Using the First Derivative Test to Determine Relative (Local) Extrema - AP Calculus AB

SECTION 1

Historical Context & Motivation

The quest to find maximum and minimum values of functions has driven mathematical innovation for centuries. Long before the formal language of calculus existed, ancient Greek mathematicians like Euclid and Apollonius solved specific optimization problems using purely geometric reasoning—determining, for instance, the shortest distance from a point to a line or the rectangle of maximum area inscribed in a given perimeter. These problems, while elegant, demanded ad hoc techniques that could not generalize across different functional forms.

The breakthrough came in the seventeenth century when Pierre de Fermat introduced a proto-calculus method of "adequality" to locate extrema. Fermat observed that near a maximum or minimum, the function's value barely changes—an insight that foreshadowed the modern condition f′(c) = 0. Building on Fermat's ideas, Isaac Newton and Gottfried Wilhelm Leibniz independently developed differential calculus in the late 1600s, providing a systematic framework for analyzing rates of change. Their work transformed optimization from a collection of clever geometric tricks into a unified, algorithmic discipline.

c. 300 BCE • Greek Geometric Optimization

Euclid and Apollonius solve specific extremal problems—such as finding shortest distances and optimal areas—using geometric constructions without any concept of a derivative.

1638 • Fermat's Method of Adequality

Fermat develops a technique for finding tangent lines and extrema by comparing function values at nearby points, effectively computing f′(x) = 0 before derivatives were formally defined.

1684–1687 • Newton and Leibniz Formalize Calculus

Leibniz publishes his differential calculus notation (1684) and Newton his Principia (1687). The derivative becomes a rigorous tool for analyzing function behavior, including sign analysis.

18th Century • Euler and the First Derivative Test

Leonhard Euler and later Joseph-Louis Lagrange systematize the classification of critical points by examining whether the derivative changes sign, establishing the First Derivative Test as a standard technique.

The central question the First Derivative Test answers is deceptively simple: given that a function has a critical point at x = c (where f′(c) = 0 or f′(c) is undefined), does the function achieve a local maximum, a local minimum, or neither at that point? Answering this requires examining how the derivative's sign behaves on either side of c—a technique that remains indispensable on the AP Calculus AB exam and throughout higher mathematics.

SECTION 2

Core Principles & Definitions

Before applying the First Derivative Test, you must have a firm grasp of several foundational concepts. The test rests on the relationship between a function's derivative and its monotonic behavior—whether the function is increasing or decreasing on an interval. A function f is increasing on an interval when f′(x) > 0 for all x in that interval, and decreasing when f′(x) < 0. The First Derivative Test exploits precisely this connection: by tracking sign changes in f′, we can determine the nature of each critical point.

Critical Point

A point x = c in the domain of f where f′(c) = 0 or f′(c) does not exist. Critical points are the only candidates for relative extrema.

Relative (Local) Maximum

f has a relative maximum at x = c if f(c) ≥ f(x) for all x sufficiently close to c. The First Derivative Test identifies this when f′ changes from positive to negative at c.

Relative (Local) Minimum

f has a relative minimum at x = c if f(c) ≤ f(x) for all x sufficiently close to c. The First Derivative Test identifies this when f′ changes from negative to positive at c.

Sign Chart of f′

A number-line diagram displaying the sign (+/−) of f′(x) on intervals between critical points. This organizational tool is central to applying the First Derivative Test systematically.

No Sign Change → No Extremum

If f′ does not change sign at a critical point (e.g., positive on both sides), then f has neither a local max nor a local min there. Example: f(x) = x³ at x = 0.

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KEY TAKEAWAY

Think of a hiker walking along a trail profiled by the graph of f. The derivative f′ tells you whether the trail slopes uphill (f′ > 0) or downhill (f′ < 0). A hilltop—a local maximum—is exactly the point where the trail transitions from uphill to downhill. A valley floor—a local minimum—is where the trail goes from downhill to uphill. If the trail flattens momentarily but continues in the same direction (like a brief plateau on a steady ascent), that's a critical point with no sign change and therefore no extremum.

SECTION 3

Visual Explanation

The following diagram illustrates how the sign of the first derivative determines whether a function is increasing or decreasing, and how sign changes at critical points correspond to relative extrema. Study the relationship between the graph of f (top) and the sign chart of f′ (bottom).

The purple curve represents f(x). At critical point c₁, f′ changes from positive to negative (the function goes from rising to falling), producing a local maximum. At c₂, f′ changes from negative to positive, producing a local minimum. At c₃, f′ changes from positive to negative again, producing another local maximum. The sign chart along the bottom encodes this information concisely.

Notice the crucial correspondence: each sign change in the bottom chart maps to a turning point on the curve above. The sign chart is not merely a bookkeeping device—it encodes the complete monotonic structure of the function, telling you on which intervals f increases and decreases. Constructing this chart accurately is the single most important skill when applying the First Derivative Test on the AP exam.

SECTION 4

Mathematical Framework

The First Derivative Test can be stated with full mathematical precision. Suppose f is continuous on an open interval containing c, and c is a critical point of f (meaning f′(c) = 0 or f′(c) does not exist). The test classifies c by examining the sign of f′ in intervals immediately to the left and right of c.

FIRST DERIVATIVE TEST — LOCAL MAXIMUM

If f′(x) > 0 for x < c and f′(x) < 0 for x > c, then f(c) is a relative maximum.

The derivative changes from positive to negative: the function rises to f(c), then falls away. The value f(c) is the highest point in a local neighborhood of c.

FIRST DERIVATIVE TEST — LOCAL MINIMUM

If f′(x) < 0 for x < c and f′(x) > 0 for x > c, then f(c) is a relative minimum.

The derivative changes from negative to positive: the function falls to f(c), then rises away. The value f(c) is the lowest point in a local neighborhood of c.

FIRST DERIVATIVE TEST — NO EXTREMUM

If f′(x) does not change sign at c (both sides + or both sides −), then f(c) is not a relative extremum.

The function either continues to increase through c or continues to decrease through c. Examples include f(x) = x³ at x = 0 (both sides: − then +? No—actually f′(x) = 3x², which is positive on both sides of 0).

Step-by-Step Procedure

Find f′(x). Differentiate the function using standard differentiation rules (power rule, product rule, quotient rule, chain rule, etc.).
Find all critical points. Set f′(x) = 0 and solve. Also identify any x-values in the domain of f where f′(x) is undefined.
Build a sign chart for f′. Place the critical points on a number line. Choose test values in each interval and evaluate the sign of f′ at those points.
Apply the First Derivative Test. At each critical point, determine whether f′ changes from + to − (local max), from − to + (local min), or does not change sign (no extremum).
State the extrema. Report the local extrema as ordered pairs (c, f(c)), substituting each critical point back into the original function.

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AP Exam Tip

On the AP Calculus AB free-response section, you must justify your classification. Simply stating "f has a local max at x = 2" without referencing the sign change of f′ will not earn full credit. Always write something like: "Because f′ changes from positive to negative at x = 2, f has a relative maximum at x = 2 by the First Derivative Test."

SECTION 5

Building and Reading Sign Charts

The sign chart (also called a sign diagram or number-line analysis) is the organizing tool that makes the First Derivative Test systematic and less error-prone. To build one, you place the critical points on a horizontal number line, dividing the domain into intervals. Within each interval, f′ maintains a constant sign—this follows from the Intermediate Value Theorem applied to f′ (which is continuous between critical points for most functions you encounter on the AP exam). You then select one convenient test value from each interval, substitute it into f′, and record whether the result is positive or negative.

Complete sign chart analysis for f(x) = 2x³ − 9x² + 12x. The factored derivative f′(x) = 6(x − 1)(x − 2) has zeros at x = 1 and x = 2. Test values in each interval reveal the sign pattern +, −, +, confirming a local maximum at (1, 5) and a local minimum at (2, 4).

A few practical tips for building sign charts on the AP exam. First, factor f′ completely whenever possible, because a factored form lets you determine the sign of each factor individually—you can then multiply signs to get the overall sign without plugging in numbers. Second, remember that critical points where f′ is undefined (such as cusps or vertical tangents) must also be included on the number line. Third, always verify that each critical point lies in the domain of the original function f; a point where f is not defined cannot be an extremum of f.

SECTION 6

Worked Example

Let us apply the First Derivative Test to a function that requires careful handling: f(x) = x⁴ − 4x³. We will identify all relative extrema.

Find All Relative Extrema of f(x) = x⁴ − 4x³

1
Step 1 — Differentiate f

Using the power rule, f′(x) = 4x³ − 12x². This derivative is defined for all real numbers, so the only critical points come from setting f′(x) = 0.

f′(x) = 4x³ − 12x²
2
Step 2 — Find Critical Points

Set f′(x) = 0: 4x³ − 12x² = 0. Factor out the greatest common factor: 4x²(x − 3) = 0. This yields x = 0 and x = 3. Both values are in the domain of f, so both are critical points.

Critical points: x = 0 and x = 3
3
Step 3 — Build the Sign Chart for f′

The critical points divide the real line into three intervals: (−∞, 0), (0, 3), and (3, ∞). Choose test values: x = −1, x = 1, and x = 4. Evaluate: f′(−1) = 4(−1)²(−1 − 3) = 4(1)(−4) = −16 < 0. f′(1) = 4(1)²(1 − 3) = 4(1)(−2) = −8 < 0. f′(4) = 4(16)(4 − 3) = 64 > 0. So the sign pattern is: −, −, +.

Sign of f′: negative on (−∞, 0), negative on (0, 3), positive on (3, ∞)
4
Step 4 — Apply the First Derivative Test

At x = 0: f′ is negative on both sides (− to −). There is no sign change, so x = 0 is neither a local maximum nor a local minimum. At x = 3: f′ changes from negative to positive (− to +). By the First Derivative Test, f has a relative minimum at x = 3.

x = 0: no extremum. x = 3: relative minimum.
5
Step 5 — Compute the Extremal Value

Find f(3) = (3)⁴ − 4(3)³ = 81 − 108 = −27. Therefore, f has a relative minimum of −27 at x = 3. The point (3, −27) is the only relative extremum of this function.

Relative minimum at (3, −27)

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Why x = 0 Is Not an Extremum

Even though f′(0) = 0, the derivative does not change sign at x = 0. The function is decreasing on both sides of x = 0, so the graph passes through (0, 0) as an inflection-like point (specifically, a point where the curve has a horizontal tangent but continues to fall). This is a common trap on the AP exam: not every critical point is an extremum.

SECTION 7

Strengths, Limitations, and Comparisons

The First Derivative Test is one of two primary methods for classifying critical points that you will encounter in AP Calculus AB—the other being the Second Derivative Test. Each method has distinct advantages and limitations, and understanding when to deploy each one is crucial for both exam efficiency and conceptual depth.

Feature	First Derivative Test	Second Derivative Test
What it requires	Sign of f′ on intervals around c	Value of f″(c)
Works when f′(c) DNE?	Yes — handles cusps, corners, vertical tangents	No — requires f′(c) = 0
Inconclusive case	Never inconclusive (always gives a definitive answer)	Inconclusive when f″(c) = 0
Computational effort	Requires evaluating f′ at multiple test points	Requires computing f″ and evaluating at one point
Also reveals…	Intervals of increase/decrease	Concavity information at c
Best used when	f′ factors easily; f′(c) may not exist; full monotonicity analysis needed	f″ is easy to compute; f′(c) = 0 and f″(c) ≠ 0

Comparison of the First and Second Derivative Tests

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KEY TAKEAWAY

Think of the First Derivative Test as a comprehensive diagnostic—like a full blood panel that always yields a result—while the Second Derivative Test is more like a targeted screening that is quicker when it works but occasionally returns 'inconclusive.' On the AP exam, the First Derivative Test is the safer, more versatile choice; reach for the Second Derivative Test when computing f″ is straightforward and you need speed.

SECTION 8

Connections to Advanced Theory

The First Derivative Test sits at the intersection of several deeper ideas in calculus that extend beyond the AP AB curriculum. Understanding these connections clarifies why the test works and previews the richer landscape of analysis you may encounter in BC Calculus or collegiate mathematics.

AP Calculus AB Concept	Advanced Extension
First Derivative Test (sign changes of f′)	Higher-order derivative tests: when f″(c) = 0, examine f‴(c), f⁽⁴⁾(c), etc., to classify c
Relative (local) extrema on open intervals	Absolute extrema on closed intervals via the Extreme Value Theorem and Candidates Test
Single-variable optimization	Multivariable optimization with gradient vectors and the Hessian matrix (Calculus III)
Sign chart analysis of f′	Mean Value Theorem guarantees f′ attains every intermediate value, justifying constant sign on intervals between zeros

From AP AB to Advanced Calculus

A particularly important connection within the AP AB curriculum itself is the relationship between the First Derivative Test and the Candidates Test (also called the Closed Interval Method). When finding absolute extrema on a closed interval [a, b], you evaluate f at every critical point within (a, b) and at both endpoints a and b, then compare. The First Derivative Test identifies where the local turning points occur, while the Candidates Test determines which of those (or the endpoints) gives the global champion. In optimization word problems—one of the most heavily tested FRQ types—you typically use derivatives to find critical points and then the closed interval context to finalize the answer.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A function f is continuous on (−∞, ∞) and has a critical point at x = 4. If f′(3) = 5 and f′(5) = −2, which of the following must be true?

A. f has an absolute maximum at x = 4
B. f has a relative maximum at x = 4
C. f has a relative minimum at x = 4
D. f has an inflection point at x = 4
E. f′(4) = 0, but no extremum exists at x = 4

Show answer (B)

The correct answer is B. By the First Derivative Test, f′ changes from positive (f′(3) = 5 > 0) to negative (f′(5) = −2 < 0) at x = 4, so f has a relative maximum at x = 4. Choice A is wrong because the First Derivative Test establishes local (relative) extrema, not absolute extrema—the function could attain larger values elsewhere. Choice C reverses the sign-change pattern: a relative minimum requires a change from negative to positive, which is the opposite of what occurs here. Choice D is incorrect because inflection points concern changes in concavity (sign changes of f″), not sign changes of f′. Choice E is wrong because the sign change from + to − definitively indicates an extremum exists.

Question 2: BASIC CALCULATION

Let g(x) = x³ − 6x² + 9x + 1. Find all relative extrema of g using the First Derivative Test.

A. Relative maximum at (1, 5); relative minimum at (3, 1)
B. Relative minimum at (1, 5); relative maximum at (3, 1)
C. Relative maximum at (3, 1) only
D. Relative maximum at (1, 5) only; no relative minimum
E. No relative extrema

Show answer (A)

The correct answer is A. Differentiate: g′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Critical points are x = 1 and x = 3. Test values: g′(0) = 9 > 0, g′(2) = 12 − 24 + 9 = −3 < 0, g′(4) = 48 − 48 + 9 = 9 > 0. The sign pattern is +, −, +. At x = 1, f′ changes + → −, so there is a relative maximum with g(1) = 1 − 6 + 9 + 1 = 5. At x = 3, f′ changes − → +, so there is a relative minimum with g(3) = 27 − 54 + 27 + 1 = 1. Choice B reverses the max/min labels. Choice C misses the extremum at x = 1 and incorrectly labels x = 3 as a maximum. Choice D ignores the sign change at x = 3. Choice E is wrong because there are two sign changes.

Question 3: INTERMEDIATE

Consider h(x) = x²/³(x − 4). Find all critical points and classify each as a relative maximum, relative minimum, or neither using the First Derivative Test.

A. Relative maximum at x = 0; relative minimum at x = 8/5
B. Relative minimum at x = 0; relative maximum at x = 8/5
C. Relative maximum at x = 0; no other extrema
D. Relative minimum at x = 0; no other extrema
E. No extremum at x = 0; relative minimum at x = 8/5

Show answer (A)

The correct answer is A. Begin by expanding h(x) = x²/³(x − 4) = x⁵/³ − 4x²/³. Differentiating term by term, h′(x) = (5/3)x²/³ − (8/3)x⁻¹/³. Factor out (1/3)x⁻¹/³ to obtain h′(x) = (1/3)x⁻¹/³(5x − 8). Critical points occur where h′(x) = 0 or where h′(x) is undefined (provided h is defined there). Setting 5x − 8 = 0 gives x = 8/5. The factor x⁻¹/³ is undefined at x = 0, and since h(0) = 0 is defined, x = 0 is also a critical point. Now construct a sign chart for h′(x) = (1/3)x⁻¹/³(5x − 8) over three intervals. For x < 0, test x = −1: (1/3)(−1)⁻¹/³(5(−1) − 8) = (1/3)(−1)(−13) = 13/3 > 0, so h′ is positive. For 0 < x < 8/5, test x = 1: (1/3)(1)⁻¹/³(5(1) − 8) = (1/3)(1)(−3) = −1 < 0, so h′ is negative. For x > 8/5, test x = 8: (1/3)(8)⁻¹/³(5(8) − 8) = (1/3)(1/2)(32) = 16/3 > 0, so h′ is positive. The sign pattern of h′ is: +, −, +. At x = 0, h′ changes from positive to negative, so by the First Derivative Test h has a relative maximum at x = 0, with h(0) = 0. At x = 8/5, h′ changes from negative to positive, so h has a relative minimum at x = 8/5, with h(8/5) = (8/5)²/³(8/5 − 4) = (8/5)²/³(−12/5), which is negative. Therefore, h has a relative maximum at x = 0 and a relative minimum at x = 8/5.

Question 4: APPLIED

5 pts

A particle moves along the x-axis with position function s(t) = t⁴ − 8t² + 3 for t ≥ 0. Using the First Derivative Test, find all times t > 0 at which the particle changes direction and determine whether the particle's position has a relative maximum or relative minimum at each such time. (a) Find s′(t) and determine all critical points for t > 0. (b) Construct a sign chart for s′(t) on the interval (0, ∞). (c) Apply the First Derivative Test to classify each critical point. (d) State the position of the particle at each extremum.

Show answer

Rubric (5 points): (a) [1 point] s′(t) = 4t³ − 16t = 4t(t² − 4) = 4t(t − 2)(t + 2). For t > 0, the critical points are t = 0 (endpoint, not interior) and t = 2. Since t + 2 > 0 for all t ≥ 0, the only interior critical point for t > 0 is t = 2. Award 1 point for correctly finding s′(t) and identifying t = 2 as the critical point. (b) [1 point] On (0, 2): test t = 1: s′(1) = 4 − 16 = −12 < 0. On (2, ∞): test t = 3: s′(3) = 108 − 48 = 60 > 0. The sign pattern is: negative on (0, 2), positive on (2, ∞). Award 1 point for a correct sign chart with appropriate test values. (c) [1 point] At t = 2, s′ changes from negative to positive, so by the First Derivative Test, s has a relative minimum at t = 2. The particle changes direction at t = 2 (from moving in the negative x-direction to moving in the positive x-direction). Award 1 point for correct classification with justification citing the sign change. (d) [1 point] s(2) = 16 − 32 + 3 = −13. The particle has a relative minimum position of −13 at t = 2. Award 1 point for the correct position value. [1 point] Interpretation/communication: The particle moves in the negative direction for 0 < t < 2, reaches its closest point to the origin (in the negative direction) at t = 2, then reverses and moves in the positive direction for t > 2. Award 1 point for clear communication connecting the derivative analysis to the physical motion.

Question 5: CRITICAL THINKING

4 pts

Let f be a differentiable function defined on all real numbers. The derivative is given by f′(x) = (x − 1)²(x + 3). (a) Find all critical points of f. (b) Use the First Derivative Test to determine whether f has a relative maximum, relative minimum, or neither at each critical point. Justify your answers. (c) Explain why the Second Derivative Test would be inconclusive at one of the critical points and how the First Derivative Test resolves the ambiguity.

Show answer

Rubric (4 points): (a) [1 point] Set f′(x) = (x − 1)²(x + 3) = 0. The solutions are x = 1 (with multiplicity 2) and x = −3. Award 1 point for both critical points. (b) [2 points — 1 point per correct classification with justification] Sign chart: Test x = −4: f′(−4) = (−5)²(−1) = −25 < 0. Test x = 0: f′(0) = (1)²(3) = 3 > 0. Test x = 2: f′(2) = (1)²(5) = 5 > 0. Sign pattern: −, +, +. At x = −3: f′ changes from − to +, so f has a relative minimum at x = −3 by the First Derivative Test. At x = 1: f′ is positive on both sides (+ to +), so f has neither a relative maximum nor a relative minimum at x = 1. Award 1 point for each correctly classified critical point with sign-change justification. (c) [1 point] At x = 1, f′(x) = (x − 1)²(x + 3), so f″(x) = 2(x − 1)(x + 3) + (x − 1)² = (x − 1)[2(x + 3) + (x − 1)] = (x − 1)(3x + 5). Then f″(1) = (0)(8) = 0, so the Second Derivative Test is inconclusive. The First Derivative Test resolves this because it examines the sign behavior of f′ in a neighborhood of x = 1, revealing no sign change (positive on both sides), which definitively shows x = 1 is not an extremum. Award 1 point for explaining the inconclusive nature of the Second Derivative Test at x = 1 and articulating how the First Derivative Test provides a definitive answer.

SUMMARY

Lesson Summary

Summary

The First Derivative Test classifies critical points — values where f′(c) = 0 or f′(c) is undefined — as relative maxima, relative minima, or neither, by examining sign changes of f′ on intervals adjacent to the critical point. When f′ transitions from positive to negative, f achieves a local maximum; when f′ transitions from negative to positive, f achieves a local minimum; and when no sign change occurs, the critical point is not an extremum.

To apply the test systematically, you first differentiate to find f′, then locate all critical points, construct a sign chart by evaluating f′ at test values in each interval, and finally read off the sign changes. The First Derivative Test is more versatile than the Second Derivative Test because it handles cases where f′(c) does not exist and is never inconclusive. On the AP exam, always justify your classification by explicitly stating the sign change; a bare conclusion without reasoning will not earn full credit.