Short on time? This page is the recap. For explanations, the graph and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Recap
- An absolute (global) maximum on an interval is the highest value of f anywhere on that interval. A local maximum is only highest nearby.
- If f is continuous on a closed interval [a, b], the Extreme Value Theorem says the absolute maximum and minimum exist.
- An interior absolute extremum is also a local extremum, so it sits at a critical point. Otherwise it sits at an endpoint. Those are the only candidates.
- The test compares values. You do not need the sign of f′ or the second derivative.
Key relationships
| Step | What you do | Watch out for |
|---|---|---|
| 1. Conditions | Confirm f is continuous on [a, b] | Open interval or a break: no guarantee |
| 2. Critical points | Solve f′(x) = 0; find where f′ does not exist | Keep only points strictly inside (a, b) |
| 3. Evaluate | Work out f at each critical point and at a and b | Use f, not f′, in the table |
| 4. Compare | Largest f = absolute max; smallest f = absolute min | A value can occur at two x’s |
Answer format: “The absolute maximum value of f on [a, b] is __, at x = __.”
Assumptions behind the method
- f is continuous on the whole closed interval, endpoints included.
- You have found every critical point in the interval, including cusps and corners where f′ does not exist.
- On a calculator question, you keep full precision until the comparison.
Mistakes to avoid
- Leaving out the endpoints.
- Missing x-values where f′ does not exist, such as x = 0 for x^(2/3).
- Keeping a critical point that lies outside the interval.
- Giving the x-value when the question asks for the maximum value.
- Assuming a local maximum is the absolute maximum.
- Using the test on an open interval, where an extremum may not exist.
Quick self-check
- Find the absolute extrema of f(x) = x³ − 3x on [0, 2]. (Candidates 0, 1, 2 give 0, −2, 2. Absolute max 2 at x = 2; absolute min −2 at x = 1.)
- f(x) = 2x³ − 9x² + 12x on [0, 3]. What is the absolute maximum value? (Critical points x = 1 and 2. Values: f(0) = 0, f(1) = 5, f(2) = 4, f(3) = 9. The absolute maximum is 9, at the endpoint x = 3, even though x = 1 is a local maximum.)
- Does x² have an absolute maximum on the open interval (−1, 2)? (No. Values approach 4 near x = 2 but never reach it.)
Next: practice questions.