# First derivative test for relative extrema

> AP Calculus BC · Unit 5: Analytical Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u5-first-derivative-test-for-relative/

This module covers critical point identification, sign analysis of the first derivative, classification of interior and endpoint relative extrema, and application to common function types for AP Calculus BC exam preparation.

**Prerequisites:** Definition of the derivative; Computing derivatives of common functions; Definition of relative (local) extrema

## Learning objectives

- Identify critical points of a function on a given domain
- Classify interior critical points as relative maxima, minima, or neither using the First Derivative Test
- Classify endpoint relative extrema on closed intervals
- Justify extrema classifications for AP Calculus BC FRQs

## Core Concepts of the First Derivative Test

The First Derivative Test (also called the test for local/relative extrema) is a core method from AP Calculus BC Unit 5, appearing on both multiple-choice and free-response sections, accounting for 4-7% of total exam score.

The test uses the geometric meaning of the first derivative: $f'(x)$ tells us if $f(x)$ is increasing ($f'(x) > 0$) or decreasing ($f'(x) < 0$). By analyzing how the sign of $f'(x)$ changes around a critical point, we can classify if the point is a relative maximum, minimum, or neither.

**Critical Point** — An interior point $c$ of the domain of $f$ where either $f'(c) = 0$ or $f'(c)$ does not exist. Critical points are the only possible locations for relative extrema.

*Notation:* $c \in \text{dom}(f)$, $c$ interior

> **tip**
>
> Unlike the Second Derivative Test, the First Derivative Test works for all critical points, including those where the second derivative is zero or undefined, making it more broadly applicable.

> **Exam tip:** On AP FRQs, you must explicitly mention the sign change of the first derivative to earn full justification credit for classifying extrema.

## Sign Analysis for Interior Critical Points

To apply the First Derivative Test to an interior critical point, split the domain of $f'$ into intervals separated by critical points, test the sign of $f'(x)$ at a point inside each interval, then use the sign change rules below to classify:

1. If $f'$ changes from positive to negative at $c$, $f$ has a **relative maximum** at $x=c$
2. If $f'$ changes from negative to positive at $c$, $f$ has a **relative minimum** at $x=c$
3. If $f'$ does not change sign at $c$, $f$ has **no relative extremum** at $x=c$

**Worked example:** Find all critical points of $f(x) = x^3 - 6x^2 + 9x + 2$ and classify each using the First Derivative Test.

1. Compute the first derivative:

   $$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$$
2. Identify critical points: $f'(x)$ is defined for all real $x$, so set $f'(x)=0$ to get critical points $x=1$ and $x=3$, both interior to the domain of $f$.
3. Sign analysis: Split the domain into $(-\infty, 1)$, $(1, 3)$, $(3, \infty)$. Test a point in each interval:
4. - At $x=0$: $f'(0)=9>0$<br>- At $x=2$: $f'(2)=-3<0$<br>- At $x=4$: $f'(4)=9>0$
5. Classify based on sign change: At $x=1$, $f'$ changes from positive to negative → relative maximum at $(1, 6)$. At $x=3$, $f'$ changes from negative to positive → relative minimum at $(3, 2)$.

> **Exam tip:** Stating only that $f'(c)=0$ is not sufficient justification for an extremum on FRQs.

## Critical Points Where $f'$ is Undefined

Critical points can occur where $f'(c)$ is undefined, as long as $c$ is in the domain of $f$. The First Derivative Test follows exactly the same process for these points as it does for points where $f'(c)=0$. This scenario is common for absolute value, root, and piecewise functions, which appear regularly on AP exams.

> **warning**
>
> If $c$ is not in the domain of $f$, it cannot be a critical point or a relative extremum, so you must discard it immediately.

**Worked example:** Let $f(x) = x^{2/3}(x - 4)$. Find all critical points and classify each using the First Derivative Test.

1. Compute $f'(x)$ using the product rule:

   $$f'(x) = \frac{2}{3}x^{-1/3}(x - 4) + x^{2/3} = \frac{5x - 8}{3x^{1/3}}$$
2. Identify critical points: $f'(x)$ is undefined at $x=0$, and $f(0)=0$ (so $x=0$ is in the domain of $f$), making it a valid critical point. Setting the numerator equal to zero gives $x=8/5$, where $f'=0$ and $f$ is defined, so $x=8/5$ is also a critical point.
3. Sign analysis across intervals $(-\infty, 0)$, $(0, 8/5)$, $(8/5, \infty)$:
4. - At $x=-1$: $f'(-1) = \frac{-13}{-3} > 0$<br>- At $x=1$: $f'(1) = \frac{-3}{3} < 0$<br>- At $x=2$: $f'(2) = \frac{2}{3 \cdot 2^{1/3}} > 0$
5. Classify: At $x=0$, $f'$ changes from positive to negative → relative maximum at $(0, 0)$. At $x=8/5$, $f'$ changes from negative to positive → relative minimum at $(1.6, \approx -3.28)$.

> **Exam tip:** Always confirm that the point where $f'$ is undefined is in the domain of $f$ before calling it a critical point.

## First Derivative Test for Endpoint Extrema

On a closed interval $[a, b]$, endpoints $x=a$ and $x=b$ can be relative extrema, since we only consider function values inside the interval when defining relative extrema. The First Derivative Test extends naturally to endpoints by only checking the sign of $f'$ on the interior side of the endpoint:

- Left endpoint $x=a$: If $f'(x) < 0$ just right of $a$, $f(a)$ is a relative maximum. If $f'(x) > 0$ just right of $a$, $f(a)$ is a relative minimum.
- Right endpoint $x=b$: If $f'(x) > 0$ just left of $b$, $f(b)$ is a relative maximum. If $f'(x) < 0$ just left of $b$, $f(b)$ is a relative minimum.

**Worked example:** Find all relative extrema of $f(x) = 2x^3 - 3x^2$ on the closed interval $[-1, 2]$, including endpoints.

1. Find interior critical points: $f'(x) = 6x^2 - 6x = 6x(x-1)$, so interior critical points at $x=0$ and $x=1$.
2. Sign analysis for interior points: Intervals $(-1, 0)$, $(0, 1)$, $(1, 2)$. Testing gives $f'(-0.5) > 0$, $f'(0.5) < 0$, $f'(1.5) > 0$. So $x=0$ is a relative maximum, $x=1$ is a relative minimum.
3. Classify endpoints: Left endpoint $x=-1$ has $f'(x)$ positive just right of $-1$, so $f$ increases away from $x=-1$ → relative minimum at $(-1, -5)$. Right endpoint $x=2$ has $f'(x)$ positive just left of $2$, so $f$ increases towards $x=2$ → relative maximum at $(2, 4)$.
4. Final result: Relative minima at $(-1, -5)$ and $(1, -1)$; relative maxima at $(0, 0)$ and $(2, 4)$.

> **Exam tip:** When asked to find all relative extrema on a closed interval, don't forget to classify endpoints—this is a common AP exam point trap.

## AP-Style Additional Worked Examples

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. Let $f(x) = \sin x + \cos x$ on the interval $[0, 2\pi]$. How many relative extrema does $f$ have on this interval, including endpoints?

   - 1
   - 2
   - 3
   - 4

   *Why:* Correct: There are 2 interior extrema plus 2 endpoint extrema, for a total of 4. If you got 2, you forgot to classify endpoints.

**Worked example:** Let $f(x) = \frac{x^2}{x - 2}$.
(a) Find all critical points of $f$. Justify your answer.
(b) Classify each critical point as a relative maximum, relative minimum, or neither using the First Derivative Test. Justify your answer.
(c) Identify all relative extrema of $f$ on the interval $[0, 4]$, including endpoints.

1. (a) Use the quotient rule to compute the first derivative:

   $$f'(x) = \frac{(2x)(x-2) - x^2(1)}{(x-2)^2} = \frac{x(x-4)}{(x-2)^2}$$
2. The domain of $f$ excludes $x=2$, so even though $f'$ is undefined at $x=2$, it is not a critical point. $f'(x)=0$ at $x=0$ and $x=4$, both in the domain of $f$, so critical points are $x=0$ and $x=4$.
3. (b) The denominator $(x-2)^2$ is always positive for $x \neq 2$, so the sign of $f'$ matches the sign of $x(x-4)$. Testing intervals: $f'>0$ for $x<0$, $f'<0$ for $0<x<2$ and $2<x<4$, and $f'>0$ for $x>4$. At $x=0$, $f'$ changes from positive to negative → relative maximum. At $x=4$, $f'$ changes from negative to positive → relative minimum.
4. (c) On $[0,4]$, endpoints are $x=0$ and $x=4$, which we already classified, and $x=2$ is not in the domain. The relative extrema are: relative maximum at $(0, 0)$ and relative minimum at $(4, 8)$.

**Worked example:** A small business models its daily profit from selling $x$ units of a product as $P(x) = -0.001x^3 + 0.12x^2 + 3x - 80$, where $P(x)$ is measured in dollars, and $0 \leq x \leq 100$. Use the First Derivative Test to find the production level that gives a relative maximum daily profit, and interpret your result.

1. Compute the first derivative:

   $$P'(x) = -0.003x^2 + 0.24x + 3$$
2. Find critical points: Set $P'(x)=0$, which simplifies to $x^2 - 80x - 1000 = 0$. Solving with the quadratic formula gives one positive critical point at $x \approx 91$, which is inside the interval $[0, 100]$.
3. Sign analysis: Test $x=10$: $P'(10) = 5.1 > 0$. Test $x=95$: $P'(95) \approx -1.28 < 0$.
4. Classification: $P'$ changes from positive to negative at $x \approx 91$, so this is a relative maximum. $P(91) \approx 482$ dollars.
5. Interpretation: Producing approximately 91 units per day gives the business a relative maximum daily profit of roughly \$482.

## Common pitfalls

- **Wrong:** Calling $x=c$ a critical point where $f'(c)$ is undefined but $f(c)$ is also undefined.
  - Why it fails: Students automatically mark any point where $f'$ is undefined as a critical point, without checking the domain of $f$.
  - Correct: For any point where $f'$ is undefined, confirm $f(c)$ exists before marking it as a critical point to test.
- **Wrong:** Justifying a relative extremum only by stating $f'(c) = 0$, without mentioning the sign change of $f'$.
  - Why it fails: Students confuse the condition for a critical point with the justification for an extremum.
  - Correct: On all FRQ problems, explicitly state the sign change of the first derivative around $c$ to earn full justification credit.
- **Wrong:** Forgetting to check for relative extrema at endpoints when working on a closed interval.
  - Why it fails: Students are taught critical points are interior points, so they ignore endpoints entirely when classifying relative extrema.
  - Correct: When the domain is given as a closed interval, always add the endpoints to your list of points to classify using the extended First Derivative Test.
- **Wrong:** Concluding no sign change because the test value of $f'$ is zero.
  - Why it fails: Students sometimes accidentally test a critical point instead of a point inside the interval, getting a zero derivative and incorrectly concluding no sign change.
  - Correct: When testing an interval between two critical points, always pick a test point strictly inside the interval, not equal to either critical point.
- **Wrong:** Using the sign of $f(test\ point)$ instead of the sign of $f'(test\ point)$ to classify extrema.
  - Why it fails: Students mix up function and derivative values when working quickly on exam day.
  - Correct: Always explicitly compute the sign of $f'(test\ point)$, not $f(test\ point)$, during sign analysis.

## Cheatsheet

| Category | Rule | Notes |
| --- | --- | --- |
| Interior Critical Point | $c$ is critical if: $f'(c)=0$ **or** ($f'(c)$ undefined and $f(c)$ defined) | Only interior domain points count as critical points |
| Interior Relative Maximum | $f'(x) > 0$ left of $c$, $f'(x) < 0$ right of $c$ | Works for all critical points, including where $f'(c)$ is undefined |
| Interior Relative Minimum | $f'(x) < 0$ left of $c$, $f'(x) > 0$ right of $c$ | Same broad applicability as the maximum rule |
| No Relative Extremum | $f'(x)$ has the same sign on both sides of $c$ | Common for horizontal inflection points like $f(x)=x^3$ at $x=0$ |
| Left Endpoint Relative Maximum | $f'(x) < 0$ just right of $a$ | Function decreases away from the endpoint |
| Left Endpoint Relative Minimum | $f'(x) > 0$ just right of $a$ | Function increases away from the endpoint |
| Right Endpoint Relative Maximum | $f'(x) > 0$ just left of $b$ | Function increases toward the endpoint |
| Right Endpoint Relative Minimum | $f'(x) < 0$ just left of $b$ | Function decreases toward the endpoint |

## What's next

The First Derivative Test is a foundational tool for all further work with extrema and curve sketching in AP Calculus BC. Mastering sign analysis and justification for extrema classifications is required to earn full credit on many free-response questions that make up a large portion of your exam score. Immediately after mastering this topic, you will move on to the Second Derivative Test for extrema, an alternative method for classifying critical points where the second derivative exists and is non-zero. You will also apply the First Derivative Test to find absolute extrema on closed intervals, a common FRQ topic, and build on this knowledge for curve sketching and optimization problems. This topic also forms the basis for analyzing motion in parametric and polar contexts later in the course.

- [Candidates Test for Absolute Extrema](https://www.owlsprep.com/study/ap-calculus-bc-u5-candidates-test-for-absolute-extrema/)
- [Determining Concavity](https://www.owlsprep.com/study/ap-calculus-bc-u5-determining-concavity/)
- [Second Derivative Test](https://www.owlsprep.com/study/ap-calculus-bc-u5-second-derivative-test/)

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