AP Calculus BC · Unit 5: Analytical Applications of Differentiation
5.5 Using the Candidates Test to Determine Absolute (Global) Extrema
What the Candidates Test finds
The Candidates Test finds absolute (global) extrema of a continuous function on a closed interval [a, b]. Compare the function's values at all critical points inside the interval (where f'(x) = 0 or f'(x) is undefined) and at the endpoints a and b. The largest value is the absolute max; the smallest is the absolute min.
Why it works
By the Extreme Value Theorem, a function continuous on a closed interval must attain an absolute max and min. These extrema occur either at endpoints or at critical points — nowhere else. So checking only these finitely many candidates guarantees you find the true global extrema without needing sign analysis.
Steps to apply it
1) Confirm f is continuous on [a, b]. 2) Find f'(x); solve f'(x) = 0 and locate where f'(x) is undefined — these are critical points inside the interval. 3) Evaluate f at each critical point and at both endpoints. 4) Compare all values: the greatest is the absolute maximum, the least is the absolute minimum.
Key terms
- Candidates Test
- Method to find absolute extrema on [a, b]: evaluate f at all critical points in the interval and at both endpoints, then compare values.
- Extreme Value Theorem
- If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and minimum on that interval.
- Critical point
- A point in the domain of f where f'(x) = 0 or f'(x) is undefined.
- Where can absolute extrema occur on [a,b]?
- Only at critical points inside the interval or at the endpoints a and b.
- Steps for the Candidates Test
- 1) Check continuity on [a,b]. 2) Find critical points. 3) Evaluate f at critical points and endpoints. 4) Compare all values to find absolute max/min.
Try an AP-style question
What are the absolute max and min of f(x) = x³ − 3x on [0, 3]?
- Max 0, min −18
- Max 18, min −2
- Max 2, min −2
- Max 18, min 0
Show the answer
B. Candidates: f(0) = 0, f(1) = −2, f(3) = 18.
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