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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]?

  1. Max 0, min −18
  2. Max 18, min −2
  3. Max 2, min −2
  4. Max 18, min 0
Show the answer

B. Candidates: f(0) = 0, f(1) = −2, f(3) = 18.

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