On a closed interval [a,b], absolute (global) extrema can only occur at endpoints or at critical numbers (where f′(x)=0 or f′ is undefined)—this follows from the Extreme Value Theorem and Fermat’s theorem (CED FUN-4.A.3). Use the Candidates Test: 1. Check that f is continuous on [a,b] (EVT applies). 2. Compute f′(x). Find all critical numbers in (a,b): solve f′(x)=0 and include x where f′ is undefined. 3. Evaluate f at each critical number and at the endpoints a and b. 4. Compare those values: the largest is the absolute maximum, the smallest is the absolute minimum. Notes: include non-differentiable interior points as candidates. You can also use first/second-derivative sign info to justify why a candidate is max/min (first derivative sign chart or second derivative test), but the Candidates Test itself is just evaluation. For examples and practice problems, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ) and Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5). Practice hundreds more problems at (https://library.fiveable.me/practice/ap-calculus).

The Candidates Test: to find absolute (global) maxima/minima on a closed interval [a,b], evaluate the function at every “candidate” and pick the largest/smallest. Candidates = critical numbers in (a,b) (where f′=0 or f′ is undefined) plus the endpoints a and b. Why it works: Extreme Value Theorem guarantees extrema exist on closed intervals, and Fermat’s theorem says interior extrema must be at critical points. How to use it (quick checklist): 1) Confirm domain and closed interval. 2) Find f′ and solve f′(x)=0; include points where f′ is undefined. 3) Evaluate f at each candidate and at endpoints. 4) The largest value = absolute max; smallest = absolute min. Use this on any closed interval question on the AP (Unit 5, FUN-4.A). For a step-by-step guide and practice, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ) and more problems (https://library.fiveable.me/practice/ap-calculus).

Relative (local) extrema are points where f(x) is higher or lower than all nearby values—you can find them with the First or Second Derivative Test at critical points (where f′ = 0 or f′ is undefined). Absolute (global) extrema are the single highest or lowest values of f on a given domain (often a closed interval). Key CED fact: on a closed interval, absolute extrema can only occur at critical points or at endpoints (FUN-4.A.3)—that’s the Candidates Test. So procedure on [a,b]: find critical numbers in (a,b); evaluate f at those and at a and b; the largest value is the absolute max, the smallest is the absolute min. Use derivative sign charts/second derivative to classify local behavior, and remember the Extreme Value Theorem guarantees absolute extrema exist on closed, continuous intervals. For more walkthroughs and examples see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ) and more practice problems (https://library.fiveable.me/practice/ap-calculus).

Step-by-step (Candidates) test for absolute extrema on a closed interval [a,b]: 1. Verify conditions: f is continuous on [a,b] (Extreme Value Theorem applies). 2. Find critical numbers: compute f′(x) and solve f′(x)=0; also include points where f′ is undefined but f is defined. (These + endpoints are your candidates per FUN-4.A.3 / Fermat’s theorem.) 3. List candidates: {a, b} ∪ {all critical numbers in (a,b)}. 4. Evaluate f at each candidate. 5. Compare values: largest value = absolute (global) maximum on [a,b]; smallest = absolute (global) minimum. 6. Justify using derivative info if asked: you can cite first-derivative sign changes or second-derivative test for local behavior, but absolute extrema must come from the candidate list. This is exactly what AP wants for Topic 5.5 (use candidates, endpoints, and justify with derivative behavior). For a quick guide and examples see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ). For more practice problems, use the unit page (https://library.fiveable.me/ap-calculus/unit-5) and general practice bank (https://library.fiveable.me/practice/ap-calculus).

Short answer: If you’re finding absolute (global) extrema on a closed interval, yes—you must always check the endpoints. Why: The CED states absolute extrema on a closed interval can only occur at critical points or at endpoints (FUN-4.A.3). So the Candidates Test is: (1) find critical numbers (where f′ = 0 or f′ undefined) inside the interval, (2) evaluate f at those points and at the two endpoints, and (3) the largest/smallest values are the global max/min. The Extreme Value Theorem guarantees at least one absolute max and min on a continuous function over a closed, bounded interval—so endpoints are essential. If the domain is open or unbounded, endpoints aren’t included; then consider limits, end behavior, and any non-differentiable points (they’re still candidates). For a quick refresher and examples, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ). For extra practice, check Fiveable’s unit practice problems (https://library.fiveable.me/practice/ap-calculus).

If you want to know whether a critical point is an absolute (global) max or min on a closed interval, use the Candidates Test (and theorems from the CED). Steps: 1) Find all critical numbers in the interval (where f′(x)=0 or f′ is undefined). Remember Fermat’s theorem and that non-differentiable points count as candidates. 2) Include the two endpoints (Extreme Value Theorem → absolute extrema on a closed interval occur only at critical points or endpoints—FUN-4.A.3). 3) Evaluate f at every candidate. The largest value is the absolute maximum; the smallest is the absolute minimum. Note: First/second derivative tests tell you local behavior (increasing/decreasing or concavity), but they don’t replace comparing values for global extrema. For practice and worked examples, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ), the Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5), and many practice problems (https://library.fiveable.me/practice/ap-calculus).

On a closed interval [a, b], the “formula” for finding absolute (global) extrema is: evaluate f at all critical points inside (a, b) and at the endpoints, then pick the largest and smallest values. In set form: - Let C = { x in (a,b) : f′(x)=0 or f′(x) does not exist }. - Then absolute max = max{ f(x) : x ∈ C ∪ {a, b} } and absolute min = min{ f(x) : x ∈ C ∪ {a, b} }. This is the Candidates Test (uses the Extreme Value Theorem + Fermat’s theorem). Remember to only include critical numbers that lie in (a,b), and to check endpoints a and b. For the AP exam make sure you show how you found critical points (solve f′=0 or note nondifferentiable points) and then list the f-values used to justify the global max/min (CED FUN-4.A.3). For a topic refresher see the study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ) and grab extra practice at (https://library.fiveable.me/practice/ap-calculus).

Use the candidates test when you need absolute (global) maxima or minima on a closed interval. The Candidates Test says check all critical numbers (where f′ = 0 or f′ doesn’t exist) AND the endpoints, then compare f-values to pick the largest and smallest (FUN-4.A.3, Extreme Value Theorem). Use the first- or second-derivative tests when you want to classify local (relative) extrema or describe local shape: - First derivative test: make a sign chart for f′ to see where f changes from + to − (local max) or − to + (local min). Good for non-differentiable critical points too. - Second derivative test: if f′(c)=0 and f″(c)>0 ⇒ local min; f″(c)<0 ⇒ local max. (Only for classifying stationary points.) In short: candidates test = find global extrema on closed intervals (compare values at critical points and endpoints). First/second derivative tests = determine whether a critical point is a local max/min. For AP review, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ) and more Unit 5 material (https://library.fiveable.me/ap-calculus/unit-5). For extra practice, use Fiveable’s problem set page (https://library.fiveable.me/practice/ap-calculus).

Think of it this way: on a closed interval [a, b] the Extreme Value Theorem guarantees a global max and min exist. Where could they hide? If an extreme value happens at an interior point c (a < c < b) and the function is differentiable there, Fermat’s theorem says f′(c) = 0—so c is a critical number (a stationary point). If the function isn’t differentiable at c, that point is still a candidate (it’s a critical point because f′ doesn’t exist). Finally, endpoints a and b are special: the function can attain its absolute extrema there even though you can’t take a two-sided derivative. So every absolute extremum on a closed interval must be either (1) an endpoint or (2) a critical point (f′(c)=0 or f′(c) undefined). That’s exactly why the Candidates Test works: check all critical points plus endpoints, then compare values. For an AP review, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ) and practice problems (https://library.fiveable.me/practice/ap-calculus).

Start by remembering the theorems: Extreme Value Theorem guarantees a continuous f on a closed interval has absolute max and min, and Fermat’s theorem (FUN-4.A) says they occur only at critical points or endpoints. Steps (Candidates Test) 1. Find critical numbers: solve f′(x)=0 and list x where f′ is undefined but f is defined. 2. Make your candidate set: all critical numbers inside the closed interval plus the two endpoints. 3. Evaluate f at every candidate. 4. The largest value is the absolute (global) maximum, the smallest is the absolute minimum. (Optional) Use first-derivative sign chart or second-derivative test to identify local behavior, but you still must compare candidate values for global extrema. This is exactly what AP asks you to justify (FUN-4.A, FUN-4.A.3). For a quick refresher use the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ) and try practice problems (https://library.fiveable.me/practice/ap-calculus).

If there are no critical points in the closed interval, don’t panic—absolute extrema can still occur, but only at the endpoints. By the Extreme Value Theorem a continuous function on a closed interval [a,b] must have a global max and min; by Fermat’s theorem (and FUN-4.A.3) interior absolute extrema can only occur at critical points (where f′=0 or f′ is undefined). So if you find no interior critical numbers, your candidate set for the Candidates Test is just {a, b}. Evaluate f(a) and f(b); the larger gives the absolute max and the smaller gives the absolute min. If the function isn’t continuous or has interior points where f′ is undefined, include those non-differentiable points as candidates too. For a quick review, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ). For extra practice, check Unit 5 and practice problems (https://library.fiveable.me/ap-calculus/unit-5 and https://library.fiveable.me/practice/ap-calculus).

Quick step-by-step for the Candidates Test (Topic 5.5, AP CED FUN-4.A): 1. Check the setting: make sure the function is continuous on a closed interval [a,b] (Extreme Value Theorem applies). 2. Find critical numbers: compute f′(x) and solve f′(x)=0; also include points in (a,b) where f′ is undefined. (Fermat’s theorem says extrema inside must be at these.) 3. Make the candidate list: all critical numbers that lie in [a,b] plus the two endpoints a and b. 4. Evaluate f at every candidate. 5. Compare values: the largest value = absolute (global) maximum; the smallest = absolute (global) minimum. 6. (Optional check) If you want justification, mention EVT ensures extremes exist on [a,b] and Fermat’s theorem locates interior candidates. Example: if f(x)=x^3−3x on [−2,2], f′=3x^2−3 → x=±1; candidates: −2, −1, 1, 2; evaluate f at each and pick biggest/smallest. For more AP-aligned practice and study, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ), the Unit 5 overview (https://library.fiveable.me/ap-calculus/unit-5), and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

Because the Extreme Value Theorem (for a continuous function on a closed interval) guarantees a global max and min exist, you only need to check the “candidates” where those could occur: critical points and the endpoints. Fermat’s theorem tells you any interior absolute extremum must be at a critical number (f ′(c)=0 or f ′ undefined). But endpoints aren’t interior—f ′ doesn’t have to be zero there—so they can still be the highest or lowest value on [a,b]. The Candidates Test just says: find all critical points in (a,b), add x=a and x=b, evaluate f at each, and pick the largest/smallest. That’s a complete, AP-aligned method (FUN-4.A.3 in the CED). For a quick review, see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ). Need practice? Try problems in the Unit 5 page (https://library.fiveable.me/ap-calculus/unit-5) or the full practice library (https://library.fiveable.me/practice/ap-calculus).

Use the Candidates Test when you need absolute (global) maxima or minima on a closed interval [a,b]. The Extreme Value Theorem guarantees an absolute max and min on a continuous function over [a,b], and Fermat’s theorem tells you they can only occur at critical points or endpoints. Steps: (1) Verify the function is continuous on [a,b]. (2) Find critical numbers in (a,b) where f′(x)=0 or f′ doesn’t exist. (3) Evaluate f at each critical number and at the endpoints a and b. (4) The largest value is the absolute maximum; the smallest is the absolute minimum. Don’t use the Candidates Test on open intervals or when you only want local extrema—use the First or Second Derivative Tests for local behavior instead. For AP alignment see Topic 5.5 in the CED and the Fiveable study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ). For unit review and extra practice problems, check (https://library.fiveable.me/ap-calculus/unit-5) and (https://library.fiveable.me/practice/ap-calculus).

Start by turning the word problem into one objective function (what you’re optimizing) and its domain. Then use the Candidates Test: 1. Write the function f(x) that models the quantity to maximize/minimize and explicitly state the closed interval or domain (use EVT—absolute extrema on a closed interval exist). 2. Compute f ′(x). By Fermat’s theorem, candidates are where f ′(x)=0 or f ′(x) is undefined, plus the interval endpoints. 3. List all candidates (critical numbers + endpoints). 4. Evaluate f at every candidate. 5. Compare the values: the largest is the global maximum, the smallest is the global minimum. (If asked, justify using sign of f ′ or f ′′ for local behavior.) Quick tips: check units and constraints (physical limits give domain). If a variable substitution simplifies the algebra, do it before differentiating. For a refresher and examples aligned to the AP CED (FUN-4.A, EVT, Fermat’s theorem), see the Topic 5.5 study guide (https://library.fiveable.me/ap-calculus/unit-5/using-candidates-test-to-determine-absolute-global-extrema/study-guide/2ONEsyKKR6nyMs3UOpOZ). For more practice problems, try Fiveable’s practice set (https://library.fiveable.me/practice/ap-calculus).