Determining intervals where a function is increasing/decreasing — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: the first derivative test for monotonic functions, identification of critical points, sign analysis of the first derivative, distinguishing open vs closed interval notation, and interpretation of increasing/decreasing function behavior on the AP exam.

You should already know: Basic derivative differentiation rules. How to solve polynomial and rational inequalities. How to find the domain of a function.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Determining intervals where a function is increasing/decreasing?

This core foundational skill is part of Unit 5: Analytical Applications of Differentiation, which accounts for 15-18% of the total AP Calculus AB exam weight per the official College Board Course and Exam Description (CED). By definition, a function is increasing on an interval if for any two inputs $x_1<x_2$ in the interval, $f(x_1)<f(x_2)$; it is decreasing if $x_1<x_2$ implies $f(x_1)>f(x_2)$. This topic connects the graphical shape of a function to its first derivative, shifting from informal graphical estimation of increasing/decreasing behavior to rigorous analytical proof using calculus.

It appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam: MCQ often asks to identify correct interval sets from a derivative graph or given function, while FRQ requires justifying increasing/decreasing intervals as part of larger curve-sketching, extrema identification, or optimization problems. Synonyms for this skill include finding monotonic intervals, identifying where a function rises or falls, and analyzing the sign of the first derivative.

2. The First Derivative Rule for Increasing/Decreasing Behavior

The core relationship between the derivative of a function and its monotonicity is derived from the Mean Value Theorem (MVT). For a function $f$ that is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$:

• If $f^{\prime }(x)>0$ for all $x$ in $(a,b)$, then $f$ is increasing on $[a,b]$
• If $f^{\prime }(x)<0$ for all $x$ in $(a,b)$, then $f$ is decreasing on $[a,b]$
• If $f^{\prime }(x)=0$ for all $x$ in $(a,b)$, then $f$ is constant on $[a,b]$

Intuition for this rule comes from the geometric meaning of the derivative: $f^{\prime }(x)$ is the slope of the tangent line to $f$ at $x$. A positive slope means the tangent line points upward as you move from left to right, so the function itself is increasing. A negative slope means the tangent points downward, so the function is decreasing. The MVT formalizes this: if every tangent slope is positive, the secant slope between any two points in the interval must also be positive, so the function satisfies the definition of increasing.

Worked Example

Problem: 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 the first derivative: $f^{\prime }(x)=3x^2-6x=3x(x-2)$.
3. The sign of $f^{\prime }(x)$ can only change where $f^{\prime }(x)=0$, so solve for critical points: $3x(x-2)=0⟹x=0,x=2$. These split the real line into three intervals: $(-\infty ,0)$, $(0,2)$, $(2,\infty )$.
4. Test the sign of $f^{\prime }(x)$ in each interval: For $(-\infty ,0)$, pick $x=-1$: $f^{\prime }(-1)=3(-1)(-3)=9>0$, so $f$ is increasing here. For $(0,2)$, pick $x=1$: $f^{\prime }(1)=3(1)(-1)=-3<0$, so $f$ is decreasing here. For $(2,\infty )$, pick $x=3$: $f^{\prime }(3)=3(3)(1)=9>0$, so $f$ is increasing here.
5. Combine intervals where $f^{\prime }(x)>0$: $f$ is increasing on $(-\infty ,0)\cup (2,\infty )$.

Exam 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 with horizontal tangent points at interval endpoints.

3. Partitioning the Domain with Critical Points and Domain Breaks

To correctly test the sign of $f^{\prime }(x)$, you first need to split (partition) the domain of $f$ into intervals where $f^{\prime }(x)$ is entirely positive or entirely negative. The only points where $f^{\prime }(x)$ can change sign are:

1. Critical points: Points $x=c$ in the domain of $f$ where $f^{\prime }(c)=0$ or $f^{\prime }(c)$ does not exist.
2. Domain breaks: Points $x=c$ not in the domain of $f$, where $f^{\prime }(x)$ is also undefined.

Derivatives satisfy the Intermediate Value Theorem, so they cannot jump from positive to negative without passing through zero or being undefined. This means no sign changes can occur inside a partitioned interval, so you only need one test per interval. A common early mistake is forgetting to include domain breaks as partition points: even though they are not in the domain, the derivative can still change sign across them, so you must test intervals on both sides.

Worked Example

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

1. First find the domain of $f$: $x-1\ne 0⟹x\ne 1$, so domain is $(-\infty ,1)\cup (1,\infty )$. The point $x=1$ is a domain break, so it will be a partition point even though it is not a critical point.
2. Compute $f^{\prime }(x)$ using the quotient rule: $$f^{\prime }(x)=\frac{(2x)(x-1)-x^2(1)}{(x-1{)}^2}=\frac{x^2-2x}{(x-1{)}^2}=\frac{x(x-2)}{(x-1{)}^2}$$
3. Find where $f^{\prime }(x)=0$ or undefined: $f^{\prime }(x)=0$ when the numerator is zero: $x=0,x=2$. $f^{\prime }(x)$ is undefined when the denominator is zero: $x=1$.
4. 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.
5. 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.

4. Sign Analysis of the First Derivative

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

1. Test-point method: Pick any value inside the interval, plug it into $f^{\prime }(x)$, and check the resulting sign. This is reliable, but slower for factored derivatives.
2. 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. This is much faster for the factored derivatives you will encounter on the AP exam.

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 $f^{\prime }(x)$ and can be ignored during sign analysis.

Worked Example

Problem: Use sign analysis on $f^{\prime }(x)=\frac{x(x-2)}{(x-1{)}^2}$ (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 $x\ne 1$, so it does not affect the sign of $f^{\prime }(x)$. We only need to find the sign of $x(x-2)$.
3. Apply the factor method to each interval:
   • $(-\infty ,0)$: $x<0⟹x$ negative; $x-2<0⟹x-2$ negative. Product: $(-)(-)=+⟹f^{\prime }(x)>0$.
   • $(0,1)$: $0<x<1⟹x$ positive; $x-2<0⟹x-2$ negative. Product: $(+)(-)=-⟹f^{\prime }(x)<0$.
   • $(1,2)$: $1<x<2⟹x$ positive; $x-2<0⟹x-2$ negative. Product: $(+)(-)=-⟹f^{\prime }(x)<0$.
   • $(2,\infty )$: $x>2⟹x$ positive; $x-2>0⟹x-2$ positive. Product: $(+)(+)=+⟹f^{\prime }(x)>0$.
4. Conclusion: $f(x)$ is increasing on $(-\infty ,0)\cup (2,\infty )$, and decreasing on $(0,1)\cup (1,2)$.

Exam tip: If you use the test-point method, pick easy integer values for testing. This avoids arithmetic errors that lead to wrong sign conclusions.

Common Pitfalls (and how to avoid them)

• Wrong move: 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: Students confuse domain breaks with critical points in the domain, so they incorrectly extend intervals to include points not in $f$'s domain. Correct move: 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 move: Claiming a function is neither increasing nor decreasing at a point where $f^{\prime }(c)=0$, so splitting the interval to exclude $c$. Why: Students confuse the derivative at a single point with monotonicity over an interval. Correct move: 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 move: Counting an even-powered linear factor (e.g., $(x-3{)}^2$) when calculating the sign of $f^{\prime }(x)$, leading to an incorrect negative sign. Why: Students forget that squaring removes the negative sign from any real number. Correct move: Cross out any even-powered factors before doing sign analysis, since they are always non-negative and do not change the sign of $f^{\prime }$.
• Wrong move: Claiming $f(x)=x^3$ is not increasing on $(-\infty ,\infty )$ because $f^{\prime }(0)=0$. Why: Students confuse zero derivative at an isolated point with zero derivative over an entire interval. Correct move: If $f^{\prime }(x)\ge 0$ on an interval and $f^{\prime }(x)=0$ only at isolated points, the function is still increasing on the entire interval.
• Wrong move: Testing the sign of the original function $f(x)$ instead of the derivative $f^{\prime }(x)$. Why: Students rush on exam questions and mix up which function determines increasing/decreasing behavior. Correct move: Double-check every problem: you test $f^{\prime }(x)$ to find the behavior of $f(x)$.
• Wrong move: Writing a combined union of disjoint intervals and claiming $f$ is increasing over the entire union, when $f$ jumps down between intervals. Why: Students are used to combining same-sign intervals and forget that monotonicity requires the inequality to hold across the entire interval. Correct move: When in doubt, list same-sign intervals separately instead of combining them with a union.

Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

The derivative of a function $f$ is given by $f^{\prime }(x)=(x+2)(x-3{)}^2$. For what values of $x$ is $f$ decreasing? A) $(-\infty ,-2)$ only B) $(-2,3)$ C) $(-\infty ,-2)\cup (3,\infty )$ D) $(-2,3)\cup (3,\infty )$

Worked Solution: A function $f$ is decreasing where its first derivative $f^{\prime }(x)<0$. First, identify the partition points at $x=-2$ and $x=3$, where $f^{\prime }(x)=0$. Next, note that $(x-3{)}^2$ is always positive for all $x\ne 3$, so the sign of $f^{\prime }(x)$ matches the sign of the linear factor $(x+2)$. The factor $(x+2)$ is negative when $x<-2$ and positive when $x>-2$, so $f^{\prime }(x)<0$ only on the interval $(-\infty ,-2)$. The correct answer is A.

Question 2 (Free Response)

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 $f(1)>f(2)$ using your results from part (b).

Worked Solution: (a) $f(x)$ is a polynomial, so its domain is all real numbers. Compute the first derivative: $f^{\prime }(x)=6x^2-18x+12=6(x^2-3x+2)=6(x-1)(x-2)$. Set $f^{\prime }(x)=0$ 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 $x=2$.

(b) The critical points split the real line into $(-\infty ,1)$, $(1,2)$, $(2,\infty )$. Since $6>0$, the sign of $f^{\prime }(x)$ matches the sign of $(x-1)(x-2)$. For $(-\infty ,1)$: $(-)(-)=+⟹f^{\prime }(x)>0$. For $(1,2)$: $(+)(-)=-⟹f^{\prime }(x)<0$. For $(2,\infty )$: $(+)(+)=+⟹f^{\prime }(x)>0$. Conclusion: $f$ is increasing on $(-\infty ,1)\cup (2,\infty )$ and decreasing on $(1,2)$.

(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.

Question 3 (Application / Real-World Style)

The daily profit a small bakery makes from selling $x$ batches of sourdough bread is given by $P(x)=-0.1x^3+1.2x^2+2x$, where $x\ge 0$ is the number of batches, and $P(x)$ is profit measured in hundreds of dollars. Determine the intervals where profit is increasing as the number of batches increases, and interpret your result in context.

Worked Solution: The domain of $P(x)$ is $x\ge 0$. Compute the derivative: $P^{\prime }(x)=-0.3x^2+2.4x+2$. Set $P^{\prime }(x)=0$, multiply through by $-10$ to get $3x^2-24x-20=0$. Use the quadratic formula: $$x=\frac{24\pm \sqrt{24^2+4(3)(20)}}{2(3)}=\frac{24\pm \sqrt{816}}{6}\approx 8.76$$ Only the positive root is in the domain, so we partition into $(0,8.76)$ and $(8.76,\infty )$. Test $x=1$: $P^{\prime }(1)=-0.3+2.4+2=4.1>0$, so $P(x)$ is increasing on $(0,8.76)$. Test $x=10$: $P^{\prime }(10)=-30+24+2=-4<0$, so $P(x)$ is decreasing for $x>8.76$.

Interpretation: The bakery's total daily profit increases as they produce up to approximately 9 batches of sourdough per day; producing more than 9 batches per day will decrease their total daily profit.

Quick Reference Cheatsheet

What's Next

This topic is the foundational prerequisite for all remaining topics in Unit 5: Analytical Applications of Differentiation. Next, you will apply your ability to find increasing/decreasing intervals to identify local extrema using the first derivative test and second derivative test, a frequent MCQ and FRQ topic on the AP exam. Without correctly identifying the sign of the first derivative on intervals, you cannot correctly classify critical points as maxima, minima, or neither, which is required for curve sketching and applied optimization problems. This topic also feeds into the larger analysis of function behavior across the entire AP course, from graphing functions to solving real-world optimization questions that appear regularly on the FRQ section.

• Finding critical points
• First derivative test for local extrema
• Curve sketching with derivatives
• Optimization problems