Unit 5 · Topic 5.7

5.7 Using the Second Derivative Test to Determine Extrema

The Second Derivative Test uses concavity at a critical point to classify it. If f′(c) = 0 and f″(c) > 0, the graph is cupped upward, so f(c) is a relative minimum; if f″(c) < 0, it's a relative maximum. If f″(c) = 0, the test tells you nothing.

Key terms

Second Derivative Test
relative extremum
inconclusive
absolute extremum

The test

Suppose f′(c) = 0 and f″ exists near c.

If f″(c) > 0, then f has a relative minimum at c.
If f″(c) < 0, then f has a relative maximum at c.
If f″(c) = 0, the test is inconclusive. Use the First Derivative Test instead.

Why it works

At a critical point with f′(c) = 0, the tangent line is horizontal. If the graph is concave up there, it curves up away from that flat tangent on both sides, so c is the bottom of a valley. If it's concave down, it curves down on both sides, a peak.

When f″(c) = 0, anything can happen. f(x) = x⁴ has a min at 0, −x⁴ has a max, and x³ has neither, yet all three have f′(0) = 0 and f″(0) = 0.

Choosing between the tests

The Second Derivative Test only needs one number, f″(c), so it's fast. But it can only handle critical points where f′(c) = 0.

Situation	Better test
f″ is easy to compute and nonzero at c	Second Derivative Test
f″(c) = 0 or f″ is messy	First Derivative Test
f′(c) doesn't exist (corner or cusp)	First Derivative Test
Only a graph of f′ is given	First Derivative Test (sign changes)

Using it with tables and context

The Second Derivative Test is handy when a table gives f′ and f″ values at the same point. If a row shows f′(4) = 0 and f″(4) = 2, f has a relative minimum at x = 4.

In context, a relative maximum of a rate has a meaning too. If R(t) is the rate water enters a tank and R′(5) = 0 with R″(5) < 0, then the inflow rate itself peaks around t = 5. That's the time water is entering fastest, not the time the tank holds the most water.

From relative to absolute: the one-critical-point idea

If f is continuous on an interval (open, closed or infinite) and has exactly one critical point there, and that critical point is a relative minimum, then it's also the absolute minimum on that interval. The same goes for a maximum.

The reason: if the function went back down somewhere else, it would have to turn around, creating another critical point. This is especially useful in optimization problems on open intervals, where the Candidates Test can't be used.

Worked examples

Try each one yourself first, then open the solution.

Example 1
Classify with the Second Derivative Test
Find and classify the critical points of f(x) = 2x³ − 3x² − 12x.
Show the solutionHide the solution
1. Step 1: f′(x) = 6x² − 6x − 12 = 6(x − 2)(x + 1). Critical points: x = 2 and x = −1.
2. Step 2: f″(x) = 12x − 6.
3. Step 3: f″(2) = 18 > 0, so a relative minimum at x = 2. f(2) = 16 − 12 − 24 = −20.
4. Step 4: f″(−1) = −18 < 0, so a relative maximum at x = −1. f(−1) = −2 − 3 + 12 = 7.
Answer: Relative maximum of 7 at x = −1 (f′(−1) = 0 and f″(−1) < 0). Relative minimum of −20 at x = 2 (f′(2) = 0 and f″(2) > 0).
Example 2
One critical point gives an absolute minimum
Find the absolute minimum value of g(x) = x + 4/x on (0, ∞). Justify.
Show the solutionHide the solution
1. Step 1: g′(x) = 1 − 4/x². Setting it to 0: x² = 4, so x = 2 (x = −2 isn't in the interval).
2. Step 2: g″(x) = 8/x³, and g″(2) = 1 > 0, so g has a relative minimum at x = 2.
3. Step 3: g is continuous on (0, ∞) and x = 2 is its only critical point there, so the relative minimum is the absolute minimum.
4. Step 4: g(2) = 2 + 2 = 4.
Answer: The absolute minimum value is 4, at x = 2.
Example 3
Trap: treating f″(c) = 0 as an answer
For f(x) = x⁴ − 2, f′(0) = 0 and f″(0) = 0. A student concludes f has no extremum at x = 0. Is that right?
Show the solutionHide the solution
1. Step 1: f″(0) = 0 means the Second Derivative Test is inconclusive. It doesn't mean “no extremum.”
2. Step 2: Use the First Derivative Test: f′(x) = 4x³ is negative for x < 0 and positive for x > 0.
3. Step 3: f′ changes from negative to positive at 0.
Answer: No. f has a relative minimum at x = 0 (in fact an absolute minimum of −2), shown by the First Derivative Test.

Common mistakes

Getting the signs reversed. f″ > 0 (concave up, cup shape) means a minimum.
Concluding “neither” when f″(c) = 0. The test just doesn't decide.
Using the Second Derivative Test at a point where f′(c) ≠ 0. It only applies at critical points with f′(c) = 0.

On the exam

Free-response justifications using this test should state both facts: “f′(c) = 0 and f″(c) < 0, so f has a relative maximum at c.”
The one-critical-point argument is a favorite for optimization justifications. State that there's only one critical point on the interval.

Connected topics

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