Second derivative test (local minimum): Infer a local minimum from curvature at a critical point

On this page: The Principle | Conditions | Failure Modes | EE Questions | Retrieval Practice | Practice Ground | Solve a Problem | Related Principles | FAQ | How This Fits

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The Principle

The move: Infer that $f$ has a local minimum at $c$ when $c$ is a critical point and the second derivative is positive there.

The invariant: Both checks refer to the same candidate point $c$: the point must be critical, and the second-derivative evidence must be positive at that same point before the minimum conclusion is available.

Pattern: $f'(c)=0,\\ f''(c)>0 \quad\Longrightarrow\quad f \text{ has a local minimum at } c$

Applies ✓	Does not apply ✗
$f'(2)=0$ and $f''(2)=5 \Rightarrow f$ has a local minimum at $x=2$	$f'(0)=0$ and $f''(0)=0 \not\Rightarrow f$ has a local minimum at $x=0$

Left: the point is critical and the second derivative is positive there, so the theorem applies. Right: this is a tempting near-miss because the critical-point check passes, but the positive-second-derivative condition fails, so the theorem gives no local-minimum conclusion.

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Conditions of Applicability

Condition: $f'(c)=0$; $f''(c)>0$

This is a sufficient-test theorem, not a generic rule that positive curvature alone proves a minimum. You need a critical point first, and then you need positive second-derivative evidence at that same point.

Before applying, check: have you verified both $f'(c)=0$ and $f''(c)>0$ at the candidate point $c$?

If the condition is violated: this local-minimum theorem is unavailable; the point might be a local maximum, might be neither, or might require the first derivative test (local minimum) or another argument.

• A positive second derivative by itself is not enough. If $f'(c)\neq 0$, then $c$ is not even a critical point, so the theorem cannot certify a local minimum there.
• If $f''(c)=0$, the test is inconclusive. That does not prove there is no minimum; it only means this theorem does not settle the question.
• The theorem depends on correct derivative evaluation, so guides such as derivative at a point and first derivative test (local maximum) help you separate theorem-ready data from a different local pattern.

Want the complete framework behind this guide? Read Masterful Learning.

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Common Failure Modes

Failure mode: treat $f''(c)>0$ as enough by itself → you declare a local minimum at a point that was never shown to satisfy $f'(c)=0$, so the theorem was not applicable.

Debug: say both checks out loud before concluding: critical point first, then positive second derivative.

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Elaborative Encoding

Use these questions to build deep understanding. (See Elaborative Encoding for the full method.)

Within the Principle

• Why does the pair $f'(c)=0$ and $f''(c)>0$ say more than either fact alone?
• Why does concave-up behavior at a critical point support a local-minimum conclusion rather than just a statement about shape?

For the Principle

• When you compute $f'(c)$ and $f''(c)$ from a formula, what evidence do you need before the second derivative test is applicable?
• If the calculation gives $f''(c)=0$, what should you conclude, and what should you avoid concluding?

Between Principles

• How does the second derivative test differ from the first derivative test (local minimum), and why can the first-derivative route still work when the second-derivative route is inconclusive?

Generate an Example

• Create a point where $f'(c)=0$ but the second derivative test does not prove a local minimum because the second derivative condition fails.

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Retrieval Practice

Answer from memory, then click to reveal and check. (See Retrieval Practice for the full method.)

State the move in one sentence: _____Infer that a function has a local minimum at c when f'(c)=0 and f''(c)>0.

Write the canonical equation: _____$f'(c)=0 \wedge f''(c)>0$

State the canonical condition: _____$f'(c)=0; f''(c)>0$

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Practice Ground

Use these exercises to build move-selection fluency. (See Self-Explanation for how to use worked examples effectively.)

Procedure Walkthrough

Starting from $f(x)=x^3-3x$, determine what the second derivative test says about $x=1$.

Step	Expression	Operation
0	$f(x)=x^3-3x$	—
1	$f'(x)=3x^2-3$	Differentiate to locate critical-point evidence
2	$f'(1)=0$	Verify the candidate point is critical
3	$f''(x)=6x$	Differentiate again to test curvature
4	$f''(1)=6>0$	Verify the concave-up condition
5	$f$ has a local minimum at $x=1$	Second derivative test (local minimum)

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Drills

Action label

What principle was used in the step below, and why is the conclusion justified?

$f'(3)=0,\ f''(3)=4 \quad\Longrightarrow\quad f \text{ has a local minimum at } x=3$

Reveal

The second derivative test for a local minimum was used. The conclusion is justified because the point is critical and the second derivative is positive there, which exactly matches the canonical condition.

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Does the theorem apply here? Name the issue before deciding.

$f'(0)=0,\ f''(0)=0 \quad\Longrightarrow\quad f \text{ has a local minimum at } x=0$

Reveal

No. This is a near-miss. The critical-point check passes, but the second derivative is not positive, so the theorem is inconclusive and does not justify a local-minimum conclusion.

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What fact makes the local-minimum conclusion available below?

$f'(2)=0,\ f''(2)=11 \quad\Longrightarrow\quad f \text{ has a local minimum at } x=2$

Reveal

The point $x=2$ satisfies both required checks: it is a critical point, and the second derivative there is positive. That is exactly the second-derivative-test pattern for a local minimum.

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Should you label this step as a valid use of the theorem? Explain why or why not.

$f'(4)=3,\ f''(4)=9 \quad\Longrightarrow\quad f \text{ has a local minimum at } x=4$

Reveal

No. Positive curvature alone is not enough. Since $f'(4)\neq 0$, the point was not shown to be critical, so the theorem cannot be applied.

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Transition identification

Which transition uses the second derivative test for a local minimum directly?

$f'(x)=3x^2-12x+9 \xrightarrow{(1)} f'(3)=0,\ f''(3)=6 \xrightarrow{(2)} f \text{ has a local minimum at } x=3$

Reveal

Transition (2) uses the theorem directly. Transition (1) gathers the derivative data; transition (2) turns the verified condition into the local-minimum conclusion.

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Which proposed applications are justified uses of the second derivative test for a local minimum?

1. $f'(c)=0,\ f''(c)>0 \Longrightarrow f$ has a local minimum at $c$
2. $f'(c)=0,\ f''(c)=0 \Longrightarrow f$ has a local minimum at $c$
3. $f'(c)\neq 0,\ f''(c)>0 \Longrightarrow f$ has a local minimum at $c$

Reveal

Only 1 is justified.

• 1 matches the theorem exactly.
• 2 is inconclusive because the second derivative is not positive.
• 3 fails the critical-point requirement, so the theorem is not available.

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At what point does the theorem become available in the chain below?

1. $f(x)=x^3-6x^2+9x$
2. $f'(x)=3x^2-12x+9$
3. $f'(3)=0$
4. $f''(x)=6x-12$
5. $f''(3)=6>0$
6. $f$ has a local minimum at $x=3$

Reveal

The theorem becomes available only after step 5. You need both checks in hand before moving to step 6.

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Which transition, if any, is a justified second-derivative-test move?

$f'(0)=0,\ f''(0)=0 \xrightarrow{(1)} f \text{ is locally concave up at } 0 \xrightarrow{(2)} f \text{ has a local minimum at } 0$

Reveal

Neither transition is justified. The data do not show $f''(0)>0$, so even the concave-up claim is unsupported here, and the local-minimum conclusion is unavailable as well.

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Forward step

Apply the second derivative test once at $c=-1$.

$f(x)=x^2+2x+5$

Reveal

$f'(x)=2x+2$, so $f'(-1)=0$. Also, $f''(x)=2$, so $f''(-1)=2>0$.

Therefore, $f$ has a local minimum at $x=-1$.

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Should you conclude a local minimum at $c=0$ from the data below? Explain your decision.

$f(x)=x^4$

Reveal

No. Here $f'(0)=0$, but $f''(0)=0$, so the theorem is inconclusive. The function does have a local minimum at $0$, but not by this test.

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Solve a Problem

Apply what you’ve learned with Problem Solving.

Problem: For $f(x)=x^3-6x^2+9x$, determine what the second derivative test says about $x=3$.

Full solution

Step	Expression	Move
0	$f(x)=x^3-6x^2+9x$	—
1	$f'(x)=3x^2-12x+9$	Differentiate to test the critical-point condition
2	$f'(3)=0$	Verify the candidate point is critical
3	$f''(x)=6x-12$	Differentiate again to test curvature
4	$f''(3)=6>0$	Verify the positive-second-derivative condition
5	$f$ has a local minimum at $x=3$	Second derivative test (local minimum)

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Related Principles

Principle	Relationship
First derivative test (local minimum)	Reaches the same conclusion by checking the sign of $f'$ on both sides instead of using $f''(c)$ at one point
First derivative test (local maximum)	Uses the same sign-chart logic, but the local pattern is $+ \to -$ instead of the local-minimum pattern
Derivative at a point	Supplies the derivative values you must compute correctly before this theorem can be applied

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FAQ

What does the second derivative test for a local minimum say?

It says that if $f'(c)=0$ and $f''(c)>0$, then $f$ has a local minimum at $c$. The point must be critical first, and the second derivative must be positive at that same point.

Is $f''(c)>0$ enough by itself?

No. Positive second derivative does not replace the critical-point check. If $f'(c)\neq 0$, the theorem does not apply.

What if $f''(c)=0$?

Then the test is inconclusive. You should not conclude a local minimum or rule one out from this theorem alone.

How is this different from the first derivative test?

The second derivative test uses values at a single point, while the first derivative test (local minimum) uses the sign of $f'$ on both sides of the point. The first-derivative route can still work when the second derivative test is inconclusive.

Does the theorem prove every local minimum?

No. It is a sufficient test, not a necessary one. Some local minima exist at points where $f''(c)=0$, so the theorem cannot certify every true minimum.

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How This Fits in Unisium

In Unisium, the second derivative test is trained as a move-selection theorem: you do not jump from concavity language to an extremum conclusion until the critical-point check is present too. That fits directly with retrieval practice, self-explanation, and the broader logic of learning to pick the right theorem under the right condition. To build that reflex with many short reps, practice directly in the Unisium app once you can check both derivative conditions without hesitation.

Explore further:

• Second derivative test (local maximum) - Contrast the local-minimum curvature condition with the matching local-maximum theorem
• First derivative test (local minimum) - Use a sign chart when pointwise second-derivative data are not enough
• Derivative at a point - Strengthen the derivative evaluations this theorem depends on
• Retrieval Practice - Make the two-condition trigger easy to recall under pressure

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