Candidates Test for Absolute Extrema

In calculus, finding the absolute maximum and absolute minimum values of a function on a closed interval is one of the most practical skills you can develop. Unlike local extrema — which are simply peaks or valleys compared to nearby points — absolute extrema are the single highest and lowest values a function achieves over an entire specified domain. For example, an engineer designing a bridge needs to know the maximum stress the structure will ever experience, not just a local spike. The Candidates Test gives a reliable, systematic procedure for locating these global extreme values. The Candidates Test (also called the Closed Interval Method) is built on a powerful theorem from calculus: the Extreme Value Theorem. This theorem guarantees that 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. Without continuity or a closed interval, such a guarantee disappears — a function with a jump discontinuity or defined on an open interval might 'escape' to infinity or simply never reach a highest point. Knowing this theorem justifies why the test works and what conditions must be met before applying it. The core insight of the Candidates Test is that absolute extrema can only occur at a small, finite set of 'candidate' points: the endpoints of the interval and any critical points interior to the interval. A critical point is a point in the domain where the derivative equals zero or does not exist. By evaluating the function at every candidate and comparing the results, you identify the absolute extrema with certainty — no graphing required. This makes the Candidates Test both rigorous and efficient, suitable for polynomials, trigonometric functions, rational functions, and many other common function families.

Key takeaways
  • Verify the Conditions
  • Find the Derivative
  • Identify All Critical Points on (a, b)
  • List All Candidates
  • Evaluate f at Each Candidate
  • Compare and Conclude

Verify the Conditions

Confirm that the function f is continuous on the closed interval [a, b]. If the function has any discontinuities (holes, jumps, or vertical asymptotes) within [a, b], the Extreme Value Theorem does not apply and the test cannot be used without modification. For most high school problems — polynomials, sinusoids, simple rational functions defined everywhere on the interval — continuity is easily confirmed.

Find the Derivative

Differentiate f(x) to obtain f'(x). Use all differentiation rules that apply: power rule, product rule, quotient rule, chain rule, or trigonometric derivatives. Simplify f'(x) as much as possible so that solving f'(x) = 0 is manageable. For example, if f(x) = 2x³ − 9x² + 12x − 4, then f'(x) = 6x² − 18x + 12 = 6(x² − 3x + 2) = 6(x − 1)(x − 2).

Identify All Critical Points on (a, b)

Solve f'(x) = 0 and also find any x-values in (a, b) where f'(x) is undefined. These are the critical points. Keep only those that lie strictly inside the open interval (a, b) — critical points outside this interval are irrelevant. In the example above, f'(x) = 0 gives x = 1 and x = 2. If the interval were [0, 3], both are valid candidates.

List All Candidates

Compile the complete list of candidates: the two endpoints x = a and x = b, plus every critical point found in Step 3. This finite list contains every location where the absolute maximum or minimum could possibly occur. For [0, 3] in our running example, the candidates are x = 0, 1, 2, 3.

Evaluate f at Each Candidate

Substitute each candidate into the original function f(x) — not f'(x). Arithmetic precision here is critical. For f(x) = 2x³ − 9x² + 12x − 4 on [0, 3]: f(0) = −4; f(1) = 2 − 9 + 12 − 4 = 1; f(2) = 16 − 36 + 24 − 4 = 0; f(3) = 54 − 81 + 36 − 4 = 5.

Compare and Conclude

Identify the largest and smallest values from your evaluations. The largest output is the absolute maximum value, and the x-value where it occurs is the absolute maximum location. The smallest output is the absolute minimum value. In the example: the values are −4, 1, 0, 5. Therefore, the absolute maximum is f(3) = 5, and the absolute minimum is f(0) = −4. Always state both the location (x-value) and the value (y-value) in your answer.

The Key Equation

f′(c)=0orf′(c) does not exist,c∈(a,b)f'(c) = 0 \quad \text{or} \quad f'(c) \text{ does not exist}, \quad c \in (a,b)
  • f'(c) — Derivative at c (dimensionless (depends on function)): The instantaneous rate of change of f evaluated at the interior point c. Setting this equal to zero locates potential peaks and valleys.
  • c — Critical Point (x-axis units): An x-value strictly inside (a, b) where f'(c) = 0 or f'(c) is undefined. These are candidate locations for the absolute extrema.
  • a, b — Endpoints (x-axis units): The left and right boundaries of the closed interval [a, b]. They are always candidates for absolute extrema, regardless of the derivative's behavior there.

Worked Example

Problem. Find the absolute maximum and absolute minimum of f(x) = x³ − 3x² − 9x + 2 on the closed interval [−2, 6].

fmax⁡=f(6)=56,fmin⁡=f(3)=−25f_{\max} = f(6) = 56, \quad f_{\min} = f(3) = -25

Answer. Absolute maximum value is 56, occurring at x = 6. Absolute minimum value is −25, occurring at x = 3.

Figure
Diagram
Candidates Test: f(x) = x³ − 3x² − 9x + 2 on [−2, 6]xf(x)−2−13603060110f(−2)=0f(−1)=7f(3)=−25f(6)=56Candidate EvaluationsAbs. Max: f(6) = 56Abs. Min: f(3) = −25Local Max: f(−1) = 7
Origins · 1 minKarl Weierstrass and how this was discovered
KW
Karl Weierstrass
1815–1897 · German

Weierstrass gave the first rigorous modern proof of the Extreme Value Theorem — the foundational guarantee that a continuous function on a closed interval must attain its absolute maximum and minimum. Before his work, mathematicians relied on geometric intuition without formal proof. He also developed the epsilon-delta definition of a limit, placing all of calculus on a rigorous logical foundation. The Candidates Test is only logically valid because of the theorem Weierstrass proved. Weierstrass did not begin his serious mathematical research career until his 40s — he spent his 20s and 30s teaching at secondary schools in remote Prussian towns. His groundbreaking papers were published while he was still a schoolteacher, shocking the Berlin mathematical establishment when they discovered who had written them.

  • Critical Points and the Derivative — The Candidates Test depends entirely on correctly identifying critical points by solving f'(x) = 0 or finding where f'(x) is undefined. Mastery of critical point analysis is a prerequisite.
  • Extreme Value Theorem — The Extreme Value Theorem is the theoretical guarantee that justifies the Candidates Test. Understanding why the test must produce a correct answer requires knowing this theorem.
  • First Derivative Test for Local Extrema — The First Derivative Test classifies critical points as local maxima or minima using sign changes in f'(x). This complements the Candidates Test, which finds absolute rather than local extrema.
  • Increasing and Decreasing Functions — Understanding where f'(x) > 0 or f'(x) < 0 helps interpret why absolute extrema appear where they do, connecting the algebraic test to the geometric shape of the curve.