The First Derivative 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” |
A trail that peaks then descends
A hiking trail climbs, reaches a peak, and then descends. Before the peak the trail gains elevation (uphill, positive slope); after the peak it loses elevation (downhill, negative slope). The peak is the moment the slope changes from positive to negative.
The First Derivative Test formalizes that picture. At a critical number, check whether $f'$ changes sign. If positive becomes negative, the function was climbing then descending: a local maximum. If negative becomes positive, it was descending then climbing: a local minimum. If the sign does not change, the function kept moving in the same direction across the critical number: no local extremum.
Quick Reference
First Derivative Test. Suppose $c$ is a critical number of a continuous function $f$.
| Sign of $f'$ left of $c$ | Sign of $f'$ right of $c$ | Conclusion at $c$ |
|---|---|---|
| $+$ | $-$ | Local maximum |
| $-$ | $+$ | Local minimum |
| $+$ | $+$ | Neither (no local extremum) |
| $-$ | $-$ | Neither (no local extremum) |
The test applies at any critical number, whether $f'(c) = 0$ or $f'(c)$ does not exist.
Key Concepts
1. The Sign-Change Principle
A local maximum is a point where the function stops increasing and starts decreasing. That is exactly when $f'$ changes from positive to negative. A local minimum is where the function stops decreasing and starts increasing: $f'$ changes from negative to positive.
If $f'$ keeps the same sign on both sides of $c$, the function was moving in the same direction through $c$ and there is no extremum there.
“Every critical number is a local extremum.” Not true. The function $f(x) = x^3$ has $f'(0) = 0$, but $f'(x) = 3x^2 > 0$ for $x \neq 0$. The sign is positive on both sides of $0$. The derivative is zero at a single point but does not change sign, so $f$ has no local extremum at $x = 0$.
2. Applying the First Derivative Test
Example 1. Find the local maxima and minima of $f(x) = x^3 - 3x + 1$.
Step 1: Critical numbers. $f'(x) = 3x^2 - 3 = 3(x-1)(x+1) = 0 \Rightarrow x = -1, 1.$
Step 2: Sign chart for $f'$.
| Interval | Sign of $f'$ | Direction |
|---|---|---|
| $x < -1$ | $+$ | Increasing |
| $-1 < x < 1$ | $-$ | Decreasing |
| $x > 1$ | $+$ | Increasing |
Step 3: Classify each critical number.
- At $x = -1$: $f'$ goes from $+$ to $-$. Local maximum. $f(-1) = -1 + 3 + 1 = 3$.
- At $x = 1$: $f'$ goes from $-$ to $+$. Local minimum. $f(1) = 1 - 3 + 1 = -1$.
Boxed answer: Local max of $3$ at $x = -1$; local min of $-1$ at $x = 1$.
Recap. The sign chart does double duty: it identifies the intervals of increase/decrease and classifies each critical number in one pass. Build the chart first; the classification follows automatically from the sign transitions.
3. A Critical Number from Non-Differentiability
The First Derivative Test applies at any critical number, including corners and cusps.
Example 2. Find the local extrema of $f(x) = x^{2/3}$.
$f'(x) = \dfrac{2}{3} x^{-1/3}$. Critical number at $x = 0$ (where $f'$ is undefined).
For $x < 0$: $x^{-1/3} < 0$, so $f'(x) < 0$ (decreasing). For $x > 0$: $x^{-1/3} > 0$, so $f'(x) > 0$ (increasing).
Sign change from $-$ to $+$ at $x = 0$: local minimum.
$f(0) = 0$.
Boxed answer: Local minimum of $0$ at $x = 0$.
Recap. The corner at $x = 0$, the point of a cusp, is still classified by the First Derivative Test. The key is the sign of $f'$ on each side, not the existence of $f'$ at the point itself.
4. No Sign Change: the Inflection-Like Case
Example 3. Find the local extrema of $f(x) = x^3$.
$f'(x) = 3x^2 \geq 0$ for all $x$, and $f'(0) = 0$.
Sign of $f'$: positive for $x < 0$, zero at $x = 0$, positive for $x > 0$. No sign change.
Boxed answer: No local extremum at $x = 0$ (the only critical number).
Recap. The graph of $y = x^3$ has a horizontal tangent at the origin but does not peak or valley there; it passes through with an inflection point instead. The absence of a sign change is the reliable indicator.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Assuming $f'(c)=0$ means a local extremum | $f(x)=x^3$ at $c=0$ | Check for a sign change of $f'$; no sign change means no extremum |
| Testing $f'(c)$ instead of nearby values | Evaluating $f'$ only at the critical number | Test $f'$ on each side of $c$, not at $c$ itself (where $f'$ might be zero or undefined) |
| Forgetting the critical numbers from non-differentiability | Skipping $x=0$ for $f(x)=\lvert x\rvert$ | Include all critical numbers in the sign chart |
| Reporting the wrong extremum value | Saying “local max at $x=-1$” without computing $f(-1)$ | State both the input and the output value: “local max of $f(-1)=3$ at $x=-1$” |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Find the local maxima and minima of $f(x) = x^3 - 6x^2 + 9x - 2$.
Show answer
$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$. Critical numbers: $x = 1, 3$.
| Interval | Sign of $f'$ |
|---|---|
| $x < 1$ | $+$ (both factors negative, product positive) |
| $1 < x < 3$ | $-$ ($x-1 > 0$, $x-3 < 0$) |
| $x > 3$ | $+$ |
At $x = 1$: $f'$ goes $+ \to -$: local max. $f(1) = 1 - 6 + 9 - 2 = 2$. At $x = 3$: $f'$ goes $- \to +$: local min. $f(3) = 27 - 54 + 27 - 2 = -2$.
Boxed answer: Local max of $2$ at $x = 1$; local min of $-2$ at $x = 3$.
Problem 2. Apply the First Derivative Test to $f(x) = x^4$.
Show answer
$f'(x) = 4x^3$. Critical number: $x = 0$.
For $x < 0$: $4x^3 < 0$ (decreasing). For $x > 0$: $4x^3 > 0$ (increasing).
Sign change from $-$ to $+$: local minimum at $x = 0$.
$f(0) = 0$.
Boxed answer: Local minimum of $0$ at $x = 0$.
Level 2 -- Multiple Steps
Problem 3. Find all local extrema of $f(x) = \dfrac{x^2}{x - 1}$ (domain: $x \neq 1$).
Show answer
Quotient rule: $f'(x) = \dfrac{2x(x-1) - x^2}{(x-1)^2} = \dfrac{2x^2 - 2x - x^2}{(x-1)^2} = \dfrac{x^2 - 2x}{(x-1)^2} = \dfrac{x(x-2)}{(x-1)^2}$.
Critical numbers where $f'(x)=0$: $x = 0$ and $x = 2$ (in the domain). $x = 1$ is not in the domain.
Sign of $f'$: the denominator $(x-1)^2 \geq 0$ is always non-negative and positive except at $x = 1$ (not in domain). So the sign of $f'$ equals the sign of $x(x-2)$.
| Interval | Sign of $x(x-2)$ |
|---|---|
| $x < 0$ | $(-)(-) = +$ |
| $0 < x < 1$ | $(+)(-) = -$ |
| $1 < x < 2$ | $(+)(-) = -$ |
| $x > 2$ | $(+)(+) = +$ |
At $x = 0$: sign goes $+ \to -$: local max. $f(0) = 0$. At $x = 2$: sign goes $- \to +$: local min. $f(2) = 4/1 = 4$.
Boxed answer: Local max of $0$ at $x = 0$; local min of $4$ at $x = 2$.
Level 3 -- Deeper Problems
Problem 4. Can a continuous function have a local maximum at a point where $f'$ does not exist? Give an example and verify it using the First Derivative Test.
Show answer
Yes. Let $f(x) = -|x|$ (negative absolute value). Then $f(0) = 0$, and for any $x \neq 0$, $f(x) = -|x| < 0 = f(0)$. So $x = 0$ is a local maximum. But $f'(0)$ does not exist (corner pointing downward).
First Derivative Test: for $x < 0$, $f(x) = -(-x) = x$, so $f'(x) = 1 > 0$. For $x > 0$, $f(x) = -x$, so $f'(x) = -1 < 0$.
Sign change from $+$ to $-$: local maximum confirmed.
Boxed answer: Yes. $f(x) = -|x|$ has a local maximum at $x = 0$ where $f'$ does not exist, confirmed by a sign change of $f'$ from positive to negative.
Mastery Checklist
Mental Model
The First Derivative Test is the direction-change detector. A local maximum is where the function switches from going up to going down. A local minimum is where it switches from going down to going up. No switch, no extremum.
The derivative is the directional indicator: positive is up, negative is down. Checking whether its sign switches at each critical number is exactly checking whether the direction changes. The sign chart records this switch or non-switch in a compact, readable form.
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