Introduction
A global maximum is often described by giving an input and a function value. These are related, but they are not the same thing. The maximum value is an output; a maximizing input is a point in the domain that produces that output. There may be several such inputs, even though the maximum value itself is a single number.
Constant functions make this distinction especially clear: every input produces the same output, so every input is both a maximizing input and a minimizing input. Nonconstant functions can also attain an absolute extreme at more than one input. We will define the value and the attaining inputs separately, then use that distinction in examples.
Absolute Values and the Inputs That Attain Them
The condition that equality occurs somewhere matters. A value that is merely an upper bound is not necessarily an absolute maximum: the function must actually attain it. Likewise, a lower bound is not an absolute minimum unless some input produces that value.
The absolute maximum value, when it exists, is unique as a number. The set of maximizing inputs, however, can contain one input or many. The same distinction applies to the minimum. In particular, it is useful to report both the value and the set of inputs where it is attained.
Why a Constant Function Attains Both Extrema Everywhere
A function is constant on \(D\) if there is a fixed number \(c\) such that \(f(x)=c\) for every \(x\in D\). If \(D\) is nonempty, then \(c\) is both an absolute maximum value and an absolute minimum value: no output is greater than \(c\) or less than \(c\), and every input gives equality.
Proof. Choose any \(x\in D\), which is possible because \(D\) is nonempty. By constancy, \(f(x)=c\) for every input in \(D\). Thus every output equals \(c\), so every output is at most \(c\) and at least \(c\). Also, equality with \(c\) occurs at every input. These facts meet the definitions of absolute maximum and absolute minimum, respectively. Since \(f(x)=c\) at every \(x\in D\), the maximizing-input set and minimizing-input set are both all of \(D\). \(\square\)
The nonempty-domain condition ensures that there is an input where the value is attained. On an empty domain there are no inputs at which an extreme value could be attained, so the definitions above do not yield an absolute maximum or minimum.
Worked Example: Both Absolute Extrema of a Constant Function
Let \(f:[-2,3]\to\mathbb{R}\) be given by \(f(x)=4\). Find its absolute maximum and minimum values and all inputs where they occur.
Step 1: Identify the outputs. For every \(x\in[-2,3]\), \(f(x)=4\). In particular, \(f(-2)=4\), \(f(0)=4\), and \(f(3)=4\); the same equality holds for every other input in the interval.
Step 2: Compare the outputs. Since every output is 4, no output is greater than 4 and no output is less than 4. The value 4 is attained, so it qualifies as both an absolute maximum and an absolute minimum.
Step 3: State the attaining inputs. Every input in the domain gives the value 4. Therefore the maximum value is 4, with maximizing-input set \([-2,3]\), and the minimum value is 4, with minimizing-input set \([-2,3]\).
Check: For every \(x\in[-2,3]\), substituting into \(f(x)=4\) gives \(f(x)=4\), confirming both equalities and both full sets of attaining inputs.
Repeated Extrema for a Nonconstant Function
A function need not be constant to have several inputs attaining the same absolute extreme. For instance, symmetry can make outputs at distinct inputs equal. The definitions do not require an extreme to occur at only one point.
Worked Example: A Nonconstant Function with Repeated Maximum and Minimum
Consider \(f:[-2,2]\to\mathbb{R}\), where \(f(x)=(x^2-1)^2\). Find its absolute maximum and minimum values and all inputs where they occur.
Step 1: Establish the range bounds. A square is nonnegative, so \(f(x)\ge 0\) for every \(x\). On the domain, \(0\le x^2\le 4\), hence \(-1\le x^2-1\le 3\). It follows that \(0\le (x^2-1)^2\le 9\).
Step 2: Find where the lower bound is attained. The value \(f(x)=0\) occurs exactly when \(x^2-1=0\). Thus \(x^2=1\), giving \(x=-1\) or \(x=1\). Both inputs are in the domain. Therefore the absolute minimum value is 0, and its minimizing-input set is \(\{-1,1\}\).
Step 3: Find where the upper bound is attained. The value 9 requires \((x^2-1)^2=9\). Within the established interval \(-1\le x^2-1\le 3\), this requires \(x^2-1=3\), so \(x^2=4\) and \(x=-2\) or \(x=2\). Therefore the absolute maximum value is 9, and its maximizing-input set is \(\{-2,2\}\).
Check: \(f(-1)=((-1)^2-1)^2=0\) and \(f(1)=(1^2-1)^2=0\). Also, \(f(-2)=((-2)^2-1)^2=9\) and \(f(2)=(2^2-1)^2=9\). The bounds \(0\le f(x)\le9\) show that none of these values can be improved.
Endpoint Ties Are Still Repeated Extrema
Repeated absolute extrema can occur at endpoints as well as at interior inputs. In a comparison over a closed interval, if two distinct inputs give the largest output, both belong in the maximizing-input set. The value itself is still reported only once.
Worked Example: Equal Minimum Values at Two Endpoints
Let \(g:[0,4]\to\mathbb{R}\) be \(g(x)=x(4-x)\). Find its absolute extrema and their attaining inputs using an algebraic comparison.
Step 1: Rewrite the function. Completing the square gives $$ g(x)=4x-x^2=4-(x-2)^2. $$
Step 2: Find the maximum. Since \((x-2)^2\ge0\), the expression \(4-(x-2)^2\) is at most 4. Equality holds exactly when \(x=2\), which belongs to \([0,4]\). The absolute maximum value is therefore 4, attained only at \(x=2\).
Step 3: Find the minimum. On \([0,4]\), the distance \(|x-2|\) is at most 2, so \(0\le(x-2)^2\le4\). Consequently \(g(x)=4-(x-2)^2\ge0\). Equality \(g(x)=0\) occurs when \((x-2)^2=4\), meaning \(x-2=-2\) or \(x-2=2\). Thus \(x=0\) or \(x=4\). The absolute minimum value is 0, attained at both endpoints.
Check: \(g(2)=2(4-2)=4\), while \(g(0)=0(4-0)=0\) and \(g(4)=4(4-4)=0\). The bounds derived above confirm that 4 is the greatest output and 0 is the least output on the interval.
Why the Extremum Value Is Unique
Although the inputs can be nonunique, an absolute maximum value cannot have two different numerical values. Suppose \(M_1\) and \(M_2\) are both absolute maximum values of the same function on the same domain. Because \(M_1\) is attained, there is an input \(a\) with \(f(a)=M_1\). Since \(M_2\) is an upper bound for every output, \(f(a)\le M_2\), so \(M_1\le M_2\). Reversing their roles gives \(M_2\le M_1\). Therefore \(M_1=M_2\). The same reasoning, with the inequalities reversed, proves uniqueness of the absolute minimum value.
This uniqueness does not imply a unique input. In the constant-function example, the value 4 is the unique maximum value and the unique minimum value, but every point of \([-2,3]\) attains each. In the polynomial example, the minimum value 0 is unique, while two inputs attain it.
A Common Mistake: Reporting an Input as the Value
A frequent error is to name a maximizing input as though it were the maximum value. For the function \(g(x)=x(4-x)\) on \([0,4]\), a mistaken report might say, “The maximum is \(x=2\).” This confuses the input with the output. Substitution gives \(g(2)=2(4-2)=4\), so 2 is the maximizing input and 4 is the maximum value. The correct report is: “The absolute maximum value is 4, attained at \(x=2\).”
Another common error is to list only one input when an extreme is attained at several points. For the same function, \(g(0)=0\) and \(g(4)=0\). Reporting only \(x=0\) omits a minimizing input. The full result is that the absolute minimum value is 0 and its minimizing-input set is \(\{0,4\}\).
When describing an extremum, check that the two parts are clear:
- Value: What is the greatest or least output?
- Attaining inputs: Which domain points produce that output?
Existence and the Role of the Domain
The definitions describe what it means for an absolute maximum or minimum to exist; they do not guarantee that one exists for every function on every domain. For example, a function may have outputs that approach an upper bound without ever reaching it. By contrast, a constant function on any nonempty domain attains its value directly, whether or not the domain is a closed interval. A continuous function on a closed, bounded interval is another important setting in which both absolute extrema are guaranteed to exist by the Extreme Value Theorem.
Once existence is known, the task is to identify the value and all the inputs where it occurs. A complete answer should not infer that an attaining input is unique unless the comparison actually shows that no other input produces the same extreme value.
Check Your Understanding
For each question, distinguish the extremum value from the inputs where it is attained.
- If \(f(x)=-3\) for every \(x\in[1,5]\), what are its absolute maximum and minimum values, and what are the corresponding sets of inputs?
- Can a function have two different absolute maximum values on the same domain? Explain briefly.
- For \(h(x)=(x^2-4)^2\) on \([-2,2]\), identify the absolute minimum value and all inputs where it is attained.
- For \(g(x)=x(4-x)\) on \([0,4]\), state separately the maximum value and the maximizing input, then do the same for the minimum.
- Why does the definition of an absolute maximum require that the proposed value be attained at some input?
