5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

An absolute maximum is the greatest value a function reaches on a specified domain, while an absolute minimum is the least value it reaches. On a closed interval, these global extrema can occur only at two kinds of locations: critical points inside the interval and the endpoints.

FUN-4.A.3: Absolute (global) extrema of a function on a closed interval can only occur at critical points or at endpoints.

The word candidate matters: a critical point or endpoint is not automatically a maximum or minimum. It is simply a location that must be tested. The Candidates Test turns that idea into a finite comparison.

The Candidates Test on a closed interval

For a function $f$ that is continuous on the closed interval $[a,b]$, use this procedure:

1. Find all critical points in $(a,b)$: values of $x$ where $f'(x)=0$ or where $f'(x)$ does not exist.
2. Keep only the critical points that lie in the interval.
3. Evaluate $f$ at every critical point and at both endpoints, $x=a$ and $x=b$.
4. Compare the resulting function values.

The largest value is the absolute maximum, and the smallest value is the absolute minimum. The theorem behind the procedure is the Extreme Value Theorem: continuity on a closed interval guarantees that absolute extrema exist, and the Candidates Test identifies where they can occur.

Worked example: comparing every candidate

Let

$$ f(x)=x^3-3x^2+1 $$

on $[-1,3]$. First differentiate:

$$ f'(x)=3x^2-6x=3x(x-2). $$

Thus the critical points are $x=0$ and $x=2$, both inside $[-1,3]$. Now evaluate all candidates:

$$ \begin{aligned} f(-1)&=-3,\ f(0)&=1,\ f(2)&=-3,\ f(3)&=1. \end{aligned} $$

The absolute maximum value is $1$, occurring at $x=0$ and $x=3$. The absolute minimum value is $-3$, occurring at $x=-1$ and $x=2$. A complete written justification is: “Because $f$ is continuous on $[-1,3]$, its absolute extrema occur at critical points or endpoints. Comparing $f(-1)$, $f(0)$, $f(2)$, and $f(3)$ gives an absolute maximum of $1$ and an absolute minimum of $-3$.”

Misconception check — “The critical point is the answer.” A critical point is only a candidate. In the example, $x=0$ is a critical point and gives the absolute maximum, but $x=2$ is also critical and gives the absolute minimum. The endpoints are equally important: $x=-1$ ties for the minimum, and $x=3$ ties for the maximum.

What changes on an open interval?

The Candidates Test above is specifically a closed-interval method. If the domain is $(a,b)$, the endpoints are not included and cannot be locations where the function attains an absolute extremum. Likewise, if the function is not continuous on $[a,b]$, the Extreme Value Theorem does not apply automatically, so the required justification must address the actual domain and behavior.

Suggested skill 1.E — Apply appropriate mathematical rules or procedures, with and without technology is explicit for Topic 5.5. A calculator may help locate zeros of $f'(x)$ or evaluate complicated expressions, but the mathematical procedure still requires identifying the interval, checking domain membership, evaluating every candidate, and comparing values. A calculator list of values without that structure is not a justification.

Retrieval check. A function $f$ is continuous on $[0,6]$, and its only critical points are $x=2$ and $x=5$. Which values must be compared to find its absolute extrema, and why is $f'(2)=0$ not enough to call $f(2)$ the absolute maximum? Answer: compare $f(0)$, $f(2)$, $f(5)$, and $f(6)$; a critical point is only a candidate until every candidate value has been compared.