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AP Calculus AB study notes

5.5.2 Applying the Candidates Test step by step

AP Syllabus focus:
‘To use the Candidates Test, list all interior critical points and the interval endpoints, evaluate the function at each candidate, and compare these values.’

The Candidates Test provides a systematic, reliable procedure for identifying absolute extrema of a continuous function on a closed interval by evaluating a carefully chosen set of key points across the domain.

Understanding the Purpose of the Candidates Test

The Candidates Test is a structured method used to determine absolute maximum and absolute minimum values of a function on a closed interval. Because absolute extrema may occur either at interior points or at the boundaries of the interval, the Candidates Test ensures students examine all necessary locations where extreme values could emerge. This subsubtopic focuses specifically on the step-by-step process, emphasizing the importance of identifying critical points and evaluating the function at each candidate.

Requirements for Applying the Candidates Test

Before starting the procedure, AP Calculus AB students must recall that the Candidates Test is valid only when the function is continuous on a closed interval. If continuity fails or the interval is not closed, the conclusions of the test may not hold. A function that is continuous on an interval is guaranteed to have both an absolute maximum and an absolute minimum somewhere on that interval, but locating these extrema requires careful evaluation of candidate points.

Step-by-Step Process of the Candidates Test

The specification emphasizes that students must list all interior critical points and the interval endpoints, evaluate the function at each candidate, and compare these values.

Pasted image

This figure shows the graph of y=cos(3πx)/xy=\cos(3\pi x)/x on the closed interval 0.1x1.10.1\le x\le 1.1, with its global and local maxima and minima clearly indicated. It illustrates how endpoints and interior critical points together form the full list of candidates for absolute extrema on a closed interval. The specific function and numerical labels extend beyond syllabus requirements but reinforce the Candidates Test procedure. Source.

Step 1: Identify and Confirm the Domain

To ensure the method can be applied correctly, begin by confirming the domain and verifying that the function is continuous on the relevant closed interval. This helps prevent overlooking any structural restrictions or undefined values that might affect analysis.

Step 2: Find All Critical Points Inside the Interval

A critical point occurs where the first derivative is zero or undefined, provided the value lies within the domain.

Critical Point: A point in the domain of a function where f(x)=0f'(x)=0 or where f(x)f'(x) does not exist.

To locate these points within the interval interior:

  • Compute the first derivative, ensuring the rules of differentiation are applied accurately.

  • Solve f(x)=0f'(x)=0 to find potential interior candidates.

  • Determine where f(x)f'(x) fails to exist but the original function is defined.

  • Retain only those points strictly inside the interval.

These interior critical points are essential candidates for absolute extrema because they represent positions where the function’s slope changes behavior or momentarily halts.

Step 3: Include the Interval Endpoints as Candidates

A central feature of the Candidates Test is the explicit inclusion of endpoints, because absolute extrema frequently occur at boundary values. Unlike local extrema, which do not rely on endpoints, absolute extrema may depend heavily on values at the boundaries. Students must therefore include:

  • The left endpoint

  • The right endpoint

No derivative conditions are required at endpoints for the purpose of this test.

Step 4: Evaluate the Function at All Candidates

Once all candidates have been identified, compute the function value at each one. This step involves evaluating the output of the original function, because absolute extrema depend on function values rather than derivative values. Students should maintain accuracy and keep track of all candidate evaluations clearly during this part of the process.

Step 5: Compare All Candidate Values

After computing values at all candidates, determine which is largest and which is smallest. These values represent the absolute maximum and absolute minimum of the function on the interval. This comparison finalizes the application of the test and provides a definitive answer supported by the systematic procedure.

Pasted image

This diagram plots y=sin(x)exp(x/5)y=\sin(x)\exp(-|x/5|) over a wide domain and labels several local maxima and minima along with the global maximum and minimum. It emphasizes that, after listing candidates, we must compare the corresponding yy-values to determine which points are absolute extrema. The wide interval shown includes extra detail not required by the syllabus, but the labeling directly illustrates how global extrema differ from local extrema. Source.

Step 6: Justifying the Extrema

Even though the process is algorithmic, students must justify their claims by stating that each candidate was evaluated, including both endpoints and interior critical points. Correct justification strengthens mathematical communication and demonstrates understanding aligned with AP expectations.

Organizational Strategies for Applying the Candidates Test

Students can make the procedure more efficient by using structured approaches:

  • Bullet lists of candidates help prevent omissions.

  • Derivative sign checks ensure that critical points were found correctly, even though the Candidates Test does not require sign analysis.

  • Clear notation for evaluations supports accuracy and logical structure.

These organizational methods support reliability in the results and align with college-level mathematical communication.

Why the Candidates Test Works

The Candidates Test relies on key properties of continuous functions on closed intervals. A continuous function on a closed interval achieves both its highest and lowest values, and these extrema must occur at either interior points where the derivative provides information or at boundary points where derivative information may be unavailable. Thus, examining both types of candidate points ensures that all possible locations for absolute extrema are considered.

FAQ

A critical point where the derivative does not exist is included only if the original function is defined at that x-value and the point lies strictly inside the interval.

If the function has a corner or cusp inside the interval, that point must be treated as an interior candidate because an absolute extremum can occur there.

Endpoints are still included separately even if the derivative is undefined at those points.

The Candidates Test does not rely on understanding how the derivative behaves; it only requires correct identification of interior points where the derivative is zero or undefined.

This makes the method reliable even when the derivative is algebraically messy, because once the candidate points are found, only function values matter.

It is a calculation-based approach rather than one requiring interpretation of derivative signs.

Students often overlook:

• Points where the derivative is undefined but the function is defined.
• Endpoints, especially when the question focuses heavily on interior behaviour.
• Restrictions in the domain that create unexpected boundary points.

Failing to check domain limitations causes many missed candidates, particularly when piecewise functions are involved.

Yes. A function that is continuous on a closed interval always has absolute extrema, even if the derivative never equals zero or is undefined only at endpoints.

In such cases, the absolute maximum and minimum both occur at endpoints.

This situation highlights why endpoints must always be included as candidates.

The First Derivative Test determines whether a point is a local maximum or minimum based on sign changes in the derivative.

The Candidates Test, however, identifies absolute extrema by comparing function values at all candidate points without needing to examine sign changes.

While the First Derivative Test gives information about local behaviour, the Candidates Test provides global information about the entire closed interval.

Practice Questions

Question 1 (1–3 marks)
A function f is continuous on the closed interval [2, 6]. The derivative f' is defined for all x in (2, 6) and satisfies f'(4) = 0.
Explain which x-values must be checked when using the Candidates Test to determine the absolute maximum of f on [2, 6].

Mark scheme:
• 1 mark for identifying x = 2 and x = 6 as required endpoints.
• 1 mark for identifying x = 4 as an interior critical point because f'(4) = 0.
• 1 mark for stating that these are all candidates for the absolute maximum.

Question 2 (4–6 marks)
Let g be a continuous function on the closed interval [0, 5]. Its derivative is given by g'(x) = x(5 - x).
(a) Find all interior critical points of g on [0, 5].
(b) List all candidate points for absolute extrema of g on [0, 5].
(c) Explain how the Candidates Test would be used to determine the absolute maximum value of g on this interval.

Mark scheme:
(a)
• 1 mark for setting g'(x) = 0.
• 1 mark for solving x(5 - x) = 0 to obtain x = 0 and x = 5, then identifying that neither is an interior point.
• 1 mark for concluding that there are no interior critical points.

(b)
• 1 mark for listing x = 0 and x = 5 as the only candidate points (endpoints).

(c)
• 1 mark for stating that the function must be evaluated at all candidate points.
• 1 mark for stating that the larger of the two resulting values will be the absolute maximum on the interval.

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