let me classify x = 1 instead:
Step 3: Test intervals around x = 1: Interval (0, 1): f'(0.5) = 3(0.5 - 1)(0.5 - 3) = 3(-0.5)(-2.5) > 0 (positive) Interval (1, 2): f'(1.5) = 3(1.5 - 1)(1.5 - 3) = 3(0.5)(-1.5) < 0 (negative)
Step 4: Apply first derivative test: f' changes from + to - at x = 1 Therefore, x = 1 is a LOCAL MAXIMUM
Answer: x = 1 is a local maximum (note: x = 2 is not a critical point)
5Problem 5hard
❓ Question:
For f(x) = x⁴ - 4x³, find all critical points and classify them using the first derivative test.
💡 Show Solution
Step 1: Find f'(x): f'(x) = 4x³ - 12x² = 4x²(x - 3)
Step 2: Find critical points: f'(x) = 0 when 4x²(x - 3) = 0 x = 0 or x = 3
Step 3: Test around x = 0: Interval (-1, 0): f'(-0.5) = 4(-0.5)²(-0.5 - 3) = 4(0.25)(-3.5) < 0 Interval (0, 1): f'(0.5) = 4(0.5)²(0.5 - 3) = 4(0.25)(-2.5) < 0 f' is negative on both sides → x = 0 is NEITHER max nor min
Step 4: Test around x = 3: Interval (2, 3): f'(2.5) = 4(2.5)²(2.5 - 3) = 4(6.25)(-0.5) < 0 Interval (3, 4): f'(3.5) = 4(3.5)²(3.5 - 3) = 4(12.25)(0.5) > 0 f' changes from - to + at x = 3 → x = 3 is LOCAL MINIMUM
Answer: x = 0 is neither (inflection point), x = 3 is local minimum
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