What Does “Local” Mean?
A function can have a highest or lowest value in one small part of its graph without having the highest or lowest value everywhere. The words local maximum and local minimum describe this nearby comparison. The word “local” is essential: values far from the point do not decide whether it is a local extremum.
To make “nearby” precise, we must also respect the function’s domain. If \(f:D\to\mathbb{R}\), then only inputs in \(D\) can be compared with \(f(c)\). A neighborhood of \(c\) within \(D\) consists of the domain points sufficiently close to \(c\).
The number \(f(c)\) is called the corresponding local maximum or minimum value. The definitions compare function values, not just the location of points. In the strict versions, the inequality excludes \(x=c\), because at that input \(f(x)=f(c)\) necessarily.
How to Read the Definitions
The condition \(|x-c|<\delta\) describes the open interval \((c-\delta,c+\delta)\). Since the definition also requires \(x\in D\), the actual comparison is made only on the intersection of that interval with the domain:
There must be one positive radius \(\delta\) that works for all domain points in this neighborhood. It is not enough that the inequality holds for a few selected inputs. On the other hand, the condition imposes no comparison with points outside this neighborhood. That is why the definition captures behavior near \(c\), rather than behavior over the entire domain.
A strict local maximum is also a non-strict local maximum: if nearby values are strictly less than \(f(c)\), they are certainly less than or equal to \(f(c)\). The same implication holds for minima. The reverse implications do not generally hold, since nearby points other than \(c\) might have the same function value.
Worked Examples with Polynomials
Worked Example: A Strict Local Minimum of a Polynomial
Show that \(f(x)=x^2-4x+7\) has a strict local minimum at \(c=2\).
First calculate the function value at the proposed point:
Now compare an arbitrary input \(x\) with this value:
A square is nonnegative, so \(f(x)-f(2)\geq 0\) for every real \(x\). If \(x\ne2\), then \((x-2)^2>0\), and therefore \(f(x)>f(2)\). In particular, these inequalities hold whenever \(|x-2|<\delta\), for example with \(\delta=1\). Thus \(c=2\) is a strict local minimum point, with local minimum value \(3\).
Check: Substituting \(x=2\) into the comparison gives \(f(2)-f(2)=(2-2)^2=0\), as required. Every other input gives a positive difference.
Worked Example: A Strict Local Maximum of a Polynomial
Show that \(g(x)=-x^2+6x-5\) has a strict local maximum at \(c=3\).
Evaluate the function at \(3\):
Subtract this value from the general function value and simplify:
Because \((x-3)^2\geq0\), it follows that \(-(x-3)^2\leq0\), so \(g(x)\leq g(3)\). If \(x\ne3\), then \((x-3)^2>0\), giving \(g(x)<g(3)\). These comparisons hold throughout the domain, so they hold in any neighborhood of \(3\); for instance, take \(\delta=1\). Hence \(3\) is a strict local maximum point, and the local maximum value is \(4\).
Check: At \(x=3\), the difference is \(-(3-3)^2=0\), so the comparison agrees with \(g(3)=4\). For every other \(x\), the difference is negative.
Non-Strict Extrema and Equal Nearby Values
Non-strict extrema allow nearby points to share the extremum’s value. A constant polynomial makes this distinction especially clear: every input has the same output. Each point is both a non-strict local maximum point and a non-strict local minimum point, but none is strict. The reason is that every neighborhood contains other inputs with equal function values.
Worked Example: A Non-Strict Minimum on a Piecewise Plateau
Consider the function \(h:\mathbb{R}\to\mathbb{R}\) defined by
For \(x<0\), \(x^2>0\). For \(0\leq x\leq1\), \(h(x)=0\). For \(x>1\), \((x-1)^2>0\). Thus \(h(x)\geq0\) for every real \(x\), and \(h(c)=0\) for every \(c\in[0,1]\).
Fix any \(c\in[0,1]\). For every \(\delta>0\), the neighborhood of \(c\) contains domain points in \([0,1]\) other than \(c\). At each such point \(x\), \(h(x)=0=h(c)\). All nearby function values are at least \(h(c)\), so \(c\) is a local minimum point. But the strict inequality \(h(x)>h(c)\) fails at those other plateau points. Therefore every point of \([0,1]\) is a non-strict local minimum point, and none is a strict local minimum point.
Check: At, for example, \(c=\tfrac12\), \(h(\tfrac12)=0\), and \(h(\tfrac34)=0\). The two points are distinct and arbitrarily close choices within the plateau give equal values, confirming that the minimum is not strict.
A Proof About Neighborhoods
The choice of radius does not need to be unique. If a comparison works for one radius, it also works for every smaller positive radius. This follows because a smaller neighborhood contains fewer points to check.
Suppose a local maximum condition holds for some \(\delta>0\), so \(f(x)\leq f(c)\) whenever \(x\in D\) and \(|x-c|<\delta\).
Let \(0<\varepsilon\leq\delta\). If \(|x-c|<\varepsilon\), then \(|x-c|<\delta\), so \(x\) also belongs to the original neighborhood.
The original condition gives \(f(x)\leq f(c)\) for every domain point in the smaller neighborhood as well. The same argument preserves the strict maximum condition and both minimum conditions.
This proves that once a neighborhood certifies an extremum, shrinking it cannot invalidate the comparison. It also explains why a local extremum is a genuinely nearby property: the definition needs only one sufficiently small neighborhood, not a comparison across the whole domain.
The Domain Changes Which Points Count
A point at the end of an interval can still have a neighborhood within the domain. That neighborhood may be one-sided because the domain itself contains inputs on only one side. The definition does not require \(c\) to have domain points both to its left and right; it requires the inequality to hold for the domain points close to \(c\).
Worked Example: Strict Extrema at the Ends of a Domain
Let \(r(x)=x\) with domain \(D=[0,2]\). First consider \(c=0\). We have \(r(0)=0\). Choose \(\delta=1\). If \(x\in[0,2]\) and \(|x-0|<1\), then \(0\leq x<1\), so \(r(x)=x\geq0=r(0)\). For every such \(x\ne0\), \(x>0\), so \(r(x)>r(0)\). Thus \(0\) is a strict local minimum point relative to \(D\).
Now consider \(c=2\). We have \(r(2)=2\). Again take \(\delta=1\). If \(x\in[0,2]\) and \(|x-2|<1\), then \(1<x\leq2\), so \(r(x)=x\leq2=r(2)\). For every such \(x\ne2\), \(x<2\), so \(r(x)<r(2)\). Thus \(2\) is a strict local maximum point relative to \(D\).
Check: The inequalities use only inputs from \([0,2]\). In particular, the comparison near \(0\) does not include negative inputs, because they are not in the domain.
Common Pitfall: Comparing Values Outside the Domain
A common mistake is to treat the neighborhood as if it automatically included inputs on both sides of \(c\). For the function \(r(x)=x\) on \(D=[0,2]\), one might incorrectly claim that \(0\) is not a local minimum because \(r(-\tfrac12)=-\tfrac12<r(0)=0\). The calculation is correct, but the input \(-\tfrac12\) is not in the domain. It cannot disqualify \(0\) under the definition.
The correction is to test the inequality only for \(x\in D\) close to \(0\). Those inputs satisfy \(x\geq0\), so \(r(x)=x\geq0=r(0)\), with strict inequality whenever \(x\ne0\). Therefore \(0\) is a strict local minimum relative to \(D\).
Another mistake is to use a distant value to decide a local question. A point may fail to be the highest value over the whole domain and still be a local maximum, provided the maximum inequality holds in some neighborhood. Conversely, a high value at one point does not establish a local maximum unless the nearby comparisons work.
Summary
To classify a local extremum, first identify the function’s domain and the point \(c\). Then ask whether there is some radius \(\delta>0\) for which the appropriate inequality holds for every domain input within distance \(\delta\). A non-strict extremum permits equal values at other nearby inputs; a strict extremum does not. Since the comparison is restricted to the domain, points at the ends of an interval are assessed using the inputs that actually belong to that domain.
Check Your Understanding
Use the neighborhood definitions and pay careful attention to the domain.
- State the inequality that defines a strict local minimum at \(c\), including which nearby inputs must satisfy it.
- For \(p(x)=(x-1)^2\) on \(\mathbb{R}\), verify whether \(x=1\) is a strict local minimum.
- For the constant function \(q(x)=4\) on \(\mathbb{R}\), classify each point as a strict or non-strict local maximum or minimum point.
- In the piecewise plateau example, why is \(c=\tfrac12\) not a strict local minimum?
- For \(r(x)=x\) on \([0,2]\), why can inputs less than \(0\) not be used to test whether \(0\) is a local minimum?
