Short on time? This page is the recap. For explanations, graphs and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- Extreme Value Theorem (EVT): if f is continuous on the closed interval [a, b], then f has at least one maximum value and at least one minimum value on [a, b].
- It is an existence theorem: it does not say where the extrema are or how many times they occur.
- Global (absolute) extremum: highest or lowest value on the whole interval. It can be at an endpoint.
- Local (relative) extremum: highest or lowest compared with points on an open interval around c. With this definition, an endpoint is not a local extremum.
- Critical point: c in the domain of f with f′(c) = 0 or f′(c) undefined.
- Every local extremum is at a critical point. Not every critical point is a local extremum.
Key relationships
| Situation | Critical point? | Extremum? | Example at x = 0 |
|---|---|---|---|
| Horizontal tangent, f′ keeps its sign | Yes (f′ = 0) | No | x³ |
| Horizontal tangent, f′ changes sign | Yes (f′ = 0) | Yes | x² |
| Corner or cusp | Yes (f′ undefined) | Can be (yes in these examples) | |x|, x^(2/3) |
| Vertical tangent, graph keeps rising | Yes (f′ undefined) | No | ∛x |
| f itself undefined | No | No | 1/x |
Assumptions behind the method
- The EVT needs both a closed interval and continuity on it. An open interval or a single break removes the guarantee.
- A failed condition means “no guarantee”, not “no extremum”.
- To find critical points, write f′ as one fraction: zeros of the top give f′ = 0; zeros of the bottom give f′ undefined (check f is defined there).
Mistakes to avoid
- Solving only f′(x) = 0 and missing points where f′ does not exist.
- Listing x-values where f is undefined as critical points.
- Calling every critical point a maximum or minimum.
- Forgetting endpoints when thinking about global extrema.
- Mixing up value and location: the maximum value is f(c); it occurs at x = c.
- Using the EVT on an open interval or across a discontinuity.
Quick self-check
- Find the critical points of f(x) = x³ − 12x. (f′(x) = 3x² − 12 = 0 gives x = −2 and x = 2.)
- Is x = 5 a critical point of f(x) = |x − 5|? (Yes: f is defined at 5 but has a corner there, so f′(5) does not exist.)
- f(x) = x³ has f′(0) = 0. Is there a local extremum at 0? (No: f(x) < 0 to the left and f(x) > 0 to the right.)
Next: practice questions.