# First Derivative Test for Relative Extrema

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

This subtopic covers how to use sign changes of the first derivative around critical points to classify them as relative maxima, minima, or neither. You will learn to handle all types of critical points, including those with undefined derivatives, for AP Calculus AB exam problems.

**Prerequisites:** [Computing first derivatives of common elementary functions](https://www.owlsprep.com/study/ap-calculus-ab-u2-basic-differentiation/); [Applying product, quotient, and chain differentiation rules](https://www.owlsprep.com/study/ap-calculus-ab-u3-differentiation-rules/); Testing the sign of a function on an open interval

## Learning objectives

- Find all critical points of a function on a domain
- Classify critical points using the sign change of the first derivative
- Handle critical points with undefined derivatives
- Apply the first derivative test to polynomial, rational, and contextual functions

## What is the First Derivative Test for Relative Extrema?

The First Derivative Test is a core technique in Unit 5 of AP Calculus AB, which makes up 15–18% of your total exam score. It is tested in both multiple-choice and free-response questions, often paired with curve sketching and optimization. The test uses the sign change of the first derivative around a critical point to classify the point.

**Relative (Local) Extremum** — A point $x=c$ where $f(c)$ is the maximum or minimum value of $f(x)$ for all $x$ in an open interval around $c$. 'Relative' and 'local' are interchangeable on the AP exam.

*Example:* A hill on a roller coaster is a relative maximum, even if there is a taller hill elsewhere on the track.

> **info**
>
> Unlike the Second Derivative Test, which only works at differentiable points with non-zero second derivatives, the First Derivative Test works for all valid critical points, making it more broadly applicable for most AP problems.

## Critical Points and Sign Interval Setup

Before applying the First Derivative Test, you must complete two preliminary steps: find all critical points, then split the domain into intervals separated by critical points to test the sign of $f'(x)$. Fermat's Theorem confirms all relative extrema occur at critical points, so no other locations need testing.

**Critical Point** — A point where $f$ is defined at $c$, and either $f'(c) = 0$ (called a stationary point) or $f'(c)$ is undefined.

*Notation:* $x=c$

*Example:* $x=0$ is a critical point of $f(x)=|x|$, since $f'(0)$ is undefined but $f(0)$ is defined.

**Worked example:** Find all critical points of $f(x) = 2x^3 - 15x^2 + 36x + 4$ and create sign intervals for the first derivative.

1. Compute the first derivative, which is a polynomial defined for all real $x$:

   $$f'(x) = 6x^2 - 30x + 36$$
2. Find critical points by setting $f'(x) = 0$ and factoring:

   $$6x^2 - 30x + 36 = 0 \implies 6(x-2)(x-3) = 0$$
3. Critical points are at $x=2$ and $x=3$. Sort them to split the domain $(-\infty, \infty)$ into three open intervals: $(-\infty, 2)$, $(2, 3)$, $(3, \infty)$.
4. Test the sign of $f'(x)$ in each interval by picking a test point inside the interval:
- For $(-\infty, 2)$, test $x=0$: $f'(0) = 36 > 0$
- For $(2, 3)$, test $x=2.5$: $f'(2.5) = -1.5 < 0$
- For $(3, \infty)$, test $x=4$: $f'(4) = 12 > 0$

> **Exam tip:** Always sort your critical points from smallest to largest before creating intervals. Skipping this step often leads to testing the wrong interval and misclassifying extrema on the AP exam.

## First Derivative Test Classification Rules

The classification rules follow directly from the relationship between derivative sign and function behavior: a positive first derivative means the function is increasing, and a negative first derivative means the function is decreasing. The sign change pattern as you move left to right across $x=c$ tells you the classification:

- If $f'(x)$ changes from **positive to negative**: $f(c)$ is a **relative maximum**
- If $f'(x)$ changes from **negative to positive**: $f(c)$ is a **relative minimum**
- If $f'(x)$ does not change sign: there is **no relative extremum** at $x=c$
- These rules work for all critical points, whether $f'(c)=0$ or $f'(c)$ is undefined

**Worked example:** Using the sign data from $f(x) = 2x^3 - 15x^2 + 36x + 4$, classify each critical point and find the coordinates of all relative extrema.

1. Recall the sign pattern: $f'(x) > 0$ on $(-\infty, 2)$, $f'(x) < 0$ on $(2, 3)$, $f'(x) > 0$ on $(3, \infty)$.
2. Classify $x=2$: $f'(x)$ changes from positive to negative moving left to right, so this is a relative maximum. Calculate $f(2)$:

   $$f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 4 = 32$$
3. Classify $x=3$: $f'(x)$ changes from negative to positive, so this is a relative minimum. Calculate $f(3)$:

   $$f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 4 = 31$$
4. The relative extrema are $(2, 32)$ (maximum) and $(3, 31)$ (minimum).

> **Exam tip:** Always compute the full function value for the extremum if the question asks for the point or value. The AP exam will dock points if you only give the x-coordinate when the question asks for the full extremum.

## Classifying Critical Points with Undefined Derivatives

The First Derivative Test works exactly the same for critical points where $f'(x)$ is undefined (corner points, cusps, vertical tangents), as long as the original function $f(x)$ is defined at the critical point. A common mistake is omitting these critical points from your list, which AP exam writers regularly test for.

**Worked example:** Classify all relative extrema of $f(x) = x^{1/3}(x - 4)$.

1. Compute the first derivative using the product rule and simplify:

   $$f'(x) = \frac{1}{3}x^{-2/3}(x - 4) + x^{1/3} = \frac{4(x - 1)}{3x^{2/3}}$$
2. Find all critical points: $f'(x)=0$ when the numerator is zero, so $x=1$. $f'(x)$ is undefined when the denominator is zero at $x=0$, and $f(0)=0$ is defined, so $x=0$ is also a critical point.
3. Sort critical points to get intervals: $(-\infty, 0)$, $(0, 1)$, $(1, \infty)$. The denominator $3x^{2/3}$ is always positive for $x \neq 0$, so sign only depends on the numerator. The sign pattern is: $f'(x) < 0$ on $(-\infty, 0)$, $f'(x) < 0$ on $(0, 1)$, $f'(x) > 0$ on $(1, \infty)$.
4. Classify each critical point: At $x=0$, $f'(x)$ does not change sign, so there is no extremum. At $x=1$, $f'(x)$ changes from negative to positive, so this is a relative minimum at $(1, -3)$.

> **Exam tip:** Never forget to add critical points where $f'(x)$ is undefined to your list. AP exam writers regularly include these to test if you remember all types of critical points.

## AP-Style Concept Check

**Check your understanding**

Use the First Derivative Test to answer the following multiple-choice question:

1. Let $f'(x) = (x+3)(x-1)(x-5)$. Which of the following correctly lists all values of $x$ where $f(x)$ has a relative maximum?

   - A) $x=-3$ only
   - B) $x=1$ only
   - C) $x=-3$ and $x=5$
   - D) $x=1$ and $x=5$

   *Answer:* B) $x=1$ only

   *Why:* Critical points sorted are $x=-3, 1, 5$. The sign pattern is $(-\infty,-3): -, (-3,1): +, (1,5): -, (5, \infty): +$. A relative maximum requires a positive to negative sign change, which occurs only at $x=1$.

## Common pitfalls

- **Wrong:** Calling any critical point automatically a maximum or minimum without testing sign change
  - Why it fails: Students confuse 'possible location of an extremum' with 'guaranteed extremum', as most introductory examples have extrema at every stationary point.
  - Correct: Always test the sign of $f'(x)$ on both sides of every critical point to confirm a sign change occurs.
- **Wrong:** Omitting critical points where $f'(x)$ is undefined from your interval setup
  - Why it fails: Students only look for solutions to $f'(x)=0$ and ignore zeros in the denominator of the derivative.
  - Correct: After computing $f'(x)$, always check where $f'(x)$ is undefined, and if $f(x)$ is defined at that point, add it to your list of critical points.
- **Wrong:** Reversing the classification rule (calling positive-to-negative change a minimum and negative-to-positive a maximum)
  - Why it fails: Students mix up the direction when moving left to right across the critical point.
  - Correct: Sketch a tiny increasing/decreasing line segment next to each interval to visualize: increasing then decreasing is a peak (maximum), decreasing then increasing is a valley (minimum).
- **Wrong:** Using a critical point as the test value for an adjacent interval
  - Why it fails: Students rush and pick the interval endpoint as the test point, which gives $f'(c)=0$ or undefined, so no valid sign.
  - Correct: Always pick a test value strictly inside the open interval to get a valid sign result.
- **Wrong:** Classifying a domain endpoint as a relative extremum using the First Derivative Test
  - Why it fails: Students forget you need derivative values on both sides of the point to check for a sign change.
  - Correct: Only classify interior critical points of the domain as relative extrema; endpoints can only be absolute extrema per AP CED conventions.
- **Wrong:** Rounding messy test values early and getting an incorrect sign
  - Why it fails: Students approximate irrational test values and round a small negative number to zero or a positive number.
  - Correct: Keep test values in exact factored form to determine the sign, do not approximate until after you confirm the sign.

## Cheatsheet

| Category | Rule/Definition | Notes |
| --- | --- | --- |
| Critical Point | $x=c$ is critical if $f$ defined at $c$, $f'(c)=0$ or $f'(c)$ undefined | Only critical points can be relative extrema |
| Relative Maximum | $f'(x)$ changes from $+$ to $-$ (left to right) | Works for stationary and undefined derivative points |
| Relative Minimum | $f'(x)$ changes from $-$ to $+$ (left to right) | Same broad applicability as maximum rule |
| No Extremum | $f'(x)$ does not change sign across $c$ | Common for stationary inflection points like $f(x)=x^3$ at $x=0$ |
| Endpoint Rule | Domain endpoints cannot be relative extrema | No derivative on both sides to test sign change |
| Sign Test Method | Pick any test point strictly inside the interval | Sign at test point holds for the entire interval |
| Test Scope | Works for all interior critical points | More general than the Second Derivative Test |

## What's next

Mastering the First Derivative Test is a non-negotiable prerequisite for all remaining topics in Unit 5, which make up a large portion of your AP Calculus AB exam score. This topic lays the groundwork for the Second Derivative Test, a complementary technique that speeds up classification of stationary points, and for analyzing concavity and inflection points. It is also required for curve sketching, finding absolute extrema, and solving optimization problems, which are common high-weight free-response questions on the exam. This topic also connects to particle motion in Unit 4, where identifying direction changes of a particle is just an application of the First Derivative Test to position functions. Without correctly finding and classifying relative extrema, you will struggle to solve many of these problems.

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

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