The Candidates Test is the AP Calculus method for finding absolute (global) extrema on a closed interval: find all critical points in the interval, evaluate the original function f at each critical point AND at both endpoints, then compare those values. The largest is the absolute max, the smallest the absolute min.
The Candidates Test is how you find the absolute (global) maximum or minimum of a continuous function on a closed interval. The logic comes straight from the CED's essential knowledge for Topic 5.5: absolute extrema on a closed interval can only occur at critical points or at endpoints. Those are your only 'candidates,' which is where the name comes from.
The procedure has three steps. First, find every critical point of f inside the interval (where f′ is zero or undefined). Second, plug each critical point and both endpoints into the original function f, not the derivative. Third, compare the outputs. The biggest f-value is the absolute maximum, and the smallest is the absolute minimum. That's it. No sign charts, no checking concavity. You're literally just lining up the candidates and picking the winner and loser.
This is the entire point of Topic 5.5 in Unit 5 (Analytical Applications of Differentiation), and it supports learning objective 5.5.A, which asks you to justify conclusions about a function's behavior using its derivatives. On the exam, 'justify' is the operative word. It's not enough to find the right answer. You have to show that you evaluated f at every critical point and every endpoint and then compared the values. Skipping an endpoint, or comparing derivative values instead of function values, costs you the justification point. The Candidates Test also shows up constantly in applied FRQ contexts (rates, accumulation, modeling problems) where a question asks for the maximum or minimum value of a quantity over a specific time interval.
Keep studying AP Calculus Unit 5
Visual cheatsheet
view galleryCritical Points (Unit 5)
Critical points are the interior candidates in the Candidates Test. You find them by setting f′(x) = 0 or finding where f′ is undefined, which is why the test is technically a derivative application even though the final step uses f itself.
First Derivative Test (Unit 5)
The First Derivative Test classifies a critical point as a local max or min by checking the sign of f′ on each side. The Candidates Test answers a different question entirely. It finds the single biggest and smallest value on a whole closed interval, and it does so by comparing f-values, not derivative signs.
Local Extrema (Unit 5)
A local max is the highest point in its neighborhood; an absolute max is the highest point on the entire interval. An absolute extremum on a closed interval is either a local extremum that wins the overall comparison or an endpoint, which is exactly why the candidate list is 'critical points plus endpoints.'
Domain (Units 1 & 5)
The Candidates Test only works as stated on a closed interval where the function is continuous. If the problem gives you an open interval or an unbounded domain, endpoints may not exist as candidates and you need a different argument, often the First Derivative Test plus reasoning about behavior at the boundary.
Multiple-choice questions test whether you know what to compare. A classic stem asks which values determine the absolute extrema, and the answer is always f evaluated at critical points and endpoints. On FRQs, the Candidates Test is a justification machine. The 2025 FRQ Q1, for example, gave a model C(t) for acres affected by an invasive species and asked about extreme values over a time interval. To earn full credit on questions like that, you must (1) find where the derivative is zero or undefined on the interval, (2) make a table or list showing the function's value at each critical point and at both endpoints, and (3) explicitly state the comparison ('the absolute maximum is f(a) because it is the greatest of these values'). Writing 'f′ = 0 so it's a max' without comparing candidates does not earn the justification point.
The First Derivative Test finds LOCAL extrema by checking whether f′ changes sign at a critical point. The Candidates Test finds ABSOLUTE extrema on a closed interval by evaluating f at critical points and endpoints and comparing the actual function values. Quick check: if the question says 'on the interval [a, b]' and asks for the absolute max or min, use the Candidates Test. If it asks whether a specific critical point is a relative max or min, use the First (or Second) Derivative Test. Mixing them up is the most common justification error on Unit 5 FRQs.
Absolute extrema on a closed interval can only occur at critical points or at endpoints, so those are the only candidates you need to check.
You evaluate the original function f at each candidate, not the derivative, and the largest output is the absolute max while the smallest is the absolute min.
Never forget the endpoints; an absolute extremum often sits at an endpoint even when the derivative is not zero there.
On FRQs, a full justification means listing f-values at all critical points and both endpoints and explicitly comparing them.
The Candidates Test finds absolute (global) extrema, while the First and Second Derivative Tests classify local extrema.
It's the method from Topic 5.5 for finding absolute (global) extrema on a closed interval. You find all critical points in the interval, evaluate f at those points and at both endpoints, and compare. The largest value is the absolute max and the smallest is the absolute min.
No. The First Derivative Test uses sign changes of f′ to classify a critical point as a local max or min. The Candidates Test compares actual f-values at critical points and endpoints to find the absolute max and min over a whole closed interval.
Into f, the original function. The derivative only helps you find which x-values are critical points. The comparison step always uses function values, because you're asking which point is literally highest or lowest.
Because the CED's essential knowledge for 5.5.A says absolute extrema on a closed interval occur only at critical points or endpoints. An endpoint can be the absolute max or min even though f′ isn't zero there, so leaving endpoints out loses the answer and the justification point.
It requires a continuous function on a closed interval. On an open or unbounded interval there may be no endpoint candidates, so you'd justify absolute extrema differently, for example by showing f′ changes sign only once on the entire domain.