Candidates Test for Global Extrema — Using the Candidates Test to Determine Absolute (Global) Extrema - Calculus 1

SECTION 1

Historical Context & Motivation

Throughout history, mathematicians and scientists have needed to find the greatest and least values that quantities can take. Whether designing the strongest arch, calculating the most efficient path for light, or maximizing a crop yield, the search for extreme values has been central to applied mathematics. The development of calculus in the 17th century finally gave us the tools to solve these problems systematically, but it took additional centuries to formalize a clean, reliable procedure for finding absolute (global) extrema on a closed interval.

1637 • Fermat's Method of Adequality

Pierre de Fermat developed a technique for finding where a function reaches its maximum or minimum by examining points where the tangent line is horizontal—an early precursor to setting the derivative equal to zero.

1684 • Leibniz Publishes Calculus

Gottfried Wilhelm Leibniz published his differential calculus, introducing the notation and rules for derivatives that made finding critical points a routine calculation.

1817 • Bolzano's Intermediate Value Work

Bernard Bolzano contributed to the rigorous foundation of continuity, a concept essential to guaranteeing that a continuous function on a closed interval actually achieves its extreme values.

1861 • Weierstrass Extreme Value Theorem

Karl Weierstrass proved the Extreme Value Theorem: any function that is continuous on a closed interval [a, b] must attain both an absolute maximum and an absolute minimum somewhere on that interval. This theorem is the theoretical backbone of the Candidates Test.

With the Extreme Value Theorem in hand, a natural question arose: if we know a continuous function must hit its highest and lowest points on a closed interval, how do we actually find them? The answer is the Candidates Test—a straightforward procedure that narrows the search down to a short list of "candidate" points and compares their function values.

SECTION 2

Core Principles & Definitions

Before diving into the procedure, you need to understand a few foundational ideas. The Candidates Test relies on the distinction between local extrema and global (absolute) extrema, the role of critical points, and the requirements that make the whole method valid.

Absolute (Global) Extrema

The absolute maximum is the single largest value f(x) achieves on the entire interval [a, b]. The absolute minimum is the single smallest value. Every continuous function on a closed interval has both.

Critical Points

A critical point is any x-value in the domain where f ′(x) = 0 or f ′(x) does not exist. These are the only interior points where extrema can occur.

Endpoints

On a closed interval [a, b], the values x = a and x = b are always candidates for global extrema—even if the function is increasing or decreasing there.

Extreme Value Theorem (EVT)

If f is continuous on [a, b], then f attains an absolute maximum and an absolute minimum on [a, b]. This theorem guarantees that the Candidates Test will always produce a valid answer.

Closed vs. Open Intervals

The Candidates Test applies only to closed intervals [a, b] (including both endpoints). On open intervals, absolute extrema are not guaranteed and the test does not apply directly.

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KEY TAKEAWAY

Think of the Candidates Test like a talent competition with a short audition list. You don't need to evaluate every single x-value—you only need to check the critical points (where the derivative is zero or undefined) and the endpoints. The winner (largest value) is the absolute max, and the lowest scorer is the absolute min. No other points can beat them.

SECTION 3

Visual Explanation

The diagram below shows a continuous function on the closed interval [1, 6]. Notice how the function has hills and valleys in the interior, plus values at the two endpoints. The candidate points are highlighted: the two endpoints and the critical points where the tangent line is horizontal (slope = 0) or where the derivative fails to exist (the cusp). By comparing the y-values at just these candidates, you can identify the global maximum and minimum without checking every point on the curve.

A continuous function on [1, 6] with three critical points (gold circles) and two endpoints (blue circles). Comparing the five y-values reveals the absolute maximum at x = 2 and the absolute minimum at x = 3.

The key insight from this diagram is that the absolute extrema occur at candidate points only—never at random interior points where the derivative exists and is nonzero. At such points, the function is either climbing or descending, so it can't be at its peak or valley there. This is why the method works: by restricting our attention to critical points and endpoints, we are guaranteed not to miss the global extrema.

SECTION 4

Mathematical Framework

The Candidates Test is a step-by-step procedure. Here is the formal statement and the equations that drive each step.

EXTREME VALUE THEOREM (PREREQUISITE)

If f is continuous on [a, b], then f attains an absolute maximum M and an absolute minimum m on [a, b].

This guarantees the answer exists. Without continuity on a closed interval, the test cannot be applied.

STEP 1 — FIND CRITICAL POINTS

Solve f ′(x) = 0 and identify where f ′(x) does not exist (for x in (a, b)).

f ′(x) is the derivative of f. Critical points are the x-values c where f ′(c) = 0 (horizontal tangent) or f ′(c) is undefined (cusp, corner, or vertical tangent), provided c is in the open interval (a, b).

STEP 2 — EVALUATE f AT ALL CANDIDATES

Compute f(a), f(c₁), f(c₂), …, f(cₙ), f(b)

Plug each endpoint and each critical point into the original function f(x) to get the corresponding y-values. The list of candidates is {a, c₁, c₂, …, cₙ, b}.

STEP 3 — COMPARE AND CONCLUDE

Absolute max = largest f-value among candidates; Absolute min = smallest f-value among candidates.

The greatest output value is the global maximum and the least output value is the global minimum. Report both the x-value and the y-value in your final answer.

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Common Mistake

Students sometimes forget to check the endpoints. A function can absolutely achieve its highest or lowest value at the boundary of the interval, not just at interior critical points. Always include x = a and x = b in your candidate list.

SECTION 5

Step-by-Step Procedure Flowchart

The flowchart below summarizes the entire Candidates Test procedure as a visual decision map. Following these steps in order ensures you never miss a candidate and always reach the correct conclusion.

Follow the five steps from top to bottom. The procedure begins by confirming continuity on a closed interval, proceeds through differentiation and evaluation, and ends by comparing outputs to identify the absolute maximum and minimum.

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Quick Tip

Before you start any calculation, double-check two things: (1) the interval is closed (written with square brackets), and (2) the function is continuous on the entire interval. If either condition fails, the Candidates Test does not apply.

SECTION 6

Worked Example

Let's apply the Candidates Test to find the absolute maximum and absolute minimum of f(x) = 2x³ − 3x² − 12x + 5 on the interval [−2, 4].

Find the absolute extrema of f(x) = 2x³ − 3x² − 12x + 5 on [−2, 4]

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Step 1 — Verify Conditions

f(x) = 2x³ − 3x² − 12x + 5 is a polynomial, so it is continuous everywhere. The interval [−2, 4] is closed. Therefore the Extreme Value Theorem guarantees absolute extrema exist, and the Candidates Test applies.
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Step 2 — Find the Derivative

Differentiate using the power rule: f ′(x) = 6x² − 6x − 12. Factor out 6 to simplify: f ′(x) = 6(x² − x − 2) = 6(x − 2)(x + 1).

f ′(x) = 6(x − 2)(x + 1)
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Step 3 — Solve for Critical Points

Set f ′(x) = 0: 6(x − 2)(x + 1) = 0, giving x = 2 and x = −1. Both values lie inside the open interval (−2, 4), so both are valid critical points. Since f ′(x) is a polynomial, it exists everywhere—there are no additional critical points from undefined derivatives.

Critical points: x = −1, x = 2
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Step 4 — Build the Candidate List and Evaluate

The candidates are the endpoints x = −2 and x = 4, plus the critical points x = −1 and x = 2. Evaluate f at each: • f(−2) = 2(−8) − 3(4) − 12(−2) + 5 = −16 − 12 + 24 + 5 = 1 • f(−1) = 2(−1) − 3(1) − 12(−1) + 5 = −2 − 3 + 12 + 5 = 12 • f(2) = 2(8) − 3(4) − 12(2) + 5 = 16 − 12 − 24 + 5 = −15 • f(4) = 2(64) − 3(16) − 12(4) + 5 = 128 − 48 − 48 + 5 = 37

f(−2) = 1, f(−1) = 12, f(2) = −15, f(4) = 37
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Step 5 — Compare and Conclude

Among the values {1, 12, −15, 37}, the largest is 37 and the smallest is −15.

Absolute maximum: f(4) = 37 at x = 4 (endpoint). Absolute minimum: f(2) = −15 at x = 2 (critical point).

Notice that the absolute maximum happened at an endpoint, not at a critical point. This is a powerful reminder that you must always check the boundary values. Without evaluating f(4), you would have incorrectly concluded that the absolute max was 12.

SECTION 7

Strengths, Limitations & Comparison with Other Tests

The Candidates Test is not the only method for analyzing extrema. You may have already learned the First Derivative Test and the Second Derivative Test, which classify local extrema. Here's how the Candidates Test compares to those approaches, and where each method shines.

Feature	Candidates Test	First Derivative Test	Second Derivative Test
What it finds	Absolute (global) max and min	Local max and min	Local max and min
Requires closed interval?	Yes — must be [a, b]	No	No
Uses endpoints?	Yes — always	No	No
Needs second derivative?	No	No	Yes
Best used when…	You need the overall highest and lowest values on a specific interval	You need to classify bumps and dips without a fixed interval	f ″ is easy to compute and you want quick local classification

Comparison of extrema-finding methods

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WHEN TO USE WHICH TEST

If the problem says "find the absolute maximum and minimum on [a, b]," reach for the Candidates Test. If the problem says "classify the critical point" or doesn't give a closed interval, use the First or Second Derivative Test instead. Think of it this way: the Candidates Test is the tool for finding the overall champion on a bounded playing field, while the other tests identify local heroes anywhere on the field.

SECTION 8

Connection to Advanced Theory

The Candidates Test you've learned in this lesson is the gateway to more sophisticated optimization techniques you'll encounter in future courses. Here's a preview of where these ideas lead.

This Lesson (Calculus 1)	What Comes Next
Functions of one variable on [a, b]	Multivariable optimization: finding extrema of f(x, y) on bounded regions in Calculus 3
Closed interval guarantees extrema exist	On open intervals or unbounded domains, limits at infinity and asymptotic behavior must be analyzed
Compare finitely many candidate values	Lagrange multipliers: optimize with constraints using the gradient, a powerful generalization
Polynomial and simple functions	Real-world modeling: cost, revenue, physics, and engineering problems requiring optimization under constraints

From Candidates Test to advanced optimization

Even in advanced settings, the core logic remains the same: narrow down the list of locations where extrema can occur, evaluate the function there, and compare. The Candidates Test instills this disciplined, systematic approach. As you move into applied optimization problems—maximizing the volume of a box, minimizing the cost of materials, or finding the fastest route—you'll see the Candidates Test appear again and again as the final step after setting up the model.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the Candidates Test cannot be used to find the absolute maximum of f(x) = 1/x on the open interval (0, 3). What condition from the Extreme Value Theorem fails?

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The Extreme Value Theorem requires two conditions: the function must be continuous on the interval, and the interval must be closed (include both endpoints). Here, the interval (0, 3) is open, meaning the endpoints are not included. Additionally, f(x) = 1/x is not defined at x = 0, so the function cannot be evaluated at the left boundary. As x approaches 0 from the right, f(x) grows without bound, so there is no finite absolute maximum. The Candidates Test relies on having a finite list of candidates, which is impossible when the function is unbounded.

Question 2: BASIC CALCULATION

Find the absolute maximum and absolute minimum of f(x) = x² − 4x + 3 on the interval [0, 5].

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First, find the derivative: f ′(x) = 2x − 4. Setting f ′(x) = 0 gives x = 2, which is inside (0, 5). The candidate list is {0, 2, 5}. Evaluate: f(0) = 0 − 0 + 3 = 3; f(2) = 4 − 8 + 3 = −1; f(5) = 25 − 20 + 3 = 8. The largest value is 8 at x = 5, so the absolute maximum is f(5) = 8. The smallest value is −1 at x = 2, so the absolute minimum is f(2) = −1. Note that both the max and min are found by simply comparing three numbers.

Question 3: INTERMEDIATE

Find the absolute extrema of f(x) = x³ − 6x² + 9x + 2 on [0, 4]. Identify whether each extremum occurs at an endpoint or a critical point.

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Differentiate: f ′(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x − 1)(x − 3). Setting f ′(x) = 0 gives x = 1 and x = 3, both in (0, 4). Candidates: {0, 1, 3, 4}. Evaluate: f(0) = 2; f(1) = 1 − 6 + 9 + 2 = 6; f(3) = 27 − 54 + 27 + 2 = 2; f(4) = 64 − 96 + 36 + 2 = 6. The absolute maximum value is 6, achieved at both x = 1 (critical point) and x = 4 (endpoint). The absolute minimum value is 2, achieved at both x = 0 (endpoint) and x = 3 (critical point). This example shows that absolute extrema can occur at multiple candidates simultaneously.

Question 4: APPLIED

A company determines that its profit (in thousands of dollars) from producing x hundred units is modeled by P(x) = −2x³ + 15x² − 36x + 40, where x is restricted to the interval [1, 5]. How many hundred units should the company produce to maximize profit, and what is the maximum profit?

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Differentiate: P ′(x) = −6x² + 30x − 36 = −6(x² − 5x + 6) = −6(x − 2)(x − 3). Critical points: x = 2 and x = 3, both in (1, 5). Candidates: {1, 2, 3, 5}. Evaluate: P(1) = −2 + 15 − 36 + 40 = 17; P(2) = −16 + 60 − 72 + 40 = 12; P(3) = −54 + 135 − 108 + 40 = 13; P(5) = −250 + 375 − 180 + 40 = −15. The absolute maximum profit is 17 thousand dollars, occurring at x = 1 (i.e., producing 100 units). The absolute minimum profit is −15 thousand dollars at x = 5 (a loss). This result shows that more production is not always better—the maximum profit occurs at the left endpoint of the feasible range.

Question 5: CRITICAL THINKING

Consider f(x) = |x² − 4| on [−3, 3]. Note that absolute value can create points where the derivative does not exist. Apply the Candidates Test to find the absolute maximum and minimum. Be sure to identify all critical points, including any where f ′(x) is undefined.

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First, analyze the expression inside the absolute value. x² − 4 = 0 when x = −2 and x = 2. For x in [−3, −2] ∪ [2, 3], x² − 4 ≥ 0 so f(x) = x² − 4 and f ′(x) = 2x. For x in (−2, 2), x² − 4 < 0 so f(x) = 4 − x² and f ′(x) = −2x. Setting f ′(x) = 0: from 2x = 0 we get x = 0 (valid, it's in (−2, 2)); from −2x = 0 we also get x = 0. At x = −2 and x = 2, the function changes formula, and the left and right derivatives disagree (e.g., at x = 2, left derivative is −4 and right derivative is +4), so f ′(x) does not exist there. These are also critical points. Candidates: {−3, −2, 0, 2, 3}. Evaluate: f(−3) = |9 − 4| = 5; f(−2) = |4 − 4| = 0; f(0) = |0 − 4| = 4; f(2) = 0; f(3) = 5. The absolute maximum is 5, occurring at both endpoints x = −3 and x = 3. The absolute minimum is 0, occurring at x = −2 and x = 2 (where f ′ is undefined). This problem illustrates that critical points where the derivative does not exist are just as important as those where f ′ = 0.

SUMMARY

Lesson Summary

Summary

The Candidates Test is a systematic procedure for finding the absolute (global) maximum and absolute (global) minimum of a continuous function on a closed interval [a, b]. It is justified by the Extreme Value Theorem, which guarantees that such extrema exist. The method works by identifying all critical points (where f ′(x) = 0 or f ′(x) does not exist) inside the interval, adding the two endpoints to the list, evaluating f at every candidate, and then comparing the resulting y-values.

The candidate with the largest function value is the absolute maximum, and the candidate with the smallest function value is the absolute minimum. Unlike the First and Second Derivative Tests, the Candidates Test finds global extrema rather than local ones. Always remember: endpoints count as candidates, the interval must be closed, and the function must be continuous on the entire interval for the test to be valid.