Increasing/Decreasing Intervals — Determining Intervals on Which a Function is Increasing or Decreasing - Calculus 1

SECTION 1

Historical Context & Motivation

Long before calculus existed, mathematicians and scientists cared deeply about how things change. Ancient Greek astronomers tracked the rising and setting of celestial bodies, noting periods when a planet appeared to speed up or slow down across the sky. Describing these changes precisely, however, required tools that wouldn't be invented for nearly two thousand years.

In the seventeenth century, Isaac Newton and Gottfried Wilhelm Leibniz independently developed calculus, giving mathematicians a rigorous way to study rates of change. The derivative became the key tool for answering a fundamental question: on which intervals does a function go up, and on which does it go down? This question is not just academic — engineers use it to optimize designs, economists use it to model profit, and physicists use it to analyze motion.

~300 BCE • Euclid's Elements

Greek mathematicians studied geometric properties of curves, but lacked algebraic tools to describe where a curve rises or falls.

1665–1676 • Newton & Leibniz Develop Calculus

Both independently created the concept of a derivative, enabling precise analysis of how functions change at every point.

1691 • Rolle's Theorem Published

Michel Rolle proved that between two roots of a function, the derivative must be zero at least once — a foundation for analyzing increasing and decreasing behavior.

1823 • Cauchy Formalizes the First Derivative Test

Augustin-Louis Cauchy rigorously connected the sign of the first derivative to whether a function is increasing or decreasing, establishing the method students use today.

The central question this lesson addresses is straightforward: given a function f(x), how do you determine exactly which intervals correspond to the function climbing upward and which correspond to it falling downward? The answer lies in analyzing the sign of the first derivative, f′(x).

SECTION 2

Core Principles & Definitions

Before diving into calculations, you need to understand the vocabulary and the logic that connects derivatives to the behavior of a function. The ideas are intuitive once you see how the slope of a tangent line tells you the direction a curve is headed.

Increasing Function

A function f is increasing on an interval if, for any two numbers a < b in that interval, f(a) < f(b). In other words, as you move right, the outputs go up.

Decreasing Function

A function f is decreasing on an interval if, for any two numbers a < b in that interval, f(a) > f(b). As you move right, the outputs go down.

Critical Numbers

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

Sign of the Derivative

If f′(x) > 0 on an interval, f is increasing there. If f′(x) < 0 on an interval, f is decreasing there. The derivative's sign is the direct indicator of the function's direction.

The First Derivative Test Process

Find f′(x), locate critical numbers, set up a sign chart by testing f′ in each sub-interval, and read off where f is increasing or decreasing.

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KEY TAKEAWAY

Think of driving on a hilly road. If you're pressing the gas and gaining speed (positive acceleration), you're going uphill — that's like f′(x) > 0. If you're coasting and losing speed, you're going downhill — that's like f′(x) < 0. The hilltops and valley bottoms are where the slope is momentarily zero — those are the critical numbers. By checking what happens between each hilltop and valley, you map out the entire ride.

SECTION 3

Visual Explanation

The diagram below shows a polynomial function along with annotations that highlight its increasing and decreasing intervals. Notice how the tangent lines tilt upward (positive slope) in the green-shaded regions and tilt downward (negative slope) in the red-shaded regions. At the critical points, the tangent line is perfectly horizontal.

The violet curve represents f(x). Green-shaded regions show intervals where f′(x) > 0 (increasing), and red-shaded regions show where f′(x) < 0 (decreasing). The amber dots mark critical points where the slope is zero.

Notice the pattern: every time the function switches from increasing to decreasing (or vice versa), it passes through a critical point. At the local maximum, the curve stops rising and starts falling. At the local minimum, the curve stops falling and starts rising. Between critical points, the derivative maintains a constant sign — either entirely positive or entirely negative.

SECTION 4

Mathematical Framework

The process for finding increasing and decreasing intervals follows a systematic series of steps. Each step builds on the previous one, and the entire method rests on a single powerful theorem.

INCREASING / DECREASING TEST

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

Here, (a, b) is an open interval. The function f must be continuous on [a, b] and differentiable on (a, b).

Step-by-Step Procedure

Step 1: Find f′(x) by differentiating f(x).
Step 2: Set f′(x) = 0 and solve. Also identify any x-values where f′(x) does not exist but f(x) does. These are your critical numbers.
Step 3: Plot the critical numbers on a number line, dividing it into sub-intervals.
Step 4: Pick a test value in each sub-interval and evaluate f′ at that test value. Record whether f′ is positive or negative.
Step 5: Conclude: f is increasing where f′ > 0 and decreasing where f′ < 0.

CRITICAL NUMBER CONDITION

f′(c) = 0 or f′(c) does not exist (c in domain of f)

Only values in the domain of f qualify as critical numbers. If c is not in the domain of f, it is not a critical number — it may be a vertical asymptote or a point of discontinuity.

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Why Test Values Work

Between consecutive critical numbers, f′(x) cannot change sign (it would have to pass through zero or be undefined, creating another critical number). So a single test point reveals the sign of f′ across the entire sub-interval. This is why the sign chart method is so efficient.

SECTION 5

Sign Charts & Classification

A sign chart (also called a sign diagram or number-line test) is the organizational tool that ties everything together. It provides a clear, visual way to track the sign of f′(x) across all sub-intervals. The diagram below demonstrates a sign chart for the function f(x) = x³ − 3x, whose derivative is f′(x) = 3x² − 3 = 3(x − 1)(x + 1).

This sign chart shows f′(x) = 3(x − 1)(x + 1) for f(x) = x³ − 3x. The amber tick marks at x = −1 and x = 1 are the critical numbers. Test values in each sub-interval confirm the sign of f′, which determines the behavior of f.

Interval	Test Value	f′(test)	Sign	f Behavior
(−∞, −1)	x = −2	3(4 − 1) = 9	+	Increasing
(−1, 1)	x = 0	3(0 − 1) = −3	−	Decreasing
(1, ∞)	x = 2	3(4 − 1) = 9	+	Increasing

Sign analysis for f′(x) = 3x² − 3

The table and diagram confirm that f(x) = x³ − 3x is increasing on (−∞, −1) ∪ (1, ∞) and decreasing on (−1, 1). Notice that the critical numbers themselves are not included in the intervals because those are the transition points — at x = −1 and x = 1, the function is neither increasing nor decreasing; it's momentarily flat.

SECTION 6

Worked Example

Let's work through a complete example with a function that has a slightly more complex derivative. We'll find the intervals on which f(x) = 2x³ + 3x² − 12x + 1 is increasing and decreasing.

Find the Increasing and Decreasing Intervals of f(x) = 2x³ + 3x² − 12x + 1

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Step 1 — Differentiate f(x)

Apply the power rule to each term: f′(x) = 6x² + 6x − 12. We can factor out a 6 to simplify: f′(x) = 6(x² + x − 2).

f′(x) = 6(x² + x − 2)
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Step 2 — Factor and Find Critical Numbers

Factor the quadratic: x² + x − 2 = (x + 2)(x − 1). Set f′(x) = 0: 6(x + 2)(x − 1) = 0. Since 6 ≠ 0, we get x + 2 = 0 or x − 1 = 0.

Critical numbers: x = −2 and x = 1
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Step 3 — Create Sub-Intervals

The critical numbers x = −2 and x = 1 divide the number line into three intervals: (−∞, −2), (−2, 1), and (1, ∞).

Sub-intervals: (−∞, −2), (−2, 1), (1, ∞)
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Step 4 — Test Each Interval

Choose test values and plug them into f′(x) = 6(x + 2)(x − 1). For x = −3: f′(−3) = 6(−3 + 2)(−3 − 1) = 6(−1)(−4) = 24 > 0. For x = 0: f′(0) = 6(0 + 2)(0 − 1) = 6(2)(−1) = −12 < 0. For x = 2: f′(2) = 6(2 + 2)(2 − 1) = 6(4)(1) = 24 > 0.

f′(−3) = +24, f′(0) = −12, f′(2) = +24
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Step 5 — State the Conclusion

Since f′(x) > 0 on (−∞, −2) and (1, ∞), the function is increasing on those intervals. Since f′(x) < 0 on (−2, 1), the function is decreasing on that interval.

Increasing on (−∞, −2) ∪ (1, ∞). Decreasing on (−2, 1).

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Bonus Insight

From this analysis, you can also identify local extrema. Since f changes from increasing to decreasing at x = −2, there is a local maximum at x = −2. Since f changes from decreasing to increasing at x = 1, there is a local minimum at x = 1. This is the essence of the First Derivative Test.

SECTION 7

Common Pitfalls & Best Practices

Even students who understand the theory can make careless mistakes in practice. The table below compares common errors with the correct approach, so you can avoid these traps on homework and exams.

Common Mistake	Why It's Wrong	Correct Approach
Forgetting to check where f′ is undefined	Points where f′ doesn't exist (cusps, vertical tangents) can also be critical numbers that separate increasing/decreasing intervals.	Always check both f′(c) = 0 and where f′(c) DNE.
Including critical numbers in the interval notation	At a critical number, the function is neither increasing nor decreasing — it's transitioning. Use open intervals.	Write (a, b), not [a, b].
Confusing f(x) values with f′(x) values	The sign of f(x) tells you if outputs are positive or negative — not whether the function is rising or falling.	Always test with f′(x), never with f(x).
Skipping the factoring step	Without factoring, you may miss critical numbers or make sign-analysis errors.	Factor f′(x) completely before setting it equal to zero.
Assuming f′(x) > 0 means f(x) > 0	The derivative's sign indicates direction (up/down), not position (above/below the x-axis). A function can be increasing while having negative y-values.	Separate the concepts: sign of f′ = direction; sign of f = position.

Common mistakes when finding increasing/decreasing intervals

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KEY TAKEAWAY

Think of a sign chart as a weather forecast for f(x). The critical numbers are the times when the weather changes. Between those times, conditions stay consistent. Just like you'd check the forecast for each block of the day, you test one point in each sub-interval to know the "weather" (positive or negative slope) for the entire stretch. The test point is your thermometer — one reading per interval is all you need.

SECTION 8

Connection to Advanced Topics

Understanding increasing and decreasing intervals is not the end of the story — it's the launching pad for several more advanced calculus ideas. The table below shows how this concept connects to what comes next in your calculus journey.

This Lesson (First Derivative Analysis)	Advanced Extension
f′(x) > 0 → f is increasing	f″(x) > 0 → f is concave up. The second derivative extends the analysis to the shape of the curve (concavity), not just its direction.
Critical numbers where f′(c) = 0	The Second Derivative Test uses f″(c) to classify critical points as local maxima or minima without a full sign chart.
Finding local maxima and minima	Optimization problems ask you to find the absolute (global) maximum or minimum on a closed interval, combining increasing/decreasing analysis with endpoint evaluation.
Sign chart for f′(x)	Curve sketching combines the sign charts for both f′ and f″ to produce a complete picture of a function's behavior without a graphing calculator.

How increasing/decreasing analysis connects to advanced calculus topics

As you continue in calculus, you'll see that the skills you build here — taking derivatives, finding critical numbers, and constructing sign charts — are reused in almost every major topic. Mastering this process now will make concavity, optimization, and curve sketching feel like natural extensions of what you already know, rather than entirely new concepts.

SECTION 9

Practice Problems

Work through these five problems in order. They start with conceptual understanding and progress toward more challenging applications. Try each one on your own before reading the answer.

Question 1: CONCEPTUAL

If f′(x) = 0 at x = 3 and f′(x) > 0 for all other x-values, is f increasing, decreasing, or neither on the interval (0, 5)? Explain your reasoning.

Show answer

The function f is increasing on the entire interval (0, 5). Even though f′(3) = 0, the derivative is positive everywhere else in the interval. A single point where the derivative equals zero does not interrupt the increasing behavior — the function simply has a horizontal tangent at x = 3 but continues to rise on both sides. A function is increasing on an interval as long as f′(x) ≥ 0 throughout and f′(x) is not identically zero on any sub-interval.

Question 2: BASIC CALCULATION

Find the intervals on which f(x) = x² − 4x + 7 is increasing and decreasing.

Show answer

First, find f′(x) = 2x − 4. Set f′(x) = 0: 2x − 4 = 0 gives x = 2. This is the only critical number. Test a value left of 2: f′(0) = −4 < 0, so f is decreasing on (−∞, 2). Test a value right of 2: f′(3) = 2 > 0, so f is increasing on (2, ∞). This makes sense because f is a parabola opening upward with its vertex at x = 2.

Question 3: INTERMEDIATE

Determine the increasing and decreasing intervals for g(x) = x⁴ − 4x³. Include a sign chart of g′(x) in your analysis.

Show answer

Differentiate: g′(x) = 4x³ − 12x² = 4x²(x − 3). Set g′(x) = 0: 4x²(x − 3) = 0 gives x = 0 and x = 3. Now build the sign chart with intervals (−∞, 0), (0, 3), and (3, ∞). Test x = −1: g′(−1) = 4(1)(−4) = −16 < 0. Test x = 1: g′(1) = 4(1)(−2) = −8 < 0. Test x = 4: g′(4) = 4(16)(1) = 64 > 0. Therefore g is decreasing on (−∞, 0) ∪ (0, 3) — equivalently, on (−∞, 3) since the sign doesn't change at x = 0 — and increasing on (3, ∞). Note that x = 0 is a critical number but not a local extremum because f′ does not change sign there.

Question 4: APPLIED

A company's revenue (in thousands of dollars) is modeled by R(t) = −t³ + 12t² − 36t + 50, where t is the number of years since the company was founded (t ≥ 0). During which years is the revenue increasing?

Show answer

Differentiate: R′(t) = −3t² + 24t − 36 = −3(t² − 8t + 12) = −3(t − 2)(t − 6). Critical numbers: t = 2 and t = 6. Test t = 0: R′(0) = −3(−2)(−6) = −36 < 0 (decreasing). Test t = 4: R′(4) = −3(2)(−2) = 12 > 0 (increasing). Test t = 7: R′(7) = −3(5)(1) = −15 < 0 (decreasing). Since t ≥ 0, the revenue is decreasing on (0, 2), increasing on (2, 6), and decreasing again on (6, ∞). So the company's revenue grows between year 2 and year 6.

Question 5: CRITICAL THINKING

Consider a continuous function h(x) defined on all real numbers whose derivative is h′(x) = (x − 1)²(x + 3). Find the intervals where h is increasing and decreasing, and explain why the critical number at x = 1 does not produce a local extremum.

Show answer

Set h′(x) = (x − 1)²(x + 3) = 0 to get x = 1 and x = −3. Build the sign chart: on (−∞, −3), test x = −4: (−5)²(−1) = −25 < 0. On (−3, 1), test x = 0: (−1)²(3) = 3 > 0. On (1, ∞), test x = 2: (1)²(5) = 5 > 0. So h is decreasing on (−∞, −3) and increasing on (−3, 1) ∪ (1, ∞), which means h is actually increasing on the entire interval (−3, ∞). At x = 1, the factor (x − 1)² makes h′ touch zero without changing sign — it's positive on both sides. Because there's no sign change, h doesn't switch between increasing and decreasing, so x = 1 is not a local extremum. Only x = −3 is a local minimum, where the sign of h′ changes from negative to positive.

SUMMARY

Lesson Summary

Summary

To determine where a function is increasing or decreasing, you follow a clear five-step process. First, compute the first derivative f′(x). Next, find all critical numbers — values of x where f′(x) = 0 or f′(x) does not exist. These critical numbers divide the number line into sub-intervals. Then, pick a test value in each sub-interval and plug it into f′(x) to determine the sign. If f′ is positive, f is increasing; if f′ is negative, f is decreasing.

This technique — organized using a sign chart — is the foundation for the First Derivative Test, which classifies critical points as local maxima or minima based on whether f′ changes from positive to negative or vice versa. Mastering this skill prepares you for concavity analysis, optimization, and complete curve sketching — the next major applications of derivatives in your calculus course.