First Derivative Test — Using the First Derivative Test to Determine Relative (Local) Extrema - Calculus 1

SECTION 1

Historical Context & Motivation

Finding the highest and lowest points on a curve is one of the oldest problems in mathematics. Ancient Greek mathematicians like Euclid studied maximum and minimum distances, but they lacked a systematic tool for locating peaks and valleys of arbitrary functions. The development of calculus in the 17th century finally gave mathematicians the language they needed. The First Derivative Test grew directly out of the realization that the slope of a curve tells you whether the function is climbing uphill or sliding downhill — and a change in that direction signals a peak or a valley.

1637 • Fermat's Method of Adequality

Pierre de Fermat developed a technique for finding maxima and minima by setting a quantity analogous to the derivative equal to zero — one of the earliest precursors to the derivative test.

1684 • Leibniz Publishes Calculus

Gottfried Wilhelm Leibniz published his foundational paper on differential calculus, introducing the notation dy/dx and formalizing the concept of the derivative as a rate of change.

1740s • Euler Systematizes Optimization

Leonhard Euler applied derivative-based tests to classify critical points of functions, laying the groundwork for the first and second derivative tests used in modern textbooks.

1797 • Lagrange's Théorie des Fonctions

Joseph-Louis Lagrange presented a rigorous algebraic treatment of derivatives and their role in identifying local extrema, helping establish the analytical framework students use today.

The central question that motivated all of this work remains the same question you'll tackle in this lesson: given a smooth function, how can you determine exactly where its relative maximum and relative minimum values occur, without having to graph every single point?

SECTION 2

Core Principles & Definitions

Before diving into the test itself, you need a few key definitions. A critical number of a function f is any value c in the domain of f where f ʹ(c) = 0 or f ʹ(c) does not exist. Critical numbers are the only candidates where relative extrema can occur. A relative (local) maximum occurs at x = c when f(c) is greater than all nearby function values, and a relative (local) minimum occurs when f(c) is less than all nearby function values.

Critical Numbers

Values of x where f ʹ(x) = 0 or f ʹ(x) is undefined. These are the only places where local extrema can happen.

Sign of f ʹ(x)

When f ʹ(x) > 0, the function is increasing. When f ʹ(x) < 0, the function is decreasing. A sign change at a critical number signals an extremum.

Relative Maximum

f ʹ changes from positive to negative at c. The function rises to the left of c and falls to the right — a hilltop.

Relative Minimum

f ʹ changes from negative to positive at c. The function falls to the left of c and rises to the right — a valley.

No Extremum

If f ʹ does not change sign at c (stays positive or stays negative), then c is a critical number but not an extremum.

✦

KEY TAKEAWAY

Think of the derivative as a road's slope indicator on your car dashboard. When the indicator switches from "uphill" to "downhill," you've just passed over the top of a hill — that's a local maximum. When it switches from "downhill" to "uphill," you've reached the bottom of a valley — that's a local minimum. If the indicator never switches, you're still on the same slope and there's no peak or valley at that point.

SECTION 3

Visual Explanation

The diagram below shows a function with two critical numbers. At the first critical number, the derivative changes from positive (increasing) to negative (decreasing), producing a relative maximum. At the second, the derivative changes from negative to positive, producing a relative minimum. Notice how the tangent lines (shown in color) rotate from positive slope to zero to negative slope around the hilltop, and from negative slope to zero to positive slope around the valley.

The curve rises (f ʹ > 0) to the left of c₁, reaches a relative maximum at c₁ where f ʹ = 0, then falls (f ʹ < 0). It continues falling until c₂, where it reaches a relative minimum, then rises again.

SECTION 4

Mathematical Framework

The First Derivative Test is stated formally as follows. Suppose c is a critical number of a continuous function f. Then we examine the sign of f ʹ(x) on each side of c to classify the critical point.

FINDING CRITICAL NUMBERS

Set f ʹ(x) = 0 or find where f ʹ(x) is undefined

Solve the equation f ʹ(x) = 0 for x. Also check where f ʹ(x) does not exist (such as cusps or corners). Each solution that lies in the domain of f is a critical number.

RELATIVE MAXIMUM CONDITION

f ʹ changes from + to − at c → f(c) is a relative maximum

If f ʹ(x) > 0 for x just to the left of c and f ʹ(x) < 0 for x just to the right of c, the function switches from increasing to decreasing. The point (c, f(c)) is a hilltop.

RELATIVE MINIMUM CONDITION

f ʹ changes from − to + at c → f(c) is a relative minimum

If f ʹ(x) < 0 for x just to the left of c and f ʹ(x) > 0 for x just to the right of c, the function switches from decreasing to increasing. The point (c, f(c)) is a valley.

NO EXTREMUM CONDITION

f ʹ does not change sign at c → no relative extremum at c

If f ʹ(x) keeps the same sign on both sides of c, the function is still increasing (or still decreasing) through c. For example, f(x) = x³ has f ʹ(0) = 0 but no extremum at x = 0.

⚠

Common Pitfall

Not every critical number produces an extremum! The function f(x) = x³ has f ʹ(x) = 3x², so f ʹ(0) = 0 and x = 0 is a critical number. But f ʹ(x) is positive on both sides of 0 (since 3x² ≥ 0), so there is no sign change and no extremum. Always check the sign on both sides.

SECTION 5

Building a Sign Chart

A sign chart (sometimes called a sign diagram) is the practical tool you'll use every time you apply the First Derivative Test. The idea is simple: once you find the critical numbers, you place them on a number line and then test a sample point in each interval to determine whether f ʹ is positive or negative there. The transitions from "+" to "−" and vice versa tell you exactly what type of extremum, if any, occurs at each critical number.

This sign chart analyzes f(x) = x³ − 6x² + 9x + 2, whose derivative factors as 3(x − 1)(x − 3). The critical numbers x = 1 and x = 3 divide the number line into three intervals. Testing one value in each interval reveals the sign of f ʹ and the resulting classification.

Step 1: Find f ʹ(x) and set it equal to zero (or locate where it is undefined) to get the critical numbers.
Step 2: Place the critical numbers on a number line, dividing it into intervals.
Step 3: Pick one test value in each interval and evaluate f ʹ at that value to determine the sign (+ or −).
Step 4: Read the sign changes: + → − means relative max; − → + means relative min; no sign change means no extremum.
Step 5: Evaluate f(c) at each critical number to find the actual y-coordinate of the extremum.

SECTION 6

Worked Example

Let's apply the First Derivative Test to find all relative extrema of f(x) = 2x³ + 3x² − 12x + 1.

Finding Relative Extrema of f(x) = 2x³ + 3x² − 12x + 1

1
Step 1 — Compute the First Derivative

Differentiate f(x) = 2x³ + 3x² − 12x + 1 using the power rule. Each term is differentiated independently: d/dx(2x³) = 6x², d/dx(3x²) = 6x, d/dx(−12x) = −12, and d/dx(1) = 0.

f ʹ(x) = 6x² + 6x − 12
2
Step 2 — Find the Critical Numbers

Set f ʹ(x) = 0 and solve: 6x² + 6x − 12 = 0. Factor out 6 to get 6(x² + x − 2) = 0, then factor the quadratic: 6(x + 2)(x − 1) = 0. The solutions are x = −2 and x = 1. Since f ʹ is a polynomial, it exists everywhere, so these are the only critical numbers.

Critical numbers: x = −2 and x = 1
3
Step 3 — Build the Sign Chart

The critical numbers divide the number line into three intervals: (−∞, −2), (−2, 1), and (1, ∞). Choose test points x = −3, x = 0, and x = 2. Evaluate: f ʹ(−3) = 6(9) + 6(−3) − 12 = 54 − 18 − 12 = 24 > 0. f ʹ(0) = 0 + 0 − 12 = −12 < 0. f ʹ(2) = 6(4) + 6(2) − 12 = 24 + 12 − 12 = 24 > 0.

Signs: +, −, + across the three intervals
4
Step 4 — Apply the First Derivative Test

At x = −2, f ʹ changes from positive to negative (+ → −), so f has a relative maximum at x = −2. At x = 1, f ʹ changes from negative to positive (− → +), so f has a relative minimum at x = 1.

Relative max at x = −2; relative min at x = 1
5
Step 5 — Find the y-values

Substitute back into the original function. f(−2) = 2(−8) + 3(4) − 12(−2) + 1 = −16 + 12 + 24 + 1 = 21. f(1) = 2(1) + 3(1) − 12(1) + 1 = 2 + 3 − 12 + 1 = −6.

Relative maximum value: f(−2) = 21. Relative minimum value: f(1) = −6.

SECTION 7

Strengths & Limitations

The First Derivative Test is powerful, but like any tool it has specific strengths and limitations. Understanding these will help you choose the best strategy for any given problem.

Feature	Strengths	Limitations
Applicability	Works even when f ʹ(c) does not exist (cusps, corners), as long as f is continuous at c.	Requires continuity at the critical number. If f has a discontinuity at c, the test doesn't apply directly.
Information Provided	Tells you whether f is increasing or decreasing on each interval — gives a full picture of the function's behavior.	Only identifies relative (local) extrema, not absolute (global) extrema. You need the Closed Interval Method for absolute extrema on [a, b].
Ease of Use	Only requires the first derivative — no need to compute f ʹʹ(x), which can be complicated for some functions.	Requires testing a point in every interval, which can be tedious if there are many critical numbers.
Conclusiveness	Always gives a definitive answer at each critical number — max, min, or neither.	Doesn't provide concavity information. For that, you need the Second Derivative Test or additional analysis.

Comparing the strengths and limitations of the First Derivative Test

🔑

WHEN TO USE WHICH TEST

Use the First Derivative Test when you want the full increasing/decreasing picture, or when the second derivative is hard to compute. Use the Second Derivative Test as a shortcut when f ʹʹ is easy to find — just remember it's inconclusive when f ʹʹ(c) = 0, and you'll have to fall back on the First Derivative Test anyway.

SECTION 8

Connection to Advanced Topics

The First Derivative Test is your entry point into a family of optimization techniques that grow more sophisticated as you advance in mathematics. Here's how it connects to what comes next.

First Derivative Test	Advanced Extension
Uses sign changes in f ʹ to classify local extrema.	The Second Derivative Test uses the sign of f ʹʹ(c) to classify critical points without testing intervals, though it can be inconclusive.
Applies to single-variable functions f(x).	In multivariable calculus, the gradient ∇f = 0 locates critical points, and the Hessian matrix (second-order partial derivatives) classifies them.
Identifies relative extrema on open intervals.	The Extreme Value Theorem and the Closed Interval Method extend the analysis to find absolute extrema on closed intervals [a, b].
Answers: where does f have local highs and lows?	Optimization word problems combine the First Derivative Test with modeling to answer real-world questions like "What dimensions maximize area?"

How the First Derivative Test connects to more advanced calculus topics

Mastering the First Derivative Test now gives you a rock-solid foundation for all of these topics. Every optimization problem you'll encounter in AP Calculus begins with the same core steps you've learned here: find the derivative, locate critical numbers, and analyze sign changes.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Suppose f is continuous and f ʹ(x) changes from negative to positive at x = 5. What can you conclude about f at x = 5, and why?

Show answer

By the First Derivative Test, f has a relative minimum at x = 5. The sign change from negative to positive means f is decreasing just to the left of x = 5 (f ʹ < 0) and increasing just to the right (f ʹ > 0). This means the function reaches a valley at x = 5, making f(5) smaller than all nearby function values. This is the definition of a relative minimum.

Question 2: BASIC CALCULATION

Find the critical numbers of f(x) = x³ − 3x + 4 and use the First Derivative Test to classify each one.

Show answer

First, f ʹ(x) = 3x² − 3 = 3(x² − 1) = 3(x − 1)(x + 1). Setting f ʹ(x) = 0 gives x = −1 and x = 1. Test x = −2: f ʹ(−2) = 3(4 − 1) = 9 > 0. Test x = 0: f ʹ(0) = 3(0 − 1) = −3 < 0. Test x = 2: f ʹ(2) = 3(4 − 1) = 9 > 0. At x = −1, f ʹ changes from + to −, so f(−1) = (−1)³ − 3(−1) + 4 = 6 is a relative maximum. At x = 1, f ʹ changes from − to +, so f(1) = 1 − 3 + 4 = 2 is a relative minimum.

Question 3: INTERMEDIATE

Find all relative extrema of f(x) = x⁴ − 4x³ using the First Derivative Test.

Show answer

Compute f ʹ(x) = 4x³ − 12x² = 4x²(x − 3). Setting f ʹ(x) = 0 gives x = 0 and x = 3. Test x = −1: f ʹ(−1) = 4(1)(−4) = −16 < 0. Test x = 1: f ʹ(1) = 4(1)(−2) = −8 < 0. Test x = 4: f ʹ(4) = 4(16)(1) = 64 > 0. At x = 0, f ʹ goes from − to − (no sign change), so x = 0 is NOT an extremum. At x = 3, f ʹ goes from − to +, so f has a relative minimum at x = 3 with f(3) = 81 − 108 = −27. This problem shows that not every critical number yields an extremum.

Question 4: APPLIED

A company's profit function is P(x) = −2x³ + 15x² − 36x + 50, where x is in thousands of units. Use the First Derivative Test to find the production level that gives a relative maximum profit.

Show answer

Compute P ʹ(x) = −6x² + 30x − 36 = −6(x² − 5x + 6) = −6(x − 2)(x − 3). Setting P ʹ(x) = 0 gives x = 2 and x = 3. Test x = 1: P ʹ(1) = −6(−1)(−2) = −12 < 0. Test x = 2.5: P ʹ(2.5) = −6(0.5)(−0.5) = 1.5 > 0. Test x = 4: P ʹ(4) = −6(2)(1) = −12 < 0. At x = 2, P ʹ changes from − to +, so there is a relative minimum. At x = 3, P ʹ changes from + to −, so there is a relative maximum. P(3) = −54 + 135 − 108 + 50 = 23. The relative maximum profit of 23 (in profit units) occurs at a production level of 3 thousand units.

Question 5: CRITICAL THINKING

Consider f(x) = x²ʹ³(x − 4). Note that x²ʹ³ means

x^{2/3}

. Find all critical numbers and use the First Derivative Test to classify each. Be sure to account for points where f ʹ(x) does not exist.

Show answer

Rewrite f(x) =

x^{2/3}

(x − 4) =

x^{5/3}

−

4x^{2/3}

. Then f ʹ(x) = (5/3)

x^{2/3}

− (8/3)

x^{−1/3}

= (1/(

3x^{1/3}

))(5x − 8). The derivative equals zero when 5x − 8 = 0, i.e., x = 8/5. The derivative is undefined when x = 0, and since f(0) = 0 exists, x = 0 is also a critical number. Test x = −1: f ʹ(−1) = (1/(3(−1)))(−5 − 8) = (−1/3)(−13) = 13/3 > 0. Test x = 1: f ʹ(1) = (1/3)(5 − 8) = −1 < 0. Test x = 2: f ʹ(2) = (1/(3·

2^{1/3}

))(10 − 8) = positive > 0. At x = 0, f ʹ changes from + to −, so f(0) = 0 is a relative maximum. At x = 8/5, f ʹ changes from − to +, so there is a relative minimum. This problem illustrates that the First Derivative Test handles critical numbers where f ʹ is undefined, which the Second Derivative Test cannot.

SUMMARY

Lesson Summary

Summary

The First Derivative Test is a systematic method for classifying critical numbers as relative maxima, relative minima, or neither. The process starts by computing f ʹ(x) and solving f ʹ(x) = 0 (or finding where f ʹ is undefined) to locate all critical numbers. You then build a sign chart by testing a sample point in each interval between critical numbers to determine whether f ʹ is positive (increasing) or negative (decreasing).

A sign change from positive to negative at a critical number means f has a relative maximum there, while a change from negative to positive signals a relative minimum. If there is no sign change, the critical number is not an extremum. This test works even at points where the derivative does not exist, making it more versatile than the Second Derivative Test in many situations.