The Extreme Value Theorem is a central result in calculus that tells us when a function is guaranteed to have an absolute maximum and an absolute minimum. It applies to any function that is continuous on a closed interval [a,b]. This matters because many real situations, such as height, profit, temperature, and distance, need a highest or lowest value over a fixed range.
The theorem gives a guarantee before we even calculate the values.
The two conditions are essential: the graph must have no breaks, jumps, or holes, and the interval must include both endpoints. To find the absolute extrema, evaluate the function at critical numbers inside the interval and at the endpoints. Then compare all those function values.
The largest value is the absolute maximum, and the smallest value is the absolute minimum.
Key Facts
- Extreme Value Theorem: If f is continuous on [a,b], then f has an absolute maximum and an absolute minimum on [a,b].
- A closed interval [a,b] includes both endpoints a and b.
- Continuity on [a,b] means the graph has no holes, jumps, breaks, or vertical asymptotes on the interval.
- Critical numbers occur where f'(x) = 0 or where f'(x) does not exist, as long as f(x) is defined.
- To find absolute extrema on [a,b], test endpoints and all critical numbers in (a,b).
- Absolute maximum = largest tested f(x) value, and absolute minimum = smallest tested f(x) value.
Vocabulary
- Extreme Value Theorem
- A theorem stating that a continuous function on a closed interval must attain both an absolute maximum and an absolute minimum.
- Absolute Maximum
- The greatest function value on a specified domain or interval.
- Absolute Minimum
- The least function value on a specified domain or interval.
- Closed Interval
- An interval [a,b] that includes both endpoint values a and b.
- Critical Number
- A number c in the domain of f where f'(c) = 0 or f'(c) does not exist.
Common Mistakes to Avoid
- Ignoring the endpoints, which is wrong because absolute extrema on a closed interval can occur at x = a or x = b.
- Assuming f'(x) = 0 is the only place to check, which is wrong because critical numbers can also occur where the derivative does not exist.
- Applying the theorem on an open interval, which is wrong because an open interval may approach a highest or lowest value without ever reaching it.
- Forgetting to verify continuity, which is wrong because a discontinuous function on [a,b] may fail to attain a maximum or minimum.
Practice Questions
- 1 Find the absolute maximum and absolute minimum of f(x) = x^2 - 4x + 1 on [0,5].
- 2 Find the absolute maximum and absolute minimum of f(x) = x^3 - 3x on [-2,3].
- 3 Explain why f(x) = 1/x on (0,1) does not satisfy the Extreme Value Theorem, and describe what happens to its values on that interval.