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

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The Increasing/Decreasing Theorem for differentiable functions, identification of critical points, construction of derivative sign charts, interval notation conventions, and applying the first derivative to find intervals of increase/decrease for explicit, implicit, and parametric functions.

You should already know: How to compute first derivatives including chain rule, quotient, and product 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 BC 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 topic is part of Unit 5: Analytical Applications of Differentiation, which counts for 15–18% of the total AP Calculus BC exam score per the official CED. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, almost always as a required step in larger problems about curve sketching, extrema classification, optimization, or contextual analysis.

By definition, a function $f(x)$ is increasing on an interval $I$ if for any two points $x_1<x_2$ in $I$, $f(x_1)<f(x_2)$. A function is decreasing on $I$ if $x_1<x_2$ implies $f(x_1)>f(x_2)$. A function that is entirely increasing or decreasing on an interval is called monotonic, a term you may see on the exam. Unlike graphical estimation, this topic uses analytical differentiation to rigorously identify monotonic intervals, which is what the AP exam expects for full credit on justifications. The standard AP convention asks for open intervals, unless you are explicitly told to include endpoints.

2. The Increasing/Decreasing Theorem

The foundational rule that links derivative slope to function monotonicity is the Increasing/Decreasing Theorem, which states:

If $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then:

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

Intuition for this theorem is straightforward: the first derivative $f^{\prime }(x)$ gives the slope of the tangent line to $f(x)$ at any point. A positive slope means the function rises as you move left to right, while a negative slope means it falls. The theorem only requires differentiability on the open interval (since endpoints only need continuity to be included), but the AP exam almost always asks for open intervals, so you only need to evaluate the sign of $f^{\prime }(x)$ on the open interval to answer the question.

Worked Example

Find all open intervals where $f(x)=x^3-3x^2+2x-1$ is increasing and decreasing.

1. Compute the first derivative using the power rule: $f^{\prime }(x)=3x^2-6x+2$.
2. Find where $f^{\prime }(x)=0$ (it is defined everywhere as a polynomial): $$x=\frac{6\pm \sqrt{36-24}}{6}=1\pm \frac{\sqrt{3}}{3}$$
3. These critical points split the domain of $f$ (all real numbers) into three open intervals: $\left(-\infty ,1-\frac{\sqrt{3}}{3}\right)$, $\left(1-\frac{\sqrt{3}}{3},1+\frac{\sqrt{3}}{3}\right)$, $\left(1+\frac{\sqrt{3}}{3},\infty \right)$.
4. Test the sign of $f^{\prime }$ in each interval: $f^{\prime }(0)=2>0$, $f^{\prime }(1)=-1<0$, $f^{\prime }(2)=2>0$.
5. Final result: Increasing on $\left(-\infty ,1-\frac{\sqrt{3}}{3}\right)\cup \left(1+\frac{\sqrt{3}}{3},\infty \right)$; decreasing on $\left(1-\frac{\sqrt{3}}{3},1+\frac{\sqrt{3}}{3}\right)$.

Exam tip: On AP FRQs, you must explicitly state that $f^{\prime }(x)>0$ (or $<0$) to justify your interval — you will lose points if you only give the interval without referencing the derivative sign.

3. Critical Points and Sign Chart Construction

The only points where the sign of $f^{\prime }(x)$ can change are critical points and domain breaks. By definition, a critical point of $f$ is a point $x=c$ that is in the domain of the original function $f$ where $f^{\prime }(c)=0$ or $f^{\prime }(c)$ is undefined. Domain breaks (points not in the domain of $f$ where $f^{\prime }$ is undefined) do not count as critical points, but they still split the domain into separate intervals, so they must be included in your sign chart.

The standard method for finding intervals of increase/decrease is:

1. Find the domain of the original function $f$.
2. Compute and fully factor $f^{\prime }(x)$.
3. List all critical points and domain breaks in order from left to right.
4. Split the domain into open intervals between consecutive ordered points.
5. Test the sign of $f^{\prime }(x)$ in each interval, then assign increasing/decreasing based on the sign.

Worked Example

Find all open intervals where $f(x)=\frac{x^2}{x-2}$ is increasing and decreasing.

1. Domain of $f$: $x\ne 2$, so $(-\infty ,2)\cup (2,\infty )$.
2. Compute $f^{\prime }(x)$ via quotient rule: $$f^{\prime }(x)=\frac{(2x)(x-2)-x^2(1)}{(x-2{)}^2}=\frac{x^2-4x}{(x-2{)}^2}=\frac{x(x-4)}{(x-2{)}^2}$$
3. Identify split points: $f^{\prime }(x)=0$ at $x=0$ and $x=4$ (both in domain, so critical points). $f^{\prime }(x)$ is undefined at $x=2$, which is not in domain, so it is a domain break, not a critical point. Ordered split points: $0,2,4$.
4. Split into intervals: $(-\infty ,0),(0,2),(2,4),(4,\infty )$.
5. Test sign: $f^{\prime }(-1)=5/9>0$, $f^{\prime }(1)=-3<0$, $f^{\prime }(3)=-3<0$, $f^{\prime }(5)=5/9>0$.
6. Final result: Increasing on $(-\infty ,0)\cup (4,\infty )$; decreasing on $(0,2)\cup (2,4)$.

Exam tip: Always factor $f^{\prime }(x)$ completely before building your sign chart. Factoring makes sign testing trivial by letting you evaluate the sign of each term separately, instead of calculating the entire derivative from scratch for each test point.

4. Intervals of Increase/Decrease for Parametric Functions

AP Calculus BC requires you to apply this skill to parametric functions, which are common on the exam. For a parametric curve defined by $x(t)$ and $y(t)$, the slope of the curve with respect to $x$ is $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$. The sign of $\frac{dy}{dx}$ follows the same rule as for explicit functions: $\frac{dy}{dx}>0$ means the curve is increasing as a function of $x$, and $\frac{dy}{dx}<0$ means it is decreasing.

If $x(t)$ is strictly monotonic (always increasing or decreasing), every interval of $t$ maps one-to-one to an interval of $x$, so you can convert your interval from $t$ to $x$ by substituting the endpoints of the $t$-interval into $x(t).Ifthequestionasksforintervalsof$t$wherethecurveisincreasing/decreasing,youcanleaveyouranswerintermsof$t.

Worked Example

Given the parametric curve $x(t)=e^t$, $y(t)=t^3-3t$ defined for all real $t$, find all intervals of $x$ where $y(x)$ is increasing.

1. Compute derivatives: $\frac{dx}{dt}=e^t>0$ for all $t$, so $x(t)$ is strictly increasing, so one-to-one mapping between $t$ and $x$.
2. Compute $\frac{dy}{dx}$: $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3t^2-3}{e^t}=\frac{3(t^2-1)}{e^t}$$
3. Find where $\frac{dy}{dx}=0$: numerator is zero at $t=-1$ and $t=1$; denominator $e^t$ is always positive, so the sign of $\frac{dy}{dx}$ matches the sign of $t^2-1$.
4. Split into intervals of $t$: $(-\infty ,-1)$, $(-1,1)$, $(1,\infty )$. $\frac{dy}{dx}>0$ on $(-\infty ,-1)$ and $(1,\infty )$.
5. Convert to intervals of $x$: $x=e^t$, so when $t<-1$, $x<e^{-1}=1/e$; when $t>1$, $x>e$.
6. Final result: $y(x)$ is increasing on $\left(-\infty ,\frac{1}{e}\right)\cup (e,\infty )$.

Exam tip: Always read the question carefully to confirm whether it asks for intervals of $t$ or intervals of $x$. Forgetting to convert from $t$ to $x$ when required is a common easy mistake that costs points on the exam.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Calling $x=2$ a critical point for $f(x)=\frac{x^2}{x-2}$ because $f^{\prime }(2)$ is undefined. Why: Students forget the critical point definition requires the point to be in the domain of the original function. Correct move: Always check if any point where $f^{\prime }$ is undefined is in the domain of $f$ before labeling it a critical point, and include all domain breaks in your sign chart even if they are not critical.
• Wrong move: Testing the sign of $f^{\prime }(x)$ at the split point itself instead of inside the interval. Why: Students rush and test the split point, which gives $f^{\prime }(x)=0$, leading to incorrect sign assignment. Correct move: Always pick a test point strictly inside each open interval when checking the sign of $f^{\prime }$.
• Wrong move: Combining non-adjacent intervals of increase into a single connected interval, e.g., writing increasing on $(-\infty ,4)$ instead of $(-\infty ,0)\cup (4,\infty )$ for the earlier rational function example. Why: Students forget critical points split the domain and incorrectly combine intervals when the derivative sign is the same on both sides of a middle decreasing interval. Correct move: Only combine adjacent intervals that have the same sign of $f^{\prime }$.
• Wrong move: Justifying an interval of increase by describing the graph as "going up" instead of referencing the derivative sign on FRQs. Why: Students rely on graphical intuition instead of the analytical justification AP requires. Correct move: Always justify with the statement "$f^{\prime }(x)>0$ for all $x$ in the interval" for increasing, and "$f^{\prime }(x)<0$ for all $x$ in the interval" for decreasing.
• Wrong move: Leaving $(x-2{)}^2$ unfactored and incorrectly assigning a negative sign to it in the denominator. Why: Students forget that any squared real term is non-negative, leading to sign errors. Correct move: Always factor $f^{\prime }(x)$ fully into linear terms, and note that squared terms are always positive (except at the root) so they do not change the sign of $f^{\prime }$.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

Which of the following gives all open intervals where $f(x)=x{e}^{-2x}$ is increasing? A) $\left(0,\frac{1}{2}\right)$ B) $\left(-\infty ,\frac{1}{2}\right)$ C) $\left(\frac{1}{2},\infty \right)$ D) $f(x)$ is never increasing

Worked Solution: First, compute $f^{\prime }(x)$ using the product rule: $f^{\prime }(x)=e^{-2x}+x(-2e^{-2x})=e^{-2x}(1-2x)$. The term $e^{-2x}$ is always positive for all real $x$, so the sign of $f^{\prime }(x)$ depends only on $(1-2x)$. Setting $f^{\prime }(x)=0$ gives $x=1/2$. For all $x<1/2$, $1-2x>0$, so $f^{\prime }(x)>0$, meaning $f(x)$ is increasing. For $x>1/2$, $f^{\prime }(x)<0$, so $f(x)$ is decreasing. The correct answer is B.

Question 2 (Free Response)

Let $f(x)=\frac{\ln x}{x}$ for $x>0$. (a) Find $f^{\prime }(x)$. (b) Find all open intervals where $f(x)$ is increasing and all open intervals where it is decreasing. (c) Justify your answer from part (b) using the Increasing/Decreasing Theorem.

Worked Solution: (a) Use the quotient rule to differentiate: $$f^{\prime }(x)=\frac{\left(\frac{1}{x}\right)(x)-(\ln x)(1)}{x^2}=\frac{1-\ln x}{x^2}$$

(b) $f(x)$ is defined and continuous for all $x>0$, and $f^{\prime }(x)$ is defined for all $x>0$. Set $f^{\prime }(x)=0$, so numerator $1-\ln x=0⟹x=e$. Split the domain into $(0,e)$ and $(e,\infty )$. Testing sign: $f^{\prime }(1)=1>0$ on $(0,e)$, $f^{\prime }(e^2)=-1/e^4<0$ on $(e,\infty )$. Final answer: Increasing on $(0,e)$, decreasing on $(e,\infty )$.

(c) $f(x)=\frac{\ln x}{x}$ is continuous on $(0,\infty )$ and differentiable on all open intervals within $(0,\infty )$, so the Increasing/Decreasing Theorem applies. $f^{\prime }(x)>0$ for all $x\in (0,e)$, so $f$ is increasing on $(0,e)$. $f^{\prime }(x)<0$ for all $x\in (e,\infty )$, so $f$ is decreasing on $(e,\infty )$.

Question 3 (Application / Real-World Style)

A local coffee shop models its daily profit from selling lattes as $P(s)=-s^3+12s^2-36s+20$, where $P(s)$ is profit in US dollars, and $s$ is the price per latte in dollars, for $2<s<8$. Find all intervals of price where profit increases as the price per latte increases, and interpret your result.

Worked Solution: First, compute the derivative of profit with respect to price: $$P^{\prime }(s)=-3s^2+24s-36=-3(s^2-8s+12)=-3(s-2)(s-6)$$ The domain is $2<s<8$, so the only critical point in the domain is $s=6$. Split into intervals $(2,6)$ and $(6,8)$. Testing sign: $P^{\prime }(3)=-3(1)(-3)=9>0$ on $(2,6)$, $P^{\prime }(7)=-3(5)(1)=-15<0$ on $(6,8)$. Profit is increasing for prices between $\$2$ and $\$6$ per latte.

Interpretation: Increasing the price per latte between $\$2$ and $\$6$ will increase the coffee shop's total daily profit, but increasing the price beyond $\$6$ per latte will cause total daily profit to decrease.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for all further analysis of function behavior in Unit 5, and mastering it is required to correctly solve almost all later analytical differentiation problems. Next, you will apply your ability to find intervals of increase and decrease to the first derivative test for classifying local extrema; without correctly identifying the sign of the derivative on either side of a critical point, you cannot correctly classify extrema, which will cost you points on both MCQ and FRQ. This topic also introduces the sign-chart method that you will reuse directly to find intervals of concavity and points of inflection, and it is the first step in all constrained optimization problems. In the bigger picture, this skill is core to all contextual problems where you need to analyze how a function changes over its domain, from physics motion problems to economic profit models.

• Finding critical points and extrema
• First derivative test for classification of extrema
• Determining intervals of concavity
• Curve sketching with derivatives