Short on time? This page is the recap. For explanations, the graphs and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- Concave up on an open interval: the slopes increase, so f′ is increasing. The graph bends like a cup.
- Concave down on an open interval: the slopes decrease, so f′ is decreasing. The graph bends like a cap.
- Concavity is separate from increasing/decreasing. A falling graph can be concave up.
- A point of inflection is a point on the graph where concavity changes.
- Candidates for inflection points: f″ = 0 or f″ undefined. Then check the sign change and that f(c) exists.
Key relationships
| You know… | On that open interval, f is… |
|---|---|
| f′ increasing, or f″ > 0 | concave up |
| f′ decreasing, or f″ < 0 | concave down |
| f″ changes sign at c, f(c) defined | point of inflection at (c, f(c)) |
| Graph of f′ has a local max or min at c | point of inflection of f at x = c |
| Graph of f′ crosses the x-axis at c | local extremum of f (Topic 5.4), not concavity |
| f′ > 0 and f″ < 0 (in context) | increasing at a decreasing rate |
Assumptions behind the method
- Concavity is decided on open intervals between the candidates.
- The sign of f″ is constant between consecutive candidates, so one test value per interval is enough (when f″ is continuous there).
- An inflection point needs f to be defined at c. A gap in the domain (such as x = 0 for 1/x) is never one.
Mistakes to avoid
- “f″(c) = 0, so inflection point.” Check the sign change: x⁴ has none at 0.
- Confusing zeros of f′ with inflection points.
- Reading a graph of f′ as a graph of f.
- Missing points where f″ does not exist, such as x = 0 for ∛x.
- Giving only x when the question asks for the point of inflection.
- Vague context wording. Say “increasing at a decreasing rate”, not “slowing down”.
Quick self-check
- Where is f(x) = x³ − 3x² + 1 concave down, and what is its point of inflection? (f″ = 6x − 6 < 0 for x < 1, so concave down on (−∞, 1). Inflection point (1, −1).)
- Find the inflection points of f(x) = x⁴ − 6x². (f″ = 12(x − 1)(x + 1) changes sign at ±1. Points (−1, −5) and (1, −5).)
- The graph of f′ crosses the x-axis at x = 2 and has a local maximum at x = 5. Which x-value gives an inflection point of f? (x = 5. x = 2 gives a local extremum of f.)
Next: practice questions.