The First Derivative Test uses the sign of fโฒ(x) to determine whether a critical point is a local maximum, local minimum, or neither.
๐ก 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
๐ Practice Problems
1Problem 1medium
โ Question:
Use the First Derivative Test to find all local extrema of f(x)=x4โ.
๐ Real-World Applications: The First Derivative Test
See how this math is used in the real world
๐ Worked Example: Related Rates โ Expanding Circle
Problem:
A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 2 cm/s. How fast is the area of the circle increasing when the radius is 10 cm?
Using the derivative to classify critical points as maxima, minima, or neither
How can I study The First Derivative Test effectively?โพ
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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What course covers The First Derivative Test?โพ
The First Derivative Test is part of the AP Calculus AB course on Study Mondo, specifically in the Applications of Derivatives section. You can explore the full course for more related topics and practice resources.
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fโฒ(x)
fโฒ(x)>0: Function is increasing (going up)
fโฒ(x)<0: Function is decreasing (going down)
fโฒ(x)=0: Function has a horizontal tangent
Think of it like driving a car:
fโฒ(x)>0: driving uphill โฌ๏ธ
fโฒ(x)<0: driving downhill โฌ๏ธ
fโฒ(x)=0: at the top or bottom of a hill
The First Derivative Test (Formal Statement)
Let c be a critical point of f (so fโฒ(c)=0 or fโฒ(c) undefined).
Test the sign of fโฒ(x) on intervals around c:
Case 1: Local Maximum
If fโฒ changes from positive to negative at c:
fโฒ(x)>0 for x<c (increasing before)
fโฒ(x)<0 for x>c (decreasing after)
Then f(c) is a local maximum ๐โก๏ธ๐
Case 2: Local Minimum
If fโฒ changes from negative to positive at c:
fโฒ(x)<0 for x<c (decreasing before)
fโฒ(x)>0 for x>c (increasing after)
Then f(c) is a local minimum ๐โก๏ธ๐
Case 3: Neither
If fโฒ does not change sign at c:
Then f(c) 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 fโฒ(x)=0 and find where fโฒ(x) is undefined)
Step 2: Create a sign chart for fโฒ(x)
Mark critical points on a number line
Test the sign of fโฒ(x) in each interval
Step 3: Analyze the sign changes
Positive โ Negative = Local MAX
Negative โ Positive = Local MIN
No change = Neither
Step 4: State your conclusions
Example 1: Complete Analysis
Classify the critical points of f(x)=x3โ3x2โ9x+5
Step 1: Find critical points
fโฒ(x)=3x2โ6xโ9=3(x2โ2xโ3)=3(xโ3)(x+1)
Critical points: x=โ1 and x=3
Step 2: Create sign chart
Test intervals: (โโ,โ1), (โ1,3), (3,โ)
Interval (โโ,โ1): Test x=โ2fโฒ(โ2)=3(โ2โ3)(โ2+1)=3(โ5)(โ1)=15>0 โ POSITIVE
Interval (โ1,3): Test x=0fโฒ(0)=3(0โ3)(0+1)=3(โ3)(1)=โ9<0 โ NEGATIVE
Interval (3,โ): Test x=4fโฒ(4)=3(4โ3)(4+1)=3(1)(5)=15>0 โ POSITIVE
Step 3: Sign chart
-1 3
++++ | ---- | ++++
โผ โฒ
Step 4: Conclusions
At x=โ1: fโฒ goes from + to โ โ LOCAL MAXIMUM
The First Derivative Test also tells us where f is increasing or decreasing:
From the Sign Chart
Where fโฒ(x)>0: f is increasing
Where fโฒ(x)<0: f is decreasing
Example from above:
f is increasing on (โโ,โ1) and (3,โ)
f is decreasing on (โ1,3)
Special Cases
Case 1: Inflection Point with Horizontal Tangent
If fโฒ(c)=0 but the sign doesn't change, it's usually an inflection point.
Example: f(x)=x3 at x=0
fโฒ(x)=3x2, so fโฒ(0)=0
fโฒ(x)>0 for all x๎ =0 (no sign change!)
x=0 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: f(x)=โฃxโฃ at x=0
fโฒ(x)<0 for x<0
fโฒ(x)>0 for x>0
Changes from โ to + โ local minimum at x=0 โ
โ ๏ธ Common Mistakes
Mistake 1: Testing AT the Critical Point
โ Don't test fโฒ(c) where c is the critical point
โ Test points on either SIDE of c
Mistake 2: Wrong Sign Interpretation
The sign of fโฒ, not f, 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 fโฒ(x) completely to make sign analysis easier!
First Derivative Test vs. Other Methods
When to Use the First Derivative Test
โ Always works (if derivative exists on either side)
โ Good for understanding overall behavior
โ Tells you increasing/decreasing intervals
โ Works when second derivative is hard to compute
When to Use the Second Derivative Test (coming next!)
โ Faster for simple cases
โ Only need to evaluate at one point
โ Fails if fโฒโฒ(c)=0
โ Doesn't give increasing/decreasing info
Quick Reference Chart
Sign Change
Result
+ โ โ
Local MAX โฌ๏ธโฌ๏ธ
โ โ +
Local MIN โฌ๏ธโฌ๏ธ
+ โ +
Neither (increasing through)
โ โ โ
Neither (decreasing through)
๐ Practice Strategy
Find and factorfโฒ(x) completely
Mark critical points on a number line
Test one point in each interval (use easy numbers)
Record signs carefully: + for positive, โ for negative
Look for sign changes: + to โ is max, โ to + is min
State intervals: Where increasing? Where decreasing?
Calculate f values at extrema for complete answer
4
x3
๐ก Show Solution
Step 1: Find the derivative
f(x)=x4โ4x3
fโฒ(x)=4x3โ12x2=4x2(xโ
Step 2: Find critical points
4x2(xโ3)=0
x=0 or x=3
Step 3: Create sign chart
Test intervals: (โโ,0), (0,3), (3,โ)
Test x=โ1 (in (โโ,0)):
โ NEGATIVE
Test x=1 (in (0,3)):
f โ NEGATIVE
Test x=4 (in (3,โ)):
f โ POSITIVE
Step 4: Analyze sign chart
0 3
---- | ---- | ++++
โฒ
At x=0: โ to โ (no change) โ NEITHER (inflection point)
At x=3: โ to + โ LOCAL MINIMUM
Step 5: Calculate the minimum value
f(3)=34โ4(3)3=81โ
Answer: Local minimum of โ27 at x=3. No local maximum. (x=0 is an inflection point with horizontal tangent.)
2Problem 2hard
โ Question:
Find the intervals where g(x)=x+x4โ is increasing and decreasing. Identify all local extrema.
๐ก Show Solution
Step 1: Find the derivative
g(x)=x+4xโ1
3Problem 3expert
โ Question:
Use the First Derivative Test on h(x)=x2/3(xโ4) to find and classify all critical points.
๐ก Show Solution
Step 1: Expand and find derivative
h(x)=x2/3(xโ4)=
4Problem 4medium
โ Question:
Use the first derivative test to classify the critical point at x = 2 for f(x) = xยณ - 6xยฒ + 9x.
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 2: Find critical points:
f'(x) = 0 when 4xยฒ(x - 3) = 0
x = 0 or x = 3
Step 3: Test around x = 0:
Interval (-1, 0): f'(-0.5) = 4(-0.5)ยฒ(-0.5 - 3) = 4(0.25)(-3.5) < 0
Interval (0, 1): f'(0.5) = 4(0.5)ยฒ(0.5 - 3) = 4(0.25)(-2.5) < 0
f' is negative on both sides โ x = 0 is NEITHER max nor min
Step 4: Test around x = 3:
Interval (2, 3): f'(2.5) = 4(2.5)ยฒ(2.5 - 3) = 4(6.25)(-0.5) < 0
Interval (3, 4): f'(3.5) = 4(3.5)ยฒ(3.5 - 3) = 4(12.25)(0.5) > 0
f' changes from - to + at x = 3 โ x = 3 is LOCAL MINIMUM
Answer: x = 0 is neither (inflection point), x = 3 is local minimum
Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.
9
(
3
)
+
5=
27โ
27โ
27+
5=
โ22
3)
fโฒ
(
โ
1
)
=
4(โ1)2(โ1โ
3)=
4(1)(โ4)=
โ16<
0
โฒ
(
1
)
=
4(1)2(1โ
3)=
4(1)(โ2)=
โ8<
0
โฒ
(
4
)
=
4(4)2(4โ
3)=
4(16)(1)=
64>
0
108
=
โ27
gโฒ(x)=1โ4xโ2=1โx24โ
Step 2: Find critical points
Set gโฒ(x)=0:
1โx24โ=0
x24โ=1
x2=4
x=ยฑ2
Also, gโฒ(x) is undefined at x=0, but g(0) is also undefined, so x=0 is NOT a critical point (not in domain).
Critical points: x=โ2 and x=2
Step 3: Create sign chart
Test intervals: (โโ,โ2), (โ2,0), (0,2), (2,โ)
Note: x=0 is not in the domain but divides the intervals.
Test x=โ3:
gโฒ(โ3)=1โ94โ=95โ>0 โ POSITIVE
Test x=โ1:
gโฒ(โ1)=1โ14โ=โ3<0 โ NEGATIVE
Test x=1:
gโฒ(1)=1โ14โ=โ3<0 โ NEGATIVE
Test x=3:
gโฒ(3)=1โ94โ=95โ>0 โ POSITIVE
Step 4: Sign chart
-2 0 2
++++ | -- | -- | ++++
โผ โฒ
Step 5: Classify critical points
At x=โ2: + to โ โ LOCAL MAXIMUM
At x=2: โ to + โ LOCAL MINIMUM
Calculate values:
g(โ2)=โ2+โ24โ=โ2โ2=โ4
g(2)=2+24โ=2+2
Answer:
Increasing on (โโ,โ2) and (2,โ)
Decreasing on (โ2,0) and (0,2)
Local maximum of โ4 at x=โ2
Local minimum of 4 at x=2
x
5/3
โ
4x2/3
hโฒ(x)=35โx2/3โ4โ 32โxโ1/3
hโฒ(x)=35โx2/3โ38โxโ1/3
Step 2: Combine using common denominator
hโฒ(x)=3x1/35x2/3โ x1/3โ8โ=3x1/35xโ8โ
Step 3: Find critical points
hโฒ(x)=0: 5xโ8=0 โ x=58โ
hโฒ(x) undefined: x1/3=0 โ x=0
Critical points: x=0 and x=58โ
Step 4: Test signs
Test x=โ1 (in (โโ,0)):
hโฒ(โ1)=3(โ1)1/35(โ1)โ8โ= โ POSITIVE
Test x=1 (in (0,58โ)):
hโฒ(1)=3(1)1/35(1)โ8โ= โ NEGATIVE
Test x=2 (in (58โ,โ)):
hโฒ(2)=3(2)1/35(2)โ8โ= โ POSITIVE
Step 5: Sign chart
0 8/5
++++ | ---- | ++++
โผ โฒ
At x=0: + to โ โ LOCAL MAXIMUM
At x=58โ: โ to + โ LOCAL MINIMUM
Step 6: Calculate values
h(0)=02/3(0โ4)=0
h(58โ)=(58โ)2/3(58โโ4)
=(58โ)2/3(โ512โ)
=โ512โโ 52/34โโโ3.07
Answer:
Local maximum of 0 at x=0
Local minimum of โ512โ(58โ)2/3โโ3.07 at x=58โ