Why Endpoints Need a One-Sided Test
At an interior point of an interval, a function has domain values on both sides. At an endpoint, values in the domain lie on only one side. That difference matters: an endpoint can be a local maximum or minimum even when the usual two-sided tests do not apply.
The key is to compare the endpoint’s function value with nearby values that actually belong to the domain. A boundary point excluded from the domain is different: the function has no value there, so it cannot be a local extremum.
Local Extrema Relative to the Domain
The requirement \(x\in D\) is essential. Only values in the function’s domain are compared. A local extremum is therefore defined relative to the domain, not relative to all real numbers.
These definitions allow equality: nearby values may equal \(f(c)\). If the inequalities are strict for every nearby \(x\in D\) with \(x\ne c\), the extremum is called a strict local maximum or minimum.
For an interval domain \([a,b]\), a neighborhood of \(a\) contains domain points only to the right of \(a\); a neighborhood of \(b\) contains domain points only to the left of \(b\). Thus, the endpoint tests are one-sided comparisons. An included endpoint is eligible to be a local extremum, but it is not automatically one.
Derivative Signs Near an Included Endpoint
Derivative signs can provide a convenient sufficient test. The signs must be considered on the interval immediately inside the domain; a two-sided derivative test at the endpoint is not required.
| Endpoint and nearby behavior | Conclusion |
|---|---|
| Left endpoint \(a\); \(f\) increases to the right | Local minimum at \(a\) |
| Left endpoint \(a\); \(f\) decreases to the right | Local maximum at \(a\) |
| Right endpoint \(b\); \(f\) increases toward \(b\) | Local maximum at \(b\) |
| Right endpoint \(b\); \(f\) decreases toward \(b\) | Local minimum at \(b\) |
For example, if \(f'(x)>0\) just to the right of an included left endpoint \(a\), then \(f\) increases as \(x\) moves into the domain. Nearby values are at least \(f(a)\), so \(a\) is a local minimum. At a right endpoint \(b\), the same positive derivative means nearby values to the left are at most \(f(b)\), so \(b\) is a local maximum.
The corresponding conclusions reverse when the derivative is negative. A derivative sign that is merely nonnegative or nonpositive gives a weak local extremum under the conditions below; strict signs give strict monotonicity and strict extrema.
Why the One-Sided Derivative Test Works
The justification follows from the Mean Value Theorem. To use it on an interval from an endpoint to a nearby point, the function must be continuous on the closed interval and differentiable in its interior. The derivative sign then determines the ordering of the endpoint value and nearby values.
Let \(a\) be an included left endpoint, and choose \(x>a\) sufficiently close to \(a\). Assume \(f\) is continuous on \([a,x]\) and differentiable on \((a,x)\).
There is a number \(c\in(a,x)\) such that \(f(x)-f(a)=f'(c)(x-a)\).
Since \(x-a>0\), if \(f'(t)\geq 0\) throughout the interval just to the right of \(a\), then \(f(x)-f(a)\geq 0\). Thus \(f(x)\geq f(a)\) for all sufficiently close domain points \(x\), which is exactly the local-minimum condition.
If \(f'(t)\leq 0\) to the right of \(a\), the same equation gives \(f(x)\leq f(a)\), so \(a\) is a local maximum. At a right endpoint, apply the theorem to \([x,b]\); the same comparison classifies the endpoint from the values to its left.
This argument explains why the endpoint itself need not have a derivative. The comparison uses derivatives only at interior points between the endpoint and a nearby domain point.
Worked Examples
Worked Example: Classify Both Endpoints of a Quadratic
Let \(f(x)=(x-1)^2\) on the domain \([1,4]\). Determine whether either endpoint is a local extremum relative to this domain.
First differentiate:
For \(1<x\leq 4\), \(x-1>0\), so \(f'(x)>0\). The function increases as \(x\) moves from the left endpoint toward the right endpoint.
At the left endpoint,
For any \(x\in[1,4]\), \((x-1)^2\geq 0\). In particular, every domain point sufficiently close to \(1\) has \(f(x)\geq f(1)\). Therefore \(x=1\) is a local minimum. It is strict because \((x-1)^2>0\) whenever \(x>1\).
At the right endpoint,
For every \(x\in[1,4]\), \(0\leq x-1\leq 3\), so \((x-1)^2\leq 3^2=9\). Thus nearby domain values satisfy \(f(x)\leq f(4)\), and \(x=4\) is a strict local maximum.
Check: the derivative is positive throughout the interval’s interior, consistent with the endpoint values increasing from \(f(1)=0\) to \(f(4)=9\).
Worked Example: A Decreasing Function on a Closed Interval
Let \(g(x)=3-x\) on \([-1,2]\). Classify the two endpoints.
The derivative is
Since \(g'(x)<0\), the function decreases from left to right. At the left endpoint,
Every domain point \(x\) satisfies \(x\geq -1\), so \(3-x\leq 3-(-1)=4\). In particular, nearby domain values are no greater than \(g(-1)\). The left endpoint \(-1\) is a strict local maximum.
At the right endpoint,
Every domain point \(x\) satisfies \(x\leq 2\), so \(3-x\geq 3-2=1\). Nearby domain values are no less than \(g(2)\). The right endpoint \(2\) is a strict local minimum.
Check: the endpoint values change from \(4\) to \(1\), matching the negative derivative and the decreasing behavior.
Worked Example: An Excluded Boundary Point
Let \(h(x)=x\) on the domain \((0,3]\). Explain why \(0\) is not a local extremum, and classify the included endpoint \(3\).
The point \(0\) is not in the domain \((0,3]\). Therefore \(h(0)\) is not defined as a value of this function. Since a local extremum must occur at a point \(c\) in the domain, \(0\) cannot be a local maximum or minimum of \(h\).
Although values of \(h(x)\) approach \(0\) as \(x\) approaches \(0\) from the right, that limiting value does not make \(0\) a function value. The range is \((0,3]\): values are positive and can be arbitrarily close to zero, but zero is not attained.
The included endpoint has value
For \(x\in(0,3]\), \(h(x)=x\leq 3=h(3)\). Thus, all domain values near \(3\) are at most \(h(3)\), and \(3\) is a strict local maximum. The derivative check agrees: \(h'(x)=1>0\), so the function increases toward the right endpoint.
Check: \(3\) belongs to the domain and has a value; \(0\) does not. Only the former can be classified as a local extremum.
An Included Endpoint Is Not Automatically an Extremum
Being included makes an endpoint eligible for classification; it does not guarantee either type of extremum. For instance, define \(q(0)=0\) and \(q(x)=x^2\sin(1/x)\) for \(0<x\leq 1\). The included endpoint \(0\) is neither a local maximum nor a local minimum.
To see why, for positive integers \(n\) consider
Both sequences consist of positive domain points and approach \(0\). Since \(\sin(1/x_n)=1\), we have \(q(x_n)=x_n^2>0=q(0)\). Since \(\sin(1/y_n)=-1\), we have \(q(y_n)=-y_n^2<0=q(0)\). Thus every neighborhood of \(0\) contains domain values both above and below \(q(0)\). Neither local-extremum inequality can hold.
Common Pitfalls
Pitfall 1: Treating an endpoint like an interior critical point. For \(f(x)=x\) on \([0,2]\), a student might say, “\(f'(0)=1\), not \(0\), so \(0\) is not an extremum.” This conclusion incorrectly uses a stationary-point test as a requirement. In fact, \(f(x)=x\geq 0=f(0)\) for all \(x\in[0,2]\), so \(0\) is a strict local minimum relative to the domain.
Pitfall 2: Treating an excluded boundary limit as an attained value. For \(h(x)=x\) on \((0,3]\), it is incorrect to claim that \(0\) is a minimum because values approach \(0\). The correction is that \(0\notin(0,3]\), so it is not a point at which \(h\) can have an extremum. The function has infimum \(0\), but does not attain that value.
Check Your Understanding
For each question, pay attention to which points belong to the stated domain.
- For \(f(x)=x^2\) on \([0,2]\), is \(0\) a local minimum, local maximum, both, or neither?
- For \(g(x)=5-x\) on \([1,4]\), classify the included right endpoint using its one-sided behavior.
- Why does \(f'(a)\) not need to exist for an included endpoint \(a\) to be a local extremum?
- For \(h(x)=x\) on \((0,2]\), can \(0\) be a local minimum? Explain the distinction between a limit and a function value.
- What conditions allow the Mean Value Theorem to justify a derivative-sign test near an endpoint?
