The Increasing/Decreasing Test
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.5: “Derivatives and the Shape of a Graph” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-5-derivatives-and-the-shape-of-a-graph |
| Textbook used in class | Stewart, Calculus, Section 3.3: “What Derivatives Tell Us about the Shape of a Graph” |
Opening Scenario
Look at a speedometer during a drive. When the needle is in the positive range, the car is moving forward -- position is increasing. When the needle drops below zero (backing up), position is decreasing. The derivative of position is velocity. Positive velocity means the position function is increasing; negative velocity means it is decreasing.
That is the whole idea. The sign of the derivative tells you the direction of change.
Quick Reference
Increasing/Decreasing Test.
- If $f'(x) > 0$ for all $x$ in an interval $I$, then $f$ is increasing on $I$.
- If $f'(x) < 0$ for all $x$ in an interval $I$, then $f$ is decreasing on $I$.
Procedure for finding intervals:
- Find all critical numbers of $f$ (where $f' = 0$ or $f'$ does not exist).
- These critical numbers divide the domain into subintervals.
- Choose one test value in each subinterval and determine the sign of $f'$ there.
- State whether $f$ is increasing or decreasing on each subinterval.
Key Concepts
1. Why the Sign of $f'$ Determines Direction
The Mean Value Theorem makes this precise. Suppose $f'(x) > 0$ on $(a, b)$ and take any two points $x_1 < x_2$ in $(a, b)$. The MVT gives a $c$ with $$f(x_2) - f(x_1) = f'(c)(x_2 - x_1).$$ Since $f'(c) > 0$ and $x_2 - x_1 > 0$, the right side is positive. So $f(x_2) > f(x_1)$: the function value increases as the input increases. That is the definition of increasing.
The same reasoning with $f'(x) < 0$ gives $f(x_2) < f(x_1)$: the function decreases.
2. Finding Intervals of Increase and Decrease
Example 1. Find the intervals on which $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ is increasing or decreasing.
(This parallels Stewart 3.3, Example 1.)
Step 1: Find $f'$ and critical numbers.
$f'(x) = 12x^3 - 12x^2 - 24x = 12x(x^2 - x - 2) = 12x(x-2)(x+1).$
Setting $f'(x) = 0$: $x = 0, 2, -1$. The derivative is a polynomial, so it exists everywhere.
Step 2: Divide the domain and test.
Critical numbers $-1, 0, 2$ create four subintervals:
| Interval | Test value | $f'$ at test | Direction |
|---|---|---|---|
| $(-\infty, -1)$ | $x = -2$ | $12(-2)(-4)(-1) = -96 < 0$ | Decreasing |
| $(-1, 0)$ | $x = -0.5$ | $12(-0.5)(-2.5)(0.5) \cdot \text{sign} = +$ (verify below) | Increasing |
| $(0, 2)$ | $x = 1$ | $12(1)(-1)(2) = -24 < 0$ | Decreasing |
| $(2, \infty)$ | $x = 3$ | $12(3)(1)(4) = 144 > 0$ | Increasing |
Verification for $x = -0.5$: $12(-0.5)(-0.5-2)(-0.5+1) = 12(-0.5)(-2.5)(0.5) = 12 \times 0.625 = 7.5 > 0$.
Boxed answer: $f$ is increasing on $(-1, 0)$ and $(2, \infty)$; decreasing on $(-\infty, -1)$ and $(0, 2)$.
Recap. The sign chart -- one test value per interval -- is the efficient method. No need to analyze the entire expression; one evaluation per region is enough since $f'$ cannot change sign without passing through zero.
3. Reading the Sign Chart Directly
An efficient way to track the sign of a factored derivative: list the factors and record their signs in each interval.
For $f'(x) = 12x(x-2)(x+1)$:
| Interval | $12x$ | $x-2$ | $x+1$ | Product |
|---|---|---|---|---|
| $x < -1$ | $-$ | $-$ | $-$ | $-$ |
| $-1 < x < 0$ | $-$ | $-$ | $+$ | $+$ |
| $0 < x < 2$ | $+$ | $-$ | $+$ | $-$ |
| $x > 2$ | $+$ | $+$ | $+$ | $+$ |
The factor signs flip at each root. This is faster than substituting a numerical test value.
4. A Function with a Non-Differentiable Point
Example 2. Find the intervals on which $f(x) = x^{2/3}$ is increasing or decreasing.
$f'(x) = \dfrac{2}{3} x^{-1/3}$. This is undefined at $x = 0$, so $x = 0$ is a critical number.
For $x > 0$: $f'(x) = \dfrac{2}{3x^{1/3}} > 0$, so $f$ is increasing. For $x < 0$: $x^{1/3} < 0$, so $f'(x) < 0$, so $f$ is decreasing.
Boxed answer: $f$ is decreasing on $(-\infty, 0)$ and increasing on $(0, \infty)$.
Recap. Always include non-differentiable points as potential dividers of sign intervals. The derivative sign can change at a corner even though no zero of $f'$ creates the division.
$f' > 0$ means the function value is positive, not that the function is increasing.
This is the height-vs-slope error. The sign of $f'(x)$ tells you about the SLOPE of the graph at $x$, not the HEIGHT. For $f(x) = x^2 - 9$: on the interval $(3, \infty)$, $f'(x) = 2x > 0$, so the function is increasing there even though $f(3) = 0$ and $f(4) = 7$ -- the heights are positive. But on $(-3, 0)$, $f'(x) = 2x < 0$ even though $f(-1) = -8$ is negative. A negative $f'$ means the graph is falling; it says nothing about whether the graph is above or below the x-axis. The sign of $f$ gives the height; the sign of $f'$ gives the direction of change.
checking that $f'$ is positive (or negative) at one point tells you the sign everywhere.
This is the action-view-of-function error applied to sign analysis. Students sometimes compute $f'$ at a single convenient value -- say $x = 0$ -- and conclude that $f'$ has the same sign everywhere. But $f'$ is itself a function that can change sign. For $f(x) = x^3 - 3x$: $f'(x) = 3x^2 - 3$, which equals $-3$ at $x=0$ (negative) and $+9$ at $x=2$ (positive). Checking only $x=0$ would miss the increasing behavior for $x > 1$. The correct approach is to find where $f'$ equals zero or is undefined, divide the real line into intervals at those points, and test one value per interval -- because $f'$ cannot change sign without passing through zero.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting non-differentiable critical numbers | Only solving $f'(x) = 0$ for $f(x) = \lvert x\rvert$ | Also include points where $f'$ does not exist |
| Testing the wrong region | Evaluating $f'$ at a critical number itself | Critical numbers are the boundaries; choose a test value strictly inside each interval |
| Concluding a critical number is a local extremum | Saying “the function changes at $x=2$, so there is a local min there” | A sign change of $f'$ gives a local extremum; sign does not automatically change at every critical number |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Find the intervals on which $f(x) = x^3 - 3x$ is increasing or decreasing.
Show answer
$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$. Critical numbers: $x = 1, -1$.
| Interval | Test | Sign of $f'$ | |
|---|---|---|---|
| $x < -1$ | $x = -2$: $3(3)(-3) < 0$ | $-$ (decreasing) | |
| $-1 < x < 1$ | $x = 0$: $3(-1)(1) < 0$ | Wait: $f'(0) = -3 < 0$ | $-$ (decreasing) |
| $x > 1$ | $x = 2$: $3(1)(3) > 0$ | $+$ (increasing) |
Actually, let us redo: $f'(x) = 3(x^2 - 1)$. For $x \in (-1, 1)$: $x^2 < 1$, so $x^2 - 1 < 0$, so $f'(x) < 0$.
Boxed answer: Increasing on $(-\infty, -1)$... wait. $f'(-2) = 3(4-1) = 9 > 0$. Increasing on $(-\infty, -1)$ and $(1, \infty)$; decreasing on $(-1, 1)$.
Problem 2. Find the intervals on which $g(x) = x^4 - 8x^2$ is increasing or decreasing.
Show answer
$g'(x) = 4x^3 - 16x = 4x(x^2 - 4) = 4x(x-2)(x+2)$. Critical numbers: $x = 0, 2, -2$.
Sign chart:
| Interval | $4x$ | $x-2$ | $x+2$ | Product |
|---|---|---|---|---|
| $x < -2$ | $-$ | $-$ | $-$ | $-$ |
| $-2 < x < 0$ | $-$ | $-$ | $+$ | $+$ |
| $0 < x < 2$ | $+$ | $-$ | $+$ | $-$ |
| $x > 2$ | $+$ | $+$ | $+$ | $+$ |
Boxed answer: Increasing on $(-2, 0)$ and $(2, \infty)$; decreasing on $(-\infty, -2)$ and $(0, 2)$.
Level 2 -- Multiple Steps
Problem 3. Find the intervals on which $f(x) = e^x / (1 + e^x)$ is increasing or decreasing.
Show answer
Using the quotient rule: $f'(x) = \dfrac{e^x(1 + e^x) - e^x \cdot e^x}{(1 + e^x)^2} = \dfrac{e^x}{(1 + e^x)^2}$.
Since $e^x > 0$ and $(1 + e^x)^2 > 0$ for all $x$, we have $f'(x) > 0$ for all $x$.
Boxed answer: $f$ is increasing on $(-\infty, \infty)$. There are no intervals of decrease.
Mastery Checklist
Mental Model
The derivative is the slope of the tangent line at each point. Positive slope means the graph goes uphill as you move left to right; negative slope means downhill. Critical numbers are the flat spots (slope zero) or sharp corners where the slope is undefined -- either can be where the hill changes direction.
The sign of $f'$ does not change inside a subinterval (since $f'$ is continuous there and would have to pass through zero or a discontinuity to change sign). So one test per interval is enough to settle the whole interval.
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