The Second 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” |
Opening Scenario
At a critical number, the tangent line is horizontal. The question is whether that flat spot is the bottom of a bowl (a local minimum) or the top of a hill (a local maximum). The concavity tells you immediately: if the curve is concave up at a flat spot, the flat spot is a bowl bottom -- a local minimum. If the curve is concave down, it is a hilltop -- a local maximum.
The Second Derivative Test reads the concavity at the critical number itself, bypassing the need for a full sign chart.
Quick Reference
Second Derivative Test. Suppose $f'(c) = 0$ and $f''(c)$ exists.
| $f''(c)$ | Conclusion |
|---|---|
| $f''(c) > 0$ | Local minimum at $c$ |
| $f''(c) < 0$ | Local maximum at $c$ |
| $f''(c) = 0$ | Test is inconclusive; use the First Derivative Test instead |
The test applies only when $f'(c) = 0$. It cannot be used at a critical number where $f'$ does not exist.
Key Concepts
1. Why the Test Works
If $f'(c) = 0$ and $f''(c) > 0$, then $f'$ is increasing at $c$. Just to the left of $c$, $f'$ is smaller than $f'(c) = 0$, so $f' < 0$ there. Just to the right, $f'$ is larger than $0$, so $f' > 0$ there. The sign of $f'$ goes from negative to positive: that is the signature of a local minimum.
Similarly, $f''(c) < 0$ means $f'$ is decreasing at $c$: $f'$ goes from positive to negative, the signature of a local maximum.
When $f''(c) = 0$, $f'$ might not be changing at all at $c$, and the test provides no information about the sign change on either side.
2. Applying the Test
Example 1. Find and classify the local extrema of $f(x) = x^3 - 3x^2 + 2$.
Step 1: Critical numbers. $f'(x) = 3x^2 - 6x = 3x(x - 2) = 0 \Rightarrow x = 0$ or $x = 2$.
Step 2: Compute $f''$ at each critical number. $f''(x) = 6x - 6$.
$f''(0) = 6(0) - 6 = -6 < 0$: local maximum at $x = 0$.
$f''(2) = 6(2) - 6 = 6 > 0$: local minimum at $x = 2$.
Step 3: Evaluate $f$ at the critical numbers. $f(0) = 2$. $f(2) = 8 - 12 + 2 = -2$.
Boxed answer: Local maximum of $2$ at $x = 0$; local minimum of $-2$ at $x = 2$.
Recap. The Second Derivative Test is faster here than the First Derivative Test: two substitutions into $f''$ classify both critical numbers without building a full sign chart for $f'$.
3. When the Test Fails
Example 2. Apply the Second Derivative Test to $f(x) = x^4$ at $x = 0$.
$f'(x) = 4x^3$. $f'(0) = 0$: critical number.
$f''(x) = 12x^2$. $f''(0) = 0$: the test is inconclusive.
Now use the First Derivative Test. $f'(x) = 4x^3$ is negative for $x < 0$ and positive for $x > 0$: sign goes from $-$ to $+$, so $x = 0$ is a local minimum.
Example 3. Now try $f(x) = x^3$ at $x = 0$.
$f'(0) = 0$, $f''(0) = 0$: inconclusive again.
First Derivative Test: $f'(x) = 3x^2 > 0$ on both sides of $x = 0$. No sign change: no local extremum.
“$f''(c) = 0$ means an inflection point.” No. In Example 2, $f''(0) = 0$ yet $x = 0$ is a local minimum, not an inflection point. In Example 3, $f''(0) = 0$ and $x = 0$ is an inflection point. The sign of $f''$ around $c$ determines whether there is an inflection point, not the value of $f''$ at $c$.
4. Choosing Between the Two Tests
| Situation | Preferred test |
|---|---|
| $f'(c) = 0$ and $f''(c) \neq 0$ | Second Derivative Test (fast: one substitution) |
| $f'(c) = 0$ and $f''(c) = 0$ | First Derivative Test (required: SDT is inconclusive) |
| $f'(c)$ does not exist | First Derivative Test (SDT cannot be used) |
Neither test is universally better. The Second Derivative Test is faster when $f''$ is easy to compute and non-zero; the First Derivative Test is always available and works in every case.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Applying SDT when $f'(c)$ does not exist | Using $f''(0)$ to classify the corner of $\lvert x\rvert$ | SDT requires $f'(c) = 0$; use the First Derivative Test for corners |
| Concluding “inconclusive = no extremum” | Saying $x=0$ is not an extremum of $x^4$ because $f''(0)=0$ | $f''(c) = 0$ means inconclusive; there might still be an extremum -- use the FDT to check |
| Forgetting to evaluate $f(c)$ | Reporting only “local min at $x=2$” | Also state the local minimum value: $f(2) = -2$ |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Use the Second Derivative Test to classify the critical points of $f(x) = 2x^3 - 3x^2 - 12x + 1$.
Show answer
$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$. Critical numbers: $x = 2, -1$.
$f''(x) = 12x - 6$.
$f''(-1) = -12 - 6 = -18 < 0$: local maximum. $f(-1) = -2 - 3 + 12 + 1 = 8$.
$f''(2) = 24 - 6 = 18 > 0$: local minimum. $f(2) = 16 - 12 - 24 + 1 = -19$.
Boxed answer: Local max of $8$ at $x = -1$; local min of $-19$ at $x = 2$.
Problem 2. Apply the Second Derivative Test to $f(x) = \sin x$ on $(0, 2\pi)$.
Show answer
$f'(x) = \cos x = 0$ on $(0, 2\pi)$ at $x = \pi/2$ and $x = 3\pi/2$.
$f''(x) = -\sin x$.
$f''(\pi/2) = -\sin(\pi/2) = -1 < 0$: local maximum. $f(\pi/2) = 1$.
$f''(3\pi/2) = -\sin(3\pi/2) = -(-1) = 1 > 0$: local minimum. $f(3\pi/2) = -1$.
Boxed answer: Local max of $1$ at $x = \pi/2$; local min of $-1$ at $x = 3\pi/2$.
Level 2 -- Multiple Steps
Problem 3. For $f(x) = x^4 - 4x^3 + 10$, the Second Derivative Test may be inconclusive at some critical numbers. Find all critical numbers, apply the SDT where possible, and fall back on the FDT where needed.
Show answer
$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$. Critical numbers: $x = 0$ and $x = 3$.
$f''(x) = 12x^2 - 24x = 12x(x - 2)$.
At $x = 0$: $f''(0) = 0$. Inconclusive. Use FDT: $f'(x) = 4x^2(x-3)$. For $x < 0$: $4x^2 > 0$, $x - 3 < 0$, so $f'(x) < 0$. For $0 < x < 3$: $4x^2 > 0$, $x - 3 < 0$, so $f'(x) < 0$. No sign change: neither a max nor a min.
At $x = 3$: $f''(3) = 12(3)(1) = 36 > 0$. Local minimum. $f(3) = 81 - 108 + 10 = -17$.
Boxed answer: No local extremum at $x = 0$; local minimum of $-17$ at $x = 3$.
Mastery Checklist
Mental Model
The Second Derivative Test is the cup-or-cap check. At a horizontal tangent ($f'(c) = 0$), the curve is momentarily flat. The second derivative tells you which way the curve bends away from that flat point: upward like a cup (local minimum) or downward like a cap (local maximum).
When $f''(c) = 0$, the curve is neither clearly cup-shaped nor cap-shaped at that scale, and you need more information. The First Derivative Test -- which looks at signs over entire intervals rather than at a single point -- always provides that extra information.
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