# Determining intervals where a function is increasing/decreasing

> AP Calculus AB · Unit 5: Analytical Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u5-determining-intervals-where-a-function/

This guide covers how to use the first derivative to find intervals where a function is increasing or decreasing, a core AP Calculus AB Unit 5 skill tested on both multiple-choice and free-response questions.

**Prerequisites:** Basic differentiation rules; Solving polynomial and rational inequalities; Finding the domain of a function

## Learning objectives

- Define increasing and decreasing functions on intervals
- Apply the first derivative rule to determine function monotonicity
- Identify partition points (critical points and domain breaks)
- Perform sign analysis of the first derivative
- Avoid common exam pitfalls when identifying intervals

## Core Definitions and Exam Context

This foundational skill makes up 15-18% of the total AP Calculus AB exam weight, connecting the graphical shape of a function to the sign of its first derivative. It appears in MCQ (identifying correct intervals from functions or derivative graphs) and FRQ (justifying behavior for curve sketching and optimization problems).

**Increasing and Decreasing Functions** — A function $f$ is **increasing** on an interval if for any two inputs $x_1 < x_2$ in the interval, $f(x_1) < f(x_2)$. A function is **decreasing** if $x_1 < x_2$ implies $f(x_1) > f(x_2)$. Monotonic functions are functions that are always increasing or always decreasing on an interval.

> **info**
>
> Increasing/decreasing is a property of intervals, not individual points. This is one of the most common points of confusion on AP exams.

> **Exam tip:** Always justify your interval conclusions by referencing the sign of the first derivative on FRQ questions to earn full credit.

## The First Derivative Rule for Monotonicity

The relationship between derivative sign and function behavior is formalized by the Mean Value Theorem. For a function $f$ continuous on closed interval $[a,b]$ and differentiable on open interval $(a,b)$:

- 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 $[a,b]$
- If $f'(x) = 0$ for all $x \in (a,b)$, then $f$ is constant on $[a,b]$

Intuitively, the derivative equals the slope of the tangent line at a point: positive slope means the function rises from left to right, and negative slope means it falls.

**Worked example:** Use the first derivative rule to find where $f(x) = x^3 - 3x^2 + 2$ is increasing.

1. Confirm $f(x)$ is a polynomial, so it is continuous and differentiable for all real $x$, so the first derivative rule applies everywhere.
2. Compute and factor the first derivative:
3. $$f'(x) = 3x^2 - 6x = 3x(x-2)$$
4. The sign of $f'(x)$ can only change where $f'(x) = 0$, so solve for critical points: $3x(x-2) = 0 \implies x=0, x=2$. These split the real line into three intervals: $(-\infty, 0)$, $(0, 2)$, $(2, \infty)$.
5. Test the sign of $f'(x)$ in each interval:
6. - For $(-\infty,0)$, test $x=-1$: $f'(-1) = 9 > 0 \implies f$ is increasing
- For $(0,2)$, test $x=1$: $f'(1) = -3 < 0 \implies f$ is decreasing
- For $(2, \infty)$, test $x=3$: $f'(3) = 9 > 0 \implies f$ is increasing
7. Combine intervals where $f'(x) > 0$: $f$ is increasing on $(-\infty, 0) \cup (2, \infty)$.

> **tip**
>
> On the AP exam, you will almost always use open intervals for increasing/decreasing, even though the theorem allows closed intervals. AP graders accept both, but open intervals avoid confusion at endpoints with horizontal tangents.

## Partitioning the Domain with Critical Points and Domain Breaks

To correctly test the sign of $f'(x)$, you first split (partition) the domain of $f$ into intervals where $f'(x)$ is entirely positive or entirely negative. The derivative can only change sign at two types of points:

- **Critical points**: Points $x=c$ in the domain of $f$ where $f'(c) = 0$ or $f'(c)$ does not exist
- **Domain breaks**: Points $x=c$ not in the domain of $f$, where $f'(x)$ is also undefined

Derivatives satisfy the Intermediate Value Theorem, so they cannot change sign inside a partitioned interval, meaning you only need one test per interval. A common early mistake is forgetting to include domain breaks as partition points.

**Worked example:** Find all partition points for $f(x) = \frac{x^2}{x-1}$ to find increasing/decreasing intervals.

1. First find the domain of $f$: $x-1 \neq 0 \implies x \neq 1$, so domain is $(-\infty, 1) \cup (1, \infty)$. $x=1$ is a domain break, so it will be a partition point even though it is not a critical point.
2. Compute $f'(x)$ using the quotient rule:
3. $$f'(x) = \frac{(2x)(x-1) - x^2(1)}{(x-1)^2} = \frac{x(x-2)}{(x-1)^2}$$
4. Find where $f'(x) = 0$ or undefined: $f'(x) = 0$ when the numerator is zero: $x=0, x=2$. $f'(x)$ is undefined when the denominator is zero: $x=1$.
5. Check which points are critical points: $x=0$ and $x=2$ are in the domain of $f$, so they are critical points. $x=1$ is not in the domain, so it is only a domain break/partition point.
6. Order the partition points from smallest to largest: $x=0, x=1, x=2$, splitting the domain into $(-\infty, 0)$, $(0,1)$, $(1,2)$, $(2, \infty)$.

> **exam_tip**
>
> Always sort your partition points from smallest to largest on a number line before testing signs. This eliminates ordering errors that lead to wrong interval conclusions.

## Sign Analysis of the First Derivative

Once you have your ordered partition points, you perform sign analysis to determine if $f'(x)$ is positive or negative in each interval, which tells you if $f$ is increasing or decreasing. There are two common, AP-accepted methods:

**Comparing methods**

Both methods are fully acceptable on the AP exam, but the factor method is faster for the factored derivatives you will encounter on most exam questions.

- **Test-point Method** — Pick any value inside the interval, plug it into $f'(x)$, and check the resulting sign of the output.
  - Pros: Simple, low conceptual overhead for beginners
  - Cons: Slower, more prone to arithmetic errors

- **Factor Method** — For a factored derivative, count the number of negative linear factors. An even number of negative factors gives a positive product, an odd number gives a negative product.
  - Pros: Fast, no arithmetic required for fully factored derivatives
  - Cons: Requires the derivative to be fully factored first

A key shortcut for the factor method: any linear factor raised to an even power is always non-negative (it can never be negative), so it does not affect the sign of $f'(x)$ and can be ignored during sign analysis.

**Worked example:** Use sign analysis on $f'(x) = \frac{x(x-2)}{(x-1)^2}$ (from the previous example) to find where $f(x)$ is increasing and decreasing.

1. We already have partitioned intervals: $(-\infty, 0)$, $(0,1)$, $(1,2)$, $(2, \infty)$.
2. Simplify for sign analysis: $(x-1)^2$ is always positive for $x \neq 1$, so it does not affect the sign of $f'(x)$. We only need to find the sign of $x(x-2)$.
3. Apply the factor method to each interval:
4. - $(-\infty, 0)$: $x$ negative, $x-2$ negative. Product: $(-)(-) = + \implies f'(x) > 0$
- $(0,1)$ and $(1,2)$: $x$ positive, $x-2$ negative. Product: $(+)(-) = - \implies f'(x) < 0$
- $(2, \infty)$: $x$ positive, $x-2$ positive. Product: $(+)(+) = + \implies f'(x) > 0$
5. Conclusion: $f(x)$ is increasing on $(-\infty, 0) \cup (2, \infty)$, and decreasing on $(0,1) \cup (1,2)$.

**Check your understanding**

Test your understanding with this AP-style multiple choice question:

1. The derivative of a function $f$ is given by $f'(x) = (x + 2)(x - 3)^2$. For what values of $x$ is $f$ decreasing?

   - A) $(-\infty, -2)$ only
   - B) $(-2, 3)$
   - C) $(-\infty, -2) \cup (3, \infty)$
   - D) $(-2, 3) \cup (3, \infty)$

   *Why:* $(x-3)^2$ is always positive for all $x \neq 3$, so the sign of $f'(x)$ matches the sign of the linear factor $(x+2)$. $f'(x) < 0$ only when $x < -2$, so the correct answer is A.

**Worked example:** Let $f(x) = 2x^3 - 9x^2 + 12x + 1$. (a) Find all critical points of $f(x)$. (b) Determine the intervals where $f(x)$ is increasing and decreasing. (c) Justify why $f(1) > f(2)$.

1. (a) $f(x)$ is a polynomial, so its domain is all real numbers. Compute and factor the first derivative:
2. $$f'(x) = 6x^2 - 18x + 12 = 6(x-1)(x-2)$$
3. Set $f'(x) = 0$ to get critical points at $x=1$ and $x=2$, both of which are in the domain of $f$. The critical points are $x=1$ and $x=2$.
4. (b) The critical points split the real line into $(-\infty, 1)$, $(1, 2)$, $(2, \infty)$. Since $6 > 0$, the sign of $f'(x)$ matches the sign of $(x-1)(x-2)$:
5. - $(-\infty,1)$: $(-)(-) = + \implies f'(x) > 0$
- $(1,2)$: $(+)(-) = - \implies f'(x) < 0$
- $(2, \infty)$: $(+)(+) = + \implies f'(x) > 0$
6. Conclusion: $f$ is increasing on $(-\infty, 1) \cup (2, \infty)$ and decreasing on $(1, 2)$.
7. (c) By definition, if $f$ is decreasing on the interval $[1, 2]$, then for any $x_1 < x_2$ in $[1,2]$, $f(x_1) > f(x_2)$. Since $1 < 2$ and $f$ is decreasing on $[1,2]$, this directly implies $f(1) > f(2)$, as required.

> **Exam tip:** If you use the test-point method, pick easy integer values for testing to avoid arithmetic errors.

## Common pitfalls

- **Wrong:** Including a domain break (e.g., $x=1$ for $f(x) = \frac{x^2}{x-1}$) as an endpoint of an increasing/decreasing interval
  - Why it fails: Students confuse domain breaks with critical points in the domain, incorrectly extending intervals to include points not in $f$'s domain
  - Correct: Always check if a partition point is in the domain of the original function before including it in an interval; never include points not in the domain.
- **Wrong:** Claiming a function is neither increasing nor decreasing at a point where $f'(c) = 0$, so splitting the interval to exclude $c$
  - Why it fails: Students confuse the derivative at a single point with monotonicity over an interval
  - Correct: Increasing/decreasing is a property of intervals, not individual points; only use isolated zero-derivative points as partition points if the sign changes across them.
- **Wrong:** Counting an even-powered linear factor (e.g., $(x-3)^2$) when calculating the sign of $f'(x)$, leading to an incorrect negative sign
  - Why it fails: Students forget that squaring removes the negative sign from any real number
  - Correct: Cross out any even-powered factors before doing sign analysis, since they are always non-negative and do not change the sign of $f'$.
- **Wrong:** Claiming $f(x) = x^3$ is not increasing on $(-\infty, \infty)$ because $f'(0) = 0$
  - Why it fails: Students confuse zero derivative at an isolated point with zero derivative over an entire interval
  - Correct: If $f'(x) \geq 0$ on an interval and $f'(x) = 0$ only at isolated points, the function is still increasing on the entire interval.
- **Wrong:** Testing the sign of the original function $f(x)$ instead of the derivative $f'(x)$
  - Why it fails: Students rush on exam questions and mix up which function determines increasing/decreasing behavior
  - Correct: Double-check every problem: you test $f'(x)$ to find the behavior of $f(x)$.
- **Wrong:** Claiming $f$ is increasing over the entire union of disjoint intervals when $f$ jumps down between intervals
  - Why it fails: Students are used to combining same-sign intervals and forget that monotonicity requires the inequality to hold across the entire interval
  - Correct: When in doubt, list same-sign intervals separately instead of combining them with a union.

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Increasing function definition | $x_1 < x_2$ on $I \implies f(x_1) < f(x_2)$ | Applies to intervals only, not single points |
| Decreasing function definition | $x_1 < x_2$ on $I \implies f(x_1) > f(x_2)$ | Applies to intervals only, not single points |
| First Derivative Rule (increasing) | $f$ increasing on $I$ if $f'(x) > 0$ for all interior $x$ of $I$ | Isolated zero derivatives do not break this rule |
| First Derivative Rule (decreasing) | $f$ decreasing on $I$ if $f'(x) < 0$ for all interior $x$ of $I$ | Requires $f$ to be continuous on $I$ |
| Critical Point Definition | $c$ in domain of $f$ where $f'(c) = 0$ or $f'(c)$ DNE | Only points in the domain of $f$ count |
| Partition Point Rule | $f'(x)$ can only change sign at critical points or domain breaks | Both types of points must be included when splitting the domain |
| Even-powered Factor Sign Rule | $(x-a)^n$ (even $n$) is non-negative for all $x \neq a$ | Never changes the sign of $f'(x)$ |
| AP Interval Convention | Use open intervals $(a,b)$ for increasing/decreasing | AP accepts closed intervals but open intervals avoid grading errors |

## What's next

This topic is the foundational prerequisite for all remaining topics in AP Calculus AB Unit 5: Analytical Applications of Differentiation. You will next apply your ability to find increasing/decreasing intervals to identify local extrema using the first and second derivative tests, a frequent MCQ and FRQ topic on the AP exam. Correct sign analysis of the first derivative is required for all downstream topics including curve sketching and applied optimization, which make up a large portion of exam points.

- [First Derivative Test for Relative Extrema](https://www.owlsprep.com/study/ap-calculus-ab-u5-first-derivative-test-for-relative/)
- [Candidates test for absolute extrema](https://www.owlsprep.com/study/ap-calculus-ab-u5-candidates-test-for-absolute-extrema/)
- [Determining Concavity](https://www.owlsprep.com/study/ap-calculus-ab-u5-determining-concavity/)

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