Determining Intervals on Which a Function is Increasing or Decreasing - AP Calculus AB

SECTION 1

Historical Context & Motivation

The ability to determine where a function increases or decreases lies at the heart of calculus and its applications. Long before formal derivatives existed, natural philosophers and mathematicians sought systematic ways to describe the behavior of changing quantities—whether the trajectory of a projectile, the growth of a population, or the profit of a merchant's enterprise. The story of how we arrived at the modern first derivative test for monotonicity is inseparable from the invention of calculus itself, and it represents one of the most elegant bridges between algebraic computation and geometric intuition.

1637 • Fermat's Method of Adequality

Pierre de Fermat developed a technique for finding maxima and minima of polynomial expressions by setting a difference quotient to zero—an early precursor to differentiation that implicitly relied on understanding where a function changes from increasing to decreasing.

1684 • Leibniz Publishes the Differential Calculus

Gottfried Wilhelm Leibniz published his foundational paper introducing the notation dy/dx. His framework made it algebraically routine to compute rates of change, enabling systematic analysis of a function's increasing and decreasing behavior.

1691 • Johann Bernoulli and the Brachistochrone

The brachistochrone problem challenged mathematicians to find the curve of fastest descent, requiring careful analysis of where speed increases and decreases along different paths. This catalyzed the use of derivatives for optimization.

1823 • Cauchy Formalizes the Derivative

Augustin-Louis Cauchy rigorously defined the derivative as a limit, placing the connection between the sign of f′(x) and the monotonic behavior of f(x) on a firm logical foundation for the first time.

The central question these developments address is deceptively simple: given a function f, on which intervals does f(x) grow larger as x moves to the right, and on which intervals does it shrink? The answer, as we will see, reduces to a systematic analysis of the sign of the first derivative. This technique is foundational not only for curve sketching but also for optimization, motion analysis, and modeling throughout the AP Calculus AB curriculum.

SECTION 2

Core Principles & Definitions

Before diving into computations, it is essential to establish precise definitions. A function f is said to be increasing on an interval (a, b) if, for every pair of numbers x₁ and x₂ in (a, b) with x₁ < x₂, we have f(x₁) < f(x₂). Similarly, f is decreasing on an interval (a, b) if x₁ < x₂ implies f(x₁) > f(x₂). These definitions are purely algebraic—they say nothing about derivatives. The power of calculus is that it converts this comparison-based definition into a simple sign check on f′(x).

Increasing ⇔ f′(x) > 0

If f′(x) > 0 for every x in an open interval, then f is increasing on that interval. A positive derivative means the tangent line slopes upward, so the function climbs from left to right.

Decreasing ⇔ f′(x) < 0

If f′(x) < 0 for every x in an open interval, then f is decreasing on that interval. A negative derivative means the tangent line slopes downward, so the function falls from left to right.

Critical Numbers

A critical number c in the domain of f is a value where f′(c) = 0 or f′(c) does not exist. These are the only points where the function can transition between increasing and decreasing behavior.

Sign Chart Method

Partition the domain using critical numbers and points where f is undefined. Test one value in each sub-interval to determine the sign of f′(x), then conclude the monotonic behavior on each piece.

✦

KEY TAKEAWAY

Think of the derivative as a speedometer for a car traveling along a number line. When the speedometer reads positive, the car moves to the right (the function increases); when it reads negative, the car backs up (the function decreases). The moments the speedometer reads exactly zero—or breaks entirely—are the critical numbers, and those are the only places where the car can switch direction.

SECTION 3

Visual Explanation

The diagram below shows a generic differentiable function f(x) alongside its derivative f′(x). Observe how the regions where f′(x) is above the x-axis (positive) correspond precisely to the intervals where f(x) is climbing, and the regions where f′(x) is below the x-axis (negative) correspond to the intervals where f(x) is falling. The critical numbers—where f′(x) crosses or touches zero—mark the boundaries between increasing and decreasing intervals.

Top: the original function f(x) (purple curve) with increasing intervals shaded green and decreasing intervals shaded pink. Bottom: the derivative f′(x) (cyan curve). The dashed yellow vertical lines mark the critical numbers c₁ and c₂ where f′(x) = 0. Notice that f has a local maximum at c₁ (where f′ changes from positive to negative) and a local minimum at c₂ (where f′ changes from negative to positive).

SECTION 4

Mathematical Framework

The theoretical backbone for this topic is the Increasing/Decreasing Test, which follows directly from the Mean Value Theorem. If f is continuous on [a, b] and differentiable on (a, b), the MVT guarantees the existence of some c in (a, b) such that f(b) − f(a) = f′(c)(b − a). Because (b − a) > 0, the sign of f(b) − f(a) matches the sign of f′(c). This reasoning, extended to every sub-interval, yields the following results.

INCREASING TEST

If f′(x) > 0 for all x ∈ (a, b), then f is increasing on (a, b).

A positive derivative everywhere on an interval ensures that f(x₂) > f(x₁) whenever x₂ > x₁ within that interval.

DECREASING TEST

If f′(x) < 0 for all x ∈ (a, b), then f is decreasing on (a, b).

A negative derivative everywhere on an interval ensures that f(x₂) < f(x₁) whenever x₂ > x₁ within that interval.

CRITICAL NUMBER DEFINITION

c is a critical number of f ⟺ f(c) is defined and (f′(c) = 0 or f′(c) DNE).

Critical numbers are the only candidates for boundary points between increasing and decreasing intervals. They include horizontal tangent points and cusps/corners.

The Sign-Chart Algorithm

Step 1. Find f′(x) using differentiation rules (power, product, quotient, chain).
Step 2. Solve f′(x) = 0 and identify where f′(x) does not exist to find all critical numbers.
Step 3. Place the critical numbers on a number line, partitioning the domain into open intervals.
Step 4. Choose a test value in each interval and evaluate the sign of f′ at that test value.
Step 5. Conclude: f is increasing where f′ > 0 and decreasing where f′ < 0.

SECTION 5

Sign Charts & Interval Classification

The sign chart (or sign diagram) is the workhorse tool for this analysis. Consider the function f(x) = x³ − 3x. Its derivative is f′(x) = 3x² − 3 = 3(x − 1)(x + 1). Setting f′(x) = 0 yields critical numbers x = −1 and x = 1. These two values divide the real line into three intervals: (−∞, −1), (−1, 1), and (1, ∞). By testing one point in each interval—say x = −2, x = 0, and x = 2—we determine f′(−2) = 9 > 0, f′(0) = −3 < 0, and f′(2) = 9 > 0. The sign chart below visualizes this classification.

Sign chart for f′(x) = 3(x − 1)(x + 1). The critical numbers x = −1 and x = 1 partition the number line. The bottom section shows how to determine the sign of f′(x) by analyzing the sign of each linear factor and multiplying. Since the constant factor 3 is always positive, only the factors (x + 1) and (x − 1) matter.

⚠

Common Pitfall

Do not forget to include points where f′(x) is undefined as partition points in your sign chart. For example, if f(x) = x^2/3, then f′(x) = (2/3)x^−1/3 is undefined at x = 0 even though f(0) is defined. The derivative changes sign at x = 0 (negative to positive), so x = 0 is a critical number that must appear on the sign chart.

SECTION 6

Worked Example

Let us walk through a complete example using a rational function, which often appears on the AP exam. Determine the intervals on which the function f(x) = x / (x² + 4) is increasing and decreasing.

Finding Intervals of Increase and Decrease for f(x) = x / (x² + 4)

1
Step 1 — Differentiate Using the Quotient Rule

Apply the quotient rule: f′(x) = [(x² + 4)(1) − x(2x)] / (x² + 4)². Simplify the numerator: x² + 4 − 2x² = 4 − x² = (2 − x)(2 + x).

f′(x) = (4 − x²) / (x² + 4)²
2
Step 2 — Find the Critical Numbers

Set the numerator equal to zero: 4 − x² = 0, so x² = 4, giving x = −2 and x = 2. The denominator (x² + 4)² is always positive (never zero), so f′(x) is defined for all real numbers. Thus the only critical numbers are x = −2 and x = 2.

Critical numbers: x = −2, x = 2
3
Step 3 — Partition the Domain

The domain of f is all real numbers. The critical numbers divide the real line into three intervals: (−∞, −2), (−2, 2), and (2, ∞).

Intervals: (−∞, −2), (−2, 2), (2, ∞)
4
Step 4 — Test the Sign of f′(x) in Each Interval

Since the denominator (x² + 4)² is always positive, we only need to check the sign of the numerator 4 − x². Test x = −3: 4 − 9 = −5 < 0, so f′(−3) < 0. Test x = 0: 4 − 0 = 4 > 0, so f′(0) > 0. Test x = 3: 4 − 9 = −5 < 0, so f′(3) < 0.

f′ < 0 on (−∞, −2); f′ > 0 on (−2, 2); f′ < 0 on (2, ∞)
5
Step 5 — State the Conclusion

By the Increasing/Decreasing Test: f is decreasing on (−∞, −2), increasing on (−2, 2), and decreasing on (2, ∞). Additionally, f has a local minimum at x = −2 (f changes from decreasing to increasing) and a local maximum at x = 2 (f changes from increasing to decreasing).

f is increasing on (−2, 2) and decreasing on (−∞, −2) ∪ (2, ∞).

SECTION 7

Strengths, Limitations & Common Errors

Aspect	Strengths	Limitations / Pitfalls
Applicability	Works for any differentiable function on open intervals; extends to piecewise functions interval by interval.	Requires differentiability—functions with infinitely many oscillations (e.g., x sin(1/x)) demand more careful treatment.
Algebraic simplicity	Factoring f′(x) makes sign analysis nearly mechanical; each linear or irreducible quadratic factor is easy to track.	If f′(x) cannot be factored easily (e.g., transcendental functions), numerical or graphical methods may be needed.
Local extrema detection	The method naturally identifies local maxima and minima where f′ changes sign.	f′(c) = 0 does not guarantee an extremum—if the sign does not change (e.g., f(x) = x³ at x = 0), there is no local max or min.
Interval notation	Open-interval notation aligns with the differentiability requirement and is standard in AP scoring.	Students sometimes incorrectly use closed brackets for the interval endpoints, but the derivative test applies on open intervals.

Strengths and limitations of the first derivative test for monotonicity

✦

KEY TAKEAWAY

The sign chart is like a diagnostic panel in engineering: it gives you a complete readout of the derivative's behavior across the domain. Just as an engineer checks multiple sensor readings before declaring a system stable, you must test each sub-interval—one positive test value does not guarantee positivity everywhere. The strength of the method is its completeness; its limitation is that it relies on your ability to differentiate and factor accurately.

SECTION 8

Connection to the First & Second Derivative Tests

Determining intervals of increase and decrease is the essential first step in the broader analytical toolkit of AP Calculus AB. Once you have classified these intervals, you can directly apply the First Derivative Test for Local Extrema: if f′ changes from positive to negative at a critical number c, then f(c) is a local maximum; if f′ changes from negative to positive, f(c) is a local minimum. This test is logically downstream from the increasing/decreasing analysis—it is simply a reading of the sign chart at each critical number.

Feature	Increasing/Decreasing Analysis	Second Derivative Test
What it determines	Intervals where f rises or falls; sign changes reveal local extrema	Concavity at a critical point; classifies as local max/min without building a full sign chart
Information required	f′(x) and its sign on every sub-interval of the domain	f″(c) at each critical number c where f′(c) = 0
When it fails	Rarely—only when f′ cannot be computed or when the domain has unusual structure	When f″(c) = 0 (inconclusive); must revert to the first derivative test
AP exam utility	Essential for justifying behavior on an interval; required for many FRQ justifications	Quick classification at a single point; often used in optimization FRQs

Comparing the first-derivative interval analysis with the second derivative test

Looking ahead, the same sign-chart technique extends to the second derivative f″(x) for concavity analysis: where f″ > 0 the graph is concave up, and where f″ < 0 it is concave down. Combining both sign charts gives you a complete qualitative picture of f—its shape, its turning points, and its inflection points—which is the foundation of accurate curve sketching without a calculator.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

If f is a differentiable function and f′(x) > 0 for all x in the interval (2, 7), which of the following must be true?

A. f(7) > f(2)
B. f has a local maximum at x = 7
C. f is concave up on (2, 7)
D. f(x) > 0 for all x in (2, 7)
E. f′ is increasing on (2, 7)

Show answer (A)

A is correct. If f′(x) > 0 on (2, 7), then f is increasing on that interval, so f(7) > f(2). B is wrong because the first derivative test for local extrema requires a sign change at x = 7, which is not guaranteed—f could continue increasing beyond 7. C is wrong because concavity depends on the sign of f″, not f′. D is wrong because the sign of f′ tells us about the direction of change, not the sign of f itself—a function can be increasing and negative. E is wrong because f′ > 0 tells us nothing about whether f′ is itself increasing; that would require f″ > 0.

Question 2: BASIC CALCULATION

Let g(x) = x³ − 12x + 5. On which interval is g decreasing?

A. (−∞, −2)
B. (−2, 2)
C. (2, ∞)
D. (−∞, 0)
E. (0, ∞)

Show answer (B)

B is correct. g′(x) = 3x² − 12 = 3(x² − 4) = 3(x − 2)(x + 2). Setting g′(x) = 0 gives critical numbers x = −2 and x = 2. Testing x = 0 in the interval (−2, 2): g′(0) = −12 < 0, so g is decreasing on (−2, 2). A is wrong because testing x = −3 gives g′(−3) = 27 − 12 = 15 > 0, so g is increasing on (−∞, −2). C is wrong because testing x = 3 gives g′(3) = 27 − 12 = 15 > 0, so g is increasing on (2, ∞). D and E are wrong because they do not correspond to the natural partition created by the critical numbers and incorrectly split the domain at x = 0, which is not a critical number.

Question 3: INTERMEDIATE

Let h(x) = xe^−x for all real x. The function h is increasing on which of the following intervals?

A. (−∞, 0)
B. (−∞, 1)
C. (1, ∞)
D. (−∞, −1)
E. (0, ∞)

Show answer (B)

B is correct. Using the product rule, h′(x) =

e^{−x}

+ x(−

e^{−x}

) =

e^{−x}

(1 − x). Since

e^{−x}

> 0 for all x, the sign of h′(x) depends entirely on the factor (1 − x). We have 1 − x > 0 when x < 1 and 1 − x < 0 when x > 1. Therefore h is increasing on (−∞, 1). A is wrong because h is also increasing on (0, 1), not just to the left of 0. C is wrong because h′(x) < 0 for x > 1, meaning h is decreasing there. D is wrong because it excludes the interval (−1, 1) where h is still increasing. E is wrong because h begins decreasing at x = 1, not increasing.

Question 4: APPLIED

5 pts

A particle moves along the x-axis so that its position at time t ≥ 0 is given by x(t) = t³ − 6t² + 9t + 2. (a) Find x′(t). (b) Determine all critical numbers of x(t) for t ≥ 0. (c) Determine the intervals on which the particle is moving to the right and the intervals on which it is moving to the left. (d) At what time(s) does the particle change direction? Justify your answer.

Show answer

Rubric (5 points): (a) [1 point] x′(t) = 3t² − 12t + 9 = 3(t² − 4t + 3) = 3(t − 1)(t − 3). Award 1 point for the correct derivative. (b) [1 point] Setting x′(t) = 0: 3(t − 1)(t − 3) = 0, so t = 1 and t = 3. Both are in the domain t ≥ 0. Award 1 point for identifying both critical numbers. (c) [2 points] The domain t ≥ 0 is partitioned into (0, 1), (1, 3), and (3, ∞). Test t = 0.5: x′(0.5) = 3(−0.5)(−2.5) = 3.75 > 0. Test t = 2: x′(2) = 3(1)(−1) = −3 < 0. Test t = 4: x′(4) = 3(3)(1) = 9 > 0. The particle moves to the right (x′ > 0) on (0, 1) and (3, ∞). The particle moves to the left (x′ < 0) on (1, 3). Award 1 point for the correct sign analysis and 1 point for correct interval statements with justification. (d) [1 point] The particle changes direction at t = 1 and t = 3 because x′(t) changes sign at each of these values (positive to negative at t = 1 and negative to positive at t = 3). Award 1 point for both times with correct justification that the velocity changes sign.

Question 5: CRITICAL THINKING

3 pts

Let f be a continuous function on [0, 5] that is differentiable on (0, 5). The derivative f′ is given by f′(x) = (x − 2)²(x − 4). Without finding f(x), determine all intervals on which f is increasing and all intervals on which f is decreasing on (0, 5). Then explain why x = 2 is not a local extremum of f, even though f′(2) = 0.

Show answer

Rubric (3 points): [1 point] Correct sign analysis: f′(x) = (x − 2)²(x − 4). Since (x − 2)² ≥ 0 for all x, the sign of f′(x) depends on (x − 4). For 0 < x < 4 (and x ≠ 2): (x − 2)² > 0 and (x − 4) < 0, so f′(x) < 0. For 4 < x < 5: (x − 2)² > 0 and (x − 4) > 0, so f′(x) > 0. Award 1 point for correctly determining that f′(x) < 0 on (0, 2) ∪ (2, 4) and f′(x) > 0 on (4, 5). [1 point] Correct monotonicity conclusion: f is decreasing on (0, 4) and increasing on (4, 5). Note that even though f′(2) = 0, the derivative does not change sign at x = 2—it is negative on both sides—so f is decreasing throughout (0, 4). Award 1 point for correctly stating both intervals. [1 point] Explanation for why x = 2 is not a local extremum: At x = 2, f′(2) = 0, so x = 2 is indeed a critical number. However, f′(x) < 0 on both (0, 2) and (2, 4), meaning f is decreasing on both sides of x = 2. Since f′ does not change sign at x = 2 (it remains negative), the first derivative test confirms that x = 2 is not a local extremum. The graph of f has a horizontal tangent at x = 2 but continues to decrease through it (similar to an inflection point with a horizontal tangent). Award 1 point for a correct justification based on the absence of a sign change.

SUMMARY

Lesson Summary

Summary

To determine where a function is increasing or decreasing, compute the first derivative f′(x) and identify all critical numbers—values where f′(x) = 0 or f′(x) does not exist. Use these critical numbers to partition the domain into open intervals. By testing the sign of f′(x) in each interval, you classify the function's behavior: f′ > 0 means increasing, and f′ < 0 means decreasing.

This analysis forms the foundation for the First Derivative Test for Local Extrema: a sign change from positive to negative at a critical number indicates a local maximum, while a sign change from negative to positive indicates a local minimum. If there is no sign change, the critical number is neither. Master the sign chart method and you will have one of the most powerful tools in the AP Calculus AB toolkit for curve sketching, optimization, and motion problems.