Why Endpoints Need One-Sided Comparisons
An interior point of an interval has domain points on both sides. An endpoint has domain points on only one side. Consequently, deciding whether an endpoint is a local maximum or minimum requires comparisons only with nearby points that actually belong to the domain.
This is not a special exception to the definition of a local extremum. It is an application of that definition relative to the function's domain. The same distinction matters when considering absolute extrema: local behavior near an endpoint does not, by itself, settle comparisons with points farther away.
Local Extrema Relative to an Interval
Let \(f:[a,b]\to\mathbb{R}\), where \(a<b\). The point \(a\) is a local minimum of \(f\) on \([a,b]\) if there is some \(\delta>0\) such that \(f(a)\leq f(x)\) for every \(x\in[a,b]\) with \(a\leq x<a+\delta\). It is a strict local minimum if the inequality is strict whenever \(a<x<a+\delta\).
At the right endpoint \(b\), the corresponding comparison uses \(b-\delta<x\leq b\). Thus, \(b\) is a local minimum if \(f(b)\leq f(x)\) for all such \(x\), and a strict local minimum if \(f(b)<f(x)\) for all \(x<b\) sufficiently close to \(b\).
For a local maximum, reverse the inequalities: the endpoint's function value must be greater than or equal to nearby values in the domain. Strict local maxima use a strict inequality. The neighborhoods can always be chosen small enough to stay within \([a,b]\).
By contrast, \(a\) is an absolute minimum on \([a,b]\) only if \(f(a)\leq f(x)\) for every \(x\in[a,b]\). An absolute maximum requires \(f(a)\geq f(x)\) for every \(x\) in the interval. The same definitions apply if the proposed absolute extremum is at \(b\).
What One-Sided Derivatives Tell Us
At \(a\), the relevant derivative is the right-hand derivative, if it exists: \(f'_+(a)=\lim_{h\to0^+}\frac{f(a+h)-f(a)}{h}\). At \(b\), it is the left-hand derivative: \(f'_-(b)=\lim_{h\to0^-}\frac{f(b+h)-f(b)}{h}\). These definitions use increments that point into the interval. A two-sided derivative at an endpoint is not required.
If the relevant one-sided derivative exists and is nonzero, its sign determines the strict local classification. The sign rule reverses between the left and right endpoints because the increment \(h\) is positive at the left endpoint and negative at the right.
Why the One-Sided Derivative Test Works
The derivative is a limit of difference quotients. If that limit is nonzero, then sufficiently close to the endpoint, the difference quotients have the same sign as the limit. Converting that sign into a comparison of function values gives the result.
For \(h>0\), the quotient is \(\frac{f(a+h)-f(a)}{h}\). If its limit is positive, then it is positive for all sufficiently small positive \(h\). Since \(h>0\), this implies \(f(a+h)-f(a)>0\), so \(a\) is a strict local minimum. A negative limit gives \(f(a+h)-f(a)<0\), so \(a\) is a strict local maximum.
For \(h<0\), the quotient is \(\frac{f(b+h)-f(b)}{h}\). A positive quotient implies \(f(b+h)-f(b)<0\), because the denominator is negative. Nearby domain values are therefore less than \(f(b)\), making \(b\) a strict local maximum. A negative quotient implies nearby values are greater, making \(b\) a strict local minimum.
This argument also explains the limitation of the test: if a one-sided derivative is zero, the quotient's limit alone does not specify whether nearby function values are above, below, or equal to the endpoint value.
Worked Example: One Endpoint Is Local but Not Absolute
Worked Example: One Endpoint Is Local but Not Absolute
Classify the left endpoint of \(f(x)=x(1-x)\) on \([0,2]\), and determine whether it is an absolute minimum.
First, \(f(0)=0(1-0)=0\). For \(x>0\) close to \(0\), \(f(x)=x(1-x)>0\), since both \(x>0\) and \(1-x>0\) when \(0<x<1\). Thus \(f(x)>f(0)\) for nearby domain points, so \(0\) is a strict local minimum.
The derivative confirms the one-sided direction: \(f'(x)=1-2x\), so \(f'_+(0)=1-2(0)=1>0\). But the value at the other endpoint is \(f(2)=2(1-2)=-2\), which is less than \(f(0)=0\). Therefore \(0\) is not an absolute minimum.
As a check, the endpoint comparison at \(2\) can be made across the whole interval: \(f(x)-f(2)=x-x^2+2=(2-x)(x+1)\geq0\) for \(0\leq x\leq2\). Hence \(2\) is an absolute minimum, consistent with \(f(2)=-2\) being below \(f(0)\).
Worked Example: A Left Endpoint Maximum
Worked Example: A Left Endpoint Maximum
For \(g(x)=x^2-x\) on \([0,2]\), classify the left endpoint locally and decide whether it is an absolute maximum.
At the endpoint, \(g(0)=0^2-0=0\). For \(0<x<1\), \(g(x)=x(x-1)<0\), so nearby domain values are less than \(g(0)\). This proves directly that \(0\) is a strict local maximum.
The derivative gives the same classification: \(g'(x)=2x-1\), and \(g'_+(0)=2(0)-1=-1<0\). However, \(g(2)=2^2-2=2>0=g(0)\). Thus \(0\) is not an absolute maximum.
Indeed, the other endpoint is an absolute maximum: for every \(x\in[0,2]\), \(g(2)-g(x)=2-(x^2-x)=2+x-x^2=(2-x)(x+1)\geq0\). This comparison checks the conclusion over the full domain, not just near an endpoint.
Worked Example: A Zero One-Sided Derivative Is Inconclusive
Worked Example: A Zero One-Sided Derivative Is Inconclusive
Compare the endpoint behavior of \(p(x)=x^2\) at \(0\) on \([0,1]\) with that of \(q(x)=x^3\) at \(0\) on \([-1,0]\).
For \(p\), \(p(0)=0\), and \(p'_+(0)=\lim_{h\to0^+}\frac{h^2-0}{h}=\lim_{h\to0^+}h=0\). The derivative test with a nonzero limit does not apply. But for every \(x\in(0,1]\), \(p(x)-p(0)=x^2>0\), so \(0\) is a strict local minimum (and, in this case, an absolute minimum).
For \(q\), \(q(0)=0\), and \(q'_-(0)=\lim_{h\to0^-}\frac{h^3-0}{h}=\lim_{h\to0^-}h^2=0\). Yet for every \(x\in[-1,0)\), \(q(x)-q(0)=x^3<0\). Thus \(0\) is a strict local maximum on this interval.
The checks show why a zero one-sided derivative does not settle the classification: one function has a minimum at the endpoint and the other has a maximum.
Local Behavior Does Not Settle Absolute Behavior
The first two examples show that an endpoint can be a strict local extremum without being the corresponding absolute extremum. The reason is built into the definitions. A local comparison covers only a small part of the domain; an absolute comparison covers the entire domain.
When a function is increasing throughout an interval, every value to the right is at least as large as every value to its left. In that situation the left endpoint is an absolute minimum and the right endpoint an absolute maximum. When it is decreasing throughout, the roles reverse. The earlier material on monotonicity explains how derivative signs can establish such behavior. A derivative sign only immediately next to an endpoint, however, supports only a local conclusion.
Another frequent error is applying an interior sign-change rule at an endpoint as though the function had domain points on both sides. At the left endpoint there is no domain to the left, and at the right endpoint there is no domain to the right. Use the relative definition and the one-sided comparison instead.
Check Your Understanding
For each question, use comparisons restricted to the stated domain.
- State the definition of a strict local minimum at the left endpoint of \([a,b]\).
- If \(f'_+(a)<0\), what strict local classification follows, assuming the one-sided derivative exists?
- If \(f'_-(b)>0\), what strict local classification follows, assuming the one-sided derivative exists?
- Why does a zero one-sided derivative fail to determine whether an endpoint is a local maximum or minimum?
- What additional scope of comparison is required to conclude that an endpoint is an absolute minimum?
