Short on time? This page is the recap. For explanations, figures and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- At a critical point with f′(c) = 0, the tangent is horizontal. The concavity at c decides the type of point.
- Concave up (f″(c) > 0): the curve is a cup above its tangent, so a relative minimum.
- Concave down (f″(c) < 0): the curve is a cap below its tangent, so a relative maximum.
- The test is a shortcut to the first derivative test: f″(c) > 0 means f′ is increasing through 0, so f′ changes from negative to positive.
- One-critical-point rule: if f is continuous on an interval with exactly one critical point c there, and c is a relative minimum (maximum), then f(c) is the absolute minimum (maximum) on that interval.
Key relationships
| Situation at x = c | Conclusion |
|---|---|
| f′(c) = 0, f″(c) > 0 | relative minimum |
| f′(c) = 0, f″(c) < 0 | relative maximum |
| f′(c) = 0, f″(c) = 0 | inconclusive: use the first derivative test |
| f′(c) does not exist | test does not apply: use the first derivative test |
| only one critical point on the interval, and it is a relative extremum | it is also the absolute extremum on that interval |
Model justification: “f has a relative minimum at x = 4 because f′(4) = 0 and f″(4) > 0.”
Assumptions behind the method
- f′(c) = 0 (not merely “c is a critical point”).
- f″(c) exists and you know its sign.
- For the one-critical-point rule: f is continuous on a single interval and has no other critical point in it.
Mistakes to avoid
- Swapping the cases. Positive f″ gives a minimum, not a maximum.
- Reading f″(c) = 0 as “no extremum” or as “point of inflection”. It means “no conclusion”.
- Leaving out f′(c) = 0 from the justification.
- Using the test at a cusp or corner, where f′(c) does not exist.
- Calling a relative extremum absolute when the interval has other critical points.
- Giving the x-value when the question asks for the value of f, or the reverse.
Quick self-check
- Let h(x) = x³ − 6x² + 2. Classify the critical points with the second derivative test. (h′ = 3x(x − 4). h″ = 6x − 12. h″(0) = −12, so relative maximum at x = 0 (h = 2). h″(4) = 12, so relative minimum at x = 4 (h = −30).)
- f′(2) = 0 and f″(2) = 0. What can you conclude? (Nothing yet. Check the sign of f′ on each side of x = 2.)
- Does k(x) = x⁴ + 2 have an extremum at x = 0, even though k″(0) = 0? (Yes. k′ = 4x³ changes from negative to positive, so there is a relative minimum, k(0) = 2.)
Next: practice questions.