Why Missing Endpoints Matter
A function can approach a value at the edge of its domain without ever taking that value. On an open interval, neither endpoint belongs to the domain; on a half-open interval, one endpoint belongs and the other does not. This distinction can decide whether a function has a maximum or minimum, even when its largest or smallest possible value is easy to identify.
To analyze global behavior on these domains, compare two kinds of candidates: the values at interior critical points and the limits at missing endpoints. Then ask a separate question: does the function actually take the candidate value at some point in its domain? A limiting value can determine a bound without being an attained extremum.
Bounds and Attained Extrema
Every maximum is a supremum, and every minimum is an infimum. The converse need not hold. The supremum or infimum describes a bound on the values of the function; a maximum or minimum also requires that the function attain that value at a point in its domain.
For example, on \(D=(0,1)\), the function \(f(x)=x\) has supremum \(1\), but it has no maximum: \(1\) is approached as \(x\) approaches the excluded endpoint, but \(f(x)\) is never \(1\) for \(x\in D\). This is the essential issue in endpoint analysis.
Endpoint Limits and Interior Candidates
Let \(f\) be defined on an interval with finite endpoints \(a<b\), such as \((a,b)\), \([a,b)\), or \((a,b]\). At an endpoint excluded from the domain, a finite one-sided limit may still exist. For example, \(\lim_{x\to a^+}f(x)=L\) means that values of \(f(x)\) approach \(L\) as \(x\) approaches \(a\) from within the interval. It does not mean that \(f(a)=L\), since \(a\) may not be in the domain.
For a differentiable function, Fermat’s theorem says that if it has a local maximum or minimum at an interior point, then its derivative there is zero. Thus interior critical points are candidates for interior extrema. Endpoint limits are also candidates for the global bounds, but a limit at an excluded endpoint is not itself an attained value.
A Candidate Principle for Open and Half-Open Intervals
The following principle makes the endpoint comparison precise. Suppose \(f\) is continuous on an interval with finite endpoints, and has a finite one-sided limit at each endpoint missing from its domain. Define a function \(F\) on the closed interval \([a,b]\) by keeping the original values in the interior and assigning each missing endpoint its corresponding one-sided limit. At an endpoint already included in the domain, assign its original function value.
The resulting \(F\) is continuous on \([a,b]\). The extreme value theorem therefore guarantees that \(F\) has a maximum and a minimum on \([a,b]\). If an extreme occurs at an interior point where \(f\) is differentiable, Fermat’s theorem implies that the derivative is zero there. If the derivative does not exist there, the point is critical under the definition above. Thus the candidates for the extrema of \(F\) are its endpoint values and its interior critical points.
Here is why the bounds agree. Every value of \(f\) on the original domain is also a value of \(F\), so the maximum of \(F\) is an upper bound for \(f\). If that maximum occurs only at a missing endpoint, values of \(f\) arbitrarily close to the endpoint approach the maximum. No smaller number can then be an upper bound for \(f\). If the maximum occurs at a point of the original domain, it is attained there. The same reasoning, with inequalities reversed, applies to the minimum.
This principle separates two tasks that are easy to confuse: finding the bound, and deciding whether it is achieved. Also, an endpoint limit equal to a bound does not by itself show that the bound is unattained. The same value might occur at an interior point or at an included endpoint.
Worked Examples
Worked Example: Both Bounds Are Approached but Not Attained
Find the supremum, infimum, maximum, and minimum of \(f(x)=x^2\) on \(D=(0,2)\).
First compute the endpoint limits:
The derivative in the interior is \(f'(x)=2x\). Solving \(2x=0\) gives \(x=0\), which is not in \((0,2)\). There are no interior critical numbers. The continuous extension to \([0,2]\) is \(F(x)=x^2\), whose minimum is \(0\) and maximum is \(4\) at the endpoints.
For every \(x\in(0,2)\), \(0<x^2<4\). Values approach \(0\) as \(x\to0^+\), and approach \(4\) as \(x\to2^-\). Therefore \(\inf f=0\) and \(\sup f=4\), but neither is attained: \(f\) has no minimum and no maximum on this domain.
Check: The inequalities show that \(0\) and \(4\) are bounds; the endpoint limits show that no larger lower bound or smaller upper bound is possible. Since neither endpoint belongs to the domain and no interior point gives either value, neither bound is an extremum.
Worked Example: An Interior Maximum and an Unattained Infimum
Find the global bounds of \(f(x)=x(2-x)\) on \(D=(0,2)\), and determine which are attained.
The endpoint limits are
Differentiate and solve for interior critical numbers:
At this critical point, \(f(1)=1(2-1)=1\). The endpoint-limit candidates are both \(0\). Hence the extension to \([0,2]\) has maximum \(1\) and minimum \(0\).
To check attainment on the original domain, rewrite the function as \(f(x)=1-(x-1)^2\). For \(x\in(0,2)\), \((x-1)^2\geq0\), so \(f(x)\leq1\), with equality at \(x=1\). Also \(x>0\) and \(2-x>0\), so \(f(x)>0\) throughout the domain. Thus the maximum \(1\) is attained at \(x=1\), while the infimum \(0\) is not attained.
Check: Substitution gives \(f(1)=1\), confirming the maximum. The factorization \(x(2-x)\) is strictly positive on \((0,2)\), while its endpoint limits are zero, confirming that zero is the infimum but not a minimum.
Worked Example: A Half-Open Interval with an Unattained Supremum
Find the extrema and bounds of \(f(x)=x\) on \(D=[0,1)\).
The left endpoint is included, so its actual function value is \(f(0)=0\). The right endpoint is excluded, but its limit exists:
The derivative is \(f'(x)=1\), so there are no interior critical numbers. The continuous extension to \([0,1]\) has minimum \(0\) and maximum \(1\). On the original domain, \(0\leq x<1\), so \(0\leq f(x)<1\). The minimum \(0\) is attained at the included endpoint \(x=0\). Values approach \(1\) as \(x\to1^-\), but \(1\) is not in the range, so the supremum is \(1\) and there is no maximum.
Check: \(f(0)=0\) verifies the minimum. For any \(x\in[0,1)\), \(f(x)=x<1\), while values can be made arbitrarily close to \(1\); this verifies the supremum and its nonattainment.
Worked Example: A Missing Endpoint Does Not Prevent Attainment Elsewhere
Find the extrema and bounds of \(f(x)=4x(1-x)\) on \(D=[0,1)\).
At the included endpoint, \(f(0)=0\). At the excluded endpoint,
Differentiate:
The critical-point value is
For \(0\leq x<1\), both factors \(x\) and \(1-x\) are nonnegative, so \(f(x)\geq0\). Also, \(4x(1-x)=1-(2x-1)^2\leq1\), with equality at \(x=\frac12\). Thus the maximum \(1\) is attained at \(x=\frac12\), and the minimum \(0\) is attained at \(x=0\). The fact that the missing endpoint also has limiting value \(0\) does not remove the minimum: the included endpoint already attains it.
Check: Direct substitution gives \(f(0)=0\) and \(f(\frac12)=1\), agreeing with the inequalities. The limit at \(1\) is \(0\), but \(1\) is excluded; the minimum is nevertheless attained at the distinct domain point \(0\).
A Common Mistake: Treating a Limit as a Function Value
A frequent error is to assign an excluded endpoint’s limiting value to the original function and then call it a maximum or minimum. For instance, on \((0,2)\), the limits of \(x^2\) at the endpoints are \(0\) and \(4\). It would be wrong to conclude that the function has minimum \(0\) and maximum \(4\): those values occur at \(x=0\) and \(x=2\) in the closed extension, not at any point in the original domain.
The correction is to keep the extension and the original function distinct. Endpoint limits help determine the supremum and infimum. To claim a maximum or minimum, name a point \(c\) in the original domain and verify that \(f(c)\) equals the proposed bound. If no such point exists, the bound is not attained.
When the Endpoint Method Applies
The continuous-extension method requires finite one-sided limits at the missing endpoints and continuity on the interval. If a limit is infinite, there is no finite endpoint value to use in the extension. For example, \(f(x)=1/x\) on \((0,1)\) grows without bound as \(x\to0^+\), so it has no finite upper bound. Other situations may require checking continuity or the behavior of interior critical points carefully; endpoint limits alone do not replace those conditions.
When the method does apply, an efficient analysis is:
Use the actual function value at an included endpoint and a one-sided limit at a missing endpoint.
Check where the derivative is zero or does not exist inside the interval.
Use the endpoint values or limits together with the function values at interior critical numbers to find the bounds.
A candidate at a missing endpoint is not an attained extremum unless the same value occurs at another point that belongs to the domain.
Why the Distinction Matters
Finding a supremum or infimum answers how far the function values extend, including values approached arbitrarily closely. Finding a maximum or minimum answers whether the function actually reaches that bound. These are different questions, especially on open and half-open domains.
In applications, the distinction can affect whether an optimal value is achievable under the stated constraints. A value may be approached as closely as desired without being attained by any permitted input. Checking endpoint inclusion prevents an analysis from reporting an impossible optimizer.
Check Your Understanding
For each question, distinguish the value of a bound from whether the function attains it.
- For \(f(x)=x\) on \((2,5)\), what are the infimum and supremum? Does either bound give a minimum or maximum?
- For \(f(x)=-(x-1)^2\) on \((0,2)\), find the endpoint limits and the interior critical-point value. Which bounds are attained?
- For \(f(x)=3-x\) on \([1,4)\), identify the included endpoint value and the missing-endpoint limit. Which extremum is not attained?
- Can an endpoint limit equal the infimum even when the function has a minimum? Explain using the half-open interval example in this tutorial.
- What additional check is needed before calling an endpoint-limit candidate a maximum or minimum?
