Difficulty: Intermediate | Prerequisites: Basic differentiation rules (power rule, quotient rule), factoring polynomials, understanding of function notation.
This topic sits at the heart of Chapter 3, where you move from "how do I take a derivative?" to "what does the derivative actually tell me about the original function?" The first derivative reveals where a function is climbing and where it is falling, which is the foundation for optimisation problems later in the course. If you are comfortable differentiating polynomials and rational functions and can factor a polynomial to find its roots, you are ready for this material. If not, revisit the power rule and basic factoring first.
A function is increasing wherever its first derivative is positive and decreasing wherever its first derivative is negative. To find the intervals, differentiate, set the derivative equal to zero, solve for the critical points, then test the sign of the derivative in each interval between those points.
Increasing function
A function f(x) is increasing on an interval if, for any two numbers x₁ and x₂ in that interval where x₂ > x₁, we have f(x₂) > f(x₁). In simple terms, the graph goes uphill as you move from left to right.
Decreasing function
A function f(x) is decreasing on an interval if, for any two numbers x₁ and x₂ in that interval where x₂ > x₁, we have f(x₂) < f(x₁). Think of it as the graph going downhill as you read left to right.
Critical point (CP)
A value of x where f'(x) = 0 or f'(x) is undefined. These are the only places where the function can switch from increasing to decreasing (or vice versa). Think of them as the potential hilltops and valley floors of the graph.
Sign chart (sign diagram)
A number-line sketch that records the sign (+/−) of f'(x) in each interval between critical points. It is the main working tool for reading off where the function rises and falls.
Vertical asymptote (VA)
A value of x where the denominator of a rational function equals zero and the function is undefined. In the sign chart, a VA acts like a critical point: you must test the sign of f'(x) on both sides of it, but the function does not actually touch or pass through that x-value.
Let f(x) be defined on an interval a < x < b, and let x₁ and x₂ be any two numbers in that interval.
f is increasing on the interval if f(x₂) > f(x₁) whenever x₂ > x₁.
f is decreasing on the interval if f(x₂) < f(x₁) whenever x₂ > x₁.
The first derivative provides the shortcut: if f'(x) > 0 on an interval, the function is increasing there; if f'(x) < 0, it is decreasing.
Differentiate the function to obtain f'(x).
Set f'(x) = 0 and solve. The solutions are the critical points.
Identify any x-values where f'(x) is undefined (e.g. vertical asymptotes for rational functions). These also partition the number line.
Build a sign chart. Place all critical points and undefined points on a number line, then pick a test value in each resulting interval and evaluate the sign of f'(x).
Read off the intervals. Positive sign → increasing. Negative sign → decreasing.
f(x) = 2x³ + 3x² − 12x − 7
f'(x) = 6x² + 6x − 12
Factor: 6(x² + x − 2) = 6(x + 2)(x − 1)
Set equal to zero: x = −2, x = 1 (the critical points)
Sign chart test values: pick x = −3 (left of −2), x = 0 (between −2 and 1), x = 2 (right of 1)
f'(−3) = 6(−1)(−4) > 0 → positive
f'(0) = 6(2)(−1) < 0 → negative
f'(2) = 6(4)(1) > 0 → positive
Increasing: (−∞, −2) ∪ (1, ∞)
Decreasing: (−2, 1)
f(x) = 2x⁴ − 4x² + 3
f'(x) = 8x³ − 8x
Factor: 8x(x² − 1) = 8x(x + 1)(x − 1)
Critical points: x = 0, x = −1, x = 1
Sign chart (test x = −2, −0.5, 0.5, 2):
f'(−2) = 8(−2)(−1)(−3) < 0 → negative
f'(−0.5) = 8(−0.5)(0.5)(−1.5) > 0 → positive
f'(0.5) = 8(0.5)(1.5)(−0.5) < 0 → negative
f'(2) = 8(2)(3)(1) > 0 → positive
Decreasing: (−∞, −1) ∪ (0, 1)
Increasing: (−1, 0) ∪ (1, ∞)
f(x) = x² / (x − 2)
Use the quotient rule:
f'(x) = [2x(x − 2) − x²·1] / (x − 2)² = (2x² − 4x − x²) / (x − 2)² = (x² − 4x) / (x − 2)²
Numerator: x² − 4x = x(x − 4). Set to zero: x = 0, x = 4.
Denominator: (x − 2)² = 0 gives a vertical asymptote at x = 2.
The sign chart now has partitions at x = 0, 2, and 4.
Test values: x = −1, x = 1, x = 3, x = 5
f'(−1) = (−1)(−5) / 9 > 0 → positive
f'(1) = (1)(−3) / 1 < 0 → negative
f'(3) = (3)(−1) / 1 < 0 → negative
f'(5) = (5)(1) / 9 > 0 → positive
Increasing: (−∞, 0) ∪ (4, ∞)
Decreasing: (0, 2) ∪ (2, 4)
Note that x = 2 is excluded from both intervals because the function is not defined there.
First Derivative Test for Increase/Decrease
If f'(x) > 0 for all x in an interval, then f is increasing on that interval.
If f'(x) < 0 for all x in an interval, then f is decreasing on that interval.
If f'(x) = 0 for all x in an interval, then f is constant on that interval.
Quotient Rule (used in Example 3)
If f(x) = g(x) / h(x), then f'(x) = [g'(x)·h(x) − g(x)·h'(x)] / [h(x)]²
In business calculus, identifying where a function increases or decreases lets you answer questions like "over what price range does revenue grow?" or "at what production level do costs start falling?" Any time you model profit, cost, or revenue as a function and want to know where things improve or deteriorate, the sign of the first derivative is the tool you reach for.
Students often write the critical points themselves as the intervals. The critical points are boundaries, not intervals. The answer is the open interval between (or beyond) those boundaries.
A common error is forgetting to check where the derivative is undefined, not just where it equals zero. For rational functions, the denominator's zeros partition the number line just as critical points do.
Some students assume a function must alternate between increasing and decreasing at every critical point. It does not. The sign chart might show the same sign on both sides of a critical point (in which case there is no direction change there).
Confusing the sign of f(x) with the sign of f'(x). You are testing the derivative, not the original function. Plugging test values into f(x) instead of f'(x) is a surprisingly frequent mistake.
⚠️ "Find the intervals of increase and decrease" is one of the most commonly tested question types in business calculus exams. Expect it as a standalone question or as the first step inside an optimisation problem.
⚠️ Show all working on the sign chart. Partial credit depends on the examiner seeing that you tested each interval, not just that you wrote the final answer.
⚠️ Use interval notation correctly: round brackets (parentheses) for open intervals, union symbol ∪ to join disjoint intervals. Writing [−2, 1] instead of (−2, 1) is a common point deduction.
⚠️ For rational functions, always state that the function is undefined at the vertical asymptote. Do not include x = 2 (or whatever the VA is) in your interval.
True or false: If f'(x) = 0 at x = 3, the function must change from increasing to decreasing (or vice versa) at x = 3.
Fill in the blank: To find intervals of increase, we look for where f'(x) is ______.
True or false: A vertical asymptote should be included in the intervals of increase or decrease.
Fill in the blank: The solutions to f'(x) = 0 are called ______.
True or false: If f'(x) > 0 on (a, b), then f(x₂) > f(x₁) for any x₁ < x₂ in that interval.
Answers: 1. False (the sign might not change). 2. Positive. 3. False (the function is undefined there). 4. Critical points. 5. True.
Q: Given f(x) = x³ − 6x² + 9x + 1, find the intervals where f is increasing and decreasing.
A: f'(x) = 3x² − 12x + 9 = 3(x − 1)(x − 3). Critical points at x = 1, x = 3. Sign chart: f' > 0 on (−∞, 1), f' < 0 on (1, 3), f' > 0 on (3, ∞). Increasing on (−∞, 1) ∪ (3, ∞). Decreasing on (1, 3).
Q: For g(x) = x⁴ − 4x³, find the intervals of increase and decrease.
A: g'(x) = 4x³ − 12x² = 4x²(x − 3). Critical points at x = 0, x = 3. Sign chart: g' < 0 on (−∞, 0), g' < 0 on (0, 3), g' > 0 on (3, ∞). Decreasing on (−∞, 3) (note: no sign change at x = 0). Increasing on (3, ∞).
Q: Find the intervals of increase and decrease for h(x) = (x + 1) / (x − 3).
A: Using the quotient rule, h'(x) = [(1)(x − 3) − (x + 1)(1)] / (x − 3)² = −4 / (x − 3)². The numerator is always negative and the denominator is always positive (a square), so h'(x) < 0 everywhere it is defined. Decreasing on (−∞, 3) ∪ (3, ∞). There is a vertical asymptote at x = 3.
Q: Suppose f'(x) = (x − 2)²(x + 5). On what intervals is f increasing?
A: f'(x) = 0 at x = 2 and x = −5. Sign chart: for x < −5, (x − 2)² > 0 and (x + 5) < 0, so f' < 0. For −5 < x < 2, (x − 2)² > 0 and (x + 5) > 0, so f' > 0. For x > 2, both factors positive, f' > 0. Increasing on (−5, ∞). (No sign change at x = 2 because of the squared factor.)
This material feeds directly into the First Derivative Test for relative extrema (covered in the second set of notes from this chapter). Once you know where f is increasing and decreasing, you can identify relative maxima and minima at the transition points. It also sets the stage for the Second Derivative Test and concavity analysis in later sections. In applied problems, the same workflow underpins optimisation of profit, cost, and revenue functions throughout the rest of the course.
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