AP® Calculus AB review sheet from Aim for Five (aimforfive.com/calc-ab/units/5/5-4)
Unit 5 · Topic 5.4
5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
The First Derivative Test decides what happens at a critical point by checking how f′ changes sign there. Positive to negative is a relative maximum, negative to positive is a relative minimum, and no sign change means neither. On the exam, the sign change is the justification.
Key terms
- First Derivative Test
- relative maximum
- relative minimum
- sign change
- critical point
The test
Let c be a critical point of a function f that is continuous at c.
- If f′ changes from positive to negative at c, f has a relative maximum at c. (Rising, then falling: a peak.)
- If f′ changes from negative to positive at c, f has a relative minimum at c. (Falling, then rising: a valley.)
- If f′ doesn't change sign at c, f has no relative extremum at c.
Why it works
If f is increasing just left of c and decreasing just right of c, then f(c) is higher than all nearby values. That's exactly a relative max. The same reasoning flipped gives a min. The test works even where f′(c) doesn't exist, like at a corner, as long as f is continuous at c.
Using a graph of f′
A very common exam setup gives only the graph of f′. Then relative extrema of f are where the graph of f′ crosses the x-axis:
| Graph of f′ at x = c | What f does at c |
|---|---|
| Crosses from above to below the x-axis | Relative maximum |
| Crosses from below to above the x-axis | Relative minimum |
| Touches the x-axis without crossing | No extremum |
From a table of f′ values
Sometimes you only get a few values of f′, like f′(1) = 2, f′(2) = 0 and f′(3) = −1. If f′ is continuous, the change from positive to negative tells you f′ has a zero somewhere between 1 and 3, and the table shows one at x = 2.
But a table can't show what f′ does between the listed points. Without more information, such as f′ being decreasing or a sign chart, you can't be sure f′ is positive on all of (1, 2) and negative on all of (2, 3), so you can't fully justify a maximum at x = 2. If a question tells you the sign of f′ on each side of c, that is enough to apply the test.
Justification language
A full-credit justification names the sign change of f′: “f has a relative maximum at x = −2 because f′ changes from positive to negative there.” Saying “because f′(−2) = 0” is not enough, since f′ = 0 can happen without an extremum.
If the question asks for the value of the max or min, plug c into f to get the y-value.
Relative extrema and endpoints
Textbooks disagree about whether an endpoint of a closed interval can be a relative extremum, so AP questions about relative extrema are written so the answer doesn't depend on it. Focus on the critical points inside the interval for relative extrema. Endpoints always matter for absolute extrema, which is why the Candidates Test in 5.5 checks them.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Classify each critical point
Find and classify the critical points of f(x) = x⁴ − 4x³.
Show the solutionHide the solution
- Step 1: f′(x) = 4x³ − 12x² = 4x²(x − 3). Critical points: x = 0 and x = 3.
- Step 2: x < 0: 4x² > 0 and x − 3 < 0, so f′ < 0.
- Step 3: 0 < x < 3: still 4x² > 0 and x − 3 < 0, so f′ < 0.
- Step 4: x > 3: f′ > 0.
- Step 5: At x = 0, f′ stays negative: no extremum. At x = 3, f′ goes from negative to positive: relative minimum. f(3) = 81 − 108 = −27.
Answer: Relative minimum of −27 at x = 3 (f′ changes from negative to positive). No extremum at x = 0 (f′ doesn't change sign).
- Example 2
Reading signs from a factored f′
f′(x) = (x + 2)(x − 1)²(x − 4). At which x-values does f have a relative maximum or minimum? Justify.
Show the solutionHide the solution
- Step 1: Critical points: x = −2, 1, 4.
- Step 2: x < −2: (−)(+)(−) = positive.
- Step 3: −2 < x < 1: (+)(+)(−) = negative.
- Step 4: 1 < x < 4: (+)(+)(−) = negative.
- Step 5: x > 4: (+)(+)(+) = positive.
Answer: Relative max at x = −2 (f′ changes from positive to negative). Relative min at x = 4 (f′ changes from negative to positive). No extremum at x = 1 (f′ is negative on both sides).
- Example 3
Trap: justifying with f′ = 0 instead of a sign change
A student writes: “f has a relative minimum at x = 3 because f′(3) = 0 and f(3) = −27 is low.” Why won't this earn the justification point, and what should they write?
Show the solutionHide the solution
- Step 1: f′(3) = 0 only says x = 3 is a critical point. It could still be a max, a min or neither (like x = 0 in the first example).
- Step 2: “f(3) is low” compares only one value and isn't a calculus reason.
- Step 3: The test needs the behavior of f′ on both sides.
Answer: Write: “f has a relative minimum at x = 3 because f′ changes from negative to positive at x = 3.”
Common mistakes
- Justifying an extremum with f′(c) = 0 alone.
- Mixing up the sign patterns. Positive to negative is a max (up, then down).
- Forgetting critical points where f′ doesn't exist, which can also be extrema.
On the exam
- This justification appears on almost every exam, usually with a graph or formula of f′. Name the sign change in words.
- When a question asks for absolute extrema on a closed interval, the endpoints must be checked too (5.5). For relative extrema, work with the critical points inside the interval.
Connected topics
- Unit 55.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points
- Unit 55.3 Determining Intervals on Which a Function Is Increasing or Decreasing
- Unit 55.7 Using the Second Derivative Test to Determine Extrema
- Unit 55.9 Connecting a Function, Its First Derivative, and Its Second Derivative
Videos
Check yourself
4 questions on 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema. Pick an answer to see if you got it, and why.
The derivative of a function f is f′(x) = (x + 2)(x − 1)²(x − 4). At which values of x does f have a relative minimum?
The derivative of a function f is given by f′(x) = 3 sin x − x. On the interval 0 < x < 4, at what value of x does f have a relative maximum?
The function f is defined for all x ≠ 0, and f′(x) = (x² − 4)/x. At which values of x does f have a relative minimum?
The function f is differentiable on (1, 5). Also, f′(3) = 0, f′(x) > 0 for 1 < x < 3 and f′(x) < 0 for 3 < x < 5. Which of the following is the best justification that f has a relative maximum at x = 3 ?
0 of 4 answered