Introduction
An absolute maximum or minimum must be a value the function actually takes. On an open or unbounded interval, a function may approach a boundary value without ever reaching it; it may also grow without bound. Either behavior can prevent an absolute extremum from existing.
Local behavior and global behavior are different. A function can have a local maximum or minimum while failing to have either absolute extremum on its whole domain. It can also be monotone and bounded without attaining its bounds. The key is to examine both the values inside the domain and what happens near excluded endpoints or far out along an unbounded interval.
Absolute Extrema and Bounds
Let \(f:A\to\mathbb{R}\), where \(A\) is its domain. An absolute maximum occurs at \(c\in A\) if \(f(x)\leq f(c)\) for every \(x\in A\). An absolute minimum occurs at \(c\in A\) if \(f(x)\geq f(c)\) for every \(x\in A\). The requirement \(c\in A\) matters: a value approached at a point outside the domain is not an extremum.
A number \(M\) is an upper bound for the values of \(f\) if \(f(x)\leq M\) for all \(x\in A\). The least such upper bound, when it exists, is the supremum of the function’s values. Similarly, a greatest lower bound is the infimum. A supremum is an absolute maximum only if the function takes the value \(M\) at some point of its domain; an infimum is an absolute minimum only if it is attained.
A local maximum or minimum is defined by comparison with points sufficiently close to a domain point, rather than with every point in the domain. Consequently, a local extremum does not by itself establish an absolute extremum.
Why an Open Endpoint Can Prevent Attainment
Consider a continuous function on an open bounded interval. Unlike a closed interval, the domain does not include its endpoints. Values near an excluded endpoint may approach a bound, but the endpoint itself cannot supply a point where that value is reached. The Extreme Value Theorem guarantees absolute extrema for a continuous function on a closed, bounded interval; an open interval does not meet that theorem’s hypotheses.
Worked Example: A Bounded Function on an Open Interval
Find the absolute extrema, if any, of \(f(x)=x^2\) on \(A=(0,1)\).
For every \(x\in(0,1)\), we have \(0<x<1\). Since \(x\) is positive, squaring gives
Thus \(0\) is a lower bound and \(1\) is an upper bound for the function values. To see that these are the greatest lower bound and least upper bound, take \(x_n=1/n\) for integers \(n\geq 2\). Then \(x_n\in(0,1)\) and \(f(x_n)=1/n^2\), which approaches \(0\). Also take \(y_n=1-1/n\) for \(n\geq 2\). These points lie in \((0,1)\), and their values \((1-1/n)^2\) approach \(1\). So the infimum is \(0\) and the supremum is \(1\).
Neither bound is attained: the equation \(x^2=0\) requires \(x=0\), and \(0\notin A\); the equation \(x^2=1\) requires \(x=1\) or \(x=-1\), neither of which lies in \(A\). Therefore \(f\) has no absolute minimum and no absolute maximum on \((0,1)\). As a check, the values \(f(1/2)=1/4\) and \(f(3/4)=9/16\) satisfy \(0<1/4<9/16<1\), consistent with both bounds being approached but not reached.
This example is bounded and continuous, but boundedness and continuity alone do not guarantee that an open interval contains points attaining the bounds. The missing ingredient is not a calculation at a critical point; it is the exclusion of the boundary points where the limiting values occur.
Behavior on an Unbounded Interval
An unbounded interval has no finite endpoint at one or both ends. As \(x\) moves farther along the domain, function values might grow without bound, or they might approach a finite number without reaching it. Both possibilities need to be checked. An unbounded domain does not automatically mean that the function is unbounded, and a bounded function on such a domain need not attain its bounds.
Worked Example: A Bounded Monotone Function on a Half-Line
Consider \(f(x)=\dfrac{x}{1+x}\) on \(A=(0,\infty)\). Determine whether it has absolute extrema.
For every \(x>0\), the denominator \(1+x\) is positive. Also \(x>0\), and \(x<1+x\). Dividing the last inequality by the positive number \(1+x\) gives
These inequalities show that all function values lie strictly between \(0\) and \(1\). The derivative is
So the function is increasing on its domain. The bounds are approached: at \(x=1/n\), \(f(x)=1/(n+1)\), which approaches \(0\); at \(x=n\), \(f(x)=n/(n+1)\), which approaches \(1\), as \(n\) grows through the positive integers. Thus the infimum is \(0\) and the supremum is \(1\). The strict inequalities show that neither is attained, so there is no absolute minimum or maximum.
For a direct check of the increasing behavior, if \(0<u<v\), then
In particular, every point has other domain points with smaller and larger function values: one can choose a smaller positive input and a larger input. That agrees with the conclusion that neither bound is attained.
Here the function is bounded despite its unbounded domain. Its two limiting bounds correspond to opposite ends of the domain: values near \(x=0\) approach \(0\), while values for large \(x\) approach \(1\). Neither behavior supplies an input where the limiting value is reached.
Local Extrema Need Not Be Absolute
A function may turn locally inside its domain and still have larger or smaller values elsewhere. On an open bounded interval, the values near the excluded endpoints can exceed every value at a local maximum, or fall below every value at a local minimum. The following example shows both local extrema and failure of both absolute extrema.
Worked Example: Local Turning Points but No Absolute Extrema
Let \(f(x)=x^3-3x\) on \(A=(-3,3)\). First find the critical points and classify the local behavior.
Differentiate and factor:
The derivative is zero at \(x=-1\) and \(x=1\). It is positive on \((-3,-1)\), negative on \((-1,1)\), and positive on \((1,3)\). Therefore \(f\) increases, then decreases, then increases. It has a local maximum at \(x=-1\) and a local minimum at \(x=1\). Substitution checks their values:
Now examine the excluded endpoints. The polynomial extends continuously to \([-3,3]\), and
These are limiting values as \(x\) approaches \(-3\) from within the domain and \(3\) from within the domain. The derivative signs show that values rise from near \(-18\) to \(2\), fall to \(-2\), then rise toward \(18\). Thus every value on \((-3,3)\) is strictly between \(-18\) and \(18\). The endpoints are excluded, so neither boundary value is attained.
For an explicit check that the bounds are approached, choose \(x_n=-3+1/n\) and \(y_n=3-1/n\) for integers \(n\geq 1\). Both sequences lie in \((-3,3)\), and continuity gives \(f(x_n)\to -18\) and \(f(y_n)\to18\). Therefore the infimum is \(-18\) and the supremum is \(18\), but neither is attained. The local maximum \(2\) and local minimum \(-2\) are not absolute extrema.
A Useful Way to Analyze These Domains
On an open or unbounded interval, critical points alone do not settle the global question. A critical point can identify a local turning point, but it does not account for values approached near excluded endpoints or far out on the domain. A careful analysis separates these sources of behavior.
Record which endpoints belong to the domain and which are excluded. Note whether the interval extends without bound.
Use critical points and any points where the function is not differentiable to investigate local behavior and candidate values.
For each excluded finite endpoint, examine the function values as inputs approach it from within the domain. On an unbounded end, examine what happens as the input grows in magnitude in that direction.
Determine whether the largest or smallest value is actually achieved at a point in the domain. A limiting value or bound by itself is not an absolute extremum.
This process keeps two questions separate: what are the greatest and least bounds of the function’s values, and does the function attain either bound? The first question concerns how values are distributed; the second depends on whether a point in the domain realizes the bound.
When Including an Endpoint Changes the Answer
Adding an endpoint can turn an unattained bound into an absolute extremum. For example, on \([0,\infty)\), the function \(g(x)=x^2\) has an absolute minimum at \(x=0\), since \(x^2\geq0\) for every \(x\geq0\), and \(g(0)=0\). It has no absolute maximum: for any \(x\geq0\), the domain also contains \(x+1\), and
So there is always a larger function value. Including \(0\) supplies the minimum, but the unbounded end still prevents a maximum.
Check Your Understanding
For each question, distinguish a bound from a value attained by the function.
- For \(f(x)=1-x\) on \((0,1)\), what are the infimum and supremum? Does either become an absolute extremum?
- Why does a finite supremum not, by itself, establish an absolute maximum?
- For \(g(x)=x^2\) on \([0,\infty)\), identify the absolute extremum and explain why the other one does not exist.
- In the cubic example on \((-3,3)\), what are the local extrema, and why are they not absolute extrema?
- What changes about the minimum of \(x^2\) when its domain changes from \((0,1)\) to \([0,1]\)?
