Increasing and Decreasing Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 1.1: “Review of Functions” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First: Reading the Graph
Picture the graph of $f(x) = x^3 - 3x$ (a cubic with a local max near $x = -1$ and local min near $x = 1$).
Before any algebra:
- On what intervals is the graph rising as you move from left to right?
- On what intervals is it falling?
Write your intervals. Then compute $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$ and check: where is $f' > 0$? Does it match the rising intervals?
Quantity-First Framing
A function is increasing on an interval when moving to the right on the interval moves you upward on the graph. It is decreasing when moving right moves you downward. These are purely geometric observations that can also be read off the formula or derivative.
Prerequisite Check
Quick Reference
Definitions.
$f$ is increasing on an interval $I$ if: for all $x_1, x_2 \in I$ with $x_1 < x_2$: $f(x_1) < f(x_2)$.
$f$ is decreasing on an interval $I$ if: for all $x_1, x_2 \in I$ with $x_1 < x_2$: $f(x_1) > f(x_2)$.
Shortcut using derivatives (for differentiable $f$):
- $f'(x) > 0$ on $I$ $\Rightarrow$ $f$ is increasing on $I$.
- $f'(x) < 0$ on $I$ $\Rightarrow$ $f$ is decreasing on $I$.
Key Concepts
1. The Formal Definitions
Increasing: as $x$ grows, so does $f(x)$. Decreasing: as $x$ grows, $f(x)$ shrinks.
The formal statements use universal quantifiers: “for ALL $x_1 < x_2$ in $I$.” It is not enough that $f$ rises at some points -- it must rise at EVERY pair of points in the interval.
Example 1. Is $f(x) = x^2$ increasing on $[-2, 2]$?
No. $f(-2) = 4$ and $f(0) = 0$: even though $-2 < 0$, $f(-2) > f(0)$. The function is decreasing on $[-2, 0]$ and increasing on $[0, 2]$.
2. Two Representations: Graph and Formula
Graph. Scan the curve from left to right:
- Rising: increasing.
- Falling: decreasing.
- Flat (horizontal tangent): neither (momentarily), but the function can resume increasing or decreasing on either side.
Formula. For $f(x) = x^3 - 3x$: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$.
| Interval | Sign of $f'$ | $f$ is... |
|---|---|---|
| $(-\infty, -1)$ | $+$ | Increasing |
| $(-1, 1)$ | $-$ | Decreasing |
| $(1, \infty)$ | $+$ | Increasing |
Translation prompt. From the table: $f$ decreases on $(-1, 1)$. The graph should have a local max at $x = -1$ (transitions from rising to falling) and a local min at $x = 1$ (transitions from falling to rising). Verify: $f(-1) = -1 + 3 = 2$ (local max), $f(1) = 1 - 3 = -2$ (local min).
3. Monotone Functions
A function that is increasing on its entire domain is called monotone increasing (or strictly increasing). Similarly for monotone decreasing.
Examples.
- $f(x) = e^x$: monotone increasing on $(-\infty, \infty)$.
- $f(x) = \ln x$: monotone increasing on $(0, \infty)$.
- $f(x) = -x$: monotone decreasing on $(-\infty, \infty)$.
- $f(x) = 1/x$: decreasing on $(-\infty, 0)$ and on $(0, \infty)$, but NOT monotone on its full domain (because it jumps from negative to positive).
4. Ask Why: Why Does $f' > 0$ Imply Increasing?
If the derivative is positive at every point of an interval, the function is always “tilted upward” -- every infinitesimally small step to the right moves you up. This is the geometric meaning of positive slope. The Mean Value Theorem makes this rigorous: for $x_1 < x_2$ in $I$, $f(x_2) - f(x_1) = f'(c)(x_2 - x_1)$ for some $c \in (x_1, x_2)$; since $f'(c) > 0$ and $x_2 - x_1 > 0$, $f(x_2) - f(x_1) > 0$.
Named Misconception: height-vs-slope
Increasing does not mean the function value is positive -- it means the function value is getting larger. $f(x) = x - 100$ is increasing on all of $\mathbb{R}$ even though $f(x) < 0$ for $x < 100$.
Similarly, a function can be positive but decreasing (e.g., $f(x) = 1/x$ for $x > 0$: always positive, always decreasing).
Height (the value $f(x)$) and slope (the direction of change) are independent.
Common Errors
| Error | Specific example | Correction |
|---|---|---|
| Confusing height with direction | “Since $f(x) > 0$, it must be increasing” | Increasing is about the direction, not the height |
| Local behavior vs. interval | “Since $f'(0) = 0$, $f$ is not increasing anywhere near 0” | Need to check the sign of $f'$ on an interval, not just at one point |
Leveled Practice
Level 1 -- Identifying Behavior
Problem 1. For $f(x) = -3x + 7$: state whether $f$ is increasing or decreasing, and on what interval.
Show answer
$f'(x) = -3 < 0$ everywhere. $f$ is decreasing on $(-\infty, \infty)$.
Problem 2. For $f(x) = x^2 - 4x + 3$: find the intervals of increase and decrease.
Show answer
$f'(x) = 2x - 4 = 2(x-2)$.
$f' > 0$ when $x > 2$: increasing on $(2, \infty)$.
$f' < 0$ when $x < 2$: decreasing on $(-\infty, 2)$.
Level 2 -- More Complex Functions
Problem 3. For $f(x) = x^3 - 6x^2 + 9x - 2$: find all intervals of increase and decrease.
Show answer
$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$.
$f' > 0$ when $x < 1$ or $x > 3$: increasing on $(-\infty, 1) \cup (3, \infty)$.
$f' < 0$ when $1 < x < 3$: decreasing on $(1, 3)$.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 4 (Extension).
(a) (Floor) Without computing a derivative, argue from the graph of $\sin x$ that it is increasing on $(-\pi/2, \pi/2)$.
(b) (Mid) Show that $f(x) = e^x$ is increasing on all of $\mathbb{R}$ using the definition (not the derivative): for $x_1 < x_2$, show $e^{x_1} < e^{x_2}$.
(c) (Ceiling) Can a function be both increasing and decreasing on an interval? Argue from the definition whether such a function can exist.
Show answer
(a) $\sin x$ rises from $-1$ at $x = -\pi/2$ to 0 at $x = 0$ to 1 at $x = \pi/2$. For any $x_1 < x_2$ in $(-\pi/2, \pi/2)$, the graph goes from left to right without any dip, so $\sin x_1 < \sin x_2$.
(b) Since $x_1 < x_2$: $x_2 - x_1 > 0$. Then $e^{x_2}/e^{x_1} = e^{x_2 - x_1} > e^0 = 1$ (since $e^t > 1$ for $t > 0$). So $e^{x_2} > e^{x_1}$. Increasing.
(c) No. If $f$ were both increasing and decreasing on $I$, take any $x_1 < x_2$ in $I$. Increasing requires $f(x_1) < f(x_2)$; decreasing requires $f(x_1) > f(x_2)$. Both cannot hold simultaneously. Contradiction. So no such function exists.
Common Misconceptions
a function is increasing wherever its graph is “high.” A function is increasing where its output rises as the input increases, not where it takes large values. The function $f(x) = -x^2 + 100$ has very large values near $x = 0$ but is decreasing for $x > 0$. A mountain peak is at its highest value exactly where it stops increasing. Height and rate of change are different things.
“increasing” and “positive” mean the same thing. A function can be increasing while taking negative values (e.g., $f(x) = x - 5$ is increasing for all $x$ but is negative for $x < 5$), and it can be positive while decreasing (e.g., $f(x) = 10 - x$ is positive for $x < 10$ but always decreasing). Increasing refers to the direction of change of the output, not to whether the output is above zero.
Mastery Checklist
Mental Model
Increasing and decreasing are purely about direction: as you walk right along the $x$-axis, does the graph go up (increasing) or down (decreasing)?
The derivative is the slope at each point. A positive slope means going uphill; negative means going downhill. Zero slope means momentarily flat -- not strictly increasing or decreasing at that point, though the function can resume either on either side.
This direction analysis is the first step in curve sketching: know where the function rises and falls to understand its shape.
Connections
Within MATH161
- Derivative sign: The connection between $f' > 0$ and increasing is the first application of derivatives to graph analysis. The full curve-sketching procedure in Chapter 3 uses this systematically.
- Optimization: The maximum and minimum values of $f$ occur where it transitions from increasing to decreasing or vice versa -- at critical points where $f' = 0$ or is undefined.