Determining Intervals Where a Function Is Increasing/Decreasing

AP Calculus BC· AP Calculus BC CED — Analytical Applications of Differentiation· 14 min read

1. Core Definitions and the Increasing/Decreasing Theorem★★☆☆☆⏱ 3 min

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 that may appear on the AP exam.

📘 Definition

Increasing/Decreasing Theorem

If $f (x)$ is continuous on a closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , then: 1. If $f^{'} (x) > 0$ for all $x \in (a, b)$ , then $f (x)$ is increasing on $[a, b]$ ; 2. If $f^{'} (x) < 0$ for all $x \in (a, b)$ , then $f (x)$ is decreasing on $[a, b]$ .

Example:

A positive derivative means the tangent slope is positive, so the function rises moving left to right.

📐 Worked Example

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

1
Compute the first derivative using the power rule:
$f'(x) = 3x^2 - 6x + 2$
2
Find where $f^{'} (x) = 0$ (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.
4
Test the sign of $f^{'}$ in each interval: $f^{'} (0) = 2 > 0$ , $f^{'} (1) = - 1 < 0$ , $f^{'} (2) = 2 > 0$ .
5
Final result: Increasing on $(- \infty, 1 - \frac{3}{3}) \cup (1 + \frac{3}{3}, \infty)$ ; decreasing on $(1 - \frac{3}{3}, 1 + \frac{3}{3})$ .

Exam tip:

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

2. Critical Points and Sign Chart Construction★★★☆☆⏱ 4 min

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

Find the domain of the original function $f$ .
Compute and fully factor $f^{'} (x)$ .
List all critical points and domain breaks in order from left to right.
Split the domain into open intervals between consecutive ordered points.
Test the sign of $f^{'} (x)$ 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 $f$ : $x \neq = 2$ , so $(- \infty, 2) \cup (2, \infty)$ .
2
Compute $f^{'} (x)$ via quotient rule, then factor:
$f'(x) = \frac{(2x)(x-2) - x^2(1)}{(x-2)^2} = \frac{x^2 - 4x}{(x-2)^2} = \frac{x(x-4)}{(x-2)^2}$
3
Identify split points: $f^{'} (x) = 0$ at $x = 0$ and $x = 4$ , both in the domain of $f$ so they are critical points. $f^{'} (x)$ is undefined at $x = 2$ , which is not in the domain of $f$ , 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 of $f^{'}$ in each interval: $f^{'} (- 1) = 5/9 > 0$ , $f^{'} (1) = - 3 < 0$ , $f^{'} (3) = - 3 < 0$ , $f^{'} (5) = 5/9 > 0$ .
6
Final result: Increasing on $(- \infty, 0) \cup (4, \infty)$ ; decreasing on $(0, 2) \cup (2, 4)$ .

Exam tip:

Always factor $f^{'} (x)$ completely before building your sign chart. Factoring makes sign testing trivial by letting you evaluate the sign of each term separately, instead of recalculating the entire derivative for each test point.

3. Intervals for Parametric Functions★★★★☆BC only⏱ 3 min

AP Calculus BC requires applying this method to parametric curves. For a parametric curve defined by $x (t)$ and $y (t)$ , the slope of $y$ with respect to $x$ is $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ . The same increasing/decreasing rule applies: $\frac{d y}{d x} > 0$ means the curve is increasing as a function of $x$ , and $\frac{d y}{d x} < 0$ 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 $x (t)$ . If the question asks for intervals of $t$ , you can leave your answer in terms of $t$ .

📐 Worked Example

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

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

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 mistake that costs points.

4. AP-Style Worked Practice Examples★★★☆☆⏱ 4 min

📐 Worked Example

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

1
Compute $f^{'} (x)$ using the product rule:
$f'(x) = e^{-2x} + x(-2e^{-2x}) = e^{-2x}(1 - 2x)$
2
The term $e^{- 2 x}$ is always positive for all real $x$ , so the sign of $f^{'} (x)$ depends only on $(1 - 2 x)$ . Setting $f^{'} (x) = 0$ gives $x = 1/2$ .
3
For all $x < 1/2$ , $1 - 2 x > 0$ , so $f^{'} (x) > 0$ , meaning $f (x)$ is increasing. For $x > 1/2$ , $f^{'} (x) < 0$ , so $f (x)$ is decreasing. The correct answer is B.

📐 Worked Example

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

1
(a) Use quotient rule to differentiate:
$f'(x) = \frac{\left(\frac{1}{x}\right)(x) - (\ln x)(1)}{x^2} = \frac{1 - \ln x}{x^2}$
2
(b) $f (x)$ is defined and continuous for all $x > 0$ , and $f^{'} (x)$ is defined for all $x > 0$ . Set $f^{'} (x) = 0$ , so $1 - ln x = 0 ⟹ x = e$ . Split the domain into $(0, e)$ and $(e, \infty)$ . Testing sign gives $f^{'} (1) = 1 > 0$ on $(0, e)$ and $f^{'} (e^{2}) = - 1/ e^{4} < 0$ on $(e, \infty)$ . Final answer: Increasing on $(0, e)$ , decreasing on $(e, \infty)$ .
3
(c) $f (x) = \frac{l n x}{x}$ is continuous on $(0, \infty)$ and differentiable on all open intervals within $(0, \infty)$ , so the Increasing/Decreasing Theorem applies. $f^{'} (x) > 0$ for all $x \in (0, e)$ , so $f$ is increasing on $(0, e)$ . $f^{'} (x) < 0$ for all $x \in (e, \infty)$ , so $f$ is decreasing on $(e, \infty)$ .

5. Common Pitfalls

Wrong move:

Calling $x = 2$ a critical point for $f (x) = \frac{x ^{2}}{x - 2}$ because $f^{'} (2)$ is undefined.

Why:

The definition of a critical point requires the point to be in the domain of the original function.

Correct move:

Always check if any point where $f^{'}$ 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 $f^{'} (x)$ at the split point itself instead of inside the interval.

Why:

Split points have $f^{'} (x) = 0$ or undefined, so testing here gives no information about the sign of $f^{'}$ on the interval.

Correct move:

Always pick a test point strictly inside each open interval when checking the sign of $f^{'}$ .

Wrong move:

Combining non-adjacent intervals of increase into a single connected interval, e.g., writing increasing on $(- \infty, 4)$ instead of $(- \infty, 0) \cup (4, \infty)$ .

Why:

Critical points and domain breaks split the domain, and the derivative sign can change between non-adjacent intervals.

Correct move:

Only combine adjacent intervals that have the same sign of $f^{'}$ .

Wrong move:

Justifying an interval of increase by describing the graph as 'going up' instead of referencing the derivative sign on FRQs.

Why:

AP exams require analytical justifications, not graphical intuition, for full credit.

Correct move:

Always justify with the statement ' $f^{'} (x) > 0$ for all $x$ in the interval' for increasing, and ' $f^{'} (x) < 0$ 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:

Any squared real term is non-negative, so it does not change the sign of the derivative.

Correct move:

Always factor $f^{'} (x)$ fully into linear terms, and note that squared terms are always positive (except at the root) so they do not change the sign of $f^{'}$ .

6. Quick Reference Cheatsheet

Category	Rule/Formula	Notes
Increasing function definition	$x_{1} < x_{2} \in I ⟹ f (x_{1}) < f (x_{2})$	Applies to any interval, any function
Decreasing function definition	$x_{1} < x_{2} \in I ⟹ f (x_{1}) > f (x_{2})$	Applies to any interval, any function
Increasing/Decreasing Theorem	$f^{'} (x) > 0 ⟹ f$ increasing; $f^{'} (x) < 0 ⟹ f$ decreasing	Requires $f$ continuous on $I$ , differentiable on interior
Critical point definition	$x = c$ is critical if $c \in dom (f)$ and $f^{'} (c) = 0$ or $f^{'} (c)$ undefined	Non-domain points are not critical, but still split domain
Parametric curve slope	$\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$	$\frac{d y}{d x}$ positive when $d y / d t, d x / d t$ have same sign
Sign testing rule	Test $f^{'}$ sign at one point per open interval	$f^{'}$ can only change sign at split points
FRQ Justification (increasing)	State $f^{'} (x) > 0$ for all $x$ in interval	Graphical descriptions not accepted for full credit
FRQ Justification (decreasing)	State $f^{'} (x) < 0$ for all $x$ in interval	Graphical descriptions not accepted for full credit

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · BC MCQ
Find intervals for rational function
2022 · BC FRQ
Parametric intervals of increase

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.